Class 9 Maths Chapter 6 NCERT Solutions: Measuring Space: Perimeter and Area

Sections 6.1 to 6.5 (Perimeter of Shapes and Circles)

This chapter develops perimeter of circles and circular arcs (including athletics-track staggers), then moves on to area of parallelograms, triangles (including Heron’s formula), cyclic quadrilaterals (Brahmagupta), and sectors and segments of circles.

Section 6.1: Think and Reflect (Fig. 6.1)

Question

In Fig. 6.1, you see athletes assembled at the start of a $4 \times 100 m$ relay race. Notice that the athletes are not at the same starting line. Those in the outer lanes seem to be starting ahead of those in the inner lanes while the finish line is the same for all of them.

What could be the reason for this?
Do you think the stagger gives anyone (those in the outer lanes or in the inner lanes) an unfair advantage? Why or why not?
On what basis can the organisers work out the length of the stagger between lanes?

Fig. 6.1: Athletes at the start of a 4 × 100 m relay race

Solution

Reason for different starting lines: An athletics track consists of two straight sections and two curved semicircular turns. As you move from the inner lanes to the outer lanes, the radius of the semicircular curves increases. This means an outer lane is physically longer than an inner lane around the curve. To ensure that every athlete runs the exact same total distance (400 meters), athletes in outer lanes must start further ahead.
Fairness: No, the stagger does not give anyone an unfair advantage. It simply compensates for the extra distance on the outer curves, making the total running distance identical for every lane.
Basis for calculation: Organizers calculate the stagger by finding the difference between the perimeters (circumferences) of the curved sections of adjacent lanes. If a lane has a width of $𝑤$ meters, the radius of each semicircle increases by $𝑤$ . The extra distance run around a full circle of turns is $2 𝜋 𝑤$ , which is the exact stagger needed per full lap. (Note: Here, stagger is first defined as the distance between the starting points of adjacent lanes.)

Section 6.1: Think and Reflect (Page 118)

Question

In my school, the playground is too small to have a 400 m track, so the school constructed a 200 m track instead. Does this mean that we need a smaller stagger for the race tracks in my school (i.e., smaller than the stagger used in the Olympics), for the same $4 \times 100 m$ relay race?

Solution

No, you do not need a smaller stagger per lap; in fact, for a $4 \times 100 m$ relay, you will need a larger total stagger if runners stay in their lanes! Here is the step-by-step reasoning:

The stagger for completing one full set of turns ( $360 \circ$ ) depends only on the width of the lanes, not on the radius of the track. Mathematically, the difference in circumference between two circles with radii $𝑟 1$ and $𝑟 2$ is $2 𝜋 (𝑟 2 - 𝑟 1)$ , which depends solely on the lane width $(𝑟 2 - 𝑟 1)$ .
On a standard 400 m track, a $4 \times 100 m$ relay race takes exactly one lap (one full circle of turns).
On a 200 m track, a 400 m race requires athletes to run two full laps (two full circles of turns).
Since the runners go around the curves twice as many times on a 200 m track, the outer lane runners experience the curve penalty twice. Therefore, if they stay in their lanes for the entire race, the total stagger required is double that of a 400 m track.

Section 6.1: Think and Reflect (Page 119)

Question

Here we see a circle with radius $𝑟 u n i t s$ . What is its perimeter? How do we find out? What is the connection between this question and the one about the 400 m athletics track?

Fig. 6.3: Circle of radius

𝑟

Solution

Perimeter of the circle: The perimeter (called the circumference) of a circle with radius $𝑟$ is $2 𝜋 𝑟 u n i t s$ .
How we find out: We find it by discovering that the ratio of a circle's circumference ( $𝐶$ ) to its diameter ( $𝐷 = 2 𝑟$ ) is always a constant value, approximately $3.1416$ , represented by the Greek letter $𝜋$ . Thus, $𝐶 = 𝜋 𝐷 = 2 𝜋 𝑟$ .
Connection to the athletics track: The two curved portions of an athletics track form two semicircles that fit together to make one complete circle. To calculate how much longer an outer lane is compared to an inner lane (the stagger), we must use the circle perimeter formula $2 𝜋 𝑟$ for the different radii of the lanes.

Section 6.2: Home Measurement Activity (Page 120)

Activity / Task

Take a cotton reel with thin thread around it. Measure the diameter $𝐷$ of the reel as accurately as possible. Unwrap and then tightly wrap the thread around the reel 20 times. Unwrap it again; measure its length $𝐿$ , and calculate $𝐿 20 𝐷$ . Do you get a ratio between 3 and 4? Between 3.1 and 3.2? It is also possible to estimate the $𝐶 / 𝐷$ ratio using pure geometry, i.e., without any measurements at all! Can you imagine how?

Solution

Experimental Result: When you perform this experiment, wrapping the thread 20 times reduces measurement errors. The total length $𝐿$ equals 20 circumferences ( $20 𝐶$ ). Dividing by $20 𝐷$ computes the ratio $𝐶 𝐷$ . Because the true value of $𝜋$ is approximately $3.14159$ , your measured result will reliably fall between 3.1 and 3.2.
Estimation using pure geometry: You can estimate the ratio without physical measurements by drawing a circle and inscribing (drawing inside) and circumscribing (drawing outside) regular polygons with straight sides, such as hexagons. Because straight line segments can be calculated using the Baudhāyana-Pythagoras theorem, we can calculate the exact perimeters of the inner and outer polygons. The circle's circumference must lie trapped between the smaller inner polygon's perimeter and the larger outer polygon's perimeter.

Section 6.2: In-Text Questions (Pages 121–122)

Question 1: Can you see why Fig. 6.6 shows that $𝜋 > 3$ ?

Fig. 6.6: Regular hexagon inscribed in a circle (r = 1)

Solution

In Fig. 6.6, a regular hexagon is inscribed inside a circle of radius $𝑟 = 1$ . A regular hexagon can be divided from its center into 6 equilateral triangles, so each straight side of the hexagon is equal to the radius $𝑟 = 1$ . The total perimeter of the hexagon is $6 \times 1 = 6 u n i t s$ . Since the curved circle fully surrounds the straight-edged hexagon, the circumference $𝐶$ of the circle must be greater than $6$ . Since diameter $𝐷 = 2 𝑟 = 2$ , the ratio $𝜋 = 𝐶 𝐷$ must be greater than $62 = 3$ .

Question 2: Can you see why Fig. 6.7 tells us that $𝜋$ is between $3$ and $2 \sqrt 3$ ? (Hint: Use the Baudhāyana-Pythagoras Theorem.)

Fig. 6.7: Inscribed and circumscribed regular hexagons

Solution

Inner Hexagon (Lower Bound): As shown above, a regular hexagon inscribed in a circle of radius $𝑟 = 1$ has side length $1$ and perimeter $6$ . Therefore, $𝜋 > 62 = 3$ .
Outer Hexagon (Upper Bound): For a regular hexagon circumscribed outside a circle of radius $𝑟 = 1$ , the radius of the circle touches the midpoint of each side at a right angle ( $90 \circ$ ), forming the height ( $ℎ = 1$ ) of an equilateral triangle with side $𝑎$ .

By the Baudhāyana-Pythagoras theorem on half of this triangle:

12 + (𝑎 2) 2 = 𝑎 2 ⟹ 1 = 3 𝑎 2 4 ⟹ 𝑎 2 = 43 ⟹ 𝑎 = 2 \sqrt 3

The perimeter of this outer hexagon is $6 \times 2 \sqrt 3 = 12 \sqrt 3 = 4 \sqrt 3$ .

Conclusion: Since the circle lies between the inner and outer hexagons, its circumference $𝐶$ is between $6$ and $4 \sqrt 3$ . Dividing by the diameter ( $𝐷 = 2$ ), we find that $𝜋$ is trapped between $62$ and $4 \sqrt 3 2$ , meaning $3 < 𝜋 < 2 \sqrt 3$ (where $2 \sqrt 3 \approx 3.46$ ).

Section 6.4: A Closer Look at a 400 m Athletics Track — Think and Reflect (Page 127)

Question

What is the difference in radius between the first and second lanes?
Use Fig. 6.11 to find the stagger needed by the runner in the second lane.
Will an equal stagger be needed between the third and second lanes?

Fig. 6.11: Schematic diagram of a 400 m athletics track (straight sections 84.39 m each; innermost curve radius 36.5 m; lane width 1.22 m)

Solution

Difference in radius: The width of each lane is given as $1.22 m$ . Therefore, the difference in radius between the first and second lanes is exactly $1.22 m$ .
Stagger for second lane: A runner in lane 1 runs at radius $𝑟 1 = 36.5 + 0.3 = 36.8 m$ . A runner in lane 2 runs at radius $𝑟 2 = 36.8 + 1.22 = 38.02 m$ . The difference in running distance over the two semicircular curves (one full circle) is:

S t a g g e r = 2 𝜋 𝑟 2 - 2 𝜋 𝑟 1 = 2 𝜋 (𝑟 2 - 𝑟 1) = 2 \times 3.1416 \times 1.22 \approx 𝟕 . 𝟔 𝟕 m

Equal stagger for third lane: Yes, an equal stagger of $7.67 m$ will be needed between the third and second lanes because all lanes have the same constant width ( $1.22 m$ ), making the radius difference identical for every adjacent pair of lanes.

Section 6.5: Worked Examples (Redone Step-by-Step)

Example 1 (Page 127–128)

Two circles of equal radius are located such that each circle passes through the centre of the other circle (Fig. 6.12). Given that the radius of each circle is $𝑟 u n i t s$ , find the perimeter of the shape formed by the two circles in terms of $𝑟 u n i t s$ . (Ignore the dotted portions that lie within the circles.)

Fig. 6.12: Two congruent circles, each through the other’s centre

Solution

Identify the geometric triangle: Let the centers of the two circles be $𝐴$ and $𝐵$ , and their intersection points be $𝐶$ and $𝐷$ . Consider triangle $△ 𝐴 𝐵 𝐶$ .
Find side lengths: Since point $𝐵$ lies on circle $𝐴$ , the distance $𝐴 𝐵 = 𝑟$ . Since $𝐶$ lies on circle $𝐴$ , $𝐴 𝐶 = 𝑟$ . Since $𝐶$ lies on circle $𝐵$ , $𝐵 𝐶 = 𝑟$ .
Determine central angles: Because $𝐴 𝐵 = 𝐴 𝐶 = 𝐵 𝐶 = 𝑟$ , triangle $△ 𝐴 𝐵 𝐶$ is an equilateral triangle. Therefore, angle $∠ 𝐶 𝐴 𝐵 = 60 \circ$ . By identical reasoning below the center line, triangle $△ 𝐴 𝐵 𝐷$ is also equilateral, so $∠ 𝐵 𝐴 𝐷 = 60 \circ$ .
Find the angle of the inner dotted arc: The total angle subtended by the inner dotted arc $𝐶 𝐷$ at center $𝐴$ is $∠ 𝐶 𝐴 𝐷 = 60 \circ + 60 \circ = 120 \circ$ .
Find the angle of the outer red arc: A full circle has $360 \circ$ . The outer visible border (red arc) of circle $𝐴$ subtends the remaining angle:

A n g l e = 360 \circ - 120 \circ = 240 \circ

In terms of fractions, this outer arc is $240 \circ 360 \circ = 23$ of the full circumference.

Calculate total perimeter: The combined shape consists of two such outer arcs (one from each circle).

L e n g t h o f o n e r e d a r c = 23 \times (2 𝜋 𝑟) = 43 𝜋 𝑟

T o t a l P e r i m e t e r = 2 \times (43 𝜋 𝑟) = 𝟖 𝟑 𝜋 𝐫 u n i t s

Example 2 (Page 128)

In Fig. 6.13, we see points $𝑃$ and $𝑄$ and two paths connecting them. The first path is made up of the semicircle $𝑎$ . The other path is made up of three semicircles ( $𝑏$ , $𝑐$ and $𝑑$ ). Which path is longer? Choose one: (i) Path $𝑎$ is longer. (ii) Path $𝑏 + 𝑐 + 𝑑$ is longer. (iii) The two paths have equal length.

Fig. 6.13: Two paths from P to Q (semicircle a vs b+c+d)

Solution

Define radii: Let the radius of semicircle $𝑎$ be $𝑟 𝑎$ , and let the radii of smaller semicircles $𝑏, 𝑐, 𝑑$ be $𝑟 𝑏, 𝑟 𝑐, 𝑟 𝑑$ respectively.
Express lengths of semicircular arcs: The length of any semicircle of radius $𝑟$ is $𝜋 𝑟$ .

Length of Path 1 (semicircle $𝑎$ ) $= 𝜋 𝑟 𝑎$ .
Length of Path 2 (semicircles $𝑏 + 𝑐 + 𝑑$ ) $= 𝜋 𝑟 𝑏 + 𝜋 𝑟 𝑐 + 𝜋 𝑟 𝑑 = 𝜋 (𝑟 𝑏 + 𝑟 𝑐 + 𝑟 𝑑)$ .

Relate the diameters along the straight line: All semicircles lie along the same straight baseline connecting $𝑃$ and $𝑄$ . The diameter of the large semicircle ( $2 𝑟 𝑎$ ) is exactly equal to the sum of the diameters of the three smaller semicircles:

2 𝑟 𝑎 = 2 𝑟 𝑏 + 2 𝑟 𝑐 + 2 𝑟 𝑑 ⟹ 𝑟 𝑎 = 𝑟 𝑏 + 𝑟 𝑐 + 𝑟 𝑑

Compare path lengths: Substituting $𝑟 𝑎$ into the equation for Path 1 gives:

L e n g t h o f P a t h 1 = 𝜋 (𝑟 𝑏 + 𝑟 𝑐 + 𝑟 𝑑) = L e n g t h o f P a t h 2

Conclusion: Therefore, (iii) The two paths have equal length.

Section 6.5: Exercise Set 6.1 (Pages 129–130)

Note: Unless stated otherwise, use the approximation

227

for

𝜋

Question 1

The perimeter of a circle is $44 c m$ . What is its radius?

Solution

We use the circumference formula $𝐶 = 2 𝜋 𝑟$ :

44 = 2 \times 227 \times 𝑟

44 = 447 \times 𝑟

Multiply both sides by $7$ and divide by $44$ :

𝑟 = 44 \times 7 44 = 𝟕 c m

Question 2

Calculate, correct to 3 significant figures, the circumference of a circle with: (i) radius $7 c m$ (ii) radius $10 c m$ (iii) radius $12 c m$ .

Solution

(i) Radius $𝑟 = 7 c m$ :

𝐶 = 2 \times 227 \times 7 = 44 c m ⟹ 𝟒 𝟒 . 𝟎 c m (t o 3 s i g . f i g s .)

(ii) Radius $𝑟 = 10 c m$ :

𝐶 = 2 \times 227 \times 10 = 4407 = 62.8571 . . . \approx 𝟔 𝟐 . 𝟗 c m

(iii) Radius $𝑟 = 12 c m$ :

𝐶 = 2 \times 227 \times 12 = 5287 = 75.4285 . . . \approx 𝟕 𝟓 . 𝟒 c m

Question 3

Calculate the length of the arc of a circle if: (i) the radius is $3.5 c m$ and the angle at the centre is $60 \circ$ , and (ii) the radius is $6.3 m$ and the angle at the centre is $120 \circ$ .

Solution

The formula for arc length is $𝑙 = 2 𝜋 𝑟 \times 𝜃 360 \circ$ .

(i) $𝑟 = 3.5 c m$ , $𝜃 = 60 \circ$ :

𝑙 = 2 \times 227 \times 3.5 \times 60360 = 22 \times 16 = 113 \approx 𝟑 . 𝟔 𝟕 c m (o r 323 c m)

(ii) $𝑟 = 6.3 m$ , $𝜃 = 120 \circ$ :

𝑙 = 2 \times 227 \times 6.3 \times 120360 = 44 \times 0.9 \times 13 = 44 \times 0.3 = 𝟏 𝟑 . 𝟐 m

Question 4

Find the perimeter of a sector (i.e., the curved portion as well as the two straight portions) of a circle of radius $14 c m$ and sector angle $75 \circ$ .

Solution

Find curved arc length ( $𝑙$ ):

𝑙 = 2 \times 227 \times 14 \times 75360 = 88 \times 524 = 553 \approx 18.33 c m

Add the two straight radius edges:

T o t a l P e r i m e t e r = 𝑙 + 2 𝑟 = 553 + 14 + 14 = 553 + 28 = 𝟒 𝟔 . 𝟑 𝟑 c m (o r 4613 c m)

Question 5

Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14i to 6.14ix):

Fig. 6.14: Shapes (i)–(ix) for perimeter

Solution

Correction: The shapes below follow the actual Fig. 6.14 in the textbook (the draft markdown had mis-identified several figures). Use

𝜋 = 227

(i) Stadium: rectangular length $80 m$ , semicircular ends of diameter $60 m$ . $P e r i m e t e r = 2 \times 80 + 𝜋 \times 60 = 160 + 13207 = 𝟐 𝟒 𝟒 𝟎 𝟕 m \approx 𝟑 𝟒 𝟖 . 𝟓 𝟕 m$
(ii) Semicircle on diameter $12 c m$ (so $𝑟 = 6 c m$ ; the printed interior “ $8 c m$ ” does not match a semicircle on a $12 c m$ diameter — the drawn shape is a semicircle, so we use $𝑟 = 6$ ): $P e r i m e t e r = 𝜋 𝑟 + 2 𝑟 = 6 𝜋 + 12 = 𝟐 𝟏 𝟔 𝟕 c m \approx 𝟑 𝟎 . 𝟖 𝟔 c m$
(iii) Square side $10 c m$ + 4 outward semicircles: $P e r i m e t e r = 4 \times (𝜋 \times 5) = 20 𝜋 = 𝟒 𝟒 𝟎 𝟕 c m \approx 𝟔 𝟐 . 𝟖 𝟔 c m$
(iv) Three semicircles on the sides of an equilateral triangle of side $12 c m$ : $P e r i m e t e r = 3 \times (𝜋 \times 6) = 18 𝜋 = 𝟑 𝟗 𝟔 𝟕 c m \approx 𝟓 𝟔 . 𝟓 𝟕 c m$
(v) Square side $14 c m$ + 4 outward semicircles: $P e r i m e t e r = 4 \times (𝜋 \times 7) = 28 𝜋 = 𝟖 𝟖 c m$
(vi) Large semicircle diameter $28 c m$ + two lower semicircles diameter $14 c m$ each: $P e r i m e t e r = 𝜋 \times 14 + 2 \times (𝜋 \times 7) = 28 𝜋 = 𝟖 𝟖 c m$
(vii) Circle diameter $8 c m$ with external semicircle diameter $6 c m$ : $P e r i m e t e r = 2 𝜋 \times 4 + 𝜋 \times 3 = 11 𝜋 = 𝟐 𝟒 𝟐 𝟕 c m \approx 𝟑 𝟒 . 𝟓 𝟕 c m$
(viii) Large semicircle diameter $12 c m$ over three small semicircles of diameter $4 c m$ : $P e r i m e t e r = 𝜋 \times 6 + 3 \times (𝜋 \times 2) = 12 𝜋 = 𝟐 𝟔 𝟒 𝟕 c m \approx 𝟑 𝟕 . 𝟕 𝟏 c m$
(ix) S-curve / double lobe with two $10 c m$ segments (two large semicircles of radius $10$ plus two small of radius $5$ along the divider): $O u t e r + d i v i d e r = 2 \times (2 𝜋 \times 10) / 2 + 2 \times (2 𝜋 \times 5) / 2 = 20 𝜋 + 10 𝜋 = 30 𝜋 = 𝟔 𝟔 𝟎 𝟕 c m \approx 𝟗 𝟒 . 𝟐 𝟗 c m$

Question 6

If the diameter of a car tyre is $56 c m$ , then: (i) How far does the car need to travel for the tyre to complete one revolution? (ii) How many revolutions does the tyre make if the car travels $10 k m$ ?

Solution

(i) Distance for one revolution: This equals the circumference of the tyre:

𝐶 = 𝜋 𝑑 = 227 \times 56 = 𝟏 𝟕 𝟔 c m

(ii) Number of revolutions in $10 k m$ : First, convert $10 k m$ into centimeters:

10 k m = 10 \times 1, 000 m = 10, 000 \times 100 c m = 1, 000, 000 c m

R e v o l u t i o n s = T o t a l D i s t a n c e C i r c u m f e r e n c e = 1, 000, 000 176 = 5681.818 . . . \approx 𝟓, 𝟔 𝟖 𝟐 r e v o l u t i o n s

Question 7

Find the total perimeter of all the petals in each of the given flowers: (i) Fig. 6.15A: Square of side $14 c m$ . The centres of the arcs are the midpoints of the sides of the square. (ii) Fig. 6.15B: Regular hexagon of side $42 c m$ . The centres of the arcs are the vertices of the hexagon.

Solution

(i) Square flower (Fig. 6.15A):

Since the arc centers are at the midpoints of the square's sides ( $14 c m$ ), the radius of every circular arc is $𝑟 = 142 = 7 c m$ .
Each corner of the square forms a $90 \circ$ angle, meaning each arc is a quarter circle ( $90 \circ 360 \circ = 14$ ).
There are 4 petals, and each petal is bounded by 2 such quarter-circle arcs, making a total of $4 \times 2 = 8$ quarter circles.
Eight quarter circles equal 2 complete circles of radius $7 c m$ :

T o t a l P e r i m e t e r = 2 \times (2 𝜋 𝑟) = 4 \times 227 \times 7 = 𝟖 𝟖 c m

(ii) Hexagon flower (Fig. 6.15B):

The centers of the arcs are the vertices of the hexagon, and the radius extends along the side length, so $𝑟 = 42 c m$ .
In a regular hexagon, the interior angle of the equilateral triangles forming the petals from each vertex is $60 \circ$ . Thus, each petal arc subtends $60 \circ$ , which is $60 \circ 360 \circ = 16$ of a circle.
There are 6 petals, bounded by $6 \times 2 = 12$ such arcs.
Twelve one-sixth circular arcs equal 2 complete circles of radius $42 c m$ :

T o t a l P e r i m e t e r = 2 \times (2 𝜋 𝑟) = 4 \times 227 \times 42 = 4 \times 22 \times 6 = 𝟓 𝟐 𝟖 c m

Question 8

The ratio of the perimeters of two circles is $5 : 4$ . What is the ratio of their radii?

Solution

Let the two circles have radii $𝑟 1$ and $𝑟 2$ , and perimeters $𝑃 1$ and $𝑃 2$ . The formula for perimeter is $𝑃 = 2 𝜋 𝑟$ .

𝑃 1 𝑃 2 = 2 𝜋 𝑟 1 2 𝜋 𝑟 2 = 𝑟 1 𝑟 2

Since the constant $2 𝜋$ cancels out, the ratio of the radii is identical to the ratio of the perimeters. Therefore, the ratio of their radii is $5 : 4$ .

Sections 6.6 to 6.9 (Parallelograms, Triangles, and Special Formulas)

Section 6.7: Area of a Parallelogram — Think and Reflect (Page 131, Top)

Question

What happens if the parallelogram is 'thin' (Fig. 6.18) and the foot of the perpendicular from $𝐶$ to $𝐴 𝐷$ does not lie on side $𝐴 𝐷$ ? The construction then does not seem to work. How do we fix this 'gap'?

Fig. 6.18: A thin parallelogram where the perpendicular from C falls outside AD — Fig. 6.18: A “thin” parallelogram where the perpendicular from $𝐶$ falls outside $𝐴 𝐷$

Solution

When a parallelogram is very slanted or "thin," dropping a perpendicular straight down from top vertex $𝐶$ lands outside the bottom base $𝐴 𝐷$ . To fix this gap and still convert the shape into a rectangle:

Extend the baseline $𝐴 𝐷$ in both directions.
Mark a point $𝐷'$ on segment $𝐴 𝐷$ close to vertex $𝐷$ , and mark a corresponding point $𝐴'$ on the extended line outside the shape such that distance $𝐴' 𝐴 = 𝐷' 𝐷$ .
If you slice off the right-angled triangle $△ 𝐶 𝐷 𝐷'$ from the right side and move it to the left position $△ 𝐵 𝐴 𝐴'$ , the two triangles match perfectly because they are congruent ( $△ 𝐶 𝐷 𝐷' ≅ △ 𝐵 𝐴 𝐴'$ ).
This cutting and shifting creates a new parallelogram $𝐴' 𝐵 𝐶 𝐷'$ that is less slanted than the original, without changing the total area.
If the new parallelogram is still too slanted, repeat this slicing and shifting step as many times as needed until the perpendicular falls inside the top side, allowing you to form a standard rectangle.

Section 6.7: Area of a Parallelogram — Think and Reflect (Page 131, Bottom)

Question

The area of a rectangle can be found when we know the lengths of its sides. Is the same true for a parallelogram? That is, can we find the area of a parallelogram when we know the lengths of its sides? Why or why not? (Hint: What happens to the area of a parallelogram if we decrease or increase the angle between the adjacent sides while keeping the lengths fixed?)

Solution

No, we cannot find the area of a parallelogram knowing only the lengths of its sides.

Reasoning: Imagine four wooden strips joined by hinges at the corners to form a parallelogram. If you keep the side lengths completely fixed, you can still push or pull the frame to tilt it at different angles.
As you lean the parallelogram over and flatten it, the perpendicular height ( $ℎ$ ) gets smaller and smaller, approaching zero. Because the area is calculated as $b a s e \times h e i g h t$ , the area shrinks toward zero as the height shrinks, even though the side lengths never change. Therefore, to find the area, you must know either the perpendicular height or the angle between the sides.

Section 6.8: Area of a Triangle — Think and Reflect (Page 132)

Question

What would we do if angle $𝐸 𝐹 𝐺$ is obtuse and the triangle were shaped like triangle $𝐸 𝐹 𝐺$ in Fig. 6.20B? Please work out the answer to this question.

Fig. 6.20A–B: Area of a triangle (acute and obtuse)

Solution

In Fig. 6.20B, triangle $△ 𝐸 𝐹 𝐺$ has an obtuse (wide) angle at vertex $𝐹$ , which causes top vertex $𝐸$ to lean far outside the base segment $𝐹 𝐺$ . We can still prove that its area is $12 \times b a s e \times h e i g h t$ using a rectangle:

Extend the baseline $𝐹 𝐺$ to the left and drop a perpendicular line from top vertex $𝐸$ down to meet the extended baseline at point $𝐻$ . This vertical line is the height of the triangle, $𝐸 𝐻 = ℎ$ .
Draw a complete bounding rectangle $𝐻 𝐼 𝐽 𝐺$ around the entire shape, with total base length $𝐻 𝐺 = 𝑏 1 + 𝑏 2$ and height $ℎ$ . Here, $𝑏 1$ is the outside segment $𝐻 𝐹$ , and $𝑏 2$ is the triangle's actual base $𝐹 𝐺$ .
The large right-angled triangle $△ 𝐸 𝐻 𝐺$ takes up exactly half of the large rectangle:

A r e a (△ 𝐸 𝐻 𝐺) = 12 (𝑏 1 + 𝑏 2) ℎ

The empty outside right-angled triangle $△ 𝐸 𝐻 𝐹$ takes up half of the smaller left section:

A r e a (△ 𝐸 𝐻 𝐹) = 12 𝑏 1 ℎ

Subtracting the outside area from the large triangle leaves the area of our obtuse triangle $△ 𝐸 𝐹 𝐺$ :

A r e a (△ 𝐸 𝐹 𝐺) = 12 (𝑏 1 + 𝑏 2) ℎ - 12 𝑏 1 ℎ = 12 𝑏 1 ℎ + 12 𝑏 2 ℎ - 12 𝑏 1 ℎ = 𝟏 𝟐 𝐛 𝟐 𝐡

Thus, the standard formula $12 \times b a s e \times h e i g h t$ holds universally true for obtuse triangles as well.

Section 6.8: Area of a Triangle — Think and Reflect (Page 133)

Question

Since $△ 𝐴 𝐵 𝐷$ and $△ 𝐴 𝐶 𝐷$ have equal area, you may wonder—Can we divide $△ 𝐴 𝐵 𝐷$ using straight cuts into two or more pieces that we can then rearrange to exactly cover $△ 𝐴 𝐶 𝐷$ ? What do you think? Is it possible?

Solution

Yes, it is absolutely possible! In geometry, a fundamental rule called the Bolyai-Gerwien Theorem proves that any two polygons (straight-sided flat shapes) that have the exact same area can always be sliced into a finite number of triangular pieces and reassembled to form one another. Since median $𝐴 𝐷$ divides the main triangle into two smaller triangles ( $△ 𝐴 𝐵 𝐷$ and $△ 𝐴 𝐶 𝐷$ ) of equal area, you can always cut one of them with straight lines and jigsaw-puzzle the pieces together to perfectly cover the other.

Section 6.8: Area of a Triangle — Think and Reflect (Page 134, Top)

Question

Suppose we are given two polygons $𝑃$ and $𝑄$ with equal area. Will it always be possible to divide one of them using straight cuts into two or more pieces and then rearrange the pieces to exactly cover the other polygon? Try this out for familiar shapes, e.g.,

A square and non-square rectangle with equal area,
Two triangles with different shapes but equal area,
A triangle and a square with equal area.

Formulate a conjecture of your own about this.

Solution

1. Square and rectangle of equal area: Yes. By stepping through Baudhāyana's ancient squaring method (shown in Section 6.9), you can geometrically cut any rectangle and rearrange its sections into a square of identical area.
2. Two differently shaped triangles of equal area: Yes. You can slice each triangle through its height to form a rectangle of half the height. Since both rectangles will have equal area, you can convert one into the other, and thus reassemble the second triangle.
3. A triangle and a square of equal area: Yes. By cutting the triangle horizontally at half its height, you can hinge the top piece down to form a rectangle, and then cut that rectangle to form a square.
My Conjecture: "Any two simple two-dimensional polygons having the same area can be dissected into a finite number of polygonal pieces and reassembled into each other." (As mentioned above, this conjecture is mathematically proven and known as the Bolyai-Gerwien Theorem).

Section 6.8: Area of a Triangle — Think and Reflect (Page 134, Middle)

Question

Think of various rectangles with perimeter $40 u n i t s$ (the sides do not have to be integers).

How many such rectangles are there?
Among them, is there one whose area is the largest? What are its dimensions?
Among all these rectangles, is there one whose area is the smallest? What are its dimensions? Do either of these answers come as a surprise to you?

Solution

Number of rectangles: There are infinitely many such rectangles. If length is $𝑙$ and width is $𝑤$ , the perimeter formula gives $2 (𝑙 + 𝑤) = 40$ , which simplifies to $𝑙 + 𝑤 = 20$ . Any pair of positive decimal numbers that add up to 20 forms a valid rectangle (e.g., $10$ and $10$ , $15$ and $5$ , $19.5$ and $0.5$ ).
Largest area: Yes, there is a maximum. The area is expressed as $𝐴 = 𝑙 \times 𝑤 = 𝑙 (20 - 𝑙) = 20 𝑙 - 𝑙 2$ . This mathematical expression reaches its absolute peak when the two dimensions are equal ( $𝑙 = 𝑤 = 10$ ). Therefore, a square of dimensions $10 u n i t s \times 10 u n i t s$ gives the largest possible area, which is $100 s q . u n i t s$ .
Smallest area: No, there is no smallest rectangle with an area greater than zero. If you make the length extremely close to $20$ (say, $19.999$ ) and the width extremely close to $0$ (say, $0.001$ ), the perimeter is still $40$ , but the area shrinks to $0.01999 s q . u n i t s$ . You can keep making the width thinner without ever hitting an exact minimum.

Surprise factor: It often surprises people that shapes with the exact same boundary length (40 units) can enclose vastly different amounts of space—ranging from virtually empty ( $0$ area) all the way up to $100 s q . u n i t s$ !

Section 6.8.1: Worked Examples (Heron's Formula Redone Step-by-Step)

Example 3 (Pages 134–135)

Use Heron's formula to find the area of an equilateral triangle with side $𝑎 u n i t s$ , and check your answer against the standard 'half base times height' formula.

Solution

Find semi-perimeter ( $𝑠$ ): Since all three sides equal $𝑎$ , the perimeter is $𝑎 + 𝑎 + 𝑎 = 3 𝑎$ . The semi-perimeter is:

𝑠 = 3 𝑎 2

Apply Heron's formula:

A r e a = \sqrt 𝑠 (𝑠 - 𝑎) (𝑠 - 𝑏) (𝑠 - 𝑐) = \sqrt 3 𝑎 2 (3 𝑎 2 - 𝑎) (3 𝑎 2 - 𝑎) (3 𝑎 2 - 𝑎)

A r e a = \sqrt 3 𝑎 2 \times 𝑎 2 \times 𝑎 2 \times 𝑎 2 = \sqrt 3 𝑎 4 16

Taking the square root of the top and bottom:

A r e a = \sqrt 𝟑 𝟒 𝐚 𝟐 s q . u n i t s

Check using $12 \times b a s e \times h e i g h t$ :

In an equilateral triangle, dropping a vertical height $ℎ$ splits the bottom base $𝑎$ exactly in half ( $𝑎 2$ ).
By the Baudhāyana-Pythagoras theorem on the right-angled half-triangle:

ℎ 2 + (𝑎 2) 2 = 𝑎 2 ⟹ ℎ 2 = 𝑎 2 - 𝑎 2 4 = 3 𝑎 2 4 ⟹ ℎ = \sqrt 3 2 𝑎

Calculating area:

A r e a = 12 \times b a s e \times h e i g h t = 12 \times 𝑎 \times \sqrt 3 2 𝑎 = \sqrt 𝟑 𝟒 𝐚 𝟐 s q . u n i t s

Both formulas yield the exact same result!

Example 4 (Page 135)

Use Heron's formula to find the area of an isosceles triangle with equal sides $𝑎 u n i t s$ and base $2 𝑏 u n i t s$ , and check your answer against the standard formula.

Solution

Find semi-perimeter ( $𝑠$ ): The three side lengths are $𝑎$ , $𝑎$ , and $2 𝑏$ .

𝑠 = 𝑎 + 𝑎 + 2 𝑏 2 = 2 𝑎 + 2 𝑏 2 = 𝑎 + 𝑏

Apply Heron's formula:

A r e a = \sqrt (𝑎 + 𝑏) (𝑎 + 𝑏 - 𝑎) (𝑎 + 𝑏 - 𝑎) (𝑎 + 𝑏 - 2 𝑏)

A r e a = \sqrt (𝑎 + 𝑏) (𝑏) (𝑏) (𝑎 - 𝑏)

Rearrange the terms to group $(𝑎 + 𝑏)$ and $(𝑎 - 𝑏)$ :

A r e a = \sqrt 𝑏 2 (𝑎 + 𝑏) (𝑎 - 𝑏)

Using the difference-of-two-squares identity $(𝑎 + 𝑏) (𝑎 - 𝑏) = 𝑎 2 - 𝑏 2$ :

A r e a = \sqrt 𝑏 2 (𝑎 2 - 𝑏 2) = 𝐛 \sqrt 𝐚 𝟐 - 𝐛 𝟐 s q . u n i t s

Check using $12 \times b a s e \times h e i g h t$ :

In an isosceles triangle, the perpendicular height $ℎ$ splits the base ( $2 𝑏$ ) into two equal segments of length $𝑏$ .
By the Baudhāyana-Pythagoras theorem:

ℎ 2 + 𝑏 2 = 𝑎 2 ⟹ ℎ 2 = 𝑎 2 - 𝑏 2 ⟹ ℎ = \sqrt 𝑎 2 - 𝑏 2

Calculating area:

A r e a = 12 \times b a s e \times h e i g h t = 12 \times 2 𝑏 \times \sqrt 𝑎 2 - 𝑏 2 = 𝐛 \sqrt 𝐚 𝟐 - 𝐛 𝟐 s q . u n i t s

The results match perfectly!

Example 5 (Page 136)

Use Heron's formula to find the area of a triangle with sides $3 u n i t s$ , $4 u n i t s$ and $5 u n i t s$ , and check your answer against the standard formula.

Solution

Find semi-perimeter ( $𝑠$ ):

𝑠 = 3 + 4 + 5 2 = 122 = 6 u n i t s

Apply Heron's formula:

A r e a = \sqrt 6 (6 - 3) (6 - 4) (6 - 5) = \sqrt 6 \times 3 \times 2 \times 1 = \sqrt 36 = 𝟔 s q . u n i t s

Check using $12 \times b a s e \times h e i g h t$ :

Notice the relationship between the sides: $32 + 42 = 9 + 16 = 25 = 52$ .
By the converse of the Baudhāyana-Pythagoras theorem, this is a right-angled triangle with hypotenuse $5$ . Thus, the two perpendicular legs can serve as base ( $3$ ) and height ( $4$ ).
Calculating area:

A r e a = 12 \times 3 \times 4 = 122 = 𝟔 s q . u n i t s

Both methods confirm the area is 6.

Section 6.8.1: Brahmagupta's Formula — Worked Examples (Pages 138–139)

Note: Brahmagupta's formula states that for any cyclic 4-gon—a quadrilateral whose four corners touch a circle—with sides

𝑎, 𝑏, 𝑐, 𝑑

and semi-perimeter

𝑠 = 𝑎 + 𝑏 + 𝑐 + 𝑑 2

, the area is

\sqrt (𝑠 - 𝑎) (𝑠 - 𝑏) (𝑠 - 𝑐) (𝑠 - 𝑑)

Example 6 (Page 138)

Verify Brahmagupta's formula for the case of a rectangle.

Solution

Cyclic property: Every rectangle can be inscribed in a circle (its corners touch a circumcircle whose diameter is the rectangle's diagonal), so Brahmagupta's formula applies.
Define sides and find semi-perimeter ( $𝑠$ ): In a rectangle, opposite sides are equal, so the four sides are $𝑎, 𝑏, 𝑎, 𝑏$ .

𝑠 = 𝑎 + 𝑏 + 𝑎 + 𝑏 2 = 2 𝑎 + 2 𝑏 2 = 𝑎 + 𝑏

Apply Brahmagupta's formula:

A r e a = \sqrt (𝑠 - 𝑎) (𝑠 - 𝑏) (𝑠 - 𝑐) (𝑠 - 𝑑) = \sqrt (𝑎 + 𝑏 - 𝑎) (𝑎 + 𝑏 - 𝑏) (𝑎 + 𝑏 - 𝑎) (𝑎 + 𝑏 - 𝑏)

A r e a = \sqrt (𝑏) (𝑎) (𝑏) (𝑎) = \sqrt 𝑎 2 𝑏 2 = 𝐚 𝐛 s q . u n i t s

This matches the standard formula for the area of a rectangle ( $l e n g t h \times w i d t h = 𝑎 𝑏$ )!

Example 7 (Pages 138–139)

Verify Brahmagupta's formula for the case of an isosceles trapezium.

Solution

Define sides: An isosceles trapezium has two parallel bases and two equal slanted sides. Let the top base be $2 𝑎$ , the bottom base be $2 𝑏$ , and the two equal side legs be $𝑐$ and $𝑐$ .
Find semi-perimeter ( $𝑠$ ):

𝑠 = 2 𝑎 + 2 𝑏 + 𝑐 + 𝑐 2 = 2 𝑎 + 2 𝑏 + 2 𝑐 2 = 𝑎 + 𝑏 + 𝑐

Calculate the four bracketed terms for Brahmagupta's formula:

$𝑠 - t o p b a s e = (𝑎 + 𝑏 + 𝑐) - 2 𝑎 = 𝑏 + 𝑐 - 𝑎$
$𝑠 - b o t t o m b a s e = (𝑎 + 𝑏 + 𝑐) - 2 𝑏 = 𝑎 + 𝑐 - 𝑏$
$𝑠 - l e g 1 = (𝑎 + 𝑏 + 𝑐) - 𝑐 = 𝑎 + 𝑏$
$𝑠 - l e g 2 = (𝑎 + 𝑏 + 𝑐) - 𝑐 = 𝑎 + 𝑏$

Substitute into Brahmagupta's formula:

A r e a = \sqrt (𝑎 + 𝑏) (𝑎 + 𝑏) (𝑏 + 𝑐 - 𝑎) (𝑎 + 𝑐 - 𝑏) = (𝑎 + 𝑏) \sqrt (𝑐 + 𝑏 - 𝑎) (𝑐 - (𝑏 - 𝑎))

Using the difference-of-two-squares identity on the term under the square root:

A r e a = (𝐚 + 𝐛) \sqrt 𝐜 𝟐 - (𝐛 - 𝐚) 𝟐

Check against standard trapezium geometry:

Drop vertical height lines ( $ℎ$ ) from the top two vertices ( $2 𝑎$ ) down to the bottom base ( $2 𝑏$ ). This leaves a remaining extra horizontal length of $(2 𝑏 - 2 𝑎)$ split equally between the two bottom outer corners: $2 𝑏 - 2 𝑎 2 = 𝑏 - 𝑎$ .
By the Baudhāyana-Pythagoras theorem on one outer right-angled triangle:

ℎ 2 + (𝑏 - 𝑎) 2 = 𝑐 2 ⟹ ℎ = \sqrt 𝑐 2 - (𝑏 - 𝑎) 2

Standard trapezium area formula is $12 (s u m o f p a r a l l e l s i d e s) \times h e i g h t$ :

A r e a = 12 (2 𝑎 + 2 𝑏) \times ℎ = (𝑎 + 𝑏) ℎ = (𝐚 + 𝐛) \sqrt 𝐜 𝟐 - (𝐛 - 𝐚) 𝟐

Both derivations arrive at the exact same formula!

Section 6.9: Squaring a Rectangle — Think and Reflect (Page 142)

Question

What procedure would you use to square a given triangle? Here, the task is to construct a square whose area is equal to the area of some given triangle. Think carefully. How would you proceed?

Solution

To geometrically construct a square that has the exact same area as a given triangle, we combine two major constructions in sequence:

Step 1: Convert the triangle into a rectangle of equal area.

Measure the base $𝑏$ and drop a perpendicular to find the height $ℎ$ of the given triangle.
Construct a rectangle that has the same base $𝑏$ , but only half the vertical height ( $ℎ 2$ ).
Since the rectangle's area is $𝑏 \times ℎ 2 = 12 𝑏 ℎ$ , it has the exact same area as the triangle.

Step 2: Convert the rectangle into a square.

Apply Baudhāyana's ancient squaring construction (detailed in Section 6.9) to this new rectangle. By drawing circular arcs and forming right-angled triangles with side segments $𝑎 + 𝑏 2$ and $𝑎 - 𝑏 2$ , you construct a square whose side length equals $\sqrt 12 𝑏 ℎ$ . This square matches the triangle's area perfectly.

Section 6.9: Exercise Set 6.2 (Pages 142–143)

Question 1

Find the area of triangle $𝐴 𝐷 𝐸$ in Fig. 6.31.

Fig. 6.31: Rectangle ABCD with triangle ADE

Solution

Analyze Fig. 6.31: The diagram shows a rectangle $𝐴 𝐵 𝐶 𝐷$ with total length $𝐷 𝐶 = 10 c m$ and vertical width $𝐵 𝐶 = 8 c m$ .
Identify triangle base and height: The triangle $△ 𝐴 𝐷 𝐸$ has its vertical base along the left side of the rectangle, so $b a s e 𝐴 𝐷 = 8 c m$ . The opposite vertex $𝐸$ rests on the right vertical side $𝐵 𝐶$ . The perpendicular height from vertex $𝐸$ across to the base line $𝐴 𝐷$ is the horizontal length of the rectangle, $h e i g h t = 10 c m$ .
Calculate Area:

A r e a (△ 𝐴 𝐷 𝐸) = 12 \times b a s e \times h e i g h t = 12 \times 8 \times 10 = 𝟒 𝟎 c m 𝟐

Question 2

The parallel sides of a trapezium are $40 c m$ and $20 c m$ . If its non-parallel sides are both equal, each being $26 c m$ , find the area of the trapezium.

Solution

Find horizontal corner segments: Because the non-parallel sides are equal ( $26 c m$ ), this is an isosceles trapezium. Drop vertical height lines ( $ℎ$ ) from the top base ( $20 c m$ ) down to the bottom base ( $40 c m$ ). The remaining bottom length is split equally between the two side triangles:

C o r n e r b a s e = 40 - 20 2 = 202 = 10 c m

Calculate height ( $ℎ$ ): Use the Baudhāyana-Pythagoras theorem on one side triangle (hypotenuse $= 26$ , base $= 10$ ):

ℎ 2 + 102 = 262 ⟹ ℎ 2 + 100 = 676 ⟹ ℎ 2 = 576

ℎ = \sqrt 576 = 24 c m

Calculate Area:

A r e a = 12 (b a s e 1 + b a s e 2) \times ℎ = 12 (40 + 20) \times 24 = 30 \times 24 = 𝟕 𝟐 𝟎 c m 𝟐

Question 3

Find the area of a triangle, given that its sides are $8 c m$ and $11 c m$ long, and its perimeter is $32 c m$ .

Solution

Find the third side ( $𝑐$ ):

𝑐 = P e r i m e t e r - (𝑎 + 𝑏) = 32 - (8 + 11) = 32 - 19 = 13 c m

Find semi-perimeter ( $𝑠$ ):

𝑠 = 322 = 16 c m

Apply Heron's formula:

A r e a = \sqrt 16 (16 - 8) (16 - 11) (16 - 13) = \sqrt 16 \times 8 \times 5 \times 3

Break into simpler numbers to find the square root:

A r e a = \sqrt 16 \times (4 \times 2) \times 15 = \sqrt 64 \times 30 = 𝟖 \sqrt 𝟑 𝟎 c m 𝟐 (o r \approx 𝟒 𝟑 . 𝟖 𝟐 c m 𝟐)

Question 4

The sides of a triangular plot are in the ratio $3 : 5 : 7$ ; its perimeter is $300 m$ . Find its area.

Solution

Find actual side lengths: Let the ratio multiplier be $𝑥$ .

3 𝑥 + 5 𝑥 + 7 𝑥 = 300 ⟹ 15 𝑥 = 300 ⟹ 𝑥 = 20

Side $𝑎 = 3 \times 20 = 60 m$
Side $𝑏 = 5 \times 20 = 100 m$
Side $𝑐 = 7 \times 20 = 140 m$

Find semi-perimeter ( $𝑠$ ):

𝑠 = 3002 = 150 m

Apply Heron's formula:

A r e a = \sqrt 150 (150 - 60) (150 - 100) (150 - 140) = \sqrt 150 \times 90 \times 50 \times 10

A r e a = \sqrt 6, 750, 000 = \sqrt 675 \times 10, 000 = 100 \sqrt 225 \times 3 = 100 \times 15 \sqrt 3 = 𝟏 𝟓 𝟎 𝟎 \sqrt 𝟑 m 𝟐 (o r \approx 𝟐 𝟓 𝟗 𝟖 . 𝟎 𝟖 m 𝟐)

Question 5

One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area $128 c m 2$ , find the length of the shorter diagonal.

Solution

Set up diagonal relationship: Let the shorter diagonal be $𝑑$ . The longer diagonal is $2 𝑑$ .
Use rhombus area formula: The area of any rhombus is half the product of its diagonals:

A r e a = 12 \times 𝑑 1 \times 𝑑 2

128 = 12 \times (2 𝑑) \times 𝑑 ⟹ 128 = 𝑑 2

Solve for shorter diagonal ( $𝑑$ ):

𝑑 = \sqrt 128 = \sqrt 64 \times 2 = 𝟖 \sqrt 𝟐 c m (o r \approx 𝟏 𝟏 . 𝟑 𝟏 c m)

Question 6

$𝐴 𝐵 𝐶 𝐷$ is a parallelogram. $𝑃$ and $𝑄$ are any two points on side $𝐴 𝐵$ . What can you say about the ratio $a r e a (△ 𝑃 𝐶 𝐷) : a r e a (△ 𝑄 𝐶 𝐷)$ ?

(Note: In the original PDF text, an OCR/printing artifact renders the triangle symbol $△$ as the letter 'A', printing "area (APCD): area (AQCD)". This is verified by Question 9 below, which prints "AABP" for $△ 𝐴 𝐵 𝑃$ .)

Solution

Both triangles have the exact same area, so the ratio is $1 : 1$ .

Reasoning: Both triangles $△ 𝑃 𝐶 𝐷$ and $△ 𝑄 𝐶 𝐷$ share the exact same bottom base segment, $𝐶 𝐷$ . The opposite vertices $𝑃$ and $𝑄$ both lie on side $𝐴 𝐵$ , which is parallel to base $𝐶 𝐷$ in a parallelogram. Because the vertical distance between two parallel lines is constant everywhere, both triangles have the exact same perpendicular height ( $ℎ$ ).
Since $A r e a = 12 \times b a s e 𝐶 𝐷 \times h e i g h t ℎ$ , their areas are identical.

Question 7

$𝑂$ is any point on the diagonal $𝑃 𝑅$ of a parallelogram $𝑃 𝑄 𝑅 𝑆$ . Prove that the areas of triangles $𝑃 𝑆 𝑂$ and $𝑃 𝑄 𝑂$ are equal.

Solution

Identify the common base: Both triangles $△ 𝑃 𝑆 𝑂$ and $△ 𝑃 𝑄 𝑂$ share the exact same line segment $𝑃 𝑂$ as their base along the diagonal $𝑃 𝑅$ .
Compare heights: In a parallelogram $𝑃 𝑄 𝑅 𝑆$ , the diagonal $𝑃 𝑅$ divides the shape into two identical, congruent halves ( $△ 𝑃 𝑆 𝑅 ≅ △ 𝑃 𝑄 𝑅$ ). Because the two halves are identical reflections across the diagonal line, the perpendicular distance (altitude $ℎ$ ) dropped from outer vertex $𝑆$ down to diagonal line $𝑃 𝑅$ is exactly equal to the perpendicular altitude dropped from outer vertex $𝑄$ down to line $𝑃 𝑅$ .
Calculate area:

$A r e a (△ 𝑃 𝑆 𝑂) = 12 \times b a s e 𝑃 𝑂 \times a l t i t u d e ℎ$
$A r e a (△ 𝑃 𝑄 𝑂) = 12 \times b a s e 𝑃 𝑂 \times a l t i t u d e ℎ$

Conclusion: Since both the base and the perpendicular height are equal, $A r e a (△ 𝑃 𝑆 𝑂) = A r e a (△ 𝑃 𝑄 𝑂)$ .

Question 8

If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a '4-gon') are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon.

Solution

Set up the shape: Let the 4-gon be $𝐴 𝐵 𝐶 𝐷$ , and let the midpoints of sides $𝐴 𝐵, 𝐵 𝐶, 𝐶 𝐷, 𝐷 𝐴$ be $𝐸, 𝐹, 𝐺, 𝐻$ respectively. Connecting them forms inner parallelogram $𝐸 𝐹 𝐺 𝐻$ .
Draw diagonal $𝐴 𝐶$ : Consider triangle $△ 𝐴 𝐵 𝐶$ . Points $𝐸$ and $𝐹$ are midpoints of sides $𝐴 𝐵$ and $𝐵 𝐶$ . By standard geometry, a triangle formed by joining two side midpoints has $12$ the base and $12$ the height of the parent triangle. Thus:

A r e a (△ 𝐸 𝐵 𝐹) = 14 A r e a (△ 𝐴 𝐵 𝐶)

Apply to opposite corner: Similarly, in top triangle $△ 𝐴 𝐷 𝐶$ , midpoints $𝐻$ and $𝐺$ give:

A r e a (△ 𝐺 𝐷 𝐻) = 14 A r e a (△ 𝐴 𝐷 𝐶)

Sum the first pair of corners: Adding these two corner areas together:

A r e a (△ 𝐸 𝐵 𝐹) + A r e a (△ 𝐺 𝐷 𝐻) = 14 [A r e a (△ 𝐴 𝐵 𝐶) + A r e a (△ 𝐴 𝐷 𝐶)] = 14 A r e a (𝐴 𝐵 𝐶 𝐷)

Sum the second pair of corners: Draw the other diagonal $𝐵 𝐷$ . Using the exact same midpoint logic on the remaining two corners ( $△ 𝐻 𝐴 𝐸$ and $△ 𝐹 𝐶 𝐺$ ):

A r e a (△ 𝐻 𝐴 𝐸) + A r e a (△ 𝐹 𝐶 𝐺) = 14 A r e a (𝐴 𝐵 𝐶 𝐷)

Combine all four outer corners:

T o t a l O u t e r C o r n e r A r e a = 14 A r e a (𝐴 𝐵 𝐶 𝐷) + 14 A r e a (𝐴 𝐵 𝐶 𝐷) = 12 A r e a (𝐴 𝐵 𝐶 𝐷)

Conclusion: The inner parallelogram $𝐸 𝐹 𝐺 𝐻$ is simply the total 4-gon area minus the four outer corners:

A r e a (𝐸 𝐹 𝐺 𝐻) = A r e a (𝐴 𝐵 𝐶 𝐷) - 12 A r e a (𝐴 𝐵 𝐶 𝐷) = 𝟏 𝟐 A r e a (𝐀 𝐁 𝐂 𝐃)

Question 9

In $△ 𝐴 𝐵 𝐶$ , the midpoint of $𝐵 𝐶$ is $𝐷$ (Fig. 6.32). Median $𝐴 𝐷$ is drawn. $𝑃$ is any point on $𝐴 𝐷$ . Show that $a r e a (△ 𝐴 𝐵 𝑃) = a r e a (△ 𝐴 𝐶 𝑃)$ .

Solution

Use the triangle median theorem: As proven in Section 6.8, a median divides any triangle into two smaller triangles of equal area.
Analyze the large triangle: In $△ 𝐴 𝐵 𝐶$ , line $𝐴 𝐷$ is a median because $𝐷$ is the midpoint of $𝐵 𝐶$ . Therefore:

A r e a (△ 𝐴 𝐵 𝐷) = A r e a (△ 𝐴 𝐶 𝐷)

Analyze the small lower triangle: In $△ 𝑃 𝐵 𝐶$ , line $𝑃 𝐷$ is also a median to base $𝐵 𝐶$ . Therefore:

A r e a (△ 𝑃 𝐵 𝐷) = A r e a (△ 𝑃 𝐶 𝐷)

Subtract the lower section from the large section:

A r e a (△ 𝐴 𝐵 𝑃) = A r e a (△ 𝐴 𝐵 𝐷) - A r e a (△ 𝑃 𝐵 𝐷)

A r e a (△ 𝐴 𝐶 𝑃) = A r e a (△ 𝐴 𝐶 𝐷) - A r e a (△ 𝑃 𝐶 𝐷)

Since we are subtracting equal quantities from equal quantities, $A r e a (△ 𝐴 𝐵 𝑃) = A r e a (△ 𝐴 𝐶 𝑃)$ .

Question 10

Given a square $𝐴 𝐵 𝐶 𝐷$ , let $𝑃$ be a point within it. Join $𝑃 𝐴, 𝑃 𝐵, 𝑃 𝐶, 𝑃 𝐷$ (Fig. 6.33). What is the ratio of the areas of the red region ( $△ 𝑃 𝐴 𝐵$ and $△ 𝑃 𝐶 𝐷$ ) and the green region ( $△ 𝑃 𝐵 𝐶$ and $△ 𝑃 𝐷 𝐴$ )?

Fig. 6.33: Square ABCD with interior point P (red vs green)

Solution

The ratio of the areas of the red region to the green region is $1 : 1$ (they have equal area).

Step-by-Step Proof:

Let the side length of square $𝐴 𝐵 𝐶 𝐷$ be $𝑠$ . The total area of the square is $𝑠 2$ .
For the red region, triangle $△ 𝑃 𝐴 𝐵$ has horizontal base $𝐴 𝐵 = 𝑠$ . Let its vertical height from point $𝑃$ up to side $𝐴 𝐵$ be $ℎ 1$ . Its area is $12 𝑠 ℎ 1$ .
Opposite red triangle $△ 𝑃 𝐶 𝐷$ has horizontal base $𝐶 𝐷 = 𝑠$ . Let its vertical height from point $𝑃$ down to side $𝐶 𝐷$ be $ℎ 2$ . Its area is $12 𝑠 ℎ 2$ .
Because sides $𝐴 𝐵$ and $𝐶 𝐷$ are parallel outer walls of the square, the sum of the two interior vertical heights equals the full side length of the square ( $ℎ 1 + ℎ 2 = 𝑠$ ).
Add the two red triangles together:

A r e a (R e d R e g i o n) = 12 𝑠 ℎ 1 + 12 𝑠 ℎ 2 = 12 𝑠 (ℎ 1 + ℎ 2) = 12 𝑠 (𝑠) = 𝟏 𝟐 𝐬 𝟐

Since the red region occupies exactly half the area of the square, the green region must occupy the remaining half ( $12 𝑠 2$ ). Therefore, both regions are equal in area ( $1 : 1$ ).

Question 11

In $△ 𝐴 𝐵 𝐶$ , $𝐷$ is the midpoint of $𝐴 𝐵$ . $𝑃$ is any point on $𝐵 𝐶$ , and $𝑄$ is a point on $𝐴 𝐵$ such that $𝐶 𝑄 ∥ 𝑃 𝐷$ . $𝑃 𝑄$ is joined (Fig. 6.34). Prove that $A r e a (△ 𝐵 𝑃 𝑄) = 12 A r e a (△ 𝐴 𝐵 𝐶)$ .

Solution

Start with median $𝐶 𝐷$ : Because $𝐷$ is the midpoint of side $𝐴 𝐵$ , line $𝐶 𝐷$ is a median of $△ 𝐴 𝐵 𝐶$ . This means $△ 𝐵 𝐶 𝐷$ takes up exactly half the area of the large triangle:

A r e a (△ 𝐵 𝐶 𝐷) = 12 A r e a (△ 𝐴 𝐵 𝐶)

Split $△ 𝐵 𝐶 𝐷$ into two pieces: Looking at line $𝑃 𝐷$ , we can write:

A r e a (△ 𝐵 𝐶 𝐷) = A r e a (△ 𝐵 𝑃 𝐷) + A r e a (△ 𝑃 𝐶 𝐷)

Compare parallel-line triangles $△ 𝑃 𝐶 𝐷$ and $△ 𝑃 𝑄 𝐷$ :

Both triangles share the exact same base segment, $𝑃 𝐷$ .
Their opposite vertices, $𝐶$ and $𝑄$ , lie on line $𝐶 𝑄$ , which is explicitly given as parallel to base $𝑃 𝐷$ ( $𝐶 𝑄 ∥ 𝑃 𝐷$ ).
Because triangles between the same parallel lines and sharing the same base have equal area:

A r e a (△ 𝑃 𝐶 𝐷) = A r e a (△ 𝑃 𝑄 𝐷)

Assemble target triangle $△ 𝐵 𝑃 𝑄$ : Looking at the diagram, triangle $△ 𝐵 𝑃 𝑄$ is made of two sections:

A r e a (△ 𝐵 𝑃 𝑄) = A r e a (△ 𝐵 𝑃 𝐷) + A r e a (△ 𝑃 𝑄 𝐷)

Substitute our equal area from Step 3 into this equation:

A r e a (△ 𝐵 𝑃 𝑄) = A r e a (△ 𝐵 𝑃 𝐷) + A r e a (△ 𝑃 𝐶 𝐷) = A r e a (△ 𝐵 𝐶 𝐷)

Conclusion: Since we established in Step 1 that $A r e a (△ 𝐵 𝐶 𝐷) = 12 A r e a (△ 𝐴 𝐵 𝐶)$ , we have proven that $A r e a (△ 𝐵 𝑃 𝑄) = 12 A r e a (△ 𝐴 𝐵 𝐶)$ .

Section 6.10 (Area of a Circle, Sectors, Segments, and Exercise Set 6.3)

Section 6.10: Area of a Circle — Think and Reflect (Page 144)

Question

Why were human beings so fond of using circular shapes? Was this only for practical reasons, or could there have been other reasons too? What kinds of uses have human beings found for the circular shape?

Solution

Reasons for fondness (Practical & Aesthetic): Human beings favored circular shapes for both practical and symbolic reasons. Practically, a circle encloses the maximum possible area for a given perimeter (boundary length), making it the most efficient shape for building storage structures or fencing cattle. Symbolically, circles represent wholeness, symmetry, the sun, the moon, and the natural cycles of seasons and time.
Kinds of uses: Throughout history, humans have used circles in:

Architecture and Storage: Cylindrical grain towers, huts, wells, and circular garden plots.
Tools and Technology: Wheels for transport, potter's wheels, gears, pulleys, and coins.
Timekeeping and Astronomy: Sun dials, clocks, and astrolabes for charting celestial bodies.

Section 6.10: Exercise Set 6.3 (Pages 148)

Note: Unless stated otherwise, use the approximation

227

for

𝜋

Question 1

Find the area of a sector of a circle with radius $7 c m$ if the angle of the sector is $60 \circ$ .

Solution

Identify the formula: The area of a sector with central angle $𝜃$ and radius $𝑟$ is:

A r e a = 𝜋 𝑟 2 \times 𝜃 360 \circ

Substitute given values: Here, $𝑟 = 7 c m$ and $𝜃 = 60 \circ$ .

A r e a = 227 \times 72 \times 60360

Simplify step-by-step:

A r e a = 227 \times 49 \times 16 = 22 \times 7 \times 16 = 1546 = 773 c m 2

Converting to a mixed number or decimal:

A r e a = 𝟐 𝟓 𝟐 𝟑 c m 𝟐 (o r \approx 𝟐 𝟓 . 𝟔 𝟕 c m 𝟐)

Question 2

Find the area of a quadrant of a circle whose circumference is $44 c m$ .

Solution

Find the radius ( $𝑟$ ): We first use the circumference formula $𝐶 = 2 𝜋 𝑟$ :

44 = 2 \times 227 \times 𝑟

44 = 447 \times 𝑟 ⟹ 𝑟 = 44 \times 7 44 = 7 c m

Calculate area of a quadrant: A quadrant is one-fourth ( $14$ ) of a complete circle ( $𝜃 = 90 \circ$ ):

A r e a o f q u a d r a n t = 14 \times 𝜋 𝑟 2 = 14 \times 227 \times 72

Simplify step-by-step:

A r e a = 14 \times 227 \times 49 = 14 \times 154 = 772 c m 2 = 𝟑 𝟖 . 𝟓 c m 𝟐

Question 3

The length of the minute hand of a clock is $7 c m$ . Find the area swept by the minute hand in $10 m i n u t e s$ .

Solution

Find the angle swept in 10 minutes: A clock face is a full circle of $360 \circ$ , representing $60 m i n u t e s$ . In $1 m i n u t e$ , the minute hand turns $360 \circ 60 = 6 \circ$ . Therefore, in $10 m i n u t e s$ , the central angle $𝜃$ is:

𝜃 = 10 \times 6 \circ = 60 \circ

Apply the sector area formula: The length of the minute hand acts as the radius, $𝑟 = 7 c m$ .

A r e a = 𝜋 𝑟 2 \times 60360 = 227 \times 72 \times 16

Simplify step-by-step:

A r e a = 154 \times 16 = 773 c m 2 = 𝟐 𝟓 𝟐 𝟑 c m 𝟐 (o r \approx 𝟐 𝟓 . 𝟔 𝟕 c m 𝟐)

Question 4

A chord of a circle of radius $10 c m$ subtends $90 \circ$ at the centre. Find the area of the corresponding: (i) minor sector (that subtends $90 \circ$ at the centre), and (ii) major sector (that subtends $270 \circ$ at the centre). (Use $𝜋 \approx 3.14$ .)

Solution

(i) Area of the minor sector ( $𝜃 = 90 \circ$ ):

Substitute $𝑟 = 10 c m$ , $𝜃 = 90 \circ$ , and $𝜋 = 3.14$ :

A r e a (m i n o r) = 3.14 \times 102 \times 90360

Simplify:

A r e a (m i n o r) = 3.14 \times 100 \times 14 = 3144 = 𝟕 𝟖 . 𝟓 c m 𝟐

(ii) Area of the major sector ( $𝜃 = 270 \circ$ ):

The central angle for the major sector is $360 \circ - 90 \circ = 270 \circ$ .

A r e a (m a j o r) = 3.14 \times 102 \times 270360

Simplify:

A r e a (m a j o r) = 314 \times 34 = 3 \times 78.5 = 𝟐 𝟑 𝟓 . 𝟓 c m 𝟐

(Check: Total area of the circle is $314 c m 2$ . Summing the two sectors gives $78.5 + 235.5 = 314 c m 2$ , confirming accuracy.)

Question 5

A chord of a circle of radius $15 c m$ subtends an angle of $60 \circ$ at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use $𝜋 \approx 3.14$ and $\sqrt 3 \approx 1.73$ .)

Solution

Find the area of the minor sector ( $𝜃 = 60 \circ$ ):

A r e a (m i n o r s e c t o r) = 3.14 \times 152 \times 60360 = 3.14 \times 225 \times 16 = 706.5 6 = 117.75 c m 2

Find the area of the triangle formed by the chord and radii:

Since the central angle is $60 \circ$ and the two radius sides are equal ( $15 c m$ each), the triangle is an equilateral triangle with side length $𝑎 = 15 c m$ .
Using the equilateral triangle area formula from Section 6.8.1:

A r e a (t r i a n g l e) = \sqrt 3 4 \times 𝑎 2 = 1.73 4 \times 152 = 1.73 \times 225 4 = 389.25 4 = 97.3125 c m 2

Calculate the area of the minor segment:

A segment is the region bounded by an arc and its chord. Subtract the triangle's area from the sector's area:

A r e a (m i n o r s e g m e n t) = 117.75 - 97.3125 = 𝟐 𝟎 . 𝟒 𝟑 𝟕 𝟓 c m 𝟐 (o r \approx 𝟐 𝟎 . 𝟒 𝟒 c m 𝟐)

Calculate the area of the major segment:

First, find the total area of the circle:

A r e a (c i r c l e) = 𝜋 𝑟 2 = 3.14 \times 225 = 706.5 c m 2

Subtract the minor segment from the total circle area:

A r e a (m a j o r s e g m e n t) = 706.5 - 20.4375 = 𝟔 𝟖 𝟔 . 𝟎 𝟔 𝟐 𝟓 c m 𝟐 (o r \approx 𝟔 𝟖 𝟔 . 𝟎 𝟔 c m 𝟐)

Question 6

A car has two wipers which do not overlap. Each wiper has a blade of length $28 c m$ and sweeps through an angle of $120 \circ$ . Find the total area cleaned at each sweep of the blades.

Solution

Find the area swept by ONE wiper: Each wiper sweeps out a circular sector with radius $𝑟 = 28 c m$ and angle $𝜃 = 120 \circ$ .

A r e a (o n e w i p e r) = 227 \times 282 \times 120360 = 227 \times 784 \times 13

A r e a (o n e w i p e r) = 22 \times 112 \times 13 = 24643 c m 2

Find the total area cleaned by TWO wipers: Multiply by 2:

T o t a l A r e a = 2 \times 24643 = 49283 c m 2 = 𝟏 𝟔 𝟒 𝟐 𝟐 𝟑 c m 𝟐 (o r \approx 𝟏 𝟔 𝟒 𝟐 . 𝟔 𝟕 c m 𝟐)

*Question 7

A chord of a circle of radius $𝑟$ subtends an angle of $60 \circ$ at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to $𝜋 𝑟 2 (16 - \sqrt 3 4)$ .

Solution

Express the minor sector area: For central angle $𝜃 = 60 \circ$ , the sector occupies $60360 = 16$ of the circle:

A r e a (s e c t o r) = 16 𝜋 𝑟 2

Express the triangle area: Because the central angle is $60 \circ$ and the two adjacent sides are equal radii ( $𝑟$ ), the triangle is equilateral with side length $𝑟$ . Its area is:

A r e a (t r i a n g l e) = \sqrt 3 4 𝑟 2

Subtract triangle area from sector area:

A r e a (m i n o r s e g m e n t) = 16 𝜋 𝑟 2 - \sqrt 3 4 𝑟 2 = 𝐫 𝟐 (𝜋 𝟔 - \sqrt 𝟑 𝟒)

FLAGGING AN AMBIGUITY/TYPO IN THE TEXTBOOK: The textbook PDF prints this formula as $𝜋 𝑟 2 (16 - \sqrt 3 4)$ . However, if you multiply $𝜋 𝑟 2$ by the second term inside the bracket, you get an extra $𝜋$ attached to the triangle's area ( $\sqrt 3 4 𝜋 𝑟 2$ ), which is mathematically incorrect. The correct algebraic factorization is $𝑟 2 (𝜋 6 - \sqrt 3 4)$ , or alternatively, $𝜋 𝑟 2 (16 - \sqrt 3 4 𝜋)$ .

*Question 8

An equilateral triangle is inscribed in a circle of radius $𝑟$ . Show that the ratio of the area of the triangle to the area of the circle is equal to $3 \sqrt 3 4 𝜋 \approx 0.413$ .

Solution

Find side length $𝑎$ of the inscribed triangle in terms of $𝑟$ : Let the circle's center be $𝑂$ . Connecting $𝑂$ to the vertices of the equilateral triangle divides it into 3 identical central isosceles triangles, each with central angle $360 \circ 3 = 120 \circ$ . Dropping an altitude from $𝑂$ splits one side $𝑎$ into half-lengths $𝑎 2$ and forms a $30 \circ - 60 \circ - 90 \circ$ right triangle.

s i n (60 \circ) = o p p o s i t e h y p o t e n u s e = 𝑎 / 2 𝑟 ⟹ \sqrt 3 2 = 𝑎 2 𝑟 ⟹ 𝑎 = 𝑟 \sqrt 3

Calculate area of the equilateral triangle:

A r e a (t r i a n g l e) = \sqrt 3 4 𝑎 2 = \sqrt 3 4 (𝑟 \sqrt 3) 2 = \sqrt 3 4 (3 𝑟 2) = 3 \sqrt 3 4 𝑟 2

Form the ratio with the circle's area ( $𝜋 𝑟 2$ ):

R a t i o = A r e a (t r i a n g l e) A r e a (c i r c l e) = 3 \sqrt 3 4 𝑟 2 𝜋 𝑟 2 = 𝟑 \sqrt 𝟑 𝟒 𝜋

Evaluate decimal approximation:

3 \times 1.73205 4 \times 3.14159 = 5.19615 12.56637 \approx 𝟎 . 𝟒 𝟏 𝟑

*Question 9

A square is inscribed in a circle of radius $𝑟$ . Show that the ratio of the area of the square to the area of the circle is equal to $2 𝜋 \approx 0.637$ .

Solution

Find side length $𝑎$ of the inscribed square: The diagonal of an inscribed square passes straight through the center of the circle, making the diagonal equal to the circle's diameter ( $𝑑 = 2 𝑟$ ).

By the Baudhāyana-Pythagoras theorem on two adjacent sides of the square:

𝑎 2 + 𝑎 2 = 𝑑 2 ⟹ 2 𝑎 2 = (2 𝑟) 2 = 4 𝑟 2 ⟹ 𝑎 2 = 2 𝑟 2

Calculate area of the square: Since the area of a square is side squared ( $𝑎 2$ ), we have:

A r e a (s q u a r e) = 2 𝑟 2

Form the ratio with the circle's area ( $𝜋 𝑟 2$ ):

R a t i o = A r e a (s q u a r e) A r e a (c i r c l e) = 2 𝑟 2 𝜋 𝑟 2 = 𝟐 𝜋

Evaluate decimal approximation:

2 3.14159 \approx 𝟎 . 𝟔 𝟑 𝟕

*Question 10

A hexagon is inscribed in a circle of radius $𝑟$ . Show that the ratio of the area of the hexagon to the area of the circle is equal to $3 \sqrt 3 2 𝜋 \approx 0.827$ . Can you see why the answer is exactly twice the answer to Question 8?

Solution

Calculate area of an inscribed regular hexagon: As established in Section 6.2, a regular hexagon inscribed in a circle of radius $𝑟$ has a side length $𝑎$ exactly equal to the radius ( $𝑎 = 𝑟$ ). It is composed of 6 identical equilateral triangles of side $𝑟$ .

A r e a (h e x a g o n) = 6 \times (\sqrt 3 4 𝑟 2) = 6 \sqrt 3 4 𝑟 2 = 3 \sqrt 3 2 𝑟 2

Form the ratio with the circle's area ( $𝜋 𝑟 2$ ):

R a t i o = A r e a (h e x a g o n) A r e a (c i r c l e) = 3 \sqrt 3 2 𝑟 2 𝜋 𝑟 2 = 𝟑 \sqrt 𝟑 𝟐 𝜋

Evaluate decimal approximation:

3 \times 1.73205 2 \times 3.14159 = 5.19615 6.28318 \approx 𝟎 . 𝟖 𝟐 𝟕

Why is this exactly twice the answer to Question 8?

In Question 8, the inscribed equilateral triangle had an area of $3 \sqrt 3 4 𝑟 2$ . Notice that the inscribed regular hexagon has an area of $3 \sqrt 3 2 𝑟 2$ , which is exactly double ( $\times 2$ ) the area of the inscribed triangle. Because both shapes are being divided by the exact same circle area ( $𝜋 𝑟 2$ ), the resulting ratio for the hexagon ( $3 \sqrt 3 2 𝜋$ ) is naturally exactly twice the ratio for the triangle ( $3 \sqrt 3 4 𝜋$ ).

End-of-Chapter Exercises (Questions 1 to 14)

Note: Unless stated otherwise, use the approximation

227

for

𝜋

Question 1 (Page 149)

Question

Identities in algebra can sometimes be shown as area relationships. For example: The figure shown (Fig. 6.41) corresponds to the identity $(𝑎 + 𝑏) 2 = 𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ . Do you see how? Draw corresponding figures to the identities $(𝑎 + 𝑏) (𝑎 - 𝑏) = 𝑎 2 - 𝑏 2$ and $(𝑎 + 𝑏 + 𝑐) 2 = 𝑎 2 + 𝑏 2 + 𝑐 2 + 2 𝑎 𝑏 + 2 𝑏 𝑐 + 2 𝑐 𝑎$ .

Fig. 6.41: Area model of

(𝑎 + 𝑏) 2 = 𝑎 2 + 2 𝑎 𝑏 + 𝑏 2

Solution

How Fig. 6.41 shows $(𝑎 + 𝑏) 2 = 𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ :

The large outer square has a total side length of $(𝑎 + 𝑏)$ , meaning its total area is $(𝑎 + 𝑏) 2$ .
The vertical and horizontal dividing lines split this large square into 4 smaller rooms:
One square in the top-left with side $𝑎$ , giving area $𝑎 2$ .
One square in the bottom-right with side $𝑏$ , giving area $𝑏 2$ .
Two rectangles (top-right and bottom-left), each with length $𝑎$ and width $𝑏$ , giving area $𝑎 𝑏$ each.
Adding the 4 rooms together equals the total area: $𝑎 2 + 𝑎 𝑏 + 𝑎 𝑏 + 𝑏 2 = 𝑎 2 + 2 𝑎 𝑏 + 𝑏 2$ .

Figure description for $(𝑎 + 𝑏) (𝑎 - 𝑏) = 𝑎 2 - 𝑏 2$ :

To draw: Start by drawing a large square of side $𝑎$ (total area $𝑎 2$ ). In the top-right corner, draw a small square of side $𝑏$ (area $𝑏 2$ ) and shade/cut it out. The remaining L-shaped region represents the area $𝑎 2 - 𝑏 2$ .
How it proves the identity: Make a single straight horizontal cut to divide the remaining L-shape into two rectangles: a vertical rectangle on the left of dimensions $𝑎 \times (𝑎 - 𝑏)$ , and a bottom horizontal rectangle of dimensions $𝑏 \times (𝑎 - 𝑏)$ . If you take the bottom rectangle, rotate it, and attach it to the top of the vertical rectangle, the two pieces form one long single rectangle with length $(𝑎 + 𝑏)$ and width $(𝑎 - 𝑏)$ . Thus, the area $(𝑎 + 𝑏) (𝑎 - 𝑏)$ equals the L-shaped area $𝑎 2 - 𝑏 2$ .

Figure description for $(𝑎 + 𝑏 + 𝑐) 2 = 𝑎 2 + 𝑏 2 + 𝑐 2 + 2 𝑎 𝑏 + 2 𝑏 𝑐 + 2 𝑐 𝑎$ :

To draw: Draw a large square with side length $(𝑎 + 𝑏 + 𝑐)$ . Draw two vertical lines and two horizontal lines across the square to divide each side into three segments of lengths $𝑎$ , $𝑏$ , and $𝑐$ .
How it proves the identity: These grid lines slice the large square into 9 smaller regions:
3 squares along the main diagonal with side lengths $𝑎$ , $𝑏$ , and $𝑐$ , giving areas $𝑎 2, 𝑏 2,$ and $𝑐 2$ .
2 identical rectangles with dimensions $𝑎 \times 𝑏$ (area $𝑎 𝑏$ each).
2 identical rectangles with dimensions $𝑏 \times 𝑐$ (area $𝑏 𝑐$ each).
2 identical rectangles with dimensions $𝑐 \times 𝑎$ (area $𝑐 𝑎$ each).
Summing all 9 rooms gives the total area: $𝑎 2 + 𝑏 2 + 𝑐 2 + 2 𝑎 𝑏 + 2 𝑏 𝑐 + 2 𝑐 𝑎$ .

Question 2 (Page 149)

Question

An isosceles triangle has perimeter $40 c m$ ; the equal sides are $15 c m$ each. Find the area of the triangle.

Solution

Find the base ( $𝑐$ ):

𝑐 = P e r i m e t e r - (𝑎 + 𝑏) = 40 - (15 + 15) = 40 - 30 = 10 c m

Find semi-perimeter ( $𝑠$ ):

𝑠 = 402 = 20 c m

Apply Heron's formula:

A r e a = \sqrt 20 (20 - 15) (20 - 15) (20 - 10) = \sqrt 20 \times 5 \times 5 \times 10

A r e a = \sqrt 5000 = \sqrt 2500 \times 2 = 𝟓 𝟎 \sqrt 𝟐 c m 𝟐 (o r \approx 𝟕 𝟎 . 𝟕 𝟏 c m 𝟐)

Question 3 (Page 149)

Question

An isosceles triangle has base $10 c m$ , and its area is $60 c m 2$ . What are the lengths of the equal sides?

Solution

Find the perpendicular height ( $ℎ$ ): We use the basic area formula $A r e a = 12 \times b a s e \times h e i g h t$ :

60 = 12 \times 10 \times ℎ ⟹ 60 = 5 ℎ ⟹ ℎ = 605 = 12 c m

Find the equal side length ( $𝑎$ ): In an isosceles triangle, the perpendicular altitude drops straight down to cut the bottom base exactly in half ( $102 = 5 c m$ ).

By the Baudhāyana-Pythagoras theorem on one half-triangle:

𝑎 2 = ℎ 2 + (h a l f - b a s e) 2 = 122 + 52 = 144 + 25 = 169

𝑎 = \sqrt 169 = 𝟏 𝟑 c m

The lengths of the equal sides are $13 c m$ each.

Question 4 (Page 149)

Question

The area of a right-angled triangle is $54 s q . c m$ . One of its legs has length $12 c m$ . Find its perimeter.

Solution

Find the second perpendicular leg ( $𝑏$ ): In a right-angled triangle, the two legs act as the base and height.

A r e a = 12 \times l e g 1 \times l e g 2 ⟹ 54 = 12 \times 12 \times 𝑏 ⟹ 54 = 6 𝑏 ⟹ 𝑏 = 9 c m

Find the hypotenuse ( $𝑐$ ): By the Baudhāyana-Pythagoras theorem:

𝑐 2 = 122 + 92 = 144 + 81 = 225 ⟹ 𝑐 = \sqrt 225 = 15 c m

Calculate perimeter:

P e r i m e t e r = 12 + 9 + 15 = 𝟑 𝟔 c m

Question 5 (Page 149)

Question

The sides of a triangle are in the ratio $2 : 3 : 4$ , and its perimeter is $45 c m$ . Find its area.

Solution

Find side lengths: Let the ratio multiplier be $𝑥$ .

2 𝑥 + 3 𝑥 + 4 𝑥 = 45 ⟹ 9 𝑥 = 45 ⟹ 𝑥 = 5

Side $𝑎 = 2 \times 5 = 10 c m$
Side $𝑏 = 3 \times 5 = 15 c m$
Side $𝑐 = 4 \times 5 = 20 c m$

Find semi-perimeter ( $𝑠$ ):

𝑠 = 452 = 22.5 c m (o r 452)

Apply Heron's formula using fractions for exact calculation:

A r e a = \sqrt 452 (452 - 10) (452 - 15) (452 - 20) = \sqrt 452 \times 252 \times 152 \times 52

A r e a = \sqrt 45 \times 25 \times 15 \times 5 16 = 14 \sqrt (9 \times 5) \times 25 \times (3 \times 5) \times 5 = 14 \sqrt 9 \times 25 \times 25 \times 15

A r e a = 3 \times 5 \times 5 4 \sqrt 15 = 𝟕 𝟓 𝟒 \sqrt 𝟏 𝟓 c m 𝟐 (o r 𝟏 𝟖 . 𝟕 𝟓 \sqrt 𝟏 𝟓 c m 𝟐 \approx 𝟕 𝟐 . 𝟔 𝟐 c m 𝟐)

Question 6 (Page 149)

Question

The sides of a triangle have lengths $7 c m$ , $24 c m$ , $25 c m$ . Find the area of the triangle in two different ways.

Solution

Method 1 (Half base times height):

Test the sides using the Baudhāyana-Pythagoras theorem: $72 + 242 = 49 + 576 = 625$ . Since $252 = 625$ , the relation $𝑎 2 + 𝑏 2 = 𝑐 2$ holds true.
By the converse of the theorem, this is a right-angled triangle with hypotenuse $25 c m$ . The perpendicular legs ( $7 c m$ and $24 c m$ ) serve as the base and height.
$A r e a = 12 \times 7 \times 24 = 7 \times 12 = 𝟖 𝟒 c m 𝟐$ .

Method 2 (Heron's Formula):

Semi-perimeter $𝑠 = 7 + 24 + 25 2 = 562 = 28 c m$ .
Apply formula:

A r e a = \sqrt 28 (28 - 7) (28 - 24) (28 - 25) = \sqrt 28 \times 21 \times 4 \times 3

Factor into squares:

A r e a = \sqrt (7 \times 4) \times (7 \times 3) \times 4 \times 3 = \sqrt 72 \times 42 \times 32 = 7 \times 4 \times 3 = 𝟖 𝟒 c m 𝟐

. Both methods confirm the area is

84 c m 2

Question 7 (Page 149)

Question

If the wheel of a bicycle has a diameter of $60 c m$ , find how far a cyclist will have travelled after the wheel has rotated 100 times.

Solution

Find the distance travelled in ONE rotation (Circumference):

𝐶 = 𝜋 𝑑 = 227 \times 60 = 13207 c m

Multiply by 100 rotations:

T o t a l D i s t a n c e = 100 \times 13207 = 1320007 c m \approx 18857.14 c m

Convert to meters (divide by 100):

T o t a l D i s t a n c e = 𝟏 𝟖 𝟖 . 𝟓 𝟕 m (o r 18847 m)

Question 8 (Page 150)

Question

Find the area of a quadrant of a circle whose circumference is $66 c m$ .

Solution

Find the radius ( $𝑟$ ):

2 𝜋 𝑟 = 66 ⟹ 2 \times 227 \times 𝑟 = 66 ⟹ 447 𝑟 = 66

𝑟 = 66 \times 7 44 = 3 \times 7 2 = 212 c m (o r 10.5 c m)

Calculate area of a quadrant (one-fourth of a circle):

A r e a = 14 \times 𝜋 𝑟 2 = 14 \times 227 \times (212) 2 = 14 \times 227 \times 4414

Simplify step-by-step:

A r e a = 14 \times 22 \times 634 = 138616 = 6938 c m 2 = 𝟖 𝟔 . 𝟔 𝟐 𝟓 c m 𝟐 (o r 8658 c m 2)

Question 9 (Page 150)

Question

The wheel of a car has an outer radius of $28 c m$ . Calculate how far the car travels after one complete turn of the wheel, and how many times the wheel turns during a journey of $1 k m$ .

Solution

Distance in ONE complete turn: This equals the circumference of the wheel.

𝐶 = 2 𝜋 𝑟 = 2 \times 227 \times 28 = 2 \times 22 \times 4 = 𝟏 𝟕 𝟔 c m (o r 1.76 m)

Number of turns in $1 k m$ : Convert $1 k m$ to centimeters ( $1 k m = 100, 000 c m$ ).

N u m b e r o f t u r n s = T o t a l D i s t a n c e C i r c u m f e r e n c e = 100, 000 176 = 1250022 = 625011 = 568.1818 . . . \approx 𝟓 𝟔 𝟖 t u r n s (o r e x a c t l y 568211 t u r n s)

*Question 10 (Page 150)

Question

Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?

Solution

Yes, they must be congruent to each other.

Step-by-Step Mathematical Proof:

Let Rectangle 1 have dimensions $𝑙 1$ and $𝑤 1$ , and Rectangle 2 have dimensions $𝑙 2$ and $𝑤 2$ .
Because their perimeters are equal, their half-perimeters are equal: $𝑙 1 + 𝑤 1 = 𝑙 2 + 𝑤 2 = 𝑆$ .
Because their areas are equal: $𝑙 1 \times 𝑤 1 = 𝑙 2 \times 𝑤 2 = 𝑃$ .
In algebra, if you know the sum ( $𝑆$ ) and product ( $𝑃$ ) of two numbers, those two numbers are the only two solutions (roots) to the quadratic equation $𝑥 2 - 𝑆 𝑥 + 𝑃 = 0$ .
Since a quadratic equation has at most one unique pair of roots, the dimensions ${𝑙 1, 𝑤 1}$ must be identical to the dimensions ${𝑙 2, 𝑤 2}$ . Therefore, both rectangles have the exact same length and width, making them congruent!

Question 11 (Page 150)

Question

You know that the area of a parallelogram is base $\times$ height. Using this and the figure (Fig. 6.42), show that the area of a trapezium is half the sum of the parallel sides $\times$ height, i.e., $12 (𝑎 + 𝑏) ℎ$ .

Fig. 6.42: Trapezium with parallel sides a, b and height h

Solution

Analyze Fig. 6.42: The diagram shows a trapezium with top parallel side $𝑎$ , bottom parallel side $𝑏$ , and height $ℎ$ . An internal line is drawn parallel to the left slanted side. This line splits the bottom base $𝑏$ into two parts: a left section of length $𝑎$ and a remaining right section of length $(𝑏 - 𝑎)$ .
Identify the two internal shapes:

The left shaded region is a parallelogram with base $𝑎$ and height $ℎ$ .
The right shaded region is a triangle with base $(𝑏 - 𝑎)$ and height $ℎ$ .

Sum the two areas:

A r e a o f T r a p e z i u m = A r e a o f P a r a l l e l o g r a m + A r e a o f T r i a n g l e

A r e a = (𝑎 \times ℎ) + [12 \times (𝑏 - 𝑎) \times ℎ] = 𝑎 ℎ + 12 𝑏 ℎ - 12 𝑎 ℎ

A r e a = 12 𝑎 ℎ + 12 𝑏 ℎ = 𝟏 𝟐 (𝐚 + 𝐛) 𝐡

Question 12 (Page 150)

Question

By dividing a trapezium into two triangles show that its area is half the sum of the parallel sides multiplied by the height (the same formula as the one given above).

Solution

Draw a dividing diagonal: Draw a single straight line connecting opposite corners (for example, from the top-left vertex to the bottom-right vertex). This diagonal slices the trapezium into two separate triangles.
Find the area of the lower triangle: Its bottom base is the lower parallel side ( $𝑏$ ). The altitude dropped to this base is the vertical height of the trapezium ( $ℎ$ ).

A r e a (l o w e r t r i a n g l e) = 12 𝑏 ℎ

Find the area of the upper triangle: Its top base is the upper parallel side ( $𝑎$ ). The altitude dropped from the bottom vertex up to the extension of this top base is also the vertical height of the trapezium ( $ℎ$ ).

A r e a (u p p e r t r i a n g l e) = 12 𝑎 ℎ

Sum the two triangles:

T o t a l A r e a = 12 𝑏 ℎ + 12 𝑎 ℎ = 𝟏 𝟐 (𝐚 + 𝐛) 𝐡

Question 13 (Page 150)

Question

Show how we can use two identical copies of a trapezium to make a parallelogram. How will this give us the formula for the area of a trapezium?

Solution

How to make the parallelogram: Take two identical (congruent) paper trapeziums, each having top side $𝑎$ , bottom side $𝑏$ , and height $ℎ$ . Rotate the second trapezium upside down ( $180 \circ$ ) and push its slanted side tightly against the matching slanted side of the first trapezium.
Identify dimensions of the new shape: The top edge of the combined shape is now $𝑎 + 𝑏$ , and the bottom edge is also $𝑏 + 𝑎$ . Because opposite sides are equal and parallel, this combined shape is a single large parallelogram with total base length $(𝑎 + 𝑏)$ and height $ℎ$ .
Derive the trapezium formula:

The area of the combined parallelogram is $b a s e \times h e i g h t = (𝑎 + 𝑏) ℎ$ .
Since this large parallelogram is built from exactly two identical trapeziums, the area of one single trapezium must be exactly half of the total:

A r e a o f o n e t r a p e z i u m = 𝟏 𝟐 (𝐚 + 𝐛) 𝐡

Question 14 (Page 150)

Question

Show that the area of a kite is half the product of its diagonals. Show this: (i) using algebra, and (ii) using geometry.

Solution

(i) Using Algebra:

In a kite, the two diagonals cross at right angles ( $90 \circ$ ), and the main axis diagonal ( $𝑑 1$ ) acts as a line of symmetry, cutting the cross diagonal ( $𝑑 2$ ) into two equal halves of length $𝑑 2 2$ .
The main diagonal $𝑑 1$ splits the kite into two identical triangles (an upper triangle and a lower triangle), both sharing the base $𝑑 1$ .
The height of the upper triangle is $𝑑 2 2$ , and the height of the lower triangle is also $𝑑 2 2$ .
Add the areas of the two triangles:

A r e a o f K i t e = (12 \times 𝑑 1 \times 𝑑 2 2) + (12 \times 𝑑 1 \times 𝑑 2 2) = 14 𝑑 1 𝑑 2 + 14 𝑑 1 𝑑 2 = 𝟏 𝟐 𝐝 𝟏 𝐝 𝟐

(ii) Using Geometry (Visual Enclosing Rectangle):

Draw a tightly fitting box (a bounding rectangle) around the entire kite by drawing horizontal and vertical lines through its 4 outer vertices.
The length of this outer rectangle equals diagonal $𝑑 1$ , and its width equals diagonal $𝑑 2$ . Thus, the total area of the enclosing box is $𝑑 1 \times 𝑑 2$ .
The internal cross-diagonals of the kite divide the bounding box into 4 smaller rectangular window panes.
Notice that within every single window pane, the slanted outer boundary of the kite acts as a diagonal line, slicing that small pane into two identical right-angled triangles—one inside the kite, and one outside in the corner.
Because exactly half of every window pane lies inside the kite, the total area of the kite is exactly half the area of the enclosing box: $12 (𝑑 1 \times 𝑑 2)$ .

End-of-Chapter Exercises (Questions 15 to 27)

Note: Unless stated otherwise, use the approximation

227

for

𝜋

Question 15 (Pages 150–151)

Question

Three problems about fitting congruent shapes together: (i) Rectangle $𝐴 𝐵 𝐶 𝐷$ has sides $𝑎, 𝑏$ , and rectangle $𝑃 𝑄 𝑅 𝑆$ has sides $2 𝑎, 2 𝑏$ . Show that $𝑃 𝑄 𝑅 𝑆$ has 4 times the area of $𝐴 𝐵 𝐶 𝐷$ . Does this mean that 4 copies of rectangle $𝐴 𝐵 𝐶 𝐷$ will fit into rectangle $𝑃 𝑄 𝑅 𝑆$ ? Check and see! (ii) $△ 𝐴 𝐵 𝐶$ has sides $𝑎, 𝑏, 𝑐$ , and $△ 𝑃 𝑄 𝑅$ has sides $2 𝑎, 2 𝑏, 2 𝑐$ . Show that $△ 𝑃 𝑄 𝑅$ has 4 times the area of $△ 𝐴 𝐵 𝐶$ . Does this mean that 4 copies of $△ 𝐴 𝐵 𝐶$ will fit into $△ 𝑃 𝑄 𝑅$ ? Check and see! (iii) $△ 𝐴 𝐵 𝐶$ has sides $𝑎, 𝑏, 𝑐$ , and $△ 𝑃 𝑄 𝑅$ has sides $3 𝑎, 3 𝑏, 3 𝑐$ . Show that $△ 𝑃 𝑄 𝑅$ has 9 times the area of $△ 𝐴 𝐵 𝐶$ . Does this mean that 9 copies of $△ 𝐴 𝐵 𝐶$ will fit into $△ 𝑃 𝑄 𝑅$ ? Check and see!

(Note on accuracy: In the original PDF text, an OCR/printing artifact renders the triangle symbol $△$ as the letter 'A', printing "AABC" and "APQR". This is corrected below to $△ 𝐴 𝐵 𝐶$ and $△ 𝑃 𝑄 𝑅$ .)

Solution

(i) Rectangles with doubled sides:

$A r e a (𝐴 𝐵 𝐶 𝐷) = l e n g t h \times w i d t h = 𝑎 𝑏$ .
$A r e a (𝑃 𝑄 𝑅 𝑆) = (2 𝑎) \times (2 𝑏) = 4 𝑎 𝑏 = 4 \times A r e a (𝐴 𝐵 𝐶 𝐷)$ .
Yes, 4 copies fit perfectly. If you draw one horizontal line and one vertical line directly through the middle of rectangle $𝑃 𝑄 𝑅 𝑆$ , you divide it into a $2 \times 2$ grid of 4 identical rectangles, each with dimensions $𝑎 \times 𝑏$ .

(ii) Triangles with doubled sides:

For $△ 𝐴 𝐵 𝐶$ , the semi-perimeter is $𝑠 = 𝑎 + 𝑏 + 𝑐 2$ , and by Heron's formula, $A r e a = \sqrt 𝑠 (𝑠 - 𝑎) (𝑠 - 𝑏) (𝑠 - 𝑐)$ .
For $△ 𝑃 𝑄 𝑅$ with sides $2 𝑎, 2 𝑏, 2 𝑐$ , the perimeter is doubled, so its semi-perimeter is $𝑆 = 2 𝑠$ .
Applying Heron's formula to $△ 𝑃 𝑄 𝑅$ :

A r e a (△ 𝑃 𝑄 𝑅) = \sqrt 2 𝑠 (2 𝑠 - 2 𝑎) (2 𝑠 - 2 𝑏) (2 𝑠 - 2 𝑐) = \sqrt 16 𝑠 (𝑠 - 𝑎) (𝑠 - 𝑏) (𝑠 - 𝑐) = 4 \times A r e a (△ 𝐴 𝐵 𝐶)

Yes, 4 copies fit perfectly. If you mark the midpoints of the three sides of $△ 𝑃 𝑄 𝑅$ and connect them with straight line segments, you slice the large triangle into 4 identical, congruent triangles of side lengths $𝑎, 𝑏, 𝑐$ .

(iii) Triangles with tripled sides:

For $△ 𝑃 𝑄 𝑅$ with sides $3 𝑎, 3 𝑏, 3 𝑐$ , its semi-perimeter is $𝑆 = 3 𝑠$ .
Applying Heron's formula:

A r e a (△ 𝑃 𝑄 𝑅) = \sqrt 3 𝑠 (3 𝑠 - 3 𝑎) (3 𝑠 - 3 𝑏) (3 𝑠 - 3 𝑐) = \sqrt 81 𝑠 (𝑠 - 𝑎) (𝑠 - 𝑏) (𝑠 - 𝑐) = 9 \times A r e a (△ 𝐴 𝐵 𝐶)

Yes, 9 copies fit perfectly. If you divide each side of $△ 𝑃 𝑄 𝑅$ into 3 equal segments and draw grid lines parallel to the outer sides, you create a $3 \times 3$ triangular grid containing exactly 9 congruent copies of $△ 𝐴 𝐵 𝐶$ .

*Question 16 (Page 151)

Question

What fraction of the triangle in Fig. 6.43 is shaded? What fraction of the square in Fig. 6.44 is shaded?

(Note: In the textbook layout, this starred question corresponds to the two geometric fraction puzzles printed above Question 17.)

Fig. 6.43: What fraction of the triangle is shaded?

Fig. 6.44: What fraction of the square is shaded?

Solution

Fraction of the triangle shaded (Fig. 6.43):

In Fig. 6.43 the equal tick marks show that the left side is bisected (divided into $2$ equal parts) and the right side is trisected (divided into $3$ equal parts).
Label the large triangle $𝑇 𝐿 𝑅$ , with $𝑇$ the top vertex, $𝐿$ the bottom-left vertex and $𝑅$ the right-hand tip. Let $𝑀$ be the midpoint of $𝑇 𝐿$ , and let $𝑃$ and $𝑄$ be the points that divide $𝑇 𝑅$ in the ratios $1 : 2$ and $2 : 1$ respectively. The shaded region is the quadrilateral $𝑀 𝑃 𝑄 𝐿$ .
Area ratios in a triangle depend only on the fractional distances along the sides from a common vertex (they are preserved by affine transformations). Place $𝑇$ at $(0, 1)$ , $𝐿$ at $(0, 0)$ and $𝑅$ at $(1, 0)$ . Then $𝑀 = (0, 12), 𝑃 = (13, 23), 𝑄 = (23, 13) .$
The area of $△ 𝑇 𝐿 𝑅$ is $12$ . The shaded quadrilateral $𝑀 𝑃 𝑄 𝐿$ has vertices $(0, 12)$ , $(13, 23)$ , $(23, 13)$ , $(0, 0)$ . By the shoelace formula its area is $14$ .
Therefore $Shaded fraction=1/41/2=𝟏𝟐.$
(Equivalently: the unshaded top triangle $𝑇 𝑀 𝑃$ has base $13$ of $𝑇 𝑅$ and height $12$ of the large triangle, so area $13 \cdot 12 = 16$ of the whole. The unshaded bottom triangle $𝐿 𝑄 𝑅$ has base $13$ of $𝑇 𝑅$ measured from $𝑅$ and the full height of $△ 𝑇 𝐿 𝑅$ relative to side $𝑇 𝑅$ scaled by the remaining fraction, giving area $13$ of the whole. Unshaded total $= 16 + 13 = 12$ , so shaded $= 12$ .)

Fraction of the square shaded (Fig. 6.44):

In Fig. 6.44 each side of the square is bisected (equal tick marks). From every vertex, lines are drawn to the midpoints of the two sides that do not meet that vertex.
Place the square as $[0, 1] \times [0, 1]$ with midpoints $𝑀 𝑏 𝑜 𝑡 𝑡 𝑜 𝑚 = (12, 0)$ , $𝑀 𝑟 𝑖 𝑔 ℎ 𝑡 = (1, 12)$ , $𝑀 𝑡 𝑜 𝑝 = (12, 1)$ , $𝑀 𝑙 𝑒 𝑓 𝑡 = (0, 12)$ .
The eight lines intersect in a central diamond (a square rotated $45 \circ$ ) whose vertices are $(12, 14), (34, 12), (12, 34), (14, 12) .$
The diagonals of this diamond are both of length $12$ and are perpendicular, so its area is $12 \cdot 12 \cdot 12 = 18 .$
The large square has area $1$ , hence $Shaded fraction=𝟏𝟖.$

Question 17 (Page 151)

Question

What fraction of the rectangle in Fig. 6.45 is covered by the circles? What fraction of the rectangle in Fig. 6.46 is covered by the circles?

Fig. 6.45: Circles packed in a rectangle

Fig. 6.46: Circles packed in a rectangle

Solution

Fraction in Fig. 6.45 (3 circles in a row):

The rectangle is packed with 3 identical circles of radius $𝑟$ touching each other and the walls.
Height of the rectangle $= 2 𝑟$ ; length $= 6 𝑟$ .
$A r e a o f r e c t a n g l e = 6 𝑟 \times 2 𝑟 = 12 𝑟 2$ .
$A r e a o f 3 c i r c l e s = 3 𝜋 𝑟 2$ .
$F r a c t i o n = 3 𝜋 𝑟 2 12 𝑟 2 = 𝜋 4 = 𝟏 𝟏 𝟏 𝟒 (u s i n g 𝜋 = 227)$

Fraction in Fig. 6.46 (4 circles in a row):

The rectangle contains 4 identical circles of radius $𝑟$ in a single row.
Height $= 2 𝑟$ , length $= 8 𝑟$ .
$A r e a o f r e c t a n g l e = 16 𝑟 2$ ; $A r e a o f 4 c i r c l e s = 4 𝜋 𝑟 2$ .
$F r a c t i o n = 4 𝜋 𝑟 2 16 𝑟 2 = 𝜋 4 = 𝟏 𝟏 𝟏 𝟒$

Both arrangements (and any $𝑚 \times 𝑛$ grid of equal circles) cover the exact same fraction $𝜋 / 4$ of their rectangles!

Question 18 (Page 151)

Question

Use the above to make a conjecture about the area occupied by circles fitted into a rectangle in the manner shown. Test your conjecture for particular cases: 10 circles; 20 circles; 50 circles. Then prove your conjecture!

Solution

Conjecture: When identical circles are packed tightly into a grid inside a rectangle such that each circle touches its neighbors and the walls, the fraction of the rectangle's area covered by the circles is always a constant, $𝜋 4$ (or $1114$ ), regardless of how many circles are inside.
Testing particular cases:
10 circles ( $2 \times 5$ grid): As calculated in Question 17, ratio $= 10 𝜋 𝑟 2 40 𝑟 2 = 𝜋 4$ .
20 circles ( $4 \times 5$ grid): Rectangle area $= (4 \times 2 𝑟) (5 \times 2 𝑟) = 80 𝑟 2$ . Circle area $= 20 𝜋 𝑟 2$ . Ratio $= 20 𝜋 𝑟 2 80 𝑟 2 = 𝜋 4$ .
50 circles ( $5 \times 10$ grid): Rectangle area $= (5 \times 2 𝑟) (10 \times 2 𝑟) = 200 𝑟 2$ . Circle area $= 50 𝜋 𝑟 2$ . Ratio $= 50 𝜋 𝑟 2 200 𝑟 2 = 𝜋 4$ .
General Proof:

Let there be $𝑛$ total circles arranged in $𝑅$ rows and $𝐶$ columns ( $𝑛 = 𝑅 \times 𝐶$ ), where each circle has radius $𝑟$ and diameter $𝑑 = 2 𝑟$ .
The height of the bounding rectangle is $𝑅 \times 𝑑 = 2 𝑅 𝑟$ , and its width is $𝐶 \times 𝑑 = 2 𝐶 𝑟$ .
$T o t a l a r e a o f r e c t a n g l e = (2 𝑅 𝑟) \times (2 𝐶 𝑟) = 4 𝑅 𝐶 𝑟 2 = 4 𝑛 𝑟 2$ .
$T o t a l a r e a o f 𝑛 c i r c l e s = 𝑛 \times 𝜋 𝑟 2$ .
Dividing the area of the circles by the area of the rectangle:

F r a c t i o n = 𝑛 𝜋 𝑟 2 4 𝑛 𝑟 2 = 𝜋 𝟒

Because the number of circles ( $𝑛$ ) and radius ( $𝑟$ ) cancel out completely, the fraction is proven to be constant for any grid size.

*Question 19 (Page 151)

Question

The figure (Fig. 6.47) shows nine identical rectangles fitted together to make a large rectangle whose area is $72 c m 2$ . Find the perimeter of each small rectangle.

Fig. 6.47: Nine identical rectangles stacked together

Solution

Find the area of one small rectangle: Since 9 identical rectangles make a total area of $72 c m 2$ , the area of ONE small rectangle (with length $𝑙$ and width $𝑤$ ) is:

A r e a = 𝑙 \times 𝑤 = 729 = 8 c m 2

Relate length and width from Fig. 6.47: Looking at the diagram, the top block consists of 4 rectangles side-by-side with their long lengths horizontal ( $4 𝑙$ ), while the bottom block consists of 5 rectangles side-by-side with their short widths horizontal ( $5 𝑤$ ). Because top and bottom borders of the large rectangle must be equal in length:

4 𝑙 = 5 𝑤 ⟹ 𝑙 = 54 𝑤 = 1.25 𝑤

Solve for $𝑤$ and $𝑙$ : Substitute $𝑙 = 1.25 𝑤$ into the area equation:

(1.25 𝑤) \times 𝑤 = 8 ⟹ 1.25 𝑤 2 = 8 ⟹ 𝑤 2 = 8 1.25 = 6.4

𝑤 = \sqrt 6.4 = 4 \sqrt 10 5 \approx 2.53 c m

𝑙 = 1.25 \times (4 \sqrt 10 5) = \sqrt 10 \approx 3.16 c m

Calculate perimeter of one small rectangle:

P e r i m e t e r = 2 (𝑙 + 𝑤) = 2 (\sqrt 10 + 4 \sqrt 10 5) = 2 (9 \sqrt 10 5) = 𝟏 𝟖 \sqrt 𝟏 𝟎 𝟓 c m (o r \approx 𝟏 𝟏 . 𝟑 𝟖 c m)

*Question 20 (Page 152)

Question

In Fig. 6.48, lines are drawn from a vertex to the points of trisection of the opposite side. Show that the areas of the shaded blue triangle and the shaded red triangle are equal. Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle.

Fig. 6.48: Lines from a vertex to trisection points

Solution

Why the areas are equal: In $△ 𝐴 𝐵 𝐶$ , the bottom baseline is divided into 3 equal segments by the two trisection points. All three inner triangles (including the blue and red ones) share the exact same top vertex, which means they all share the same perpendicular altitude ( $ℎ$ ) dropped to the baseline. Since $A r e a = 12 \times b a s e \times ℎ$ , and their base segments are identical in length, $A r e a (B l u e) = A r e a (R e d)$ .
How to cut and rearrange:

Measure the vertical height $ℎ$ of the blue triangle and draw a straight horizontal line across it at exactly half its height ( $ℎ 2$ ).
Cut along this line to detach the top small triangle. Rotate this top piece $180 \circ$ and attach it to the slanted side of the bottom trapezoidal piece to form a rectangle of height $ℎ 2$ and width equal to the base.
Perform the exact same half-height horizontal cut on the red triangle to turn it into an identical rectangle ( $ℎ 2 \times b a s e$ ).
Because both rectangular assemblies have identical dimensions, you can directly overlay the puzzle pieces of the blue triangle to perfectly cover the red triangle!

*Question 21 (Page 152)

Question

The figure (Fig. 6.49) shows a quarter circle in a square. Its centre is at one vertex, and it passes through two adjacent vertices. There are two semicircles on two adjacent sides as diameters. They create the shaded regions A and B. Show that A and B have equal area.

Fig. 6.49: Quarter circle and two semicircles (regions A and B)

Solution

Let the side length of the square be $𝑠$ .
Area of the quarter circle: Since its radius is $𝑠$ , its area is:

A r e a (Q u a r t e r C i r c l e) = 14 𝜋 𝑠 2

Combined area of the two semicircles: Each semicircle is drawn on a square side of length $𝑠$ as its diameter (so radius is $𝑠 2$ ). The area of ONE semicircle is $12 𝜋 (𝑠 2) 2 = 18 𝜋 𝑠 2$ . The sum of TWO semicircles is:

S u m o f t w o s e m i c i r c l e s = 2 \times (18 𝜋 𝑠 2) = 14 𝜋 𝑠 2

Notice that $A r e a (Q u a r t e r C i r c l e) = S u m o f T w o S e m i c i r c l e s = 14 𝜋 𝑠 2$ . Both enclose the exact same amount of space.
Apply inclusion-exclusion: Both curves lie inside the square. Let the overlapping region where the two semicircles cross each other be called $𝑋$ . By algebraic identity:

S e m i c i r c l e 1 + S e m i c i r c l e 2 = T o t a l a r e a c o v e r e d b y s e m i c i r c l e s + O v e r l a p 𝑋

In Fig. 6.49, Region A represents the outer part of the semicircles extending outside the quarter circle, while Region B represents the inner gap of the quarter circle not covered by the semicircles. Because the total area of the semicircles equals the total area of the quarter circle, any extra outer area (A) must be exactly balanced by an equal unfilled inner gap (B). Therefore, $A r e a (𝐴) = A r e a (𝐵)$ .

*Question 22 (Page 152)

Question

In Fig. 6.50, four semicircles have been drawn within the given square whose side is $2 u n i t s$ . The centres of these semicircles are the midpoints of the sides. They create a 4-petalled flower (shown in blue). Find the perimeter and the area of this flower.

Fig. 6.50: Four semicircles forming a 4-petalled flower

Solution

Find the perimeter of the flower:

The flower has 4 petals, and each petal is bounded by 2 circular arcs, making 8 curved edges in total.
Notice that each of the 4 semicircles drawn on the sides of the square (diameter $𝑑 = 2$ , radius $𝑟 = 1$ ) provides exactly two of these petal edges.
Therefore, the total boundary of the flower is simply the combined arc lengths of all 4 semicircles:

P e r i m e t e r = 4 \times (𝜋 𝑟) = 4 \times 𝜋 (1) = 4 𝜋 u n i t s

Using $𝜋 \approx 227$ :

P e r i m e t e r = 4 \times 227 = 𝟖 𝟖 𝟕 u n i t s (o r \approx 𝟏 𝟐 . 𝟓 𝟕 u n i t s)

Find the area of the flower:

Let's find the area of the 4 unshaded (white) corner regions first.
Look at the top semicircle and bottom semicircle. Each has area $12 𝜋 (12) = 𝜋 2$ . Combined, they have area $𝜋$ .
When placed inside the square (total area $𝑠 2 = 22 = 4$ ), these two semicircles cover everything except the left and right white corner gaps!
Thus, $A r e a o f t w o w h i t e c o r n e r s = 4 - 𝜋$ .
By symmetry, the top and bottom white corners also have an area of $4 - 𝜋$ .
Adding all 4 white corners together: $T o t a l W h i t e A r e a = 2 (4 - 𝜋) = 8 - 2 𝜋$ .
Finally, subtract the white corners from the total square area to get the blue flower area:

A r e a o f f l o w e r = 4 - (8 - 2 𝜋) = 𝟐 𝜋 - 𝟒 s q . u n i t s

Using $𝜋 \approx 227$ :

A r e a = 2 (227) - 4 = 447 - 287 = 𝟏 𝟔 𝟕 s q . u n i t s (o r \approx 𝟐 . 𝟐 𝟗 s q . u n i t s)

*Question 23 (Page 152)

Question

In Fig. 6.51 we see two concentric circles with a common centre $𝑂$ . A chord $𝐵 𝐶$ of the larger circle is drawn, touching the smaller circle at $𝐴$ . The length of $𝐵 𝐶$ is $𝑙$ . Show that the area of the green region enclosed between the two circles is $14 𝜋 𝑙 2$ .

Fig. 6.51: Concentric circles; chord touches the smaller

Solution

Let the radius of the outer circle be $𝑅$ , and the radius of the inner circle be $𝑟$ .
The area of the green ring (the annulus) is the difference between the two circle areas:

A r e a (G r e e n R e g i o n) = 𝜋 𝑅 2 - 𝜋 𝑟 2 = 𝜋 (𝑅 2 - 𝑟 2)

Connect center $𝑂$ to point of tangency $𝐴$ , and to outer chord endpoint $𝐶$ .

Because chord $𝐵 𝐶$ is tangent to the inner circle at point $𝐴$ , radius $𝑂 𝐴$ meets the chord at a right angle ( $∠ 𝑂 𝐴 𝐶 = 90 \circ$ ).
In geometry, a perpendicular line from the center of a circle to a chord always cuts that chord exactly in half. Therefore, half-chord $𝐴 𝐶 = 12 𝐵 𝐶 = 𝑙 2$ .

Look at right-angled triangle $△ 𝑂 𝐴 𝐶$ : vertical leg is $𝑂 𝐴 = 𝑟$ , horizontal leg is $𝐴 𝐶 = 𝑙 2$ , and hypotenuse is outer radius $𝑂 𝐶 = 𝑅$ .

By the Baudhāyana-Pythagoras theorem:

𝑟 2 + (𝑙 2) 2 = 𝑅 2 ⟹ 𝑅 2 - 𝑟 2 = 𝑙 2 4

Substitute $(𝑅 2 - 𝑟 2) = 𝑙 2 4$ directly into our ring area equation from Step 2:

A r e a (G r e e n R e g i o n) = 𝜋 (𝑅 2 - 𝑟 2) = 𝟏 𝟒 𝜋 𝐥 𝟐

*Question 24 (Page 153)

Question

In Fig. 6.52, semicircles have been drawn on all the sides of a right-angled triangle as shown. Show that $A r e a (𝐴) + A r e a (𝐵) = A r e a (𝐶)$ .

Fig. 6.52: Semicircles on the sides of a right-angled triangle

Solution

Let the right-angled triangle have perpendicular legs $𝑎$ and $𝑏$ (which serve as the diameters of semicircles A and B), and hypotenuse $𝑐$ (which serves as the diameter of semicircle C).
By the Baudhāyana-Pythagoras theorem:

𝑎 2 + 𝑏 2 = 𝑐 2

The area of any semicircle of diameter $𝑑$ is $12 𝜋 (𝑑 2) 2 = 18 𝜋 𝑑 2$ . Therefore:

$A r e a (𝐴) = 18 𝜋 𝑎 2$
$A r e a (𝐵) = 18 𝜋 𝑏 2$
$A r e a (𝐶) = 18 𝜋 𝑐 2$

Sum the areas of the two smaller semicircles:

A r e a (𝐴) + A r e a (𝐵) = 18 𝜋 𝑎 2 + 18 𝜋 𝑏 2 = 18 𝜋 (𝑎 2 + 𝑏 2)

Since $𝑎 2 + 𝑏 2 = 𝑐 2$ , replace $(𝑎 2 + 𝑏 2)$ with $𝑐 2$ :

A r e a (𝐴) + A r e a (𝐵) = 18 𝜋 𝑐 2 = A r e a (𝐂)

*Question 25 (Page 153)

Question

Fig. 6.53 shows two circles passing through each other's centres. Find the area of the region enclosed by the two circles in terms of the common radius $𝑟$ .

Fig. 6.53: Two congruent circles through each other’s centres

Solution

Let the centers of the circles be $𝐴$ and $𝐵$ , and their top and bottom intersection points be $𝐶$ and $𝐷$ .
Because each circle passes through the other's center, distance $𝐴 𝐵 = 𝑟$ . Since both circles have radius $𝑟$ , lines $𝐴 𝐶, 𝐵 𝐶, 𝐴 𝐷, 𝐵 𝐷$ all equal $𝑟$ . This forms two equilateral triangles ( $△ 𝐴 𝐵 𝐶$ and $△ 𝐴 𝐵 𝐷$ ) sharing base $𝐴 𝐵$ .
In circle $𝐴$ , the central angle subtended by arc $𝐶 𝐷$ is $∠ 𝐶 𝐴 𝐷 = 60 \circ + 60 \circ = 120 \circ$ .

The area of this circular sector ( $120 \circ 360 \circ = 13$ of a circle) is:

A r e a (S e c t o r) = 13 𝜋 𝑟 2

The enclosed lens-shaped region is formed by two identical circular segments overlapping across chord $𝐶 𝐷$ . We can find its total area by adding the sectors from both circles and subtracting the central rhombus $𝐴 𝐶 𝐵 𝐷$ (which gets double-counted):

$A r e a o f S e c t o r f r o m C i r c l e 𝐴 = 13 𝜋 𝑟 2$ .
$A r e a o f S e c t o r f r o m C i r c l e 𝐵 = 13 𝜋 𝑟 2$ .
$A r e a o f R h o m b u s 𝐴 𝐶 𝐵 𝐷 = 2 \times (\sqrt 3 4 𝑟 2) = \sqrt 3 2 𝑟 2$ .

Combine these to find the enclosed area:

E n c l o s e d A r e a = (13 𝜋 𝑟 2) + (13 𝜋 𝑟 2) - \sqrt 3 2 𝑟 2 = 𝐫 𝟐 (𝟐 𝜋 𝟑 - \sqrt 𝟑 𝟐) s q . u n i t s

*Question 26 (Page 153)

Question

In Fig. 6.54, we see three triangles within a rectangle. The areas of the triangles are $𝐴, 𝐵, 𝐶$ , as marked. Show that the area of the rectangle is $2 (𝐴 + 𝐶) (𝐵 + 𝐶) 𝐶$ .

Fig. 6.54: Three triangles of areas A, B, C in a rectangle

Solution

Let the bounding rectangle have total horizontal length $𝐿$ and total vertical height $𝑊$ . Let the total area of the rectangle be $𝑆 = 𝐿 \times 𝑊$ .
In Fig. 6.54, notice how the three triangles meet: Triangle $𝐶$ is a right-angled triangle tucked into the top-right corner of the rectangle. Let its horizontal width be $𝑤 1$ and its vertical height be $ℎ 1$ .

A r e a (𝐶) = 12 𝑤 1 ℎ 1

Look at the combined shape of Triangle $𝐴$ and Triangle $𝐶$ along the top border: together, they span the entire horizontal length $𝐿$ of the rectangle, sharing the same vertical altitude $ℎ 1$ .

A r e a (𝐴 + 𝐶) = 12 𝐿 ℎ 1

Look at the combined shape of Triangle $𝐵$ and Triangle $𝐶$ along the right vertical border: together, they span the entire vertical height $𝑊$ of the rectangle, sharing the same horizontal altitude $𝑤 1$ .

A r e a (𝐵 + 𝐶) = 12 𝑊 𝑤 1

Multiply the expression for $(𝐴 + 𝐶)$ by the expression for $(𝐵 + 𝐶)$ :

(𝐴 + 𝐶) (𝐵 + 𝐶) = (12 𝐿 ℎ 1) \times (12 𝑊 𝑤 1) = 14 (𝐿 𝑊) (𝑤 1 ℎ 1)

Rearrange the right side by grouping $𝑆 = 𝐿 𝑊$ and $𝐶 = 12 𝑤 1 ℎ 1$ :

(𝐴 + 𝐶) (𝐵 + 𝐶) = 12 (𝐿 𝑊) \times (12 𝑤 1 ℎ 1) = 12 𝑆 \times 𝐶

Multiply both sides by 2 and divide by $𝐶$ to solve for the rectangle's area $𝑆$ :

A r e a o f R e c t a n g l e (𝑆) = 𝟐 (𝐀 + 𝐂) (𝐁 + 𝐂) 𝐂

*Question 27 (Page 153)

Question

In the figure (Fig. 6.55) we see two shaded regions formed by a quarter circle, a semicircle, and a triangle. Show that the areas of the two shaded regions are equal.

Fig. 6.55: Two shaded regions formed by a quarter circle, a semicircle, and a triangle

Solution

In Fig. 6.55, points $𝐴$ , $𝑂$ , and $𝐶$ are collinear, with $𝑂$ the midpoint of $𝐴 𝐶$ . A semicircle is drawn with diameter $𝐴 𝐶$ (centre $𝑂$ , radius $𝑅 = 𝑂 𝐴 = 𝑂 𝐶$ ). Point $𝐵$ lies on this semicircle so that $𝑂 𝐵 ⟂ 𝐴 𝐶$ . An arc with centre $𝐴$ and radius $𝐴 𝐵$ is drawn through $𝐵$ (the arc $𝐸 𝐵$ in the figure). Two regions are shaded with hatching.

Step 1 — Equal building-block areas. Since $𝐵$ is on the semicircle with diameter $𝐴 𝐶$ and $𝑂$ is the midpoint with $𝑂 𝐵 ⟂ 𝐴 𝐶$ , triangles $△ 𝐴 𝑂 𝐵$ and $△ 𝐶 𝑂 𝐵$ are congruent right-angled isosceles triangles, so $𝐴 𝐵 = 𝑅 \sqrt 2$ .

Area of the large semicircle with diameter $𝐴 𝐶$ : $12 𝜋 𝑅 2$ .
Area of the quarter circle with centre $𝐴$ and radius $𝐴 𝐵 = 𝑅 \sqrt 2$ : $14 𝜋 (𝑅 \sqrt 2) 2 = 14 𝜋 \cdot 2 𝑅 2 = 12 𝜋 𝑅 2 .$

So the large semicircle (centre $𝑂$ ) and the quarter circle (centre $𝐴$ ) have exactly the same area.

Step 2 — Equal remainders. These two equal-area regions overlap on a common unshaded (or differently hatched) middle portion. Subtracting that common overlap from each leaves the two non-overlapping remainders — precisely the two shaded regions in the figure. Equal totals minus equal overlap implies equal remainders.

Therefore the two shaded regions have equal area.

Shown: the two shaded areas are equal.