Class 9 Maths Chapter 5 NCERT Solutions: I’m Up and Down, and Round and Round

Introduction

This chapter explores circles: their definition, symmetries, chords, distances from the centre, and the angles subtended by arcs and chords.

Activity

List some objects from nature that resemble a circle.

Solution

Many natural forms take a circular shape. Examples include the sun, the full moon, the ripples that form when a raindrop hits a puddle, the cross-section of a tree trunk, and the arrangement of petals on certain flowers like sunflowers.

Think and Reflect

Jamuna has a circular piece of paper. She is trying to locate its centre. Amina gives her a suggestion. She follows the instructions and is thrilled to find that it works. Can you guess what Amina told her?

Solution

Amina told her to fold the circular paper exactly in half so the edges meet, creating a straight crease across the middle. This crease is a diameter line. Then, she should fold the paper in half a second time, making the ends of the first crease meet. This creates a second diameter line. When she unfolds the paper, the point where the two straight creases intersect is the exact centre of the circle.

5.1 Definitions

A circle is the set of all points on the plane that are equidistant from a given point (the centre). That common distance is the radius. A chord is a line segment joining two points on the circle. A diameter is a chord through the centre.

Fig. 5.3: Circle, centre

𝐴

, chord

𝐵 𝐶

(radii to

𝐵, 𝐶, 𝐷, 𝐸

)

5.2 Symmetries of a Circle

A circle has complete rotational symmetry about its centre, and every diameter is a line of reflection symmetry.

Think and Reflect

What are the rotational symmetries of a square? How many lines of reflection symmetry does it have? What about a regular pentagon? A regular hexagon?
What is the length of the longest chord in a circle of radius 5 units? Is there a smallest chord?
The locus of points at a given distance from a given point is a circle. What can we say about the locus of points equidistant from two given points?

Solution

Square: 4 rotational symmetries (

90 \circ

180 \circ

270 \circ

360 \circ

) and 4 reflection lines (2 diagonals, 2 midlines). Regular pentagon: 5 rotational symmetries (every

72 \circ

) and 5 reflection lines. Regular hexagon: 6 rotational symmetries (every

60 \circ

) and 6 reflection lines.

The longest chord is the diameter: $2 \times 5 = 10$ units. Chords can be arbitrarily short (no positive minimum).

The locus of points equidistant from two points $𝐴$ and $𝐵$ is the perpendicular bisector of segment $𝐴 𝐵$ .

5.3 How Many Circles?

Fig. 5.4: Circles through two points — centres lie on the perpendicular bisector

Think and Reflect

How many circles pass through two points on a plane?
Are there circles of all possible radii through $𝐴$ and $𝐵$ ? What is the smallest radius? Largest?
As you move away from $𝐴 𝐵$ along its perpendicular bisector, do the radii increase or decrease?
Do those circles appear more curved or less curved?
How many squares can have $𝐴$ and $𝐵$ on the boundary? How many with $𝐴$ and $𝐵$ as corners?

Solution

Infinitely many circles pass through two given points; centres lie on the perpendicular bisector of the joining segment.

Not every radius works. Minimum radius is $12 𝐴 𝐵$ (when $𝐴 𝐵$ is a diameter). There is no maximum radius.

Radii increase as the centre moves farther from $𝐴 𝐵$ .

Larger circles appear less curved (flatter).

Infinitely many squares can have $𝐴, 𝐵$ on the boundary. Exactly three squares have $𝐴$ and $𝐵$ as vertices: two with $𝐴 𝐵$ as a side, one with $𝐴 𝐵$ as a diagonal.

Three non-collinear points determine a unique circle — the circumcircle of the triangle they form. The circumcentre is the intersection of the perpendicular bisectors of the sides.

Fig. 5.5: Circumcircle of

Δ 𝐴 𝐵 𝐶

(acute) — circumcentre

𝑂

inside, at intersection of perpendicular bisectors

Fig. 5.6: Obtuse-angled triangle — circumcentre

𝑂

lies outside the triangle

Fig. 5.7: Right-angled triangle — circumcentre

𝑂

is the midpoint of the hypotenuse

Exercise Set 5.1

Question

Draw

Δ 𝐴 𝐵 𝐶

with

𝐴 𝐵 = 5

cm,

∠ 𝐴 = 70 \circ

and

∠ 𝐵 = 60 \circ

. Draw the circumcircle. Is the centre inside or outside?

Solution

$∠ 𝐶 = 180 \circ - (70 \circ + 60 \circ) = 50 \circ$ .

All angles $70 \circ$ , $60 \circ$ , $50 \circ$ are acute, so the triangle is acute-angled.

For an acute-angled triangle, the circumcentre lies inside the triangle.

Question

Draw

Δ 𝐴 𝐵 𝐶

with

𝐴 𝐵 = 5

cm,

∠ 𝐴 = 100 \circ

𝐴 𝐶 = 4

cm. Draw the circumcircle. Is the centre inside or outside?

Solution

Since $∠ 𝐴 = 100 \circ > 90 \circ$ , the triangle is obtuse-angled.

For an obtuse-angled triangle, the circumcentre lies outside the triangle.

Question

Draw

Δ 𝐴 𝐵 𝐶

with

𝐴 𝐵 = 6

cm,

𝐵 𝐶 = 7

cm,

𝐶 𝐴 = 7

cm. Measure

𝑂 𝐴

𝑂 𝐵

𝑂 𝐶

for circumcentre

𝑂

Solution

$𝐴$ , $𝐵$ , $𝐶$ lie on the circumcircle with centre $𝑂$ .

So $𝑂 𝐴$ , $𝑂 𝐵$ , $𝑂 𝐶$ are radii of the same circle: $𝑂 𝐴 = 𝑂 𝐵 = 𝑂 𝐶$ .

Question

What is the least possible radius of a circle through two points

𝐴

and

𝐵

Solution

The smallest circle through $𝐴$ and $𝐵$ has $𝐴 𝐵$ as diameter.

Least radius $= 12 𝐴 𝐵$ .

Think, Draw and Infer

Question

Three collinear points

𝐴, 𝐵, 𝐶

: can

𝑃 𝐴 = 𝑃 𝐵 = 𝑃 𝐶

? Are the perpendicular bisectors of

𝐴 𝐵

and

𝐵 𝐶

parallel? Can a circle pass through three collinear points? Can a line cut a circle in three points?

Solution

No point $𝑃$ satisfies $𝑃 𝐴 = 𝑃 𝐵 = 𝑃 𝐶$ for three collinear points.

Both perpendicular bisectors are perpendicular to the same line $𝐴 𝐵 𝐶$ , so they are parallel and never meet.

Hence no circle through three collinear points (centre would need that intersection).

A line meets a circle in at most two distinct points.

Question

Can other triangles congruent to

Δ 𝐴 𝐵 𝐶

share the same circumcircle?

Solution

Yes. Rotating $Δ 𝐴 𝐵 𝐶$ about the circumcentre yields congruent triangles on the same circumcircle.

5.4 Chords and the Angles They Subtend

Equal chords

𝐴 𝐵

and

𝐷 𝐸

with centre

𝐶

form congruent isosceles triangles

Theorem 2: Equal chords subtend equal angles at the centre. Theorem 3 (converse): Chords subtending equal central angles are equal.

Exercise Set 5.2

Question

Show that the triangle formed by a chord and the centre is isosceles.

Solution

Let chord $𝐴 𝐵$ and centre $𝐶$ form $Δ 𝐶 𝐴 𝐵$ .

$𝐶 𝐴$ and $𝐶 𝐵$ are both radii, so $𝐶 𝐴 = 𝐶 𝐵$ .

A triangle with two equal sides is isosceles.

Question

Show that if two such isosceles triangles have equal base length, they are congruent.

Solution

Let $Δ 𝐶 𝐴 𝐵$ and $Δ 𝐶 𝐷 𝐸$ arise from equal chords $𝐴 𝐵 = 𝐷 𝐸$ and the same centre $𝐶$ .

Then $𝐶 𝐴 = 𝐶 𝐵 = 𝐶 𝐷 = 𝐶 𝐸 = 𝑟$ (radii) and $𝐴 𝐵 = 𝐷 𝐸$ .

By SSS, $Δ 𝐶 𝐴 𝐵 ≅ Δ 𝐶 𝐷 𝐸$ .

5.5 Midpoints and Perpendicular Bisectors of Chords

Fig. 5.12: Line from centre

𝐶

to midpoint

𝑀

of chord

𝐴 𝐵

is perpendicular to

𝐴 𝐵

Theorem 4: The line from the centre to the midpoint of a chord is perpendicular to the chord. Theorem 5 (converse): The perpendicular from the centre to a chord bisects it.

Exercise Set 5.3

Question

Explain why the perpendicular from the centre to a chord bisects the chord.

Solution

In Fig. 5.12, $𝐶 𝑀 ⟂ 𝐴 𝐵$ . Consider $Δ 𝐶 𝑀 𝐴$ and $Δ 𝐶 𝑀 𝐵$ .

$∠ 𝐶 𝑀 𝐴 = ∠ 𝐶 𝑀 𝐵 = 90 \circ$ ; hypotenuses $𝐶 𝐴 = 𝐶 𝐵$ (radii); $𝐶 𝑀$ is common.

By RHS congruence, $Δ 𝐶 𝑀 𝐴 ≅ Δ 𝐶 𝑀 𝐵$ .

Hence $𝐴 𝑀 = 𝐵 𝑀$ : the perpendicular bisects the chord.

Question

Isosceles

Δ 𝐴 𝐵 𝐶

inscribed with

𝐴 𝐵 = 𝐴 𝐶

. Show the altitude from

𝐴

𝐵 𝐶

passes through the centre.

Solution

In isosceles $Δ 𝐴 𝐵 𝐶$ with $𝐴 𝐵 = 𝐴 𝐶$ , the altitude from $𝐴$ to $𝐵 𝐶$ is also the perpendicular bisector of chord $𝐵 𝐶$ .

The perpendicular bisector of any chord passes through the centre.

Therefore that altitude passes through the centre.

Question

Parallel chords

6

cm and

8

cm on opposite sides of the centre; radius

5

cm. Distance between midpoints?

Solution

Let $𝑑 1$ be distance to the $6$ cm chord and $𝑑 2$ to the $8$ cm chord.

Half of $6$ is $3$ : $𝑑 21 + 32 = 52 \Rightarrow 𝑑 21 + 9 = 25 \Rightarrow 𝑑 21 = 16 \Rightarrow 𝑑 1 = 4$ cm.

Half of $8$ is $4$ : $𝑑 22 + 42 = 52 \Rightarrow 𝑑 22 + 16 = 25 \Rightarrow 𝑑 22 = 9 \Rightarrow 𝑑 2 = 3$ cm.

Opposite sides of the centre: distance between midpoints $= 𝑑 1 + 𝑑 2 = 4 + 3 = 7$ cm.

5.6 Distance of Chords from the Centre

Theorem 6: Equal chords are equidistant from the centre. Theorem 7 (converse): Chords equidistant from the centre are equal. Theorem 8: Of two unequal chords, the longer is closer to the centre.

Fig. 5.15: If

𝐶 𝐸 = 𝐶 𝐻

(perpendicular distances), then chords

𝐴 𝐵 = 𝐺 𝐹

Exercise Set 5.4

Question

Use Baudhāyana–Pythagoras to show Theorem 6.

Solution

Let equal chords $𝐴 𝐵 = 𝐺 𝐹$ , centre $𝐶$ . Drop $𝐶 𝐸 ⟂ 𝐴 𝐵$ and $𝐶 𝐻 ⟂ 𝐺 𝐹$ .

Then $𝐴 𝐸 = 12 𝐴 𝐵$ and $𝐹 𝐻 = 12 𝐺 𝐹$ , so $𝐴 𝐸 = 𝐹 𝐻$ .

In right triangles: $𝐶 𝐴 2 = 𝐶 𝐸 2 + 𝐴 𝐸 2$ and $𝐶 𝐹 2 = 𝐶 𝐻 2 + 𝐹 𝐻 2$ .

Since $𝐶 𝐴 = 𝐶 𝐹$ (radii), $𝐶 𝐸 2 + 𝐴 𝐸 2 = 𝐶 𝐻 2 + 𝐹 𝐻 2$ .

With $𝐴 𝐸 = 𝐹 𝐻$ , we get $𝐶 𝐸 2 = 𝐶 𝐻 2$ , so $𝐶 𝐸 = 𝐶 𝐻$ .

Question

Fig. 5.15:

𝐶 𝐸 ⟂ 𝐴 𝐵

𝐶 𝐻 ⟂ 𝐺 𝐹

𝐶 𝐸 = 𝐶 𝐻

. Show

𝐴 𝐵 = 𝐺 𝐹

Solution

In right $Δ 𝐶 𝐸 𝐴$ and $Δ 𝐶 𝐻 𝐹$ : hypotenuses $𝐶 𝐴 = 𝐶 𝐹$ (radii), legs $𝐶 𝐸 = 𝐶 𝐻$ (given).

By RHS, $Δ 𝐶 𝐸 𝐴 ≅ Δ 𝐶 𝐻 𝐹$ , so $𝐴 𝐸 = 𝐹 𝐻$ .

Then $𝐴 𝐵 = 2 𝐴 𝐸$ and $𝐺 𝐹 = 2 𝐹 𝐻$ , hence $𝐴 𝐵 = 𝐺 𝐹$ .

Question

Solve the previous question using Baudhāyana–Pythagoras.

Solution

$𝐴 𝐸 2 = 𝐶 𝐴 2 - 𝐶 𝐸 2$ and $𝐹 𝐻 2 = 𝐶 𝐹 2 - 𝐶 𝐻 2$ .

With $𝐶 𝐴 = 𝐶 𝐹$ and $𝐶 𝐸 = 𝐶 𝐻$ : $𝐹 𝐻 2 = 𝐴 𝐸 2$ , so $𝐹 𝐻 = 𝐴 𝐸$ .

Thus $𝐴 𝐵 = 2 𝐴 𝐸 = 2 𝐹 𝐻 = 𝐺 𝐹$ .

Exercise Set 5.5

Question

Radius

7

cm, perpendicular distance

6

cm. Find chord length.

Solution

Half-chord $𝑥$ : $𝑥 2 + 62 = 72 \Rightarrow 𝑥 2 + 36 = 49 \Rightarrow 𝑥 2 = 13 \Rightarrow 𝑥 = \sqrt 13$ .

Full chord $= 2 \sqrt 13$ cm.

Question

Explain: chord length

= 2 \sqrt 𝑟 2 - 𝑑 2

Solution

The perpendicular from the centre bisects the chord. Half-chord $𝑥$ satisfies $𝑥 2 + 𝑑 2 = 𝑟 2$ .

So $𝑥 = \sqrt 𝑟 2 - 𝑑 2$ . Full chord $= 2 𝑥 = 2 \sqrt 𝑟 2 - 𝑑 2$ .

Question

If distance of

𝐴 𝐵

from centre is twice that of

𝐶 𝐷

, is

𝐶 𝐷 = 2 𝐴 𝐵

Solution

No. Let distance to $𝐶 𝐷$ be $𝑑$ ; then distance to $𝐴 𝐵$ is $2 𝑑$ .

$𝐶 𝐷 = 2 \sqrt 𝑟 2 - 𝑑 2$ and $𝐴 𝐵 = 2 \sqrt 𝑟 2 - 4 𝑑 2$ .

These are not related by the factor $2$ in general (the relation is not linear).

5.7 Angles Subtended by an Arc

Fig. 5.19: Circle with centre

𝑂

; arcs

𝐴 𝐾 𝐵

and

𝐶 𝐿 𝐷

Question

Fig. 5.19: Are arcs

𝐴 𝐾 𝐵

and

𝐶 𝐿 𝐷

minor or major?

Solution

Central angle less than $180 \circ$ means a minor arc; greater means major.

In the figure both arcs shown are the shorter routes, so both are minor arcs.

Theorem 9: Central angle is double the angle at the circumference (same arc). Angle in a semicircle is $90 \circ$ . Theorem 10: equal angles on the same side of a segment imply concyclic points. Theorem 11: opposite angles of a cyclic quadrilateral sum to $180 \circ$ .

Exercise Set 5.6

Question

Central angle

∠ 𝐴 𝑂 𝐵 = 60 \circ

, radius

12

cm. Find chord

𝐴 𝐵

Solution

In $Δ 𝐴 𝑂 𝐵$ , $𝑂 𝐴 = 𝑂 𝐵 = 12$ cm, so base angles are equal.

$∠ 𝑂 𝐴 𝐵 = ∠ 𝑂 𝐵 𝐴 = 180 \circ - 60 \circ 2 = 60 \circ$ .

All angles $60 \circ$ : equilateral triangle. Hence $𝐴 𝐵 = 12$ cm.

Question

Points

𝑋, 𝑌

on the same side of chord

𝐴 𝐵

on the circle: can

∠ 𝐴 𝑋 𝐵 \neq ∠ 𝐴 𝑌 𝐵

Solution

No. Angles in the same segment are equal (corollary of Theorem 9).

Question

∠ 𝐴 𝑋 𝐵 = ∠ 𝐴 𝑌 𝐵

(on the circle), must

𝑋

and

𝑌

lie on the same side of

𝐴 𝐵

Solution

Yes. Equal angles characterise the same segment; opposite segments give supplementary (not equal) angles in general.

Question

∠ 𝐴 𝑋 𝐵 = ∠ 𝐴 𝑌 𝐵

on the same side of

𝐴 𝐵

, does the circle through

𝐴, 𝐵, 𝑋

pass through

𝑌

Solution

Yes (Theorem 10): $𝐴, 𝐵, 𝑋, 𝑌$ are concyclic.

Fig. 5.26: Cyclic quadrilateral

𝐴 𝐵 𝐶 𝐷

inscribed in the circle;

∠ 𝐴 𝐷 𝐶 = 100 \circ

∠ 𝐴 𝐵 𝐶 = 𝑥

Question

Find

𝑥

in Fig. 5.26.

Solution

$𝐴 𝐵 𝐶 𝐷$ is a cyclic quadrilateral (all four vertices lie on the circle).

By Theorem 11, opposite angles of a cyclic quadrilateral sum to $180 \circ$ .

$∠ 𝐴 𝐵 𝐶$ and $∠ 𝐴 𝐷 𝐶$ are opposite, with $∠ 𝐴 𝐷 𝐶 = 100 \circ$ and $∠ 𝐴 𝐵 𝐶 = 𝑥$ .

So $𝑥 + 100 \circ = 180 \circ$ , hence $𝑥 = 80 \circ$ .

End-of-Chapter Exercises

Question

Chord

5

cm from centre; radius

13

cm. Chord length?

Solution

Half-chord $𝑥$ : $𝑥 2 + 52 = 132 \Rightarrow 𝑥 2 + 25 = 169 \Rightarrow 𝑥 2 = 144 \Rightarrow 𝑥 = 12$ .

Full chord $= 24$ cm.

Question

Arc subtends

70 \circ

at centre. Angle at a point on the remaining circle?

Solution

By Theorem 9: circumference angle $= 70 \circ 2 = 35 \circ$ .

Question

Diameter

26

cm; chord

24

cm. Distance from centre to chord?

Solution

$𝑟 = 13$ cm; half-chord $= 12$ cm.

$𝑑 2 + 122 = 132 \Rightarrow 𝑑 2 + 144 = 169 \Rightarrow 𝑑 2 = 25 \Rightarrow 𝑑 = 5$ cm.

Question

Radius

15

cm; distance to chord

9

cm. Chord length?

Solution

$𝑥 2 + 92 = 152 \Rightarrow 𝑥 2 + 81 = 225 \Rightarrow 𝑥 2 = 144 \Rightarrow 𝑥 = 12$ .

Full chord $= 24$ cm.

Question

Prove: perpendicular bisector of a chord passes through the centre.

Solution

Let chord $𝐴 𝐵$ , centre $𝑂$ . Then $𝑂 𝐴 = 𝑂 𝐵$ (radii).

So $𝑂$ is equidistant from $𝐴$ and $𝐵$ , hence $𝑂$ lies on the perpendicular bisector of $𝐴 𝐵$ .

Question

Diameter

𝐴 𝐵

;

𝐶

on circumference. Measure of

∠ 𝐴 𝐶 𝐵

Solution

$∠ 𝐴 𝐶 𝐵 = 90 \circ$ (angle in a semicircle).

Central angle for diameter is $180 \circ$ ; half is $90 \circ$ by Theorem 9.

Question

Cyclic

𝐴 𝐵 𝐶 𝐷

∠ 𝐴 = 75 \circ

∠ 𝐵 = 110 \circ

. Find

∠ 𝐶

∠ 𝐷

Solution

$∠ 𝐶 = 180 \circ - 75 \circ = 105 \circ$ .

$∠ 𝐷 = 180 \circ - 110 \circ = 70 \circ$ .

Question

Cyclic

𝑃 𝑄 𝑅 𝑆

∠ 𝑃 = (2 𝑥 + 10) \circ

∠ 𝑅 = (3 𝑥 - 20) \circ

. Find

𝑥

∠ 𝑃

∠ 𝑅

Solution

$(2 𝑥 + 10) + (3 𝑥 - 20) = 180 \Rightarrow 5 𝑥 - 10 = 180 \Rightarrow 5 𝑥 = 190 \Rightarrow 𝑥 = 38$ .

$∠ 𝑃 = 2 (38) + 10 = 86 \circ$ ; $∠ 𝑅 = 3 (38) - 20 = 94 \circ$ .

Check: $86 \circ + 94 \circ = 180 \circ$ .

Question

Chord

16

cm at distance

6

cm from centre. Find radius.

Solution

Half-chord $= 8$ cm.

$𝑟 2 = 82 + 62 = 64 + 36 = 100 \Rightarrow 𝑟 = 10$ cm.

Question

Cyclic quadrilateral with sides

5, 5, 12, 12

. Find its area.

Solution

This is a cyclic kite (two pairs of adjacent equal sides).

A cyclic kite has right angles where unequal sides meet.

Area of one right triangle with legs $5$ and $12$ : $12 \times 5 \times 12 = 30$ .

Total area $= 2 \times 30 = 60$ square units.

(Check: $52 + 122 = 132$ , so the diagonal between equal-side joints is a diameter of length $13$ .)

Question

*11. Without drawing the circumcircle, how can we tell if the circumcentre of a cyclic quadrilateral lies inside or outside?

Solution

Draw a diagonal, splitting the cyclic quadrilateral into two triangles that share the same circumcircle.

The circumcentre is the circumcentre of either triangle.

A triangle circumcentre is inside / on / outside according as the triangle is acute / right / obtuse.

Check whether that point lies inside both triangular halves (hence inside the quadrilateral), or outside.

Equivalently, construct perpendicular bisectors of two sides; their intersection is the centre — see where it falls.

Question

*12. Equal chords intersect inside a circle. Show corresponding segments are equal.

Solution

Let equal chords $𝐴 𝐵 = 𝐶 𝐷$ meet at $𝑃$ ; centre $𝑂$ .

Equal chords are equidistant from $𝑂$ : if $𝑂 𝐸 ⟂ 𝐴 𝐵$ and $𝑂 𝐹 ⟂ 𝐶 𝐷$ , then $𝑂 𝐸 = 𝑂 𝐹$ , and $𝐸, 𝐹$ are midpoints.

In right $Δ 𝑂 𝐸 𝑃$ and $Δ 𝑂 𝐹 𝑃$ : common hypotenuse $𝑂 𝑃$ , $𝑂 𝐸 = 𝑂 𝐹$ . By RHS, $Δ 𝑂 𝐸 𝑃 ≅ Δ 𝑂 𝐹 𝑃$ , so $𝐸 𝑃 = 𝐹 𝑃$ .

Also $𝐴 𝐸 = 𝐶 𝐹$ (halves of equal chords).

Hence the segments of the two chords match (equal or reversed order). Combined with $𝐴 𝐵 = 𝐶 𝐷$ , both pairs of segments correspond.

This is consistent with the intersecting-chords theorem $𝐴 𝑃 \cdot 𝑃 𝐵 = 𝐶 𝑃 \cdot 𝑃 𝐷$ .

Question

*13. Draw a circle in which a chord of

6

cm is

3

cm from the centre.

Solution

Half-chord $= 3$ cm, distance $= 3$ cm.

$𝑟 2 = 32 + 32 = 18 \Rightarrow 𝑟 = 3 \sqrt 2$ cm.

Construction: form a right isosceles triangle with legs $3$ cm; use the hypotenuse as radius; draw the circle; place the chord at distance $3$ cm from the centre with half-length $3$ cm each side of the foot.

Question

*14. Show a rectangle is the only parallelogram that can be inscribed in a circle.

Solution

Let parallelogram $𝐴 𝐵 𝐶 𝐷$ be cyclic.

Opposite angles of a parallelogram are equal: $∠ 𝐴 = ∠ 𝐶$ .

Opposite angles of a cyclic quadrilateral sum to $180 \circ$ : $∠ 𝐴 + ∠ 𝐶 = 180 \circ$ .

So $2 ∠ 𝐴 = 180 \circ \Rightarrow ∠ 𝐴 = 90 \circ$ .

A parallelogram with a right angle is a rectangle.

Question

*15. Rectangle inscribed in a circle: show diagonals meet at the centre.

Solution

Diagonals of a rectangle are equal and bisect each other at $𝑀$ .

Then $𝑀 𝐴 = 𝑀 𝐵 = 𝑀 𝐶 = 𝑀 𝐷$ (each is half a diagonal).

The unique point equidistant from all four vertices on the circle is the circumcentre.

Hence $𝑀$ is the centre.

Question

*16. Midpoints of all chords of a fixed length form what shape?

Solution

Equal chords lie at a fixed distance $𝑑$ from the centre (Theorem 6).

Each midpoint is at distance $𝑑$ from the centre.

So the midpoints form a concentric circle of radius $𝑑$ .

Question

*17. Congruent chords

𝐴 𝐵

and

𝐴 𝐶

; centre

𝑂

. Why does

𝑂

lie on the bisector of

∠ 𝐵 𝐴 𝐶

Solution

In $Δ 𝑂 𝐴 𝐵$ and $Δ 𝑂 𝐴 𝐶$ : $𝐴 𝐵 = 𝐴 𝐶$ (given), $𝑂 𝐵 = 𝑂 𝐶$ (radii), $𝑂 𝐴$ common.

By SSS, $Δ 𝑂 𝐴 𝐵 ≅ Δ 𝑂 𝐴 𝐶$ .

Hence $∠ 𝑂 𝐴 𝐵 = ∠ 𝑂 𝐴 𝐶$ : $𝑂 𝐴$ bisects $∠ 𝐵 𝐴 𝐶$ .

Question

Parallel chords

10

cm and

24

cm on the same side of the centre; distance between them

7

cm. Find radius.

Solution

Longer chord is closer to the centre. Let its distance be $𝑥$ ; then the $10$ cm chord is at $𝑥 + 7$ .

$𝑟 2 = 122 + 𝑥 2 = 144 + 𝑥 2$

$𝑟 2 = 52 + (𝑥 + 7) 2 = 25 + 𝑥 2 + 14 𝑥 + 49 = 𝑥 2 + 14 𝑥 + 74$

$144 + 𝑥 2 = 𝑥 2 + 14 𝑥 + 74 \Rightarrow 144 = 14 𝑥 + 74 \Rightarrow 70 = 14 𝑥 \Rightarrow 𝑥 = 5$ .

$𝑟 2 = 144 + 25 = 169 \Rightarrow 𝑟 = 13$ cm.

Question

*19. Regular hexagon inscribed in circle of radius

𝑟

. Side length? Distance of each side from centre?

Solution

Six radii to the vertices make six equilateral triangles of side $𝑟$ .

Side of hexagon $= 𝑟$ .

Distance to a side: right triangle with hypotenuse $𝑟$ and base $𝑟 2$ :

$𝑑 = \sqrt 𝑟 2 - (𝑟 / 2) 2 = \sqrt 3 𝑟 2 4 = 𝑟 \sqrt 3 2$ .

Question

Cyclic quadrilateral

𝑀 𝑁 𝑂 𝑃

with diameter

𝑀 𝑁

. What about

∠ 𝑀 𝑂 𝑃

and

∠ 𝑀 𝑁 𝑃

Solution

Since $𝑀 𝑁$ is a diameter, the angle subtended by $𝑀 𝑁$ at any point on the remaining circumference is $90 \circ$ (angle in a semicircle).

In $Δ 𝑀 𝑁 𝑂$ , the angle at $𝑂$ is $∠ 𝑀 𝑂 𝑁 = 90 \circ$ .

In $Δ 𝑀 𝑁 𝑃$ , the angle at $𝑃$ is $∠ 𝑀 𝑃 𝑁 = 90 \circ$ .

Clarification / correction: The angles forced to be

90 \circ

by the diameter theorem are the angles at the circumference subtended by arc

𝑀 𝑁

, namely

∠ 𝑀 𝑂 𝑁

and

∠ 𝑀 𝑃 𝑁

. The labels

∠ 𝑀 𝑂 𝑃

and

∠ 𝑀 𝑁 𝑃

name different angles and are not automatically

90 \circ

from

𝑀 𝑁

being a diameter alone. The intended fact is that the angles in the semicircle at the other two vertices are right angles.

Question

Cyclic

𝐴 𝐵 𝐶 𝐷

: exterior angle equals interior opposite angle (e.g.

∠ 𝐶 𝐷 𝐸 = ∠ 𝐴 𝐵 𝐶

Solution

Extend $𝐶 𝐷$ to $𝐸$ . Then $∠ 𝐴 𝐷 𝐶 + ∠ 𝐶 𝐷 𝐸 = 180 \circ$ (straight line).

By Theorem 11: $∠ 𝐴 𝐷 𝐶 + ∠ 𝐴 𝐵 𝐶 = 180 \circ$ .

Therefore $∠ 𝐶 𝐷 𝐸 = ∠ 𝐴 𝐵 𝐶$ .

Question

*22. Justify: no chord is longer than the diameter.

Solution

Chord length $= 2 \sqrt 𝑟 2 - 𝑑 2$ .

Maximised when $𝑑 = 0$ (chord through centre), giving length $2 𝑟$ (the diameter).

For $𝑑 > 0$ the chord is strictly shorter.

Question

*23. Point

𝐴

inside circle centre

𝑂

. Shortest chord through

𝐴

is perpendicular to

𝑂 𝐴

Solution

Longer chords are closer to the centre (Theorem 8). So the shortest chord through $𝐴$ maximises distance from $𝑂$ .

For a chord through $𝐴$ , if $𝑃$ is the foot of the perpendicular from $𝑂$ , then $𝑂 𝑃 \leq 𝑂 𝐴$ , with equality iff $𝑃 = 𝐴$ (chord $⟂ 𝑂 𝐴$ at $𝐴$ ).

Thus the perpendicular chord is farthest from $𝑂$ and therefore shortest.

Fig. 5.30: Angle in a semicircle —

∠ 𝐵 𝐴 𝐶 = 𝑎 + 𝑏 = 90 \circ

Question

How does Fig. 5.30 justify that the angle in a semicircle is

90 \circ

Solution

Let $𝐵 𝐶$ be a diameter, centre $𝑂$ , and $𝐴$ on the semicircle. Draw $𝑂 𝐴$ .

$Δ 𝑂 𝐴 𝐵$ is isosceles ( $𝑂 𝐴 = 𝑂 𝐵$ ); base angles equal $𝑎$ ; angle at $𝑂$ is $180 \circ - 2 𝑎$ .

$Δ 𝑂 𝐴 𝐶$ is isosceles ( $𝑂 𝐴 = 𝑂 𝐶$ ); base angles equal $𝑏$ ; angle at $𝑂$ is $180 \circ - 2 𝑏$ .

Angles at $𝑂$ on the straight diameter: $(180 \circ - 2 𝑎) + (180 \circ - 2 𝑏) = 180 \circ$ .

$360 \circ - 2 𝑎 - 2 𝑏 = 180 \circ \Rightarrow 2 𝑎 + 2 𝑏 = 180 \circ \Rightarrow 𝑎 + 𝑏 = 90 \circ$ .

But $∠ 𝐵 𝐴 𝐶 = 𝑎 + 𝑏$ , so $∠ 𝐵 𝐴 𝐶 = 90 \circ$ .

Question

*25. Chords

𝐶 𝐶'

and

𝐷 𝐷'

perpendicular to diameter

𝐴 𝐵

. Midpoints

𝑀, 𝑀'

𝐶 𝐷

and

𝐶' 𝐷'

. Show

𝑀 𝑀' ⟂ 𝐴 𝐵

Solution

Diameter $𝐴 𝐵$ is a reflection symmetry line of the circle.

Since $𝐶 𝐶' ⟂ 𝐴 𝐵$ and $𝐷 𝐷' ⟂ 𝐴 𝐵$ , reflection in $𝐴 𝐵$ swaps $𝐶$ with $𝐶'$ and $𝐷$ with $𝐷'$ .

Thus chord $𝐶 𝐷$ reflects to $𝐶' 𝐷'$ , so midpoint $𝑀$ reflects to $𝑀'$ .

The segment joining a point to its mirror image is perpendicular to the mirror line.

Therefore $𝑀 𝑀' ⟂ 𝐴 𝐵$ .

Fig. 5.31: Opposite angles of a cyclic quadrilateral sum to

180 \circ

Question

*26. How does Fig. 5.31 justify opposite angles of a cyclic quadrilateral sum to

180 \circ

Solution

Join centre $𝑂$ to $𝐴, 𝐵, 𝐶, 𝐷$ . Each of the four triangles is isosceles (two radii).

Let base angles be $𝑝$ in $Δ 𝑂 𝐴 𝐵$ , $𝑞$ in $Δ 𝑂 𝐵 𝐶$ , $𝑢$ in $Δ 𝑂 𝐶 𝐷$ , $𝑣$ in $Δ 𝑂 𝐷 𝐴$ .

Angle sum of four triangles $= 720 \circ$ . Angles at $𝑂$ total $360 \circ$ .

Base angles total $720 \circ - 360 \circ = 360 \circ$ , i.e. $2 (𝑝 + 𝑞 + 𝑢 + 𝑣) = 360 \circ$ , so $𝑝 + 𝑞 + 𝑢 + 𝑣 = 180 \circ$ .

Opposite angles of the quadrilateral: $∠ 𝐴 = 𝑝 + 𝑣$ and $∠ 𝐶 = 𝑞 + 𝑢$ .

Sum: $(𝑝 + 𝑣) + (𝑞 + 𝑢) = 𝑝 + 𝑞 + 𝑢 + 𝑣 = 180 \circ$ .

End of Chapter 5 — I'm Up and Down, and Round and Round