Circle

ICSE Class 8 Mathematics · ML Aggarwal Solutions Understanding ICSE Mathematics Class 8 · 31 solved questions

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1Exercise 15

Draw a circle with centre O and radius 2.5 cm. Draw two radii OA and OB such that ∠AOB = 60°. Measure the length of the chord AB.

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Steps of construction:

Mark a point O and with O as centre and radius 2.5 cm, draw a circle.
Take any point A on the circle and join OA.
Draw another radius OB such that ∠AOB = 60°.
Join A and B. AB is the required chord.

Draw a circle with centre O and radius 2.5 cm. Draw two radii OA and OB such that ∠AOB = 60°. Measure the length of the chord AB. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 8

In ∆OAB,

⇒ OA = OB = 2.5 cm (Radii of the circle)

⇒ ∠OAB = ∠OBA [Angles opposite to equal sides of a triangle are always equal]

Using angle sum property,

⇒ ∠OAB + ∠OBA + ∠AOB = 180°

⇒ ∠OAB + ∠OAB + 60° = 180°

⇒ 2∠OAB = 120°

⇒ ∠OAB = $\frac{120^{\circ}}{2}$ = 60°

⇒ ∠OAB = ∠OBA = 60°

So, ∆OAB is an equilateral triangle.

⇒ AB = OA = OB = 2.5 cm

On measuring the chord AB, we get AB = 2.5 cm.

Hence, the length of the chord AB = 2.5 cm.

2Exercise 15

Draw a circle of radius 3.2 cm. Draw a chord AB of this circle such that AB = 5 cm. Shade the minor segment of the circle.

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Steps of construction:

Mark a point C and with C as centre and radius 3.2 cm, draw a circle.
Take any point A on the circle.
With A as centre and radius 5 cm, draw an arc to meet the circle at B.
Join A and B. AB is the required chord.
Shade the segment which does not contain the centre C. It is the minor segment of the circle.

Draw a circle of radius 3.2 cm. Draw a chord AB of this circle such that AB = 5 cm. Shade the minor segment of the circle. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 8

3Exercise 15

Find the length of the tangent drawn to a circle of radius 3 cm, from a point at a distance 5 cm from the centre.

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Let C be the centre of the circle of radius 3 cm and PT be the tangent drawn to the circle from the point P, touching the circle at T.

Find the length of the tangent drawn to a circle of radius 3 cm, from a point at a distance 5 cm from the centre. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 8

Given, CT = 3 cm and CP = 5 cm.

We know that,

The tangent at any point of a circle and the radius through that point are perpendicular to each other.

⇒ ∠CTP = 90°

So, ∆CTP is a right angled triangle with CP as the hypotenuse.

Using Pythagoras theorem,

⇒ CT² + PT² = CP²

⇒ 3² + PT² = 5²

⇒ 9 + PT² = 25

⇒ PT² = 25 - 9

⇒ PT² = 16

⇒ PT = $\sqrt{16}$

⇒ PT = 4 cm.

Hence, the length of the tangent = 4 cm.

4Exercise 15

In the adjoining figure, PT is a tangent to the circle with centre C. Given CP = 20 cm and PT = 16 cm, find the radius of the circle.

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Given,

CP = 20 cm and PT = 16 cm.

We know that,

The tangent at any point of a circle and the radius through that point are perpendicular to each other.

⇒ ∠CTP = 90°

So, ∆CTP is a right angled triangle with CP as the hypotenuse.

Using Pythagoras theorem,

⇒ CT² + PT² = CP²

⇒ CT² + 16² = 20²

⇒ CT² + 256 = 400

⇒ CT² = 400 - 256

⇒ CT² = 144

⇒ CT = $\sqrt{144}$

⇒ CT = 12 cm.

CT is the radius of the circle.

Hence, the radius of the circle = 12 cm.

5(i)Exercise 15

In the following figure, O is the centre of the circle. Find the size of lettered angles :

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From figure,

AB is the diameter of the circle, ∠CAB = 32° and ∠ABD = 50°.

We know that,

Angle in a semicircle is a right angle.

⇒ ∠ACB = ∠ADB = 90°

In ∆ACB, using angle sum property,

⇒ ∠CAB + ∠ABC + ∠ACB = 180°

⇒ 32° + x + 90° = 180°

⇒ x + 122° = 180°

⇒ x = 180° - 122°

⇒ x = 58°

In ∆ADB, using angle sum property,

⇒ ∠BAD + ∠ABD + ∠ADB = 180°

⇒ y + 50° + 90° = 180°

⇒ y + 140° = 180°

⇒ y = 180° - 140°

⇒ y = 40°

Hence, x = 58° and y = 40°.

5(ii)Exercise 15

In the following figure, O is the centre of the circle. Find the size of lettered angles :

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From figure,

AC is the diameter of the circle, ∠DAC = 37° and AD ∥ BC.

Since, AD ∥ BC and AC is the transversal,

⇒ ∠ACB = ∠DAC (Alternate angles are equal)

⇒ x = 37°

We know that,

Angle in a semicircle is a right angle.

⇒ ∠ABC = 90°

In ∆ABC, using angle sum property,

⇒ ∠BAC + ∠ACB + ∠ABC = 180°

⇒ y + 37° + 90° = 180°

⇒ y + 127° = 180°

⇒ y = 180° - 127°

⇒ y = 53°

Hence, x = 37° and y = 53°.

5(iii)Exercise 15

In the following figure, O is the centre of the circle. Find the size of lettered angles :

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From figure,

AC is the diameter of the circle and BA = BC.

⇒ ∠BAC = ∠BCA = x [Angles opposite to equal sides of a triangle are always equal]

We know that,

Angle in a semicircle is a right angle.

⇒ ∠ABC = 90°

In ∆ABC, using angle sum property,

⇒ ∠BAC + ∠BCA + ∠ABC = 180°

⇒ x + x + 90° = 180°

⇒ 2x = 180° - 90°

⇒ 2x = 90°

⇒ x = $\frac{90^{\circ}}{2}$

⇒ x = 45°

Hence, x = 45°.

5(iv)Exercise 15

In the following figure, O is the centre of the circle. Find the size of lettered angles :

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From figure,

AC is the diameter of the circle and ∠ACD = 122°.

⇒ ∠ACB + ∠ACD = 180° (Linear pair)

⇒ ∠ACB + 122° = 180°

⇒ ∠ACB = 180° - 122°

⇒ ∠ACB = 58°

We know that,

Angle in a semicircle is a right angle.

⇒ ∠ABC = 90°

In ∆ABC, using angle sum property,

⇒ ∠BAC + ∠ACB + ∠ABC = 180°

⇒ x + 58° + 90° = 180°

⇒ x + 148° = 180°

⇒ x = 180° - 148°

⇒ x = 32°

Hence, x = 32°.

5(v)Exercise 15

In the following figure, O is the centre of the circle. Find the size of lettered angles :

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From figure,

AC is the diameter of the circle, OD ∥ CB and ∠CAB = 40°.

We know that,

Angle in a semicircle is a right angle.

⇒ ∠ABC = 90°

In ∆ABC, using angle sum property,

⇒ ∠CAB + ∠BCA + ∠ABC = 180°

⇒ 40° + x + 90° = 180°

⇒ x + 130° = 180°

⇒ x = 180° - 130°

⇒ x = 50°

Since, OD ∥ CB and AC is the transversal,

⇒ ∠AOD = ∠ACB (Corresponding angles)

⇒ ∠AOD = 50°

Since, AOC is a straight line,

⇒ ∠AOD + ∠DOC = 180° (Linear pair)

⇒ 50° + y = 180°

⇒ y = 180° - 50°

⇒ y = 130°

Hence, x = 50° and y = 130°.

5(vi)Exercise 15

In the following figure, O is the centre of the circle. Find the size of lettered angles :

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From figure,

AC is the diameter of the circle and BA = BC = CD.

In ∆ABC,

⇒ BA = BC

⇒ ∠BAC = ∠BCA = x [Angles opposite to equal sides of a triangle are always equal]

We know that,

Angle in a semicircle is a right angle.

⇒ ∠ABC = 90°

Using angle sum property,

⇒ ∠BAC + ∠BCA + ∠ABC = 180°

⇒ x + x + 90° = 180°

⇒ 2x = 180° - 90°

⇒ 2x = 90°

⇒ x = $\frac{90^{\circ}}{2}$

⇒ x = 45°

In ∆BCD,

⇒ BC = CD

⇒ ∠CBD = ∠CDB = y [Angles opposite to equal sides of a triangle are always equal]

According to exterior angle theorem,

Exterior angle of a triangle is equal to the sum of the two opposite interior angles.

⇒ ∠BCA = ∠CBD + ∠CDB

⇒ 45° = y + y

⇒ 2y = 45°

⇒ y = $\frac{45^{\circ}}{2}$

⇒ y = $22\frac{1}{2}^^{\circ}$

Hence, x = 45° and y = $\mathbf{22\frac{1}{2}}^^{\circ}$

5(vii)Exercise 15

In the following figure, O is the centre of the circle. Find the size of lettered angle :

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From figure,

AB is the diameter of the circle, ST is the tangent to the circle at B and ∠ASB = 65°.

We know that,

The tangent at any point of a circle and the radius through that point are perpendicular to each other.

⇒ OB ⊥ ST

⇒ ∠ABS = 90°

In ∆ABS, using angle sum property,

⇒ ∠BAS + ∠ASB + ∠ABS = 180°

⇒ x + 65° + 90° = 180°

⇒ x + 155° = 180°

⇒ x = 180° - 155°

⇒ x = 25°

Hence, x = 25°.

5(viii)Exercise 15

In the following figure, O is the centre of the circle. Find the size of lettered angle :

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From figure,

AB is the diameter of the circle, ST is the tangent to the circle at B and AB = BS.

We know that,

The tangent at any point of a circle and the radius through that point are perpendicular to each other.

⇒ OB ⊥ ST

⇒ ∠ABS = 90°

In ∆ABS, using angle sum property,

⇒ ∠BAS + ∠BSA + ∠ABS = 180°

⇒ x + y + 90° = 180°

⇒ x + y = 180° - 90°

⇒ x + y = 90°

Since, AB = BS

⇒ x = y [Angles opposite to equal sides of a triangle are always equal]

⇒ x + x = 90°

⇒ 2x = 90°

⇒ x = $\frac{90^{\circ}}{2}$

⇒ x = 45°

⇒ y = 45°

Hence, x = 45° and y = 45°.

5(ix)Exercise 15

In the following figure, O is the centre of the circle. Find the size of lettered angle :

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From figure,

RS is the diameter of the circle, SR is produced to Q, QT is the tangent to the circle at P and ∠Q = 36°.

We know that,

The tangent at any point of a circle and the radius through that point are perpendicular to each other.

⇒ OP ⊥ QT

⇒ ∠OPQ = 90°

In ∆OPQ, using angle sum property,

⇒ ∠OQP + ∠POQ + ∠OPQ = 180°

⇒ 36° + x + 90° = 180°

⇒ x + 126° = 180°

⇒ x = 180° - 126°

⇒ x = 54°

In ∆OPS,

⇒ OP = OS (Radii of the circle)

⇒ ∠OPS = ∠OSP = y [Angles opposite to equal sides of a triangle are always equal]

According to exterior angle theorem,

Exterior angle of a triangle is equal to the sum of the two opposite interior angles.

⇒ ∠POQ = ∠OPS + ∠OSP

⇒ 54° = y + y

⇒ 2y = 54°

⇒ y = $\frac{54^{\circ}}{2}$

⇒ y = 27°

Hence, x = 54° and y = 27°.

6(i)Exercise 15

In the following figure, O is the centre of the circle. Find the values of x and y.

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From figure,

AC is the diameter of the circle, AB = 15 cm and BC = 8 cm.

We know that,

Angle in a semicircle is a right angle.

⇒ ∠ABC = 90°

So, ∆ABC is a right angled triangle with AC as the hypotenuse.

Using Pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 15² + 8²

⇒ AC² = 225 + 64

⇒ AC² = 289

⇒ AC = $\sqrt{289}$

⇒ AC = 17 cm

⇒ x = 17

AC is the diameter and OA is the radius of the circle.

As we know that radius is half of the diameter.

⇒ y = $\frac{1}{2} \times 17$

⇒ y = 8.5

Hence, x = 17 and y = 8.5.

6(ii)Exercise 15

In the following figure, O is the centre of the circle. Find the values of x and y.

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From figure,

P is a point outside the circle from which two tangents are drawn to the circle, the radius of the circle is 5 cm and the length of one tangent is 12 cm.

We know that,

The tangent at any point of a circle and the radius through that point are perpendicular to each other.

So, the radius, the tangent and the line segment joining the centre O to the point P form a right angled triangle with OP as the hypotenuse.

Using Pythagoras theorem,

⇒ x² = 5² + 12²

⇒ x² = 25 + 144

⇒ x² = 169

⇒ x = $\sqrt{169}$

⇒ x = 13

We know that,

Tangents drawn from an exterior point to a circle are equal in length.

⇒ y = 12

Hence, x = 13 and y = 12.

6(iii)Exercise 15

In the following figure, O is the centre of the circle. Find the values of x and y.

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From figure,

P is a point outside the circle from which two tangents are drawn to the circle, the radius of the circle is 24 cm and the length of one tangent is 18 cm.

We know that,

The tangent at any point of a circle and the radius through that point are perpendicular to each other.

So, the radius, the tangent and the line segment joining the centre O to the point P form a right angled triangle with OP as the hypotenuse.

Using Pythagoras theorem,

⇒ x² = 24² + 18²

⇒ x² = 576 + 324

⇒ x² = 900

⇒ x = $\sqrt{900}$

⇒ x = 30

We know that,

Tangents drawn from an exterior point to a circle are equal in length.

⇒ y = 18

Hence, x = 30 and y = 18.

1Objective Type Questions - Mental Maths

Fill in the blanks:

(i) A chord of a circle is a line segment with its end points ... .

(ii) A diameter of a circle is a chord that ... the centre of circle.

(iii) A line meets a circle atmost in ... points.

(iv) One-half of the whole arc of a circle is called a ... .

(v) The angle subtended by an arc of a circle at the centre of the circle is called the ... by the arc.

(vi) A line which meets a circle in one and only one point is called a ... to the circle.

(vii) The tangent at any point of a circle and the radius through that point are ... to each other.

(viii) From a point outside the circle, ... tangents can be drawn to the circle.

(ix) The measure of an angle in a semicircle is ... .

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(i) A chord of a circle is a line segment with its end points on the circle.

(ii) A diameter of a circle is a chord that passes through the centre of circle.

(iii) A line meets a circle atmost in two points.

(iv) One-half of the whole arc of a circle is called a semicircle.

(v) The angle subtended by an arc of a circle at the centre of the circle is called the angle subtended by the arc.

(vi) A line which meets a circle in one and only one point is called a tangent to the circle.

(vii) The tangent at any point of a circle and the radius through that point are perpendicular to each other.

(viii) From a point outside the circle, two tangents can be drawn to the circle.

(ix) The measure of an angle in a semicircle is 90°.

2Objective Type Questions - Mental Maths

State whether the following statements are true (T) or false (F):

(i) A line segment with its end-points lying on a circle is called a radius of the circle.

(ii) Diameter is the longest chord of the circle.

(iii) The end-points of a diameter of a circle divide the circle into two parts; each part is called a semicircle.

(iv) A diameter of a circle divides the circular region into two parts; each part is called a semicircular region.

(v) The diameters of a circle are concurrent. The centre of the circle is the point common to all diameters.

(vi) Every circle has unique centre and it lies inside the circle.

(vii) Every circle has unique diameter.

(viii) From a given point in the exterior of a circle, two tangents can be drawn to it and these two tangents are equal in length.

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(i) A line segment with its end-points lying on a circle is called a chord of the circle and not a radius of the circle.

Hence, the given statement is false.

(ii) The longest chord of a circle is the one which passes through the centre of the circle and it is called the diameter.

Hence, the given statement is true.

(iii) The end-points of a diameter of a circle divide the circle into two equal arcs and each of these arcs is called a semicircle.

Hence, the given statement is true.

(iv) A diameter of a circle divides the circular region into two equal parts and each of these parts is called a semicircular region.

Hence, the given statement is true.

(v) Every diameter of a circle passes through its centre, so all the diameters of a circle are concurrent and the centre is the point common to all of them.

Hence, the given statement is true.

(vi) A circle has one and only one centre and this centre always lies in the interior of the circle.

Hence, the given statement is true.

(vii) An infinite number of diameters can be drawn in a circle as every chord passing through the centre is a diameter.

Hence, the given statement is false.

(viii) From a point lying in the exterior of a circle, exactly two tangents can be drawn to the circle and these two tangents are equal in length.

Hence, the given statement is true.

31 Mark · Objective Type Questions - Multiple Choice…

If P and Q are any two points on a circle, then the line segment PQ is called a

(a)radius of the circle
(b)diameter of the circle
(c)chord of the circle
(d)secant of the circle

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We know that,

A chord of a circle is a line segment with its end points on the circle.

Here, P and Q are two points lying on the circle.

∴ The line segment PQ is a chord of the circle.

Hence, option 3 is the correct option.

41 Mark · Objective Type Questions - Multiple Choice…

If P is a point in the interior of a circle with centre O and radius r, then

(a)OP = r
(b)OP > r
(c)OP ≥ r
(d)OP < r

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We know that,

A point lies in the interior of a circle if its distance from the centre is less than the radius of the circle.

Here, P lies in the interior of the circle with centre O and radius r.

∴ OP < r

Hence, option 4 is the correct option.

51 Mark · Objective Type Questions - Multiple Choice…

If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is

(a)6 cm
(b)8 cm
(c)10 cm
(d)20 cm

Show SolutionHide Solution↓

Given, AB = 12 cm, BC = 16 cm and AB ⊥ BC.

If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 8

Since, ∠ABC = 90°, the chord AC subtends a right angle at the point B of the circle.

So, AC is the diameter of the circle passing through the points A and C.

In ∆ABC, using Pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 12² + 16²

⇒ AC² = 144 + 256

⇒ AC² = 400

⇒ AC = $\sqrt{400}$

⇒ AC = 20 cm

As we know that radius is half of the diameter.

⇒ Radius = $\frac{1}{2} \times 20$

⇒ Radius = 10 cm

Hence, option 3 is the correct option.

61 Mark · Objective Type Questions - Multiple Choice…

In the adjoining figure, AB is a diameter of the circle. If AC = BC, then ∠CAB is equal to

(a)30°
(b)45°
(c)60°
(d)90°

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Given, AB is a diameter of the circle and AC = BC.

⇒ ∠CAB = ∠CBA [Angles opposite to equal sides of a triangle are always equal]

We know that,

Angle in a semicircle is a right angle.

⇒ ∠ACB = 90°

In ∆ACB, using angle sum property,

⇒ ∠CAB + ∠CBA + ∠ACB = 180°

⇒ ∠CAB + ∠CAB + 90° = 180°

⇒ 2∠CAB = 180° - 90°

⇒ 2∠CAB = 90°

⇒ ∠CAB = $\frac{90^{\circ}}{2}$

⇒ ∠CAB = 45°

Hence, option 2 is the correct option.

7Objective Type Questions - Statement I-II Type…

Statement I: A chord of a circle passing through its centre is called a diameter.

Statement II: Diameter = 2 × radius.

(a)Statement I is true but Statement II is false.
(b)Statement I is false but Statement II is true.
(c)Both Statement I and Statement II are true.
(d)Both Statement I and Statement II are false.

Show SolutionHide Solution↓

A chord of a circle which passes through the centre of the circle is called a diameter of the circle.

So, statement I is true.

A diameter of a circle is made up of two radii lying on the same straight line, so its length is twice the radius of the circle.

⇒ Diameter = 2 × radius

So, statement II is true.

∴ Both statement I and statement II are true.

Hence, option 3 is the correct option.

8Objective Type Questions - Statement I-II Type…

Statement I: In the adjoining figure, AB is the diameter. If the radius of the circle is 5 cm and PB = 6 cm, then AP = 7 cm.

Statement II: The arc $\overset{\frown}{APB}$ = the arc $\overset{\frown}{AQB}$.

(a)Statement I is true but Statement II is false.
(b)Statement I is false but Statement II is true.
(c)Both Statement I and Statement II are true.
(d)Both Statement I and Statement II are false.

Show SolutionHide Solution↓

Given, the radius of the circle = 5 cm and PB = 6 cm.

As we know that the diameter is twice the radius.

⇒ AB = 2 × 5

⇒ AB = 10 cm

We know that,

Angle in a semicircle is a right angle.

⇒ ∠APB = 90°

In ∆APB, using Pythagoras theorem,

⇒ AP² + PB² = AB²

⇒ AP² + 6² = 10²

⇒ AP² + 36 = 100

⇒ AP² = 100 - 36

⇒ AP² = 64

⇒ AP = $\sqrt{64}$

⇒ AP = 8 cm

So, statement I is false.

Since AB is the diameter of the circle, it divides the circle into two equal arcs and each of these arcs is a semicircle.

⇒ arc APB = arc AQB

So, statement II is true.

∴ Statement I is false but statement II is true.

Hence, option 2 is the correct option.

1Check Your Progress

Draw a circle of radius 2.7 cm. Draw a chord PQ of length 4 cm of this circle. Shade the major segment of this circle.

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Steps of construction:

Mark a point O and with O as centre and radius 2.7 cm, draw a circle.
Take any point P on the circle.
With P as centre and radius 4 cm, draw an arc to meet the circle at Q.
Join P and Q. PQ is the required chord.
Shade the segment which contains the centre O. It is the major segment of the circle.

Draw a circle of radius 2.7 cm. Draw a chord PQ of length 4 cm of this circle. Shade the major segment of this circle. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 8

2(i)Check Your Progress

Draw a circle of radius 3.2 cm and in it make a sector of angle 30°. Draw the diagram and shade the sector.

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Steps of construction:

Mark a point C and with C as centre and radius 3.2 cm, draw a circle.
Draw a radius CA of the circle.
Draw another radius CB such that ∠ACB = 30°.
Shade the sector ACB.

Draw a circle of radius 3.2 cm and in it make a sector of angle 30°. Draw the diagram and shade the sector. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 8

2(ii)Check Your Progress

Draw a circle of radius 3.2 cm and in it make a sector of angle 135°. Draw the diagram and shade the sector.

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Steps of construction:

Mark a point C and with C as centre and radius 3.2 cm, draw a circle.
Draw a radius CA of the circle.
Draw another radius CB such that ∠ACB = 135°.
Shade the sector ACB.

Draw a circle of radius 3.2 cm and in it make a sector of angle 135°. Draw the diagram and shade the sector. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 8

2(iii)Check Your Progress

Draw a circle of radius 3.2 cm and in it make a sector of angle $2\frac{2}{3}$ right angles. Draw the diagram and shade the sector.

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We know that,

1 right angle = 90°.

⇒ Angle of the sector = $2\frac{2}{3} \times 90^{\circ} = \frac{8}{3} \times 90^{\circ} = 240^{\circ}$

Steps of construction:

Mark a point C and with C as centre and radius 3.2 cm, draw a circle.
Draw a radius CA of the circle.
Draw another radius CB such that the angle ACB measured in the anticlockwise direction is 240°.
Shade the sector ACB.

Draw a circle of radius 3.2 cm and in it make a sector of angle 22/3 right angles. Draw the diagram and shade the sector. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 8

3Check Your Progress

Draw a line segment PQ = 6.4 cm. Construct a circle on PQ as diameter. Take any point R on this circle and measure ∠PRQ

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Steps of construction:

Draw a line segment PQ = 6.4 cm.
Draw the perpendicular bisector of PQ, which cuts PQ at the point O.
With O as centre and radius OP, draw a circle. It passes through P as well as through Q.
Take any point R on the circle and join PR and QR.
Measure ∠PRQ.

Draw a line segment PQ = 6.4 cm. Construct a circle on PQ as diameter. Take any point R on this circle and measure ∠PRQ. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 8

On measuring, we get ∠PRQ = 90°.

This is because PQ is a diameter of the circle and the angle in a semicircle is a right angle.

Hence, ∠PRQ = 90°.

4Check Your Progress

In the adjoining figure, the tangent to a circle of radius 6 cm from an external point P is of length 8 cm. Calculate the distance of the point P from the nearest point of the circumference.

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Given,

CT = 6 cm and PT = 8 cm.

We know that,

The tangent at any point of a circle and the radius through that point are perpendicular to each other.

⇒ ∠CTP = 90°

So, ∆CTP is a right angled triangle with CP as the hypotenuse.

Using Pythagoras theorem,

⇒ CP² = CT² + PT²

⇒ CP² = 6² + 8²

⇒ CP² = 36 + 64

⇒ CP² = 100

⇒ CP = $\sqrt{100}$

⇒ CP = 10 cm

Let R be the nearest point of the circumference from the point P.

From figure,

⇒ PR = CP - CR

As, CR = CT = 6 cm

⇒ PR = 10 - 6

⇒ PR = 4 cm

Hence, the distance of the point P from the nearest point of the circumference = 4 cm.

5Check Your Progress

In the adjoining figure, O is the centre of the circle. If ∠ABP = 35° and ∠BAQ = 65°, find

(i) ∠PAB

(ii) ∠QBA

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From figure,

AB is the diameter of the circle with centre O, ∠ABP = 35° and ∠BAQ = 65°.

(i) We know that,

Angle in a semicircle is a right angle.

⇒ ∠APB = 90°

In ∆APB, using angle sum property,

⇒ ∠PAB + ∠ABP + ∠APB = 180°

⇒ ∠PAB + 35° + 90° = 180°

⇒ ∠PAB + 125° = 180°

⇒ ∠PAB = 180° - 125°

⇒ ∠PAB = 55°

Hence, ∠PAB = 55°.

(ii) We know that,

Angle in a semicircle is a right angle.

⇒ ∠AQB = 90°

In ∆AQB, using angle sum property,

⇒ ∠BAQ + ∠QBA + ∠AQB = 180°

⇒ 65° + ∠QBA + 90° = 180°

⇒ ∠QBA + 155° = 180°

⇒ ∠QBA = 180° - 155°

⇒ ∠QBA = 25°

Hence, ∠QBA = 25°.

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