Circle — Exercise 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.

Answer

Steps of construction:

1. Mark a point O and with O as centre and radius 2.5 cm, draw a circle.

2. Take any point A on the circle and join OA.

3. Draw another radius OB such that ∠AOB = 60°.

4. Join A and B. AB is the required chord.

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 = $\dfrac{120\degree}{2}$ = 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.

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.

Answer

Steps of construction:

1. Mark a point C and with C as centre and radius 3.2 cm, draw a circle.

2. Take any point A on the circle.

3. With A as centre and radius 5 cm, draw an arc to meet the circle at B.

4. Join A and B. AB is the required chord.

5. Shade the segment which does not contain the centre C. It is the minor segment of the circle.

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.

Answer

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.

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.

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.

Answer

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.

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

Answer

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°.

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

Answer

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°.

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

Answer

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 = $\dfrac{90\degree}{2}$

⇒ x = 45°

Hence, x = 45°.

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

Answer

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°.

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

Answer

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°.

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

Answer

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 = $\dfrac{90\degree}{2}$

⇒ 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 = $\dfrac{45\degree}{2}$

⇒ y = $22\dfrac{1}{2}^\degree$

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

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

Answer

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°.

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

Answer

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 = $\dfrac{90\degree}{2}$

⇒ x = 45°

⇒ y = 45°

Hence, x = 45° and y = 45°.

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

Answer

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 = $\dfrac{54\degree}{2}$

⇒ y = 27°

Hence, x = 54° and y = 27°.

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

Answer

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 = $\dfrac{1}{2} \times 17$

⇒ y = 8.5

Hence, x = 17 and y = 8.5.

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

Answer

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.

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

Answer

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.