Circle — Check 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.

Answer

Steps of construction:

1. Mark a point O and with O as centre and radius 2.7 cm, draw a circle.

2. Take any point P on the circle.

3. With P as centre and radius 4 cm, draw an arc to meet the circle at Q.

4. Join P and Q. PQ is the required chord.

5. Shade the segment which contains the centre O. It is the major segment of the circle.

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

Answer

Steps of construction:

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

2. Draw a radius CA of the circle.

3. Draw another radius CB such that ∠ACB = 30°.

4. 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.

Answer

Steps of construction:

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

2. Draw a radius CA of the circle.

3. Draw another radius CB such that ∠ACB = 135°.

4. Shade the sector ACB.

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

Answer

We know that,

1 right angle = 90°.

⇒ Angle of the sector = $2\dfrac{2}{3} \times 90\degree = \dfrac{8}{3} \times 90\degree = 240\degree$

Steps of construction:

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

2. Draw a radius CA of the circle.

3. Draw another radius CB such that the angle ACB measured in the anticlockwise direction is 240°.

4. Shade the sector ACB.

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

Answer

Steps of construction:

1. Draw a line segment PQ = 6.4 cm.

2. Draw the perpendicular bisector of PQ, which cuts PQ at the point O.

3. With O as centre and radius OP, draw a circle. It passes through P as well as through Q.

4. Take any point R on the circle and join PR and QR.

5. Measure ∠PRQ.

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°.

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.

Answer

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.

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

(i) ∠PAB

(ii) ∠QBA

Answer

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°.