## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 15 Circle Check Your Progress

Question 1.

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.

Solution:

(i) Draw a circle of radius = 2·7 cm.

(ii) Take a point P anywhere on the circle.

(iii) With P as centre and 4 cm as radius,

draw an arc which cuts the circle at Q.

(iv) Join PQ which is the required chord.

(v) Shade the major segment.

Question 2.

Draw a circle of radius 3·2 cm and in it make a sector of angle.

(i) 30°

(ii) 135°

(iii) \(2 \frac{2}{3}\) right angles

Draw separate diagrams and shade the sectors.

Solution:

(i) 30°

Steps :

(a) Draw a circle with centre C and radius CB = 3.2 cm

(b) From C, make an angle of 30°.

(c) Shade the region enclosed in ABC.

(ii) 135°

Steps :

(a) Draw a circle with centre C and radius CB = 3.2 cm.

(b) From C, make an angle of 135°.

(c) Shade the region enclosed in ACB.

(iii) \(2 \frac{2}{3}\) right angles

Steps :

(a) Draw a circle with centre C and radius 3.2 cm.

(b) From C, make an ∠, \(2 \frac{2}{3}\) of right angle

= \(2 \frac{2}{3}\) × 90°

= \(\frac{8}{3}\) × 90° = 240°

(c) Shade the region enclosed in ACB.

= \(\frac{8}{3}\) × 90° = 240°

Question 3.

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.

Solution:

(i) Draw a line segment PQ = 6·4 cm.

(ii) Draw ⊥ bisector of PQ.

(iii) With O as centre and OP or OQ as radius draw a circle

which passes through P as well as through Q.

(iv) Take point R on the circle.

(v) Join PR and QR.

(vi) Measure ∠PRQ, we get ∠PRQ = 90°.

Question 4.

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.

Solution:

C is the centre of the circle

PT is the tangent to the circle from P.

CT is the radius

∴ CT ⊥ PT

CT = CR = 6 cm, PT = 8 cm

Now in right ∆CPT (By Pythagoras Theorem)

CP^{2} = PT^{2} + CT^{2} = (8)^{2} + (6)^{2}

= 64 + 36 = 100 = (10)^{2}

∴ CP = 10 cm

Now PR = CP – CR = 10 – 6 = 4 cm

Question 5.

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

(i) ∠PAB

(ii) ∠QBA

Solution:

In the figure,

AB is the diameter of the circle with centre O

∠ABP = 35° and ∠BAQ = 65°

(i) ∠APB = 90° (Angle in a semicircle)

In ∆APB, By ∠sum property of ∆

∠PAB + ∠P + ∠ABP = 180°

∠PAB + 90° + ∠ABP = 180°

∴ ∠PAB + ∠ABP = 90°

⇒ ∠PAB + 35° = 90° ⇒ ∠PAB = 90° – 35° ∠PAB = 55°

(ii) Similarly ∠AQB = 90° (Angle in a semicircle)

In ∆AOB, By angle sum property of ∆

∠BAQ + ∠Q + ∠QBA = 180°

∠BAQ + ∠QBA + 90° = 180°

∴ ∠BAQ + ∠QBA = 90°

⇒ 65° + ∠QBA = 90°

⇒ ∠QBA = 90°- 65° = 25°

Hence ∠PAB = 55° and ∠QBA = 25°