# ML Aggarwal Circle Check Your Progress Class 8 ICSE Ch-15 Maths Solutions

ML Aggarwal Circle Check Your Progress Class 8 ICSE Ch-15 Maths Solutions. We Provide Step by Step Answer of  Check Your Progress  Questions for Circle as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

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

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 8th Chapter-15 Circle Writer ML Aggarwal Book Name Understanding Topics Solution of Check Your Progress Questions Edition 2023-2024

ML Aggarwal Class 8 ICSE Maths Solutions

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Circle Class-8 ML Aggarwal ICSE Mathematics Solutions

#### 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.

(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.

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

(i) 30°
(ii) 135°
(iii) 2(2/3)° right angles
Draw separate diagrams and shade the sectors.

(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(2/3) right angles
Steps :
(a) Draw a circle with centre C and radius 3.2 cm.
(b) From C, make an ∠, 2(2/3) of right angle
= 2(2/3) × 90°
= 8/3 × 90° = 240°
(c) Shade the region enclosed in ACB.

= 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.

(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.

C is the centre of the circle
PT is the tangent to the circle from P.
∴ CT ⊥ PT
CT = CR = 6 cm, PT = 8 cm
Now in right ∆CPT (By Pythagoras Theorem)
CP2 = PT2 + CT2 = (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

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°

— End of Circle Check Your Progress Class 8 ICSE Maths Solutions :–

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