RS Aggarwal - Mathematics

Book: RS Aggarwal - Mathematics

Chapter: 18. Area of Circle, Sector and Segment

Subject: Maths - Class 10th

Q. No. 27 of Exercise 18B

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The side of a square is 10 cm. Find

(i) The area of the inscribed circle, and

(ii) The area of the circumscribed circle. [Take π = 3.14.]


Consider the above figure, Join PR,

Now PR = Diameter of the inscribed circle

Also, PR = BC = 10 cm.

So, PR = 10 cm

r = 5 cm

Area of inscribed circle = πr2 (put value of r in this equation)

Area of inscribed circle = π(52)

Area of inscribed circle = 78.57 cm2

The area of inscribed circle is 78.57 cm2.


Consider the above figure, O is the centre of circle and ABCD is a square inscribed. Now OB and OD are radii of circle.

Consider ∆DBC right angled at c (as C is a vertex of square)

Apply Pythagoras theorem in triangle DBC

Hypotenuse2 = Perpendicular2 + Base2

In triangle DBC, hypotenuse = DB,

perpendicular = BC and

base = DC

Put the values of BC and DC i.e. 10 cm

BD2 = 200

BD = √200

BD = 10√2 cm

Now radius of circle = half of BD

r = (10√2)/2

r = 5√2 cm

Hence Area of circumscribing circle = πr2

Area of circumscribing circle = 3.14×5√2×5√2

(put π = 3.14 and r = 5√2 cm)

Area of circumscribing circle = 3.14 × 50

Area of circumscribing circle = 157 cm2

Area of circumscribing circle is 157 cm2.

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