NCERT Solution For Class 10, Maths, Chapter 12, Areas Related To Circles Exercise 12.3 solutions are given below. These questions are very important for Class 10 maths board examination. Students of class 10 first should go through the ex. 12.1 and Ex. 12.2 then solve exercise 12.3 for better learning.
Class 10, Maths, Chapter 12, Exercise 12.3, Solutions
Q.1. Find the area of the shaded region in Figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.
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Q.2. Find the area of the shaded region in Fig. 12.20, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠ AOC = 40°.
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Q.3. Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
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Q.4. Find the area of the shaded region in Figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.
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Q.5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Figure. Find the area of the remaining portion of the square.
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Area of the remaining portion of the square
= Area of the Square (ABCD) – Area of two circles
=(Side)2 – 2×(πr2) where, side = 4 cm and r = 1 cm
=(4)2 – 2 × $\frac{22}{7}$ × 1 × 1
=16 – $\frac{44}{7}$ = $\frac{122-44}{7}=\frac{88}{7}$ cm2
Q.6. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. 12.24. Find the area of the design.
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Q.7. In Fig. 12.25, ABCD is a square of side 14 cm. With centers A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.
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Area of the shaded region
= Area of the Square (ABCD) – Area of one circle
= (Side)2 – πr2 where, side = 14 cm and r = 7 cm
= (14)2 – $\frac{22}{7}$ x 7 x 7
=196 – 154 = 42 cm2
Hence, Area of the shaded region is 42 cm2
Q.8. Fig. 12.26 depicts a racing track whose left and right ends are semicircular.
The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:
(i) the distance around the track along its inner edge
(ii) the area of the track.
Ans: We have, OB=O’C = 30 m
And, AB = CD = 10 m
∴ OA = O’D = (30+10 )= 40 m
Q.9. In Fig. 12.27, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.
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Q.10. The area of an equilateral triangle ABC is 17320.5 cm2. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and $\sqrt{3}$= 1.73205)
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Q.11. On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig. 12.29). Find the area of the remaining portion of the handkerchief.
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Q.12. In Fig. 12.30, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the
(i) Quadrant OACB,
(ii) Shaded region.
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Q.13. In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)
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Q.14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Figure. If ∠ AOB = 30°, find the area of the shaded region.
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Q.15. In Fig. 12.33, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.
Ans: Let 14 cm be the radius of the quadrant with A as the centre.
Q.16. Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each.
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