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# 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

**Solution:**

From figure, it is clear that the diameter of both the semicircles = Side of the square = 14 cm

∴ Radius of each semicircle (r) = 14/2 = 7 cm

Visually, it is clear that

Semicircles APD and BPC are drawn using sides AD and BC respectively as their diameter.

∴ Diameter of each semicircle = 14 cm

Radius of each semicircle (r) = 14/2 = 7 cm

Area of shaded region = Area of square ABCD - (Area of semicircle APD + Area of semicircle BPC)

= (side)^{2} - (1/2πr^{2} + 1/2πr^{2})

= (14)^{2} - π × (7)^{2}

= 196 cm^{2} - 22/7 × 7 cm × 7 cm

= 196 cm^{2} - 154 cm^{2}

= 42 cm^{2}

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 12

**Video Solution:**

## 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.

NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.3 Question 3

**Summary:**

The area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles is 42 cm^{2}.

**☛ Related Questions:**

- Find the area of the shaded region in Fig. 12.22, 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.
- 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 Fig. 12.23. Find the area of the remaining portion of the square.
- 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.
- In Fig. 12.25, ABCD is a square of side 14 cm. With centres 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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