# Find the area of the shaded region given in Fig. 11.20

**Solution:**

We have to find the shaded region from the given figure.

We observe ABCD is a square with a side 14 cm.

Join JK, KL, LM and JM.

So, JKLM forms a square

KEJ, JHM, MGL and LFK are four equal semicircles.

The dimension of FH = EG = 14 - 3 - 3

= 14 - 6

= 8 cm

The side of square JKLM will be 4 cm

The diameter of semicircle will be 4 cm

So, radius of the semicircle = 4/2 = 2 cm

Area of the shaded region = area of square ABCD - area of square JKLM - area of 4 semicircles

Area of square = (side)²

Area of square ABCD = (14)²

= 196 cm²

Area of square JKLM = (4)²

= 16 cm²

Area of semicircle = πr²/2

Area of semicircle KEJ = (22/7)(2)²/2

= (22/7)(2)

= 44/7

= 6.2857 cm²

Since all semicircles are equal.

Area of 4 semicircle = 4(area of one semicircle)

= 4(6.2857)

= 25.14286 cm²

Area of the shaded region = 196 - 16 - 25.14286

= 180 - 25.14286

= 154.857 cm²

Therefore, the area of the shaded region is 154.857 cm²

**✦ Try This:** ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠ DBC = 70°, ∠ BAC = 30°, find ∠ BCD. Further, if AB = BC, find ∠ ECD.

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

**NCERT Exemplar Class 10 Maths Exercise 11.4 Problem 17**

## Find the area of the shaded region given in Fig. 11.20

**Summary:**

The area of the shaded region is 154.857 cm²

**☛ Related Questions:**

- Find the number of revolutions made by a circular wheel of area 1.54 m² in rolling a distance of 176 . . . .
- Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm subten . . . .
- Find the difference of the areas of a sector of angle 120° and its corresponding major sector of a c . . . .

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