# Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle

**Solution:**

We know that, the angle in a semicircle is a right angle.

∴ ∠RPQ = 90°

Thus, ΔRQP is a right-angled triangle.

Given, PQ = 24 cm, PR = 7 cm

∴ Using Pythagoras theorem,

RQ^{2} = PR^{2 }+ PQ^{2}

RQ = √7² + 24²

= √49 + 576

= √625

Thus, PQ = 25 cm which is the diameter

∴ Radius (r) = 25/2 cm

Area of shaded region = Area of semicircle RPQ - Area of ΔRQP

= 1/2 × πr^{2 }- 1/2 × PQ × RP

= 1/2 [(22/7 × 25/2 × 25/2) - (24 × 7)]

= 1/2 [6875/14 - 168]

= 1/2 [(6875 - 2352)/14]

= 1/2 × 4523/14

= 4523/28 cm^{2}

= 161.54 cm^{2 }(approximately)

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

**Video Solution:**

## Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle

NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.3 Question 1

**Summary:**

The area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle is 161.54 cm^{2}.

**☛ Related Questions:**

- 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°.
- 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.
- 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.

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