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

**Answer
1** :

Solution:

Here, P is in the semi-circle and so,

P = 90°

So, it can be concluded that QR is hypotenuseof the circle and is equal to the diameter of the circle.

∴ QR = D

Using Pythagorean theorem,

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

Or, QR^{2 }= 7^{2}+24^{2}

QR= 25 cm = Diameter

Hence, the radius of the circle = 25/2 cm

Now, the area of the semicircle = (πR^{2})/2

= (22/7)×(25/2)×(25/2)/2 cm^{2}

= 13750/56 cm^{2 }= 245.54 cm^{2}

Also, area of the ΔPQR = ½×PR×PQ

=(½)×7×24 cm^{2}

= 84 cm^{2}

Hence, the area of the shaded region = 245.54cm^{2}-84 cm^{2}

= 161.54 cm^{2}

Find the area of theshaded region in Fig. 12.20, if radii of the two concentric circles with centreO are 7 cm and 14 cm respectively and AOC = 40°.

**Answer
2** :

Given,

Angle made by sector = 40°,

Radius the inner circle = r = 7 cm, and

Radius of the outer circle = R = 14 cm

We know,

Area of the sector = (θ/360°)×πr^{2}

So, Area of OAC = (40°/360°)×πr^{2 }cm^{2}

= 68.44 cm^{2}

Area of the sector OBD = (40°/360°)×πr^{2 }cm^{2}

= (1/9)×(22/7)×7^{2 }= 17.11 cm^{2}

Now, area of the shaded region ABDC = Area ofOAC – Area of the OBD

= 68.44 cm^{2} – 17.11 cm^{2 }=51.33 cm^{2}

Find the area of theshaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPCare semicircles.

**Answer
3** :

Find the area of theshaded region in Fig. 12.22, where a circular arc of radius 6 cm has been drawnwith vertex O of an equilateral triangle OAB of side 12 cm as centre.

**Answer
4** :

**Answer
5** :

Side of the square = 4 cm

Radius of the circle = 1 cm

Four quadrant of a circle are cut from cornerand one circle of radius are cut from middle.

Area of square = (side)^{2}= 4^{2 }=16 cm^{2}

Area of the quadrant = (πR^{2})/4 cm^{2} =(22/7)×(1^{2})/4 = 11/14 cm^{2}

∴ Total area of the 4 quadrants = 4 ×(11/14) cm^{2} =22/7 cm^{2}

Area of the circle = πR^{2 }cm^{2} =(22/7×1^{2}) = 22/7 cm^{2}

Area of the shaded region = Area of square –(Area of the 4 quadrants + Area of the circle)

= 16 cm^{2}-(22/7) cm^{2}+(22/7)cm^{2}

= 68/7 cm^{2}

**Answer
6** :

Find the area of the design.