# Mensuration-36

### From Homeworkwiki

**ABCP is a quadrant of a circle of radius 14 cm. With AC as diameter, semicircle is drawn. Find the area of the shaded**
**portion APCQA .**

**Solution:** In right angle triangle ABC,

AC^{2} = AB^{2} + BC^{2} (Pythagoras theorem)

=> AC^{2} = 14^{2} + 14^{2} = 2 x 14^{2}

=> AC = √(2 x 14) = 14 √2

=> AC / 2 = 14 √2 / 2 cm = 7 √2 cm

Area of the portion between the quadrant and the semi-circle

= (Area of region ABCQA) - Area of quadrant ABCPA

Area of region ABCQA = Area of triangle ABC + Area of semicircle ACQA

= (1/2 x 14 x 14 + 1/2 x 22/7 x 7 √2 x 7 √2) cm^{2} = (98 + 154) cm^{2}

= 252 cm^{2}

Area of quadrant ABCPA = (1/4 x 22/7 x 14 x 14) cm^{2} = 154 cm^{2}

Therefore, Area of shaded region APCQA = 252 cm^{2} - 154 cm^{2} = 98 cm^{2}