Question

Calculate the area of the designed region in figure, common between the two quadrants of circles of radius $8cm$ each.

Answer 1

i) Area of Square $ABCD$
$=a^2=(8{)}^2=64c{m}^2$
ii) Sum of Areas of Part II and Part III:
= Area of segment with centre $D$ and radius of $8cm$
$\therefore =\frac{\theta }{360}\times \pi r^2$
$=\frac{90}{360}\times \frac{22}{7}\times (8{)}^2$
$=\frac{1}{4}\times \frac{22}{7}\times 64$
$=\frac{352}{7}sq.cm$
iii) Area of part I:
=(Sum of Part I, II and III)-(Sum of area of Part II, III)
= Area of Square $ABCD-$ (Sum of area of Part II and III)
$=64c{m}^2-\frac{352}{7}c{m}^2$
$=\frac{448-352}{7}c{m}^2$
$=\frac{96}{7}c{m}^2$
Similarly, Area of Part II$=\frac{96}{7}c{m}^2$
$\therefore$ Area of Region III $=\frac{352}{7}-\frac{96}{7}$
$=\frac{256}{7}c{m}^2$