Question

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

Answer 1

Given,
AB = BC = CD = AD = 8 cm

We know that, area of $\Delta$
$=\frac{1}{2}\times \left(base\right)\times \left(height\right)$

Area of $\Delta ABC=Areaof\Delta ADC$
$=\frac{1}{2}\times 8\times 8=32c{m}^2$

Area of quadrant AECB = Area of quadrant AFCD
$=\frac{\left(\pi R^2\right)}{4}c{m}^2$
$=\frac{(\frac{22}{7}\times 8^2)}{4}c{m}^2$
$=\frac{352}{7}c{m}^2$

$\therefore$ Area of shaded region

= (Area of quadrant AECB - Area of $\Delta ABC)$ + (Area of quadrant AFCD - Area of $\Delta ADC)$

$=(\frac{352}{7}-32)+(\frac{352}{7}-32)c{m}^2$

$=2\times (\frac{352}{7}-32)c{m}^2$

$=\frac{256}{7}c{m}^2$