Question

The figure shows an equilateral $\Delta ABC$ of side 10 cm and a right angle $\Delta BDC$ whose side BD = 8 cm and $\angle D={90}^{\circ }$.

Then, find the area of the shaded portion.

Answer 1

In $\Delta BDC$,

$D{C}^2={10}^2-8^2$

$D{C}^2=6^2$
(6-8-10 Pythagoras Triplet)

Area of shaded region = Area $\Delta ABC-$ Area $\Delta BDC$

$=\frac{\sqrt{3}}{4}(10{)}^2-(\frac{1}{2}\times 6\times 8$)

$=\frac{\left(1.732\right)\left(100\right)}{4}-24$

$=43.3-24$

$=19.3c{m}^2$