Question

Find the area of a quadrilateral ABCD in which AD = 24 cm, $\angle BAD={90}^0$ and BCD forms an equilateral triangle whose each side is equal to 26 cm. (Take $\sqrt{3}=1.73)$

Answer 1

In quadrilateral ABCD, AD = 24 cm, $\angle BAD={90}^0$

BCD is an equilateral triangle with side 26 cm
In right $\Delta ABD$,
$B{D}^2=A{B}^2+A{D}^2$
$(26{)}^2=A{B}^2+(24{)}^2$
$\Rightarrow 676=A{B}^2+576$
$A{B}^2=676-576=100=(10{)}^2$
$\therefore$ AB = 10 cm
Now area of right $\Delta ABD$,
= $\frac{1}{2}base\times altitude=\frac{1}{2}\times 10\times 24$
= $120c{m}^2$
and area of equilateral $\Delta BCD$ with side 26 cm
= $\frac{\sqrt{3}}{4}\times (side{)}^2=\frac{1.73}{4}\times 26\times 26c{m}^2$
= 292.37 $c{m}^2$
$\therefore$ Total area of a quadrilateral ABCD,
= $120+292.37c{m}^2=412.37c{m}^2$