Question

Find the area of the quadrilateral ABCD in which AD = 24 cm, $\angle$ BAD = ${90}^o$ and $\Delta$ BCD is an equilateral triangle having each side equal to 26 cm. Also, find the perimeter of the quadrilateral. [Given, $\sqrt{3}$ = 1.73.]

Answer 1

ΔBDC is an equilateral triangle with side a = 26 cm

Area of ΔBDC $=\frac{\sqrt{3}{a}^2}{4}=\frac{\sqrt{3}\times {26}^2}{4}=\frac{\sqrt{3}\times 676}{4}=292.37c{m}^2$

By using Pythagoras theorem in the right-angled triangle ΔDAB, we get:

$A{D}^2+A{B}^2=B{D}^2{24}^2+A{B}^2={26}^{2}A{B}^2={26}^2-{24}^2=676-{576}^2=100AB=10cm$

Area of ΔABD $=\frac{1}{2}\times base\times height\frac{1}{2}\times 10\times 24=120c{m}^2$

Area of the quadrilateral = Area of ΔBDC + Area of ΔABD
$=292.37+120=412.37c{m}^2$

Perimeter of the quadrilateral $=AB+AC+CD+AD=24+10+26+26=86cm$