Question

Find the area of the quadrilateral ABCD in which BCD is an equilateral triangle, each of whose sides is 26 cm, AD = 24 cm, and $\angle BAD={90}^{\circ }.$ Also, find the perimeter of the quadrilateral. $(Given,\sqrt{3}=\mathrm{1.73.})$

Answer 1

AB = $\sqrt{{26}^2–{24}^2}=10$ [Pythagoras theorem]

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

Now, BCD is an equilateral triangle, the angles of the equilateral triangle with side, $a=26cm$

Area of equilateral triangle BCD $=\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}\times {26}^2=\frac{1.73}{4}\times 676=292.37c{m}^2$

Area of quadrilateral ABCD $=$ Area of triangle ABD $+$ Area of equilateral triangle BCD

$=120+292.37=412.37c{m}^2$

Perimeter of quadrilateral ABCD = AB+BC + CD+DA

$=10+26+26+24=86cm$