Question

Find the area of the quadrilateral ABCD in which AD = 24 cm, ∠BAD = 90° and BCD forms an equilateral triangle whose each side is equal to 26 cm. Also find the perimeter of the quadrilateral. Take $\sqrt{3}$ = 1.73.

Answer 1

$∆$BDC is an equilateral triangle with side a= 26 cm.
Area of $∆BDC=\frac{\sqrt{3}}{4}{a}^2$
$$\begin{gathered} =\frac{\sqrt{3}}{4}\times {26}^2 \\ =\frac{1.73}{4}\times 676 \\ =292.37{\mathrm{cm}}^2 \end{gathered}$$

By using Pythagoras' theorem in the right-angled triangle $∆$DAB, we get:
$$\begin{gathered} A{D}^2+A{B}^2=B{D}^2 \\ \Rightarrow {24}^2+A{B}^2={26}^2 \\ \Rightarrow A{B}^2={26}^2-{24}^2 \\ \Rightarrow A{B}^2=676-576 \\ \Rightarrow A{B}^2=100 \\ \Rightarrow AB=10\mathrm{cm} \end{gathered}$$

Area of $∆ABD=\frac{1}{2}\times b\times h$
$$\begin{gathered} =\frac{1}{2}\times 10\times 24 \\ =120{\mathrm{cm}}^2 \end{gathered}$$

Area of the quadrilateral = Area of $∆$BCD + Area of $∆$ABD
$=292.37+120$
= 412.37 cm²

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