ABC26GN4688 · Area
Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
The base of an isosceles is 14 cm and its perimeter is 36 cm. Find its area.
(a)$42 \sqrt{2}$ sq. cm.
(b)42 sq. cm
(c)84 sq. cm
(d)48 sq. cm [ESIC-UDC Exam, 2016]
Answer
Explanation
Let each equal side of isosceles triangle be $x$ cm Perimeter of an isosceles triangle $=36 \mathrm{cm}$ $$\begin{aligned} & \therefore x+x+14=36 \\ & \Rightarrow 2 x=36-14=22 \\ & =x=\frac{22}{2}=11 \mathrm{cm} . \end{aligned}$$ $\mathrm{BD}=\mathrm{DC}=7 \mathrm{cm}$. From $\triangle \mathrm{ABD}$. By using Pythagoras theorem $$\begin{aligned} \mathrm{AD} & =\sqrt{A B^{2}-B D^{2}}=\sqrt{11^{2}-7^{2}} \\ & =\sqrt{121-49}=\sqrt{72} \\ & =3 \times 2 \sqrt{2} \end{aligned}$$ $6 \sqrt{2}$ cm. ∴ Area of $\triangle \mathrm{ABC}$ $$\begin{aligned} & =\frac{1}{2} \times B C \times A D \\ & =\frac{1}{2} \times 14 \times 6 \sqrt{2} \\ & =42 \sqrt{2} \text { sq. } \mathrm{cm} . \end{aligned}$$
Open in whiteboard · Browse this chapter in the app