ABC26GN4477 · Area
Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
The perimeter of an isosceles right-angled triangle having an area of $162 \mathrm{cm}^{2}$ is
(a)40 cm
(b)56.5 cm
(c)61.38 cm
(d)68.2 cm
Answer
Explanation
Let the length of the base and height be $x \mathrm{cm}$ each. Then, $\frac{1}{2} x^{2}=162 \Rightarrow x^{2}=324 \Rightarrow x=\sqrt{324}=18 \mathrm{cm}$. $$\begin{aligned} \text { Hypotenuse } & =\sqrt{(18)^{2}+(18)^{2}} \mathrm{cm}=\sqrt{648} \mathrm{cm}=18 \sqrt{2} \mathrm{cm} . \\ \therefore \text { Perimeter } & =(18+18+18 \sqrt{2}) \mathrm{cm}=18(2+\sqrt{2}) \mathrm{cm} \\ & =18(2+1.41) \mathrm{cm}=(18 \times 3.41) \mathrm{cm} \\ & =61.38 \mathrm{cm} . \end{aligned}$$
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