ABC26GN4476 · Area
Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
If the perimeter of an isosceles right triangle is $(6+3 \sqrt{2}) \mathrm{m}$, then the area of the triangle is
(a)$4.5 \mathrm{m}^{2}$
(b)$5.4 \mathrm{m}^{2}$
(c)$9 \mathrm{m}^{2}$
(d)$81 \mathrm{m}^{2}$
Answer
Explanation
Let the sides be $a$ metres, $a$ metres and $b$ metres. Then, $2 a+b=6+3 \sqrt{2}$ and $b^{2}=a^{2}+a^{2}=2 a^{2} \Leftrightarrow b=\sqrt{2} a$. $$\begin{array}{ll} \therefore & 2 a+\sqrt{2} a=6+3 \sqrt{2} \Leftrightarrow a=3 . \\ \therefore & \quad \text { Area }=\left(\frac{1}{2} \times 3 \times 3\right) \mathrm{m}^{2}=4.5 \mathrm{m}^{2} . \end{array}$$
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