Question : In an isosceles right-angled triangle, the perimeter is 30 m. Find its area (in m2). (Rounded off to the nearest integral value)
Option 1: 41
Option 2: 36
Option 3: 39
Option 4: 34
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Correct Answer: 39
Solution :
The perimeter of right-angled isosceles triangle = 30 m
Let the equal sides of the isosceles triangle be $a$ m.
Hence, hypotenuse of the isosceles right angled triangle = $\sqrt{a^2+a^2}=\sqrt2a$
According to the question,
$a+a+\sqrt2a=30$
⇒ $a=\frac{30}{2+\sqrt2}$
⇒ $a=15(2-\sqrt2)$
Area of isosceles right angle triangle
= $\frac{1}{2}\times a^2$
= $\frac{1}{2}\times (15(2-\sqrt2))^2$
= $\frac{1}{2}\times 225\times (4+2-4\sqrt2)$
= $225\times(3-2\sqrt2)$
= $38.63$ m
Hence, the correct answer is 39.
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