Question : The altitude drawn to the base of an isosceles triangle is 8 cm and its perimeter is 64 cm. The area (in cm2) of the triangle is:
Option 1: 240
Option 2: 180
Option 3: 360
Option 4: 120
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Correct Answer: 120
Solution :
Let the base of the isosceles triangle as $b\operatorname{ cm }$ and each of the equal sides as $a\operatorname{ cm }$.
The altitude drawn to the base = $8\operatorname{ cm }$
The perimeter of the isosceles triangle = $64\operatorname{ cm }$
In an isosceles triangle, the altitude bisects the base, dividing it into two equal parts.
Let each part as $\frac{b}{2}$.
We can use the Pythagorean theorem,
$⇒a = \sqrt{(\frac{b}{2})^2 + 8^2}$ ____(i)
Given that the perimeter of the triangle is $64\operatorname{ cm }$.
$⇒2a + b = 64$ ____(ii)
From equation (i) and equation (ii),
$⇒2\sqrt{(\frac{b}{2})^2 + 8^2} + b = 64$
$⇒\sqrt{ b^2 + 256}=64-b$
$⇒b^2 + 256=4096-128b+b^2$
$⇒128b=3840$
$⇒b = 30\operatorname{ cm }$
From equation (ii),
$⇒a = 17\operatorname{ cm }$.
The area of the triangle $=\frac{1}{2}bh=\frac{1}{2} \times 30 \times 8 = 120\operatorname{ cm^2 }$
Hence, the correct answer is 120.
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