The altitude drawn to the base of an isosceles triangle is 8 cm and the perimete

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Question
The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. Find the area of the triangle.
Options
A. 24
B. 48
C. 60
D. 72

Correct Answer

48

Explanation

Let ABC be the isosceles triangle and AD be the altitude

Let AB = AC = x. Then, BC = (32 - 2x).

Since, in an isosceles triangle, the altitude bisects the base,
so BD = DC = (16 - x).
In triangle ADC, ${(A C)}^{2} = {(A D)}^{2} + {(D C)}^{2}$

$? x^{2} = {(8)}^{2} + {(16 - x)}^{2} ? x = 10$

BC = (32- 2x) = (32 - 20) cm = 12 cm.

Hence, required area = $\frac{1}{2} * B C * A D = \frac{1}{2} * 12 * 8 = 48 c m^{2}$

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The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. Find the area of the triangle.

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The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. Find the area of the triangle.

Area problems

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Aptitude

Engineering

Programming

Reasoning

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