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.
Solution
Correct Answer: Option B
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 , \(\left(AC\right)^2=\left(AD\right)^2+\left(DC\right)^2\)
\(\Rightarrow x^2=\left(8\right)^2+\left(16-x\right)^2\Rightarrow x=10\)
BC = (32- 2x) = (32 - 20) cm = 12 cm.
Hence, required area = \(\frac12\ast BC\ast AD=\frac12\ast12\ast8=48cm^2\)
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