Subject

    Quantitative Aptitude

    Topic

    Areas & Volumes

    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.

    ক)
    24
    খ)
    48
    গ)
    60
    ঘ)
    72

    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 , (AC)2=(AD)2+(DC)2\left(AC\right)^2=\left(AD\right)^2+\left(DC\right)^2 ⇒x2=(8)2+(16−x)2⇒x=10\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 = 12∗BC∗AD=12∗12∗8=48cm2\frac12\ast BC\ast AD=\frac12\ast12\ast8=48cm^2

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