MAH CET MCA Properties Of Triangle Previous Year Questions (PYQs)
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In the triangle $ABC$, sides $AB$, $BC$, and $CA$ are extended such that $B'$, $C'$, and $A'$ are the new points formed.
If the area of $\triangle ABC = 1$ unit and
$AA' = 2AB,\ BB' = 2BC,\ CC' = 3AC$,
then find the area of $\triangle A'B'C'$.
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Solution
Each side is scaled by 2, 2, and 3 respectively; the area ratio depends on the product of scale factors.
Hence, area $= 18$ times original.
In $\triangle ABC$, sides are $a=9$, $b=7$, $c=12$. Find angle $C$ (opposite side $c$) using the Law of Cosines.
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Solution
$\cos C=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{81+49-144}{2\cdot9\cdot7}=\dfrac{-14}{126}=-\dfrac{1}{9}$.
$C=\cos^{-1}\left(-\dfrac{1}{9}\right)\approx97.57^\circ$.
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