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Find the percentage increase in the area of a triangle if each side is increased by ^@n^@ times.
Answer:
^@100(n^2 -1) \%^@
- Consider a ^@\triangle QRS^@ with sides ^@a, b^@ and ^@c.^@ Let ^@S = \dfrac { a+b+c } {2}^@.
Area of ^@\triangle QRS, A_1 = \sqrt { S(S-a)(S-b)(S-c) } ^@ - Increasing the side of each side by ^@n^@ times, we get a new ^@\triangle XYZ^@.
^@\triangle XYZ^@ has sides ^@na, nb^@ and ^@nc^@. - By Heron's formula:
Area of new triangle ^@= \sqrt{ S_{1}(S_{1}-na)(S_{1}-nb)(S_{1}-nc) }^@
Where, ^@S_1 = \dfrac { na + nb + nc }{2} = n × \dfrac { a+b+c } {2}^@
^@\begin{align} \text { Area of } \triangle XYZ & = \sqrt{ nS(nS-na)(nS-nb)(nS-nc) } \\ & = \sqrt{ n^{4}S(S-a)(S-b)(S-c)) } \\ & = n^2 \times A_1 \end{align}^@ - Increase in area ^@ = n^2 A_1 - A_1 ^@
^@ \% ^@ Increase in area ^@ = \dfrac{ A_1(n^2 - 1) }{ \dfrac{ A_1 }{ 100 } } = ^@ ^@100(n^2 -1) \%^@.
