Triangle Similarity

when 2 traingles are similar then their areas are in the ratio of the squares of their sides so when ratio of areas is 36 ;25 their ratio of sides will be 6:5 and when we consider smaller side is 5x then x is 16 and thus 6x is 96

1) Draw a rough sketch of the two triangles.

2) Consider the two triangles as equilateral triangles. Let the sides of the triangles be a and b where a > b.

3) The areas of these triangles are: √3/4 a-square and √3/4 b-square.

4) According to question: √3/4 a2 divided by √3/4 b2 = 36x/25x; a2/b2 = 36x/25x; a/b = 6x/5x.

5) According to question: 5x = 80, hence x = 18. Therefore 6x = 6(18) = 108.

        Hence solved...

Relevant wiki: Heron's Formula

Call the sides of the $\triangle_{big}$ are $S_{1},S_{2},S_{3},\frac{S_{1}+S_{2}+S_{3}}{2}=K$ and the $\triangle_{small}$ respectively are $s_{1},s_{2},s_{3},\frac{s_{1}+s_{2}+s_{3}}{2}=k$ .Because $\triangle_{big}~is~congruent~to \triangle_{small}$ so $\frac{S_{1}}{s_{1}}=\frac{S_{2}}{s_{2}}=\frac{S_{3}}{s_{3}}=\frac{K}{k}=n$ Use Heron's formula , we have $S_{\triangle_{big}}(=\frac{36}{25} \times S_{\triangle_{small}})(1)$ $=\sqrt{K(K-S_{1})(K-S_{2})(K-S_{3})}$ $=\sqrt{nk \times n(k-s_{1}) \times n(k-s_{2}) \times n(k-s_{3})}$ $=\sqrt{n^{4} \times k(k-s_{1})(k-s_{2})(k-s_{3})}$ $=\sqrt{n^{4}} \times \sqrt{k(k-s_{1})(k-s_{2})(k-s_{3})}$ $=n^2 \times S_{\triangle_{small}}(2)$ From $(1)$ and $(2)$ , $n^{2}=\frac{36}{25}$ $\iff n=\pm \frac{6}{5}$ Because a ratio of 2 sides can't be negative $\implies n=\frac{6}{5}$ $WLOG$ , call $s_{1}=80$ $S_{1}=s_{1} \times \frac{S_{1}}{s_{1}}=80 \times \frac{6}{5}=\Large \boxed{96}$

Use theoram of Area of two similar triangle.