The perimeter of an isosceles triangle is 32 cm. If each equal side is 5/6 th of the base, find the area of the triangle.

Let each equal side of triangle be x and base be y ATQ, \(\displaystyle x+x+y=32\) \(\displaystyle 2 \mathrm{x}+\mathrm{y}=32\) Also, \(\displaystyle \mathrm{x}=\frac{5}{6} \mathrm{y}\) On solving these equations, we get \(\displaystyle \mathrm{x}=10\) and \(\displaystyle \mathrm{y}=12\) ∴ sides of the traingle are $\displaystyle 10$ cm, $\displaystyle 10$ cm, $\displaystyle 12$ cm Semi - perimeter of the triangle \(\displaystyle =16 \mathrm{~cm}\) Area of the triangle \(\displaystyle =\sqrt{16 \times 6 \times 6 \times 4}\) \(\displaystyle =48 \mathrm{~cm}^{2}\) Let first term = a and common difference = d ATQ, \(\displaystyle (\mathrm{a}+2 \mathrm{~d})+(\mathrm{a}+6 \mathrm{~d})=6\) \(\displaystyle \mathrm{a}+4 \mathrm{~d}=3\) \(\displaystyle \mathrm{a}=3-4 \mathrm{~d}\) Also, \(\displaystyle (\mathrm{a}+2 \mathrm{~d})(\mathrm{a}+6 \mathrm{~d})=8\) \(\displaystyle (3-4 d+2 d)(3-4 d+6 d)=8\) \(\displaystyle 9-4 \mathrm{~d}^{2}=8\) \(\displaystyle \mathrm{d}= \pm \frac{1}{2}\) When \(\displaystyle \mathrm{d}=\frac{1}{2} \Rightarrow \mathrm{a}=1\) \(\displaystyle \mathrm{S}_{16}=\frac{16}{2}\left[2 \times 1+15 \times \frac{1}{2}\right]\) \(\displaystyle =76\) When \(\displaystyle \mathrm{d}=-\frac{1}{2} \Rightarrow \mathrm{a}=5\) \(\displaystyle \mathrm{S}_{16}=\frac{16}{2}\left[2 \times 5+15 \times\left(-\frac{1}{2}\right)\right]\) = $\displaystyle 20$ The ages of the participants form the following AP \(\displaystyle 8,8 \frac{1}{3}, 8 \frac{2}{3}, 9, \cdots\) where first term \(\displaystyle =8\) and common difference \(\displaystyle =\frac{1}{3}\) Let the number of participants be n \(\displaystyle \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2 \times 8+(\mathrm{n}-1) \frac{1}{3}\right]=168\) \(\displaystyle \mathrm{n}^{2}+47 \mathrm{n}-1008=0\) ⇒ n = $\displaystyle 16$ ∴ the age of the eldest participant \(\displaystyle =8+15 \times \frac{1}{3}=13\) years