Question

An altitude of a triangle is five-thirds thelength of its corresponding base. If the altitude be increased by 4 cm and the base decreased by 2 cm, the area of the triangle remains the same. Find the base and the altitude of the triangle.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{base}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{be}\mathrm{x}\mathrm{cm}. \\ \mathrm{Then},\mathrm{its}\mathrm{altitude}\mathrm{will}\mathrm{be}\frac{5}{3}x\mathrm{cm}. \\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{triangle}=\frac{1}{2}\left(x\right)\left(\frac{5}{3}x\right)=\frac{5}{6}{x}^2 \\ \therefore \frac{1}{2}(x-2)(\frac{5}{3}x+4)=\frac{5}{6}{x}^2 \\ \Rightarrow \left(\frac{x-2}{2}\right)\left(\frac{5x+12}{3}\right)=\frac{5x^2}{6} \\ \Rightarrow \frac{(x-2)(5x+12)}{6}=\frac{5x^2}{6} \\ \Rightarrow \frac{5x^2+12x-10x-24}{6}=\frac{5x^2}{6} \\ \Rightarrow 5x^2+2x-24=5x^2(\mathrm{cancelling}\mathrm{the}\mathrm{denominators}\mathrm{from}\mathrm{both}\mathrm{the}\mathrm{sides}\mathrm{since}\mathrm{they}\mathrm{are}\mathrm{same}) \\ \Rightarrow 5x^2-5x^2+2x=24 \\ \Rightarrow 2x=24 \\ \Rightarrow x=\frac{24}{2}=12m \\ \mathrm{Therefore},\mathrm{the}\mathrm{bas}e\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{is}12\mathrm{m}. \\ \mathrm{Altitude}\mathrm{of}\mathrm{the}\mathrm{triangle}=\frac{5}{3}x=\frac{5}{3}\left(12\right)=20m \end{gathered}$$