Question

The height of a triangle is 3 cm more than its base. If the area of the triangle is 104cm square, find the lengths of its base and height.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{base}\mathrm{of}∆=x\mathrm{cm} \\ \mathrm{then}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{altitude}\mathrm{of}∆=\left(x+3\right)\mathrm{cm} \\ \mathrm{area}\mathrm{of}∆=104{\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{base}\times \mathrm{altitude}=104 \\ \Rightarrow \frac{1}{2}\times x\times \left(x+3\right)=104 \\ \Rightarrow x\left(x+3\right)=208 \\ \Rightarrow x^2+3x-208=0 \\ \Rightarrow x^2+16x-13x-208=0 \\ \Rightarrow x\left(x+16\right)-13\left(x+16\right)=0 \\ \Rightarrow \left(x-13\right)\left(x+16\right)=0 \\ \Rightarrow x-13=0\mathrm{or}x+16=0 \\ \Rightarrow x=13\mathrm{or}x=-16\left(\mathrm{rejected}\right) \\ \mathrm{so},\mathrm{base}\mathrm{of}∆=13\mathrm{cm} \\ \mathrm{height}\mathrm{of}∆=13+3=16\mathrm{cm} \end{gathered}$$