Question

The area of a right-triangle is 600 cm². If the base of the triangle exceeds the altitude by 10 cm, find the dimensions of the triangle.

Answer 1

Let the altitude of the triangle be x cm.
Therefore, the base of the triangle will be (x + 10) cm.
$$\begin{gathered} \text{Area of triangle =}\frac{1}{2}x(x+10)=600 \\ \Rightarrow x(x+10)=1200 \\ \Rightarrow x^2+10x-1200=0 \\ \Rightarrow x^2+(40-30)x-1200=0 \\ \Rightarrow x^2+40x-30x-1200=0 \\ \Rightarrow x(x+40)-30(x+40)=0 \\ \Rightarrow (x+40)(x-30)=0 \\ \Rightarrow x=-40\text{or}x=30 \\ \Rightarrow x=30[\because \text{Altitude cannot be negative}] \\ \text{Thus, the altitude and base of the triangle are 30 cm and (30 + 10 = 40) cm, respectively.} \\ \because {\left(\mathrm{Hypotenuse}\right)}^2={\left(\mathrm{Altitude}\right)}^2+{\left(\mathrm{Base}\right)}^2 \\ \Rightarrow {\left(\mathrm{Hypotenuse}\right)}^2={\left(30\right)}^2+{\left(40\right)}^2 \\ \Rightarrow {\left(\mathrm{Hypotenuse}\right)}^2=900+1600=2500 \\ \Rightarrow {\left(\mathrm{Hypotenuse}\right)}^2={\left(50\right)}^2 \\ \Rightarrow \left(\mathrm{Hypotenuse}\right)=50 \\ \mathrm{Thus},\mathrm{the}\mathrm{dimensions}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{are}:, \\ \mathrm{Hypotenuse}=50\mathrm{cm} \\ \mathrm{Altitude}=30\mathrm{cm} \\ \mathrm{Base}=40\mathrm{cm} \end{gathered}$$