Question

Prove that (2, -2), (-2, 1) and (5, 2) are the vertices of a right angled triangle. Find the area of the triangle and the length of the hypotenuse.

Answer 1

Let A (2, – 2), B (– 2, 1) and C (5, 2) be the vertices of ΔABC.

$AB=\sqrt{(2-(-2){)}^2+(-2-1{)}^2}=\sqrt{4^2+(-3{)}^2}=\sqrt{16+9}=\sqrt{25}=5$

$BC=\sqrt{(-2-5{)}^2+(1-2{)}^2}=\sqrt{(-7{)}^2+(-1{)}^2}=\sqrt{49+1}=\sqrt{50}=5\sqrt{2}$

$CA=\sqrt{(5-2{)}^2+(2-(-2){)}^2}=\sqrt{(3{)}^2+(4{)}^2}=\sqrt{16+9}=\sqrt{25}=5$

$A{B}^2+C{A}^2=(5{)}^2+(5{)}^2=25+25=50$

$B{C}^2=(5\sqrt{2}{)}^2=50$

$A{B}^2+C{A}^2=B{C}^2$

In ΔABC,

$A{B}^2+C{A}^2=B{C}^2$

∴ ΔABC is a right triangle right angled at $\angle A$. (Converse of Pythagoras theorem)

Length of hypotenuse, $BC=5\sqrt{2}$

Area of ΔABC

$=\frac{1}{2}\times AB\times CA=\frac{1}{2}\times 5\times 5=\frac{25}{2}=12.5sq.units$

$[Since,Areaoftriangle=\frac{1}{2}\times base\times height]$

Hence required area 12.5 sq units