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 lenght of hypotenuse.

Answer 1

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Answer 2

Answer:

It is a right angle triangle.

The area of triangle is 17.5 unit square.

Step-by-step explanation:

Given : (2,-2), (-2,1) and (5,2) are the vertices of a right angled triangle.

To find : Prove that given points are the vertices of a right angled triangle. Find the area of the triangle and the length of hypotenuse?

Solution :

Distance formula, $D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Distance between (2,-2), (-2,1)

$D=\sqrt{(-2-2)^2+(1+2)^2}$

$D=\sqrt{(-4)^2+(3)^2}$

$D=\sqrt{16+9}$

$D=\sqrt{25}$

$D=5$

Distance between (-2,1),(5,2)

$D=\sqrt{(5+2)^2+(2-1)^2}$

$D=\sqrt{(7)^2+(1)^2}$

$D=\sqrt{49+1}$

$D=\sqrt{50}$

Distance between (2,-2),(5,2)

$D=\sqrt{(5-2)^2+(2+2)^2}$

$D=\sqrt{(3)^2+(4)^2}$

$D=\sqrt{9+16}$

$D=\sqrt{25}$

$D=5$

If it is a right angle triangle,

Then square of longest side = sum of square of other side

LHS : $(\sqrt{50})^2=50$

RHS : $5^2+5^2=25+25=50$

LHS=RHS , it is a right angle triangle.

Length of hypotenuse $\sqrt{50}=7.1$

Area of triangle $A=\frac{1}{2}\times b\times h$

$A=\frac{1}{2}\times 5\times 5$

$A=\frac{25}{2}$

$A=17.5 unit^2$