Question

prove that the points A(0,0), b(0,2) and C(2,0) vertices of an isosceles right triangle area​

Answer 1

Answer:

Given:

A → (0,0).

B → (0,2).

C → (2,0).

Here,

AB = √[(0-0)²+(2-0)²] = √[0²+2²] = √[0+4] = √4 = 2 units. ----(1)

BC = √[(2-0)²+(0-2)²] = √[2²+(-2)²] = √[4+4] = √8 units. ------(2)

CA = √[(2-0)²+(0-0)²] = √[2²+0²] = √[4+0] = √4 = 2 units. -----(3)

Since, AB + CA ≠ BC, these points are NOT collinear. So, the figure formed by joining the given points is a Triangle.

(i.e., ∆ABC).

Also,

From (1) and (3), we get,

AB = CA.

Since two sides of the triangle have equal length,

→ ∆ABC is an Isosceles Triangle.

Also,

From (1), (2), and (3), we get,

AB² + CA² = BC² [i.e., (Altitude)² + (Base)² = (Hypotenuse)² ]

This is possible only in the case of a Right angled Triangle.

So, ∆ABC is a right angled triangle.

Therefore,

The points A(0,0), B(0,2) and C(2,0) vertices of an isosceles right triangle.

To find Area:

Here,

Base = 2 units.

Height/Altitude = 2 units.

So,

Area = ½ × (Base) × (Height)

= ½ × 2 × 2

= 2 sq. units.