Question

The coordinates of the vertices of a triangle are (3, 5), (9, 13), (10, 6). Prove that this triangle is isosceles. Calculate its area.

Answer 1

If A (3, 5), B (9, 13), C(10, 6) are the vertices of the triangle ABC

AB \(=\sqrt{(9-3)^2+(13-5)^2}=\sqrt{100}=10\)

BC \(=\sqrt{(10-9)^2+(6-13)^2}=\sqrt{50}\)

AC \(=\sqrt{(10-3)^2+(6-5)^2}=\sqrt{50}\)

As BC = AC the triangle is an isosceles triangle

CD = \(\sqrt{\sqrt{50^2}-5^2}\)

= \(\sqrt{50-25}\)

= \(\sqrt{25}\)

= 5

Area = \(\frac{1}{2}\times10\times5=25\) m²