Question

Find the area of the right angled triangle whose vertices are (2, –2), (–2, 1) and (5, 2).

(a) 5√2 sq. units 

(b)\(\frac{25}{2}\) sq. units 

(c) 15√2 sq. units 

(d) 10 sq. units

Answer 1

(b)  \(\frac{25}2\) sq. units

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

Then AB² = (–2 – 2)² + (1 + 2)² = 16 + 9 = 25 

BC² = (5 + 2)² + (2 – 1)² = 49 + 1 = 50 

AC² = (5 – 2)² + (2 + 2)² = 9 + 16 = 25 

⇒ AB² + AC² = BC² 

⇒ The Δ is rt ∠d at A.

⇒ Area of Δ ABC = \(\frac12\) x AC x AB

= \(\frac12\) x 5 x 5 = \(\frac{25}2\) sq. units.