The perimeter of a right triangle is 40cm and hypotenuse is 17cm Find the area of the triangle using - Maths - Heron\s Formula

Question

The perimeter of a right triangle is 40cm and hypotenuse is 17cm
Find the area of the triangle using Heron's formula

Priyanka Kedia · Accepted Answer

$L e t A B C b e a r i g h t t r i a n g l e w i t h B C = y c m a n d A B = x c m$

$S i n c e p e r i m e t e r o f t r i a n g l e = 40 c m T h e r e f o r e, x + y + 17 = 40 \Rightarrow x + y = 23 (1) S q u a r i n g b o t h s i d e, w e h a v e, {(x + y)}^{2} = {(23)}^{2} \Rightarrow x^{2} + y^{2} + 2 x y = 529 (2) U \sin g P y t h a g o r a s T h e o r e m i n Δ A B C, w e h a v e, {A C}^{2} = {A B}^{2} + {B C}^{2} \Rightarrow {(17)}^{2} = x^{2} + y^{2} \Rightarrow x^{2} + y^{2} = 289 (3) S u b t r a c t i n g e q u a t i o n (2) a n d (3), w e h a v e, 2 x y = 529 - 289 \Rightarrow x y = 120 (4) F r o m (1) a n d (4), {(x - y)}^{2} = {(x + y)}^{2} - 4 x y \Rightarrow {(x - y)}^{2} = {(23)}^{2} - 4 \times 120 = 529 - 480 = 49 \Rightarrow x - y = 7 (5) A d d i n g e q u a t i o n (1) a n d (2), w e g e t 2 x = 30 \Rightarrow x = 15 c m S u b s t i t u t e v a l u e o f x i n e q u a t i o n (1), w e g e t, y = 8 c m T h u s, A r e a o f Δ A B C = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{20 (20 - 15) (20 - 8) (20 - 17)} = \sqrt{20 \times 5 \times 12 \times 3} = 60 {c m}^{2} H e n c e, a r e a o f Δ A B C i s 60 {c m}^{2}$

The perimeter of a right triangle is 40cm and hypotenuse is 17cm. Find the area of the triangle using Heron's formula.