A right triangle has perimeter 20. If the sum of the squares of the triangle's three sides is 162,
what is the area of the triangle?


Let the three sides are a, b and c......think a and b are sides and c as the hypotenuse of the right angle traingle.

a+b+c = 20 , a^2+b^2+c^2 = 162

from phythogoreous theorem , a^2+b^2 = c^2

use below formula

(a+b+c) ^2 = a^2+b^2+c^2+2ab+2bc+2ca


20 ^ 2 = 162 + 2ab+2bc+2ca

ab+bc+ca = 400 - 162/2 = 119

since a^2+b^2 = c^2 and a^2+b^2+c^2 = 162 => 2 * c^2 = 162 => c^2 = 81 => c = 9

ab+bc+ca = 119 => ab+9b+9a (substitute the value of c from above)

ab+9(a+b) = 119

ab+9(11) = 119 ( since, a+b+c = 20 and c= 9 => a+b =11)

=> ab = 119 -99 = 20

=> ab = 20

area of triangle = 1/2* a * b = 1/2 * 20 = 10

Therefor area of triangle is 10

I did that. Took a long time. Better way?

Wow. there u go. huge difference in time.

[quote="Rahul@gurome"][quote="yellowho"]A right triangle has perimeter 20. If the sum of the squares of the triangle's three sides is 162, what is the area of the triangle?[/quote]

Say, length of sides othe triangle are a, b, and c and c is the hypotenuse. Thus, area of the triangle = (ab)/2

Hence, (a + b + c) = 20 and (a² + b² + c²) = 162
As, (a² + b²) = c²
=> (c² + c²) = 162
=> c² = 81
=> c = 9

Hence, (a + b) = 11
=> (a + b)² = (a² + 2ab + b²) = 121

As, (a² + b²) = c² = 81
=> (2ab + 81) = 121
=> 2ab = 40
=> (ab)/2 = 10

Hence area of the triangle is 10 sq. units.[/quote]