The area of a isosceles right triangle is 24 cm^2 where the hypotenuse is 12. What is the length of the other two sides.
The area of a right angle triangle is half of product of legs of triangle, such that:
`A = (c_1*c_2)/2`
Since the problem provides that the right triangle is isosceles, hence, the legs have equal lengths, such that:
`A = (x*x)/2`
The problem provides the information that the area of triangle is of `24 cm^2` , such that:
`24 = x^2/2 => x^2 = 2*24 => x = sqrt(24*2) => x = sqrt(2^4*3)`
`x = 4sqrt3`
Since the problem provides the information that the hypotenuse of triangle is of `12 cm` , you need to test if the lengths of the legs verify the Pythagorean theorem, such that:
`12^2 = (4sqrt3)^2 + (4sqrt3)^2`
`144 = 2*16*3 => 144!=96`
Hence, there exists no such a right isosceles triangle whose hypotenuse is of `12 cm` and area of `24 cm^2` .
The area of an isosceles right triangle is 24 with hypotenuse 12. Let the length of the equal sides be L, `2*L^2 = 24`
=> L = `sqrt 12`
The area of the triangle is `(1/2)*L^2 = 6`
It is not possible for an isosceles right triangle to have area 24 and hypotenuse 12