A right angled triangle has the length of its shorter sides as(2x+4) and (8x+8). if the length of hypotenuse is 10x, finds its area?

Answer 1

Remember for Right Angled Triangles.

First using Pythagoras the area(A) of triangles is equal to half the base multiplied to the perpendicular height.

side 'a' = 2x + 4

side b = 8x + 8

side h = 10x

First using Pythagoras find the value of 'x' .

Hence

(10x)^2 = ( 2x + 4)^2 + ()8x + 8)^2

Hence

100x^2 = 4x^2 + 8x + 16 + 64x^2 + 128x + 64

Collect like terms and form a quadratic equation.

100x^2 - 4x^2 - 64x^2 -8x - 128x - 16 - 64 = 0

32x^2 - 136x - 80 = 0

Factor out '8'

4x^2 - 17x - 10 = 0

Use Quadratic Eq'n

x = {--17 +/- sqrt[(-17)^2 - 4(4)(-10)]} / 2(4)

x = {17 +/- sqrt[289 + 160]} / 8

x = {17 +/- sqrt[449]} / 8

x = {17 +/- 21.189....} / 8

x = 38.189... / 8 = 4.773....

& x = - 4.189 / 8 ; Unresolved because , philosophically you cannot have a negative length.

Using 4.773... substitute for 'x' in to the two shorter sides.

Hence

A = (0.5)(2(4.773...) + 4)(8(4.773...) + 8)

A = (0.5(9.547... + 4)(38.184... + 8)

A = (0.5)(13.547...)(46.184...)

A = 312.827.... units^2