The length of the sides forming right angle of triangle are 5x and (3x-1)cm. If the area of the triangle is 60cm2, find its hypotenuse
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1/2 bh
1/2 (5x (3x-1)) = 60
1/2 (15x^2 -5x ) = 60
15x^2 - 5x = 120
15x^2 -5x - 120 = 0
3x^2 - x - 24 = 0
3x^2 + 8x - 9x - 24 = 0
x(3x + 8) -3 (3x + 8)
(x-3) (3x + 8)
x = 3 or x = -8/3
since length cannot be -ve
x = 3
so the sides of the triangle are
5(3) and 3(3) - 1
15 and 8
Pythogorean prop
8^2 + 15^2 = hypotenuse ^2
(sum of the squares on the two non - hypotenuse sides is equal to the square on the hypotenuse )
64 + 225 = 289
hypotenuse sqr = 289
hy[otenuse = Route of 289
Hyp - 17 !
1/2 (5x (3x-1)) = 60
1/2 (15x^2 -5x ) = 60
15x^2 - 5x = 120
15x^2 -5x - 120 = 0
3x^2 - x - 24 = 0
3x^2 + 8x - 9x - 24 = 0
x(3x + 8) -3 (3x + 8)
(x-3) (3x + 8)
x = 3 or x = -8/3
since length cannot be -ve
x = 3
so the sides of the triangle are
5(3) and 3(3) - 1
15 and 8
Pythogorean prop
8^2 + 15^2 = hypotenuse ^2
(sum of the squares on the two non - hypotenuse sides is equal to the square on the hypotenuse )
64 + 225 = 289
hypotenuse sqr = 289
hy[otenuse = Route of 289
Hyp - 17 !
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