Question

what is the hypotenuse of a right-angled triangle with sides 6x and 8x?

Answer 1

hey mate this is your answer

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 

By 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
hypotenuse = Root of 289

I hope it's helpful you
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Answer 2

As sides are proportional

So take 6 and 8

Hypotenuse = √ 6^2 + 8^2 = √ 36 + 64 = √ 100= 10

So it would be 10x