The hypotenuse of a garden in the shape of a | vedclass

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(D) Let the shorter side among the sides containing the right angle (i.e.,base) be $x$ metres.Then,the other side containing the right angle (i.e.,alt

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$8\,m, 15\,m, 17\,m$

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(D) Let the shorter side among the sides containing the right angle (i.e.,base) be $x$ metres.Then,the other side containing the right angle (i.e.,altitude) is $(x+7)$ metres and the hypotenuse is $(2x+1)$ metres.By Pythagoras' theorem,$(	ext{Hypotenuse})^2 = (	ext{Base})^2 + (	ext{Altitude})^2$.Therefore,$(2x+1)^2 = x^2 + (x+7)^2$.$4x^2 + 4x + 1 = x^2 + x^2 + 14x + 49$.$4x^2 + 4x + 1 = 2x^2 + 14x + 49$.$2x^2 - 10x - 48 = 0$.Dividing by $2$,we get $x^2 - 5x - 24 = 0$.$(x-8)(x+3) = 0$.So,$x=8$ or $x=-3$.Since the side of a triangle cannot be negative,$x=8$.Thus,the base is $8\,m$,the altitude is $8+7=15\,m$,and the hypotenuse is $2(8)+1=17\,m$.The sides of the garden are $8\,m, 15\,m$ and $17\,m$.

The hypotenuse of a garden in the shape of a right triangle is $1$ metre more than twice the shortest side. If the longer side containing the right angle is $7$ metres more than the shorter side,find the sides of the garden.

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