The altitude of a right triangle is 7 , cm le | vedclass

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Question

(A) Let the base of the right triangle be $x \, cm$.Its altitude is $(x - 7) \, cm$.According to the Pythagoras theorem:$	ext{Base}^2 + 	ext{Altitud

Vedclass · Accepted Answer

$5 \, cm$ and $12 \, cm$

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(A) Let the base of the right triangle be $x \, cm$.Its altitude is $(x - 7) \, cm$.According to the Pythagoras theorem:$	ext{Base}^2 + 	ext{Altitude}^2 = 	ext{Hypotenuse}^2$$x^2 + (x - 7)^2 = 13^2$$x^2 + x^2 - 14x + 49 = 169$$2x^2 - 14x - 120 = 0$Dividing by $2$:$x^2 - 7x - 60 = 0$Factoring the quadratic equation:$x^2 - 12x + 5x - 60 = 0$$x(x - 12) + 5(x - 12) = 0$$(x - 12)(x + 5) = 0$So,$x = 12$ or $x = -5$.Since the length of a side cannot be negative,we take $x = 12$.Therefore,the base is $12 \, cm$ and the altitude is $12 - 7 = 5 \, cm$.

The altitude of a right triangle is $7 \, cm$ less than its base. If the hypotenuse is $13 \, cm$,find the other two sides.

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