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The altitude of the right-angled triangle is 7cm less than its base. If the hypotenuse is 13cm. find the other two sides

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Hint: Here we have to use Pythagoras theorem to solve this question as it is mentioned that the triangle is a right-angled triangle.

“Complete step-by-step answer:”
Let altitude of right-angle triangle is $hcm$
Let base = $bcm$
Now according to question,
$h = b – 7$
Given, hypotenuse is $l$ = $13cm$
Now apply Pythagoras theorem
${l^2} = {h^2} + {b^2}$
$ \Rightarrow 13^2 = {\left( {b - 7} \right)^2} + {b^2}$
$ \Rightarrow {b^2} + {b^2} + 49 - 14b = 169$
$ \Rightarrow 2{b^2} - 14b - 120 = 0$
$ \Rightarrow {b^2} - 7b - 60 = 0$
Now factorize this equation
$ \Rightarrow \left( {b - 5} \right)\left( {b - 12} \right) = 0$
Therefore, base of right-angle triangle is 5 or 12cm
Altitude of triangle, $h = b - 7$
$ \Rightarrow h = 5 - 7 = - 2$
$ \Rightarrow h = 12 - 7 = 5cm$
$h$ cannot be negative So,
Base = 12cm and altitude =5cm

NOTE: Whenever we face such a problem the key concept is that we have to remember the Pythagoras theorem and based on the given condition we have to form an equation to get the desired value.
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