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