
The altitude of the right-angled triangle is 7cm less than its base. If the hypotenuse is 13cm. find the other two sides
Answer
510.3k+ views
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.
“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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