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Q5. 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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Let the length of the base of the triangle be $b\ cm$ .

Then, the altitude length will be: $b-7\ cm$ .

Given if hypotenuse is $13\ cm$ .

Applying the Pythagoras theorem; we get

$Hypotenuse^2 = Perpendicular^2 + Base^2$

So, $(13)^2 = (b-7)^2 +b^2$

$\Rightarrow 169 = 2b^2+49-14b$

$\Rightarrow 2b^2-14b-120 = 0$ Or $b^2-7b-60 = 0$

$\Rightarrow b^2-12b+5b-60 = 0$

$\Rightarrow b(b-12) + 5(b-12) = 0$

$\Rightarrow (b-12)(b+5) = 0$

$\Rightarrow b= 12\ or\ -5$

But, the length of the base cannot be negative.

Hence the base length will be $12\ cm$ .

Therefore, we have

Altitude length $= 12cm -7cm = 5cm$ and Base length $= 12\ cm$

Posted by

Divya Prakash Singh

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