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# Triangle has sides $5cm,{\text{ }}12cm,$ and $13cm$. Find the length of the perpendicular from the opposite vertex to the side whose length is $13cm$.

Last updated date: 19th Jul 2024
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Hint: In this question we will use the area of the right angle triangle that is half multiplied by its base and perpendicular, by considering the other base and perpendicular of the same triangle then we will equate both areas.

Let $AB = 5cm,{\text{ }}BC = 12cm,{\text{ }}CA = 13cm$
${\left( {CA} \right)^2} = {\left( {AB} \right)^2} + {\left( {BC} \right)^2} \\ {13^2} = {5^2} + {12^2} = 169 = {13^2} \\$
Let $BD = xcm$
$\Delta ABC = \dfrac{1}{2}\left( {AB} \right)\left( {BC} \right) = \dfrac{1}{2}\left( {BD} \right)\left( {AC} \right) \\ = \dfrac{1}{2}\left( 5 \right)\left( {12} \right) = \dfrac{1}{2}\left( x \right)\left( {13} \right) \\ \Rightarrow 60 = 13x \\ \Rightarrow x = \dfrac{{60}}{{13}}cm \\$
So, this is the required perpendicular distance from the opposite vertex to the side whose length is $13cm$.
Note: In such types of questions first draw the pictorial representation of the given problem, then check whether it is right angle triangle or not if it is then using the formula of area of triangle which is half multiply by perpendicular time’s base, then we can easily calculated the length of the perpendicular from the opposite vertex to the side whose length is $13cm$.