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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 $13cm$.  Answer Verified
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.

Complete step-by-step answer: Let $AB = 5cm,{\text{ }}BC = 12cm,{\text{ }}CA = 13cm$
So, these sides makes a right angle triangle because
${\left( {CA} \right)^2} = {\left( {AB} \right)^2} + {\left( {BC} \right)^2} \\ {13^2} = {5^2} + {12^2} = 169 = {13^2} \\$
Therefore ABC is a right angle triangle at B
Let BD be the perpendicular on side AC
Let $BD = xcm$
From figure the area of right angle triangle is half multiply by perpendicular time’s base
$\Delta ABC = \dfrac{1}{2}\left( {AB} \right)\left( {BC} \right) = \dfrac{1}{2}\left( {BD} \right)\left( {AC} \right) \\ = \dfrac{1}{2}\left( 5 \right)\left( {12} \right) = \dfrac{1}{2}\left( x \right)\left( {13} \right) \\ \Rightarrow 60 = 13x \\ \Rightarrow x = \dfrac{{60}}{{13}}cm \\$
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 $13cm$.
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Sides of a Triangle  Determinant to Find the Area of a Triangle  Isosceles Triangle and Equilateral Triangle  Area of Isosceles Triangle  Area of a Triangle  Area of Scalene Triangle  Area of Equilateral Triangle  Triangle and It’s Properties  Altitude and Median of a Triangle  Medians and Altitudes of a Triangle  