Question

the hypotenuse of a right angle triangle is 5 metre if one of its side is 4 metre find the length of other side.
do it fast
by explaning..... ​

Answer 1

$\huge\bf{\underline{Solution:-}}$

$\bf\underline{\underline{\blue{\:\:\:\:\: Given:-\:\:\:\:}}}$

$\sf\bullet A\:right\: triangle\\ \\ \sf\bullet hypotenuse=5m\\ \\ \sf\bullet One\:of\:its\:side=4m\\ \\ \\ \sf\:\:\:\:\:Find\:the\:length\:of\:other\:side:-$

$\sf{\purple{\:\:\:\:\:\:\:Now\:by\: Pythagoras\:\:\:}}\\ \\ \sf{\boxed{\sf{ hypotenuse {}^{2}=Base{}^{2}+perpendicular {}^{2}}}}$

$\bf\underline{\underline{\red{\:\:\:\:\: A.T.Q:-\:\:\:\:}}}$

$\sf\implies H{}^{2}=B{}^{2}+P{}^{2}\\ \\ \sf\implies (5){}^{2}=(4){}^{2}+P{}^{2}\\ \\ \sf\implies 25=16+P{}^{2}\\ \\ \sf\implies 25-16=P{}^{2}\\ \\ \sf\implies 9=P{}^{2}\\ \\ \sf\implies P=\sqrt{9}\\ \\ \sf\implies P=3m$

$\boxed{\sf{length\:of\:Other\:side=3m}}$

$\tt\underline{formula\: related\:\triangle}$

$\sf\bullet Area\:of\: right\triangle=\dfrac{1}{2}\times base\times height\\ \\ \sf\bullet Area\:of\:scalene\triangle=\sqrt{s(s-a)(s-b)(s-c)}\\ \\ \sf\bullet Area\:of\: equilateral\: triangle=\dfrac{\sqrt{3}}{4}a{}^{2}$

Answer 2

$\huge\bold\green{Question}$

The hypotenuse of a right angle triangle is 5 metre if one of its side is 4 metre find the length of other side

$\huge\bold\green{Answer}$

According to the question we have given that :-

•°• Hypotenuse = 5m

•°• One side of triangle = 4m

So , we have to find out the other side of triangle

Simply by using Pythagoras theorem we can solve it :-

$\begin{lgathered} \tt\orange{\boxed{\sf{ Hypotenuse {}^{2}=Base{}^{2}+Perpendicular {}^{2}}}}\end{lgathered}$

So, as said in question

$\begin{lgathered}\tt= H{}^{2}=B{}^{2}+P{}^{2}\\ \\ \tt= (5){}^{2}=(4){}^{2}+P{}^{2}\\ \\ \tt= 25=16+P{}^{2}\\ \\ \tt = 25-16=P{}^{2}\\ \\ \tt = 9=P{}^{2}\\ \\ \tt = P=\sqrt{9}\\ \\ \tt\green = P=3m\end{lgathered}$

Hence the length of other side is = 3 m