(10) The legs of a right triangle are in theratio 3 : 4 and its area i

1.	(10) The legs of a right triangle are in theratio 3 : 4 and its area is 1014 cm.Find the lengths of its sides and itsperimeter.
Answer» $\huge{\boxed{\mathtt{\red{Answer:-}}}}$ $\sf\underline{\blue{\:\:\: <klux>GIVEN</klux>:-\:\:\:}}$ Sides of ∆ =3:4 AREA of ∆= 1014sq.cm Let the side of ∆ be X Then 3x and 4x $\boxed{\sf{Area\:of\triangle=\dfrac{1}{2}\times base\times height}}$ $\sf\underline{\red{\:\:\: A.T.Q:-\:\:\:}}$ $\sf\implies 1014=\dfrac{1}{2}\times 3x\times 4x\\ \\ \sf\implies 1014\times 2= 12x{}^{2}\\ \\ \sf\implies 2028=12x{}^{2}\\ \\ \sf\implies \cancel{\dfrac{2028}{12}}=x{}^{2}\\ \\ \sf\implies 169=x{}^{2}\\ \\ \sf\implies \sqrt{169}=x\\ \\ \sf\implies {\blue{x=13}}$ Now sides of ∆ $\sf\bullet 3\times 13= 39cm\\ \\ \sf\bullet 4\times 13= 52cm$ We have to FIND the perimeter of ∆ .Two sides are given find the 3rd side of ∆ by Pythagoras theorm $\boxed{\sf{Hypotenuse {}^{2}=Perpendicular {}^{2}+base{}^{2}}}$ $\sf\implies H{}^{2}= (39){}^{2}+(52){}^{2}\\ \\ \sf\implies H{}^{2}=1521+2704\\ \\ \sf\implies H{}^{2}= 4225\\ \\ \sf\implies H=\sqrt{4225}\\ \\ \sf\implies H=65cm$ Now the perimeter of ∆ $\boxed{\sf{\purple{Perimeter\:of\:\triangle=Sum\:of\:All\:sides}}}$ $\sf{Given:-}\begin{cases}\sf{Sides\:of\: triangle}\\ \sf{39cm\:\:,52cm\:\:,65cm}\end{cases}$ $\sf\implies Perimeter=39+52+65\\ \\ \sf\implies Perimeter= 156cm$ $\underline{\boxed{\sf{Perimeter\:of\triangle=156cm}}}$

(10) The legs of a right triangle are in theratio 3 : 4 and its area is 1014 cm.Find the lengths of its sides and itsperimeter.

Answer»

$\huge{\boxed{\mathtt{\red{Answer:-}}}}$

$\sf\underline{\blue{\:\:\: <klux>GIVEN</klux>:-\:\:\:}}$

Sides of ∆ =3:4

AREA of ∆= 1014sq.cm

Let the side of ∆ be X

Then 3x and 4x

$\boxed{\sf{Area\:of\triangle=\dfrac{1}{2}\times base\times height}}$

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

$\sf\implies 1014=\dfrac{1}{2}\times 3x\times 4x\\ \\ \sf\implies 1014\times 2= 12x{}^{2}\\ \\ \sf\implies 2028=12x{}^{2}\\ \\ \sf\implies \cancel{\dfrac{2028}{12}}=x{}^{2}\\ \\ \sf\implies 169=x{}^{2}\\ \\ \sf\implies \sqrt{169}=x\\ \\ \sf\implies {\blue{x=13}}$

Now sides of ∆

$\sf\bullet 3\times 13= 39cm\\ \\ \sf\bullet 4\times 13= 52cm$

We have to FIND the perimeter of ∆

.Two sides are given find the 3rd side of ∆ by Pythagoras theorm

$\boxed{\sf{Hypotenuse {}^{2}=Perpendicular {}^{2}+base{}^{2}}}$

$\sf\implies H{}^{2}= (39){}^{2}+(52){}^{2}\\ \\ \sf\implies H{}^{2}=1521+2704\\ \\ \sf\implies H{}^{2}= 4225\\ \\ \sf\implies H=\sqrt{4225}\\ \\ \sf\implies H=65cm$

Now the perimeter of ∆

$\boxed{\sf{\purple{Perimeter\:of\:\triangle=Sum\:of\:All\:sides}}}$

$\sf{Given:-}\begin{cases}\sf{Sides\:of\: triangle}\\ \sf{39cm\:\:,52cm\:\:,65cm}\end{cases}$

$\sf\implies Perimeter=39+52+65\\ \\ \sf\implies Perimeter= 156cm$