Question

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

Answer 1

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

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

• 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 ∆

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

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

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

$\underline{\boxed{\sf{Perimeter\:of\triangle=156cm}}}$

Answer 2

Step-by-step explanation:

Step-by-step explanation:

Let it's leg be 3x and 4x. Then,

according to question,

Area of right angle triangle = 1/2 × 3x × 4x

1,014 = 6x^2

1,014/6 = x^2

√169 = x

13 cm = x

Hence, first side = 3×13 = 39 cm

second side = 4×13 = 52 cm.

We know that

H^2 = P^2 + B^2

H^2 = 39^2 + 52^2

H^2 = 1521 + 2704

H = √4225

H = 65 cm.

Hence, the hypotenuse is 65 cm.

Perimeter of triangle = sum of all side

so,

perimeter = 39+52+65 cm

perimeter of triangle = 156 cm