Question

Maths The perimeter of a right angled triangle is 60 cm and its area is 120 cm^2. Find the lengths of its sides. Answer is :10cm, 24 cm and 26 cm

Answer 1

Answer:

$\sf{The \ lengths \ of \ the \ sides \ are \ 24 \ cm,}$

$\sf{10 \ cm \ and \ 26 \ cm \ respectively.}$

Given:

• The perimeter of a right angled triangle is 60 cm

• Area of right angled triangle is 120 cm²

Solution:

$\sf{Let \ the}$

$\sf{Side \ 1 = x \ cm,}$

$\sf{Side \ 2= y \ cm,}$

$\sf{Hypotenuse=z \ cm.}$

$\sf{According \ to \ the \ first \ condition.}$

$\sf{x+y+z=60}$

$\sf{\therefore{z=60-x-y...(1)}}$

$\sf{According \ to \ the \ second \ condition.}$

$\sf{Here, \ x \ is \ perpendicular \ to \ y.}$

$\sf{Hence, \ x=height \ and \ y=base}$

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

$\sf{\therefore{\dfrac{xy}{2}=120}}$

$\sf{\therefore{xy=240...(2)}}$

$\sf{By \ Pythagoras \ theorem.}$

$\sf{z^{2}=x^{2}+y^{2}....(3)}$

$\sf{Substitute \ equation (1) \ in \ equation (3),}$

$\sf{we \ get}$

$\sf{(60-x-y)^{2}=x^{2}+y^{2}}$

$\sf{\therefore{3600+x^{2}+y^{2}-120(x+y)+2xy=x^{2}+y^{2}}}$

$\sf{\therefore{3600+2xy-120(x+y)=0}}$

$\sf{\therefore{3600+2xy=120(x+y)...(4)}}$

$\sf{Substitute \ equation (2) \ in \ equation (4),}$

$\sf{we \ get,}$

$\sf{3600+2(240)=120(x+y)}$

$\sf{\therefore{3600+480=120(x+y)}}$

$\sf{\therefore{x+y=\dfrac{4080}{120}}}$

$\sf{\therefore{x+y=34...(5)}}$

$\sf{By \ identity}$

$\sf{(a+b)^{2}=(a-b)^{2}+4ab}$

$\sf{...from \ equations (5) \ and \ (2)}$

$\sf{(x+y)^{2}=(x-y)^{2}+4xy}$

$\sf{\therefore{34^{2}=(x-y)^{2}-4(240)}}$

$\sf{\therefore{(x-y)^{2}=1156-960}}$

$\sf{\therefore{(x-y)^{2}=196}}$

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\sf{x-y=14...(6)}$

$\sf{Add \ equations (5) \ and \ (6), \ we \ get}$

$\sf{x+y=34}$

$\sf{+}$

$\sf{x-y=14}$

_______________

$\sf{2x=48}$

$\sf{\therefore{x=\dfrac{48}{2}}}$

$\boxed{\sf{\therefore{x=24}}}$

$\sf{Substitute \ x=24 \ in \ equation (5), \ we \ get}$

$\sf{24+y=34}$

$\sf{\therefore{y=34-24}}$

$\boxed{\sf{\therefore{y=10}}}$

$\sf{Substitute \ x=24 \ and \ y=10 \ in \ equation (1),}$

$\sf{we \ get}$

$\sf{z=60-24-10}$

$\boxed{\sf{\therefore{z=26}}}$

$\sf\purple{\tt{\therefore{The \ lengths \ of \ the \ sides \ are \ 24 \ cm,}}}$

$\sf\purple{\tt{10 \ cm \ and \ 26 \ cm \ respectively.}}$

Answer 2

Answer:

Step-by-step explanation:10cm

24cm

26cm