Question

the measure of two adjacent sides of a rectangle are in the ratio of 11:10 .if the perimeter of the rectangle is 84 cm ,find its area

Answer 1

let the sides be 11x and 10x resp.,
BTP,
PERIMETER=2(L+B)=2(11x+10x)=84
thus, x=2
thus the sides are 22 and 20
thus area= 440cm^2

Answer 2

Given,

• The Measure of two adjacent side of a rectangle are in the ratio of 11:10.
• If the Perimeter of the Rectangle is 84 cm.

To Find,

• Area of Rectangle

Solution :

$\implies$ Suppose the Adjacent side of rectangle be a

Therefore, The Adjacent Side be 11a and 10a

║Now Given,that the Perimeter of Rectangle is 84 cm and Length is 11a and Breadth is 10a, So with the help of Perimeter of Rectangle Formula We Will Find the Value of a then find Area of Rectangle║

$\longrightarrow \underline{\boxed{\sf{\pink{Perimeter \ of \ Rectangle = 2(Length + Breadth)}}}}$

$\implies \sf{84 = 2(11a + 10a)}$

$\implies \sf{84 = 22a + 20a}$

$\implies \sf{42a = 84}$

$\implies \sf{a = \dfrac{84}{42}}$

$\implies \boxed{\sf{a = 2}}$

Therefore,

$\star{\boxed{\textbf{Length = 11a = 11(2) = 22 \ cm}}}$

$\star{\boxed{\textbf{Breadth = 10a = 10(2) = 20 \ cm}}}$

$\mapsto \underline{\sf{\red{Now,\: Find\:the\:Rectangle\:of\: Area :}}}$

$\implies \sf{Area \ of \ Rectangle = (Length \ \times Breadth)}$

$\implies \sf{Area \ of \ Rectangle = 22 \times 20}$

$\undeline{\boxed{\sf{\pink{Area \ of \ Rectangle = 440 \ cm^{2}}}}}$

$\rule{200}2$