Question

the area of rectangle with perimeter 40 centimeter is 75 square meter. find the length and breadth of the rectangle? ​

Answer 1

Correct Question:

$\textsf{The area of a rectangle with perimeter 40} \\ \textsf{centimetre is 75 square centimetre. Find the} \\ \textsf{length and breadth of the rectangle.}$

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Answer:

$\sf{The \ length \ of \ the \ rectangle \ is \ 15 \ cm} \\ \sf{and \ the \ breadth \ is \ 5 \ cm}$

Given:

$\sf{\leadsto{The \ perimeter \ of \ a \ rectangle \ is \ 40 \ cm}} \\ \\ \sf{\leadsto{Area \ of \ the \ rectangle \ is \ 75 \ cm^{2}}}$

To find:

$\sf{Length \ and \ breadth \ of \ the \ rectangle.}$

Solution:

$\sf{Let \ the \ length \ of \ the \ rectangle \ be \ x \ cm} \\ \sf{and \ breadth \ of \ a \ rectangle \ be \ y \ cm} \\ \\ \boxed{\sf{Perimeter \ of \ rectangle=2(l+b)}} \\ \\ \textsf{According to the first condition.} \\ \\ \sf{40=2(x+y)} \\ \\ \sf{\therefore{x+y=20...(1)}} \\ \\ \\ \boxed{\sf{Area \ of \ rectangle=Length\times \ Breadth}} \\ \\ \textsf{According \ to \ the \ second \ condition.} \\ \\ \sf{xy=75...(2)}$

$\textsf{According to the identity} \\ \\ \boxed{\sf{(a-b)^{2}=(a+b)^{2}-4ab}} \\ \\ \sf{(x-y)^{2}=(x+y)^{2}-4xy} \\ \\ \sf{\therefore{(x-y)^{2}=20^{2}-4(75)}} \\ \\ \sf{\therefore{(x-y)^{2}=400-300}} \\ \\ \sf{\therefore{(x-y)^{2}=100}} \\ \\ \sf{\therefore{x-y=10...(3)}} \\ \\ \textsf{Add equations (1) and (3), we get} \\ \\ \sf{2x=30} \\ \\ \sf{\therefore{x=15}} \\ \\ \textsf{Substitute x=15 in equation (1), we get} \\ \\ \sf{15+y=20} \\ \\ \sf{\therefore{y=5}} \\ \\ \purple{\tt{\therefore{The \ length \ of \ the \ rectangle \ is \ 15 \ cm}}} \\ \purple{\tt{the \ breadth \ is \ 5 \ cm}}$