Question

please tell the correct answer with solution

A wire is of length 90 cm bent in the form of a rectangle such that its length is 5 more than its breadth.Find length and breadth of the rectangle so formed.​

Answer 1

Answer:

you can solve this question like that only.

Answer 2

Step-by-step explanation:

GIVEN :-

$\sf \: Length \: of \: wire \: = 90 \: cm$

TO FIND :-

$\sf \: Length \: and \: breadth \: of \: rectangle$

SOLUTION :-

$\sf \: let \: \: the \: breadth \: of \: rectangle \: is \: \: x \: cm \\ \sf \: length \: = (x + 5) \: cm$

$\sf \: According \: to \: the \: question \\ \sf \: perimeter \: of \: rectangle = length \: \: of \: wire \\ \implies \sf \: 2(length + breadth) = 90 \\ \sf \: \implies \: 2(x + 5 + x) = 90 \\ \sf \implies \: 2x + 5 = \frac{90}{2} \\ \sf \implies \: 2x + 5 = 45 \\ \sf \implies \: 2x = 45 - 5 \\ \sf \implies2x = 40 \: \: \: \: \: \: \: \: \: \\ \\ \sf \implies \: x = \frac{40}{2} = 20$

$\sf \: breadth \: \green{ \boxed { \implies x = 20 \: cm \: }}$

$\sf \: length \: \green{ \boxed{ \implies \: x + 5 \: = 20 + 5 = 25 cm}}$