Question

the perimeter of a sqaure of side 12 cm is the same as that of a 15 cm long rectangle find the area of the rectangle is solution​

Answer 1

$\huge \green{answer}$

$\pink{given}$

perimeter of sq= perimeter of rectangle

4x = 2(l+b)

4×12=2(l+b)

24 = 15+b

b= 24- 15

therefore, breadth of rectangle = 9 cm

area of rectangle =l×b

15×9

answer, 135

#be brainly

Answer 2

$\begin{gathered}\begin{gathered}\ \tt \: Given - \begin{cases} &\sf{Side_{(square)}, \: x \: = \: 12 \: cm} \\ &\sf{Length_{(rectangle)}, \: l \: = 15 \: cm }\\ &\sf{Perimeter_{(square)} = Perimeter_{(rectangle)}} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf To \: Find :- \begin{cases} &\tt{Area \: of \: rectangle} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\tt \: Formula \: used \begin{cases} &\sf{Perimeter_{(square)} = 4 \times side} \\ &\sf{Perimeter_{(rectangle)} = 2(l \: + \: b)}\\ &\sf{Area_{(rectangle)} = l \: \times \: b } \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

$\begin{gathered}\begin{gathered}\bf Let :- \begin{cases} &\sf{Breadth_{(rectangle)} \: be \: b \: cm} \end{cases}\end{gathered}\end{gathered}$

$\tt \begin{gathered}\bf\red{ \tt \: According \: to \: statement}\end{gathered}$

$\tt \: Perimeter_{(square)} \: = \: Perimeter_{(rectangle)}$

$\tt\implies \:4x = 2(l \: + \: b)$

$\tt\implies \:4 \times 12 = 2(15 + b)$

$\tt\implies \:48 = 30 + 2b$

$\tt\implies \:2b \: = 48 - 30$

$\tt\implies \:2b \: = \: 18$

$\tt\implies \:b \: = \: 9 \: cm$

$\tt \: Hence, \: Area_{(rectangle)} \: = \: length \: \times \: breadth$

$\tt\implies \:Area_{(rectangle)} = 15 \times 9 = 135 \: {cm}^{2}$