Question

the rectangle shown has a perimeter of 56 cm and the given area. it's length is 8 more than three times it's width. write and solve a system of equations to find the dimensions of the rectangle.​

Answer 1

`\qquad\qquad\huge\underline{{\sf Answer}}♨`

Let's solve ~

Assume width of rectangle be " x ", length = 3×width + 8 = 3x + 8 ~

Now, Perimeter of rectangle is :

`\qquad \sf  \dashrightarrow \:2(l + w) = 56`

`\qquad \sf  \dashrightarrow \:2(3x + 8 + x) = 56`

`\qquad \sf  \dashrightarrow \:2(4x + 8) = 56`

`\qquad \sf  \dashrightarrow \:4x + 8 = 56 / 2`

`\qquad \sf  \dashrightarrow \:4x + 8 = 28`

`\qquad \sf  \dashrightarrow \:4x = 28 - 8`

`\qquad \sf  \dashrightarrow \:x = 20/ 4`

`\qquad \sf  \dashrightarrow \:x =5 \: cm`

Hence, width = x = 5 cm

`\qquad \sf  \dashrightarrow \:l = 3w + 8`

`\qquad \sf  \dashrightarrow \:l = 3(5)+ 8`

`\qquad \sf  \dashrightarrow \:l = 15+ 8`

`\qquad \sf  \dashrightarrow \:l =23 \:cm`

And, length = 26 cm