Question

The length of a rectangle is 3 yds longer than its width.
If the perimeter of the rectangle is 70 yds, find its length and width.

Answer 1

Given

• Length of rectangle=3 yds more than Breadth
• Perimeter of Rectangle=70 yds

To Find

• Length and Breadth

Solution

Let Breadth be x

.°. Length of rectangle=3+x

We know that

$\boxed{\bf{Perimeter\ of\ rectangle= 2(length+Breadth}}$

$\sf\dashrightarrow{70= 2(3 + x + x)}$

$\sf\dashrightarrow{2 x + 3= 35}$

$\sf\dashrightarrow{2 x = 35 - 3}$

$\sf\dashrightarrow{2 x = 32}$

$\bf\dashrightarrow{x = 16}$

• Breadth of rectangle=16 yds.
• Length of rectangle= 19 yds.

$\begin{gathered}\begin{gathered}\begin{gathered} \dag \: \underline{\bf{More \: Useful \: Formula}}\\ {\boxed{\begin{array}{cc}\dashrightarrow {\sf{perimeter \: of \: rectangle = 2(l + b)}} \\ \\ \dashrightarrow \sf{area \: of \: rectangle = length \: \times breadth }\\ \\ \dashrightarrow \sf{perimeter \: of \: square = 4 \times side } \\ \\ \dashrightarrow \sf{area \: of \: square =(side) ^{2} } \\ \\ \dashrightarrow \sf{area \: of \: parallelogram = base \times height} \\ \\ \dashrightarrow \sf{area \: of \: trapezium = \frac{1}{2}sum \: of \: parallel \: de \: \times \: height }\\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}$