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. About the author Iris
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 [tex] \boxed{\bf{Perimeter\ of\ rectangle= 2(length+Breadth}}[/tex] [tex]\sf\dashrightarrow{70= 2(3 + x + x)}[/tex] [tex]\sf\dashrightarrow{2 x + 3= 35}[/tex] [tex]\sf\dashrightarrow{2 x = 35 – 3}[/tex] [tex]\sf\dashrightarrow{2 x = 32}[/tex] [tex]\bf\dashrightarrow{x = 16}[/tex] Breadth of rectangle=16 yds. Length of rectangle= 19 yds. [tex]\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}[/tex] Reply
Given
To Find
Solution
Let Breadth be x
.°. Length of rectangle=3+x
We know that
[tex] \boxed{\bf{Perimeter\ of\ rectangle= 2(length+Breadth}}[/tex]
[tex]\sf\dashrightarrow{70= 2(3 + x + x)}[/tex]
[tex]\sf\dashrightarrow{2 x + 3= 35}[/tex]
[tex]\sf\dashrightarrow{2 x = 35 – 3}[/tex]
[tex]\sf\dashrightarrow{2 x = 32}[/tex]
[tex]\bf\dashrightarrow{x = 16}[/tex]
[tex]\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}[/tex]