Question

The length of the rectangle is twice its width. The perimeter is \[60 ft\]. Find its area.

Answer 1

Hint: The relation between the length and the breadth are given. So consider one of them as a variable and then write an equation for the other in terms of the variable. The perimeter of a rectangle is the sum of the lengths of all its sides and the area is the product of any two adjacent sides. Use these relations to find out the length and breadth and solve for the area.

Complete step by step solution:
First assign a variable to the width, let us assume width to be \[x\] ft.
As given in the question length of the rectangle is twice its width, so that means the length is two times the width
\[\therefore \] length = \[2x\] ft.
Now the perimeter of the rectangle is the sum of all the sides of the rectangle.
\[\therefore \] Perimeter \[ = \] length \[ + \] length \[ + \] width \[ + \] width
                        \[ = \] 2 (length \[ + \] width)
Substituting the values of length and breadth from above, we get:
Perimeter \[ = \] \[2\left( {x + 2x} \right)\]
\[ \Rightarrow 60 = 2\left( {x + 2x} \right)\] [ Since perimeter is given in question]
Dividing both sides by \[2\],
\[ \Rightarrow 30 = \left( {x + 2x} \right)\]
\[ \Rightarrow 30 = 3x\]
Dividing both sides by \[3\],
\[ \Rightarrow x = 10\]
Hence the width of the rectangle is \[10\] ft, and the length is \[20\] ft.
Now the area of the rectangle is the product of any two adjacent sides,
\[\therefore \] Area \[ = \] length \[ \times \] width
\[ \Rightarrow \] Area \[ = \] \[20 \times 10\]
\[ \Rightarrow \] Area \[ = \] \[200\] sq. ft
Hence the area of the rectangle is \[200\] sq. ft.

Note: Note that even though both length and width are unknown, two different variables should not be assigned to them as we have only one equation to solve them. Also be careful about the units of the different parameters units of length, breadth and perimeter will be ft but that of the area will be sq. ft.