The width of a rectangle is 4x which is only 25% of its length. What a

The width of a rectangle is 4x which is only 25% of its length. What are the area and the perimeter of the rectangle respectively ?

16x² squnit and 16x unit
20x² squnit and 40x unit
32x² squnit and 64x unit
64x² squnit and 40x unit
This question was previously asked in
UPSC CAPF – 2009
Let the width of the rectangle be w and the length be l.
We are given the width: w = 4x.
We are told that the width is 25% of its length.
w = 25% of l
w = 0.25 * l
Substitute the value of w:
4x = 0.25 * l
To find the length l, divide by 0.25:
l = 4x / 0.25
l = 4x / (1/4)
l = 4x * 4
l = 16x

So, the length of the rectangle is 16x and the width is 4x.

Area of the rectangle = length * width
Area = l * w = (16x) * (4x) = 16 * 4 * x * x = 64x² sq units.

Perimeter of the rectangle = 2 * (length + width)
Perimeter = 2 * (l + w) = 2 * (16x + 4x) = 2 * (20x) = 40x units.

The area is 64x² sq units and the perimeter is 40x units.

The problem requires calculating the area and perimeter of a rectangle given its width and a relationship between its width and length. The key steps are converting the percentage relationship into an equation, solving for the length, and then applying the standard formulas for the area and perimeter of a rectangle.
Understanding that 25% is equivalent to the fraction 1/4 simplifies the calculation of the length. Area is always in square units, and perimeter is in linear units. The variable ‘x’ is treated algebraically throughout the calculations.