Question

The length of a rectangle is 11 centimeters less than its width. Its area is 20 square centimeters. Find the dimensions of the rectangle.

Answer 1

Given:

The length of a rectangle is 11 centimeters less than its width. Its area is 20 square centimeters.

To Find:

Dimensions of the rectangle

Solution:

Let the breadth of the rectangle be 'b'

So, the length of the rectangle will be 'b - 11'

We know that, Area of rectangle = Length x Breadth  

⇒ 20 = (b - 11) x b  

⇒ 20 = b² - 11 b  

⇒ b² - 11b - 20 = 0

a = 1, b = -11, c = -20

Substituting the given values in the formula, x = - b ± √ b² - 4 ac / 2a,

b = - (-11) ± √ (-11)² - 4(1)(-20) / 2(1)

b = 11 ± √201 / 2

b = 12.6 (approx)

so, l = b - 11 = (12.6 - 11) = 1.6 (approx)

Hence, the length of the rectangle is 1.6 cm and its breadth is 12.6 approximately

Answer 2

SOLUTION

According to the given conditions in the question, length is less than width for the rectangle. Which is impossible.

CORRECT QUESTION

The length of a rectangle is 11 centimeters less than four times its width. Its area is 20 square centimeters. Find the dimension of the rectangle

GIVEN

• The length of a rectangle is 11 centimeters less than four times its width.

• Its area is 20 square centimeters.

TO DETERMINE

The dimension of the rectangle

EVALUATION

Let the width of the rectangle = x cm

So Length of the rectangle = ( 4x - 11 ) cm

So by the given condition

$\sf{x(4x - 11) = 20}$

$\sf{ \implies \: 4 {x}^{2} - 11x - 20 = 0}$

$\sf{ \implies \: 4 {x}^{2} - 16x + 5x- 20 = 0}$

$\sf{ \implies \: 4x(x - 4) + 5(x - 4) = 0}$

$\sf{ \implies \: (4x + 5) (x - 4) = 0}$

$\displaystyle \: \sf{ \implies \: x = - \frac{5}{4} \:, \: 4}$

Since width of a rectangle can not be negative

$\displaystyle \: \sf{ \therefore \: \: x \ne - \frac{5}{4}}$

$\displaystyle \: \sf{ \therefore \: \: x = 4}$

Hence width of the rectangle = 4 cm

Length of the rectangle

= ( 4 × 4 - 11 ) cm

= ( 16 - 11 ) cm

= 5 cm

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