Question

The length of the whiteboard is 8 units more than the width. The area is 33 sq. units. Find the dimension of the garden.

Answer 1

SOLUTION

GIVEN

• The length of the whiteboard is 8 units more than the width.

• The area is 33 sq. units.

TO DETERMINE

The dimension of the garden.

EVALUATION

Let width of the whiteboard = x unit

Since length of the whiteboard is 8 units more than the width.

Length of the whiteboard = x + 8 unit

Now it is given that area is 33 sq. units.

So by the given condition

Length × Width = 33

$\displaystyle \sf{ \implies (x + 8)x = 33}$

$\displaystyle \sf{ \implies {x}^{2} + 8x - 33 = 0}$

$\displaystyle \sf{ \implies {x}^{2} + (11 - 3)x - 33 = 0}$

$\displaystyle \sf{ \implies {x}^{2} + 11x - 3x - 33 = 0}$

$\displaystyle \sf{ \implies x(x + 11) - 3(x + 11)= 0}$

$\displaystyle \sf{ \implies (x + 11) (x - 3)= 0}$

x + 11 = 0 gives x = - 11

x - 3 = 0 gives x = 3

Since width can not be negative

∴ x ≠ - 11

∴ x = 3

Width of the whiteboard = 3 unit

Length of whiteboard = 3 + 8 = 11 unit

FINAL ANSWER

Length of whiteboard = 11 unit

Width of the whiteboard = 3 unit

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Answer 2

Given :  The length of the whiteboard is 8 units more than the width.

The area is 33 sq. units.

To  Find :  the dimension of the whiteboard

Solution:

width = x unit

Length = x + 8 units

Area = width * length

x(x + 8)

= x² + 8x

x² + 8x  = 33

=> x² + 8x - 33 = 0

=> (x + 11)(x - 3)  = 0

=> x = 3  , - 11

Length can not be negative

Hence x = 3

so x + 8 = 11

Dimensions are 3 and 11 units

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