The length of a rectangle is 6 in. more than its width. Its area is 40 sq. in. How do you find the width of the rectangle?

Question

Kennedy Adams · Accepted Answer

The width of the rectangle is #4# inches.

We consider the width of the rectangle as #x# which will make the length #(x+6)#. Since we know the area, and the formula of a rectangle's area to be length #xx# breadth, we can write: #x xx (x+6)=40# Open the brackets and simplify. #x^2+6x=40# Subtract #40# from both sides. #x^2+6x-40=0# Factorise. #x^2+10x-4x-40=0# #x(x+10)-4(x+10)=0# #(x-4)(x+10)=0# #x-4=0# and #x+10=0# #x=4# and #x=-10# The only possibility in the above problem is that #x=4#. That will make the width #4# and the length #(x+6)# which is #10#, and the area #(4xx10)# which is #40#.

Eli Assouline · Answer

undefined Let $ x $ represent the width of the rectangle. Since the length is 6 inches more than the width, the length can be represented as $ x + 6 $.

The area of a rectangle is given by the formula: $ \text{Area} = \text{Length} \times \text{Width} $.

Given that the area is 40 square inches, we can set up the equation:

$$ (x + 6) \times x = 40 $$

Expanding the equation, we get:

$$ x^2 + 6x = 40 $$

Rearranging terms to set the equation to zero:

$$ x^2 + 6x - 40 = 0 $$

Now, we can solve this quadratic equation for $ x $ using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:

$$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $$

Where $ a = 1 $, $ b = 6 $, and $ c = -40 $.

Plugging in the values, we get:

$$ x = \frac{{-6 \pm \sqrt{{6^2 - 4 \times 1 \times (-40)}}}}{{2 \times 1}} $$

$$ x = \frac{{-6 \pm \sqrt{{36 + 160}}}}{2} $$

$$ x = \frac{{-6 \pm \sqrt{{196}}}}{2} $$

$$ x = \frac{{-6 \pm 14}}{2} $$

So, the possible values for $ x $ are:

$$ x_1 = \frac{{-6 + 14}}{2} = \frac{8}{2} = 4 $$

$$ x_2 = \frac{{-6 - 14}}{2} = \frac{-20}{2} = -10 $$

Since the width cannot be negative, we discard $ x_2 $.

Therefore, the width of the rectangle is $ x = 4 $ inches.