Question

a rectangle has an area of 40 square units. the length is 6 units greater than the width.

what are the dimensions of the rectangle?

8 by 5
10 by 4
11 by 9
13 by 7

Answer 1

A=L*W
Let w be width
L=(w+6)
40=(w+6)*w
40=$w^{2} +6w$
$w^{2} +6w-40$=0
$w^{2}+10w-4w-40=0$
w(w+10)-4(w+10)=0
(w-4)(w+10)=0
w=4
Length= 10 units
Width=4 units

Answer 2

The length and width of the rectangle is 4 and 10 respectively

Data;

• Area = 40
• Length = x
• width = y

Area of  Rectangle

The formula of area of a rectangle is given as

$A = L* W$

But the length is 6units greater than the width

Let's substitute that

$l = w + 6$

we can put the value of the length and solve the equation

$A = l * w\\l = w + 6\\ A = (w+6)* w\\A = w^2 + 6w\\40 = w^2 + 6w \\w^2 + 6w - 40 = 0$

solving the quadratic equation, we get a solution of - 10 and 4. But since the width can only have a positive value, the width of the rectangle is 4.

We can substitute this value and solve for the length

$l = w + 6\\l = 4 + 6\\l = 10$

The length and width of the rectangle is 4 and 10 respectively

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