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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

Verified Answer :

Ckaranja

A=LW
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

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

Learn more on area of rectangle here;

https://brainly.com/question/13048427

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