Question

a rectangle has the perimeter of 116 cm and its length is 1cm more than twice its width. I got 39L and 19w but I'm having problems setting up the problems

Answer 1

To answer this question, we can proceed as follows:

1. We have that the perimeter of the rectangle is equal to 116cm, then, we have:

`w+l+w+l=116\Rightarrow2w+2l=116`

We know that a rectangle is a parallelogram. Then, its opposite sides are congruent.

2. We have that the length of the rectangle is 1 cm more than twice its width. We can translate this, algebraically, as follows:

`l=2w+1`

Now, to find the measures of the length and the width of the rectangle, we can substitute this last formula into the first one, as follows:

`2w+2(2w+1)=116`

We need to apply the distributive property to find w:

`2w+4w+2=116`

Adding like terms:

`6w+2=116`

Subtracting 2 to both sides of the equation, and then dividing by 6:

`6w+2-2=116-2\Rightarrow6w=114\Rightarrow(6w)/(6)=(114)/(6)\Rightarrow w=19`

Then, the width of the rectangle is equal to 19cm. The measure of the length can be calculated using either equation above. Let us use the first equation:

`2w+2l=116\Rightarrow2\cdot19+2l=116\Rightarrow38+2l=116`

Then, using similar properties as before, we have:

`38-38+2l=116-38\Rightarrow2l=78\Rightarrow(2l)/(2)=(78)/(2)\Rightarrow l=39`

In summary, we have that the measures of the length and width of this rectangle are:

• Width, ,(w) =, 19cm

,

• Legth (l) = ,39cm