Question

A rectangular picture frame has a perimeter of 46 inches and a width 3 inches shorter than its length. Find the area of the picture frame.

Answer 1

130in squared

Explanation:

If you have a perimeter of 46 and a width that is 3 inches shorter than its length you would want to set up the equation of L=length and W=L-3. In order to find the length you would set up the equation 2(L)+2(L-3)=46 inches as the perimeter is both lengths plus both widths in this case it would equal 46 inches. So you would solve for L as you would

2L+2L-6=46

4L=52

L=13

Therefore the length is equal to 13 inches. Since the width is 3 inches shorter the width is 10 inches. The area is equal to length times width therefore the area is equal to 13in times 10in. Therefore the area is 130in squared.

Answer 2

Answer: 130 Square inches

Explanation:

Let's call the length of the rectangular picture frame "l" and the width "w". We know that:

l = w + 3 (since the width is 3 inches shorter than the length)

And we also know that the perimeter of the picture frame is 46 inches, so:

2l + 2w = 46

We can substitute the expression for l from the first equation into the second equation:

2(w + 3) + 2w = 46

Expanding the first term on the left side:

2w + 6 + 2w = 46

Combining like terms:

4w + 6 = 46

Subtracting 6 from both sides:

4w = 40

Dividing both sides by 4:

w = 10

And since l = w + 3, we can find the length:

l = 10 + 3 = 13

So the area of the picture frame is given by l * w:

A = l * w = 13 * 10 = 130 square inches.