Question

Francine has a picture whose length is 4/3 its width. She wants to enlarge the picture to have an area of 192 in2. What will the dimensions of the enlarged picture be? Model the scenario and solve. Then, explain in at least one sentence your solution and include the reasonableness of your solution.

Answer 1

L=4W/3

A=LW, using L found above in this equation we have:

A=(4W)W/3

A=4W^2/3

3A=4W^2

3A/4=W^2

W=√(3A/4), we are told that A=192 in^2

W=√(3*192/4)

W=√144

W=12 in, since L=4W/3

L=4*12/3

L=16 in

So the length is 16 inches and the width is 12 inches.

Answer 2

Answer:

The length of picture is 16 inches and width is 12 inches.

Step-by-step explanation:

Francine has a picture whose length is 4/3 its width.

Let the width of the picture be = w

Then its length would be = $\frac{4w}{3}$

She wants to enlarge the picture to have an area of 192 square inches.

This becomes: $w*\frac{4w}{3}=192$

$4w^{2}= 576$

$w^{2} =144$

This gives w = 12 inches.

We know that length*width =192

Length = $\frac{192}{12}$ = 16

Hence, the length of picture is 16 inches and width is 12 inches.