8.6.1 Finding and Reasoning Unknown Factors

Compare your answer:

To determine the area of the blue rectangle, we need to figure out it’s width and length.

The top shape is a square measuring 8 in. x 8 in. If the measurement of the longer left side is 10 inches, then the width of the blue rectangle can be an equation.

$\begin{array}{ccc}\hfill 8+W& \hfill =\hfill & 10\hfill \\ \hfill W& \hfill =\hfill & 2\hfill \end{array}$

The width of the rectangle is 2 inches.

Because the bottom of the square is 8 inches, the length of the blue rectangle can be written as an equation. But we have an unknown part that extends beyond the 8 inches of the square.

Let’s look at the measurements of the smaller rectangle. In addition to seeing that the rectangle has a length of 3 inches, we see that along the top of the rectangle, there is a measurement that includes the top length of our rectangle and a part that extends beyond the rectangle. If we call the extension part, $x$, then we can solve the equation:

$5=x+3$
$2=x$

Do you see that $x$ is the length of the blue rectangle that we were missing? So, the full length of the blue rectangle is $8+x$, or $8+2$, which is 10 inches.

Now we can determine the area of the blue rectangle.

$\text{Area}=l\cdot w$
$\text{Area}=10\cdot 2$
$\text{Area}=20$

The area of the blue rectangle is equal to 20 square inches.

Compare your answer:

You have to start with what you know to figure out the other sides.

First, select a rectangle to start with. Let’s start with the rectangle that measures 60 square inches because we know the full measure of one of its sides. The bottom length of that rectangle is labeled 12 inches. So, to find the width, we solve an area equation.

$\text{Area}=l\cdot w$
$60=12\cdot w$
$5=w$

Now that we know it has a width of 5 inches, we can combine this with some information we know about the rectangle to the right side of the figure that has an area of 48 square inches.

On this right-side rectangle, we see that there is a small part of the width that is labeled 3 inches. The rest of the width happens to match the measurement we just found from the 60 square inch rectangle - we can use the 5 inch measurement we just calculated!

That means the full width of the 48 square inch rectangle is 5 + 3 or 8 inches. This now allows us to use the area formula to determine the length of this rectangle.

$\text{Area}=l\cdot w$
$48=l\cdot 8$
$6=l$

The length of the rectangle on the right side is 6 inches.

The last measurements for the 36 square inch rectangle are a little tricky to find, so let’s review what we know.

• From the 60 square inch rectangle, we know the length is 12 inches across the bottom and the width is 5 inches.
• From the 48 square inch rectangle, we know the length is 6 inches and the width is 8 inches (which we know can be divided into a 3 inch segment and a 5 inch segment).
• Other lengths labeled on the figure include a 3 inch segment and an 11 inch segment that extend across parts of all three rectangles.

Actually, this last part may be helpful. The length of the 48 square inch rectangle is aligned with the 3 and 11 inch measurements.

From the 11 inches, if we subtract the length of the 48 square inch rectangle (which is 6 inches), then the part that extends along a portion of the 60 square inch rectangle equals 5 inches since $11-6=5$.

And, we know the full length of the 60 square inch rectangle should measure 12 inches. If label the missing length, $l$, then we know that $3+l+5=12$.

$8+l=12$
$l=4$

Now we can solve for the width of the 36 square inch rectangle because $l$ also measures its length.

$\text{Area}=l\cdot w$
$36=4\cdot w$
$9=w$

The missing width of the 36 square inch rectangle is 9 inches.