Find the area of the shaded region.
An image of an attached horizontal rectangle and a vertical rectangle is shown. The top is labeled 12, the side of the horizontal rectangle is labeled 4. The side is labeled 10, the width of the vertical rectangle is labeled 2.
Solution
The given figure is irregular, but we can break it into two rectangles. The area of the shaded region will be the sum of the areas of both rectangles.
An image of an attached horizontal rectangle and a vertical rectangle is shown. The top is labeled 12, the side of the horizontal rectangle is labeled 4. The side is labeled 10, the width of the vertical rectangle is labeled 2.
The blue rectangle has a width of
12
and a length of
4
. The red rectangle has a width of
2
, but its length is not labeled. The right side of the figure is the length of the red rectangle plus the length of the blue rectangle. Since the right side of the blue rectangle is
4
units long, the length of the red rectangle must be
6
units.
An image of a blue horizontal rectangle attached to a red vertical rectangle is shown. The top is labeled 12, the side of the blue rectangle is labeled 4. The whole side is labeled 10, the blue portion is labeled 4 and the red portion is labeled 6. The width of the red rectangle is labeled 2.
The first line says A sub figure equals A sub rectangle plus A sub red rectangle. Below this is A sub figure equals bh plus red bh. Below this is A sub figure equals 12 times 4 plus red 2 times 6. Below this is A sub figure equals 48 plus red 12. Below this is A sub figure equals 60.
The area of the figure is
60
square units.
Is there another way to split this figure into two rectangles? Try it, and make sure you get the same area.
