We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

TEXT
Unlock Solution Steps
Sign in to
Symbolab
Get full access to all Solution Steps for any math problem
By continuing, you agree to our Terms of Use
and have read our Privacy Policy

Study Guides > Prealgebra

Finding the Area of Irregular Figures

Learning Outcomes

  • Combine area of regular shapes to find the area of irregular shapes.
 

So far, we have found area for rectangles, triangles, trapezoids, and circles. An irregular figure is a figure that is not a standard geometric shape. Its area cannot be calculated using any of the standard area formulas. But some irregular figures are made up of two or more standard geometric shapes. To find the area of one of these irregular figures, we can split it into figures whose formulas we know and then add the areas of the figures.

example

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 [latex]12[/latex] and a length of [latex]4[/latex]. The red rectangle has a width of [latex]2[/latex], 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 [latex]4[/latex] units long, the length of the red rectangle must be [latex]6[/latex] 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 [latex]60[/latex] square units. Is there another way to split this figure into two rectangles? Try it, and make sure you get the same area.
 

try it

[ohm_question]146946[/ohm_question]

example

Find the area of the shaded region. A blue geometric shape is shown. It looks like a rectangle with a triangle attached to the top on the right side. The left side is labeled 4, the top 5, the bottom 8, the right side 7.

Answer: Solution We can break this irregular figure into a triangle and rectangle. The area of the figure will be the sum of the areas of triangle and rectangle. The rectangle has a length of [latex]8[/latex] units and a width of [latex]4[/latex] units. We need to find the base and height of the triangle. Since both sides of the rectangle are [latex]4[/latex], the vertical side of the triangle is [latex]3[/latex] , which is [latex]7 - 4[/latex] . The length of the rectangle is [latex]8[/latex], so the base of the triangle will be [latex]3[/latex] , which is [latex]8 - 4[/latex] . A geometric shape is shown. It is a blue rectangle with a red triangle attached to the top on the right side. The left side is labeled 4, the top 5, the bottom 8, the right side 7. The right side of the rectangle is labeled 4. The right side and bottom of the triangle are labeled 3. Now we can add the areas to find the area of the irregular figure. The top line reads A sub figure equals A sub rectangle plus A sub red triangle. The second line reads A sub figure equals lw plus one-half red bh. The next line says A sub figure equals 8 times 4 plus one-half times red 3 times red 3. The next line reads A sub figure equals 32 plus red 4.5. The last line says A sub figure equals 36.5 sq. units. The area of the figure is [latex]36.5[/latex] square units.

 

try it

[ohm_question]146949[/ohm_question]
 

example

A high school track is shaped like a rectangle with a semi-circle (half a circle) on each end. The rectangle has length [latex]105[/latex] meters and width [latex]68[/latex] meters. Find the area enclosed by the track. Round your answer to the nearest hundredth. A track is shown, shaped like a rectangle with a semi-circle attached to each side.

Answer: Solution We will break the figure into a rectangle and two semi-circles. The area of the figure will be the sum of the areas of the rectangle and the semicircles. A blue geometric shape is shown. It looks like a rectangle with a semi-circle attached to each side. The base of the rectangle is labeled 105 m. The height of the rectangle and diameter of the circle on the left is labeled 68 m. The rectangle has a length of [latex]105[/latex] m and a width of [latex]68[/latex] m. The semi-circles have a diameter of [latex]68[/latex] m, so each has a radius of [latex]34[/latex] m. The top line reads A sub figure equals A sub rectangle plus A sub semicircles. The second line reads A sub figure equals bh plus red 2 times (in parentheses) red 1/2pi times r squared. The next line says A sub figure approximately equals 105 times 68 plus red 2 times (in parentheses) red 1/2 times 3.14 times 34 squared. The next line reads A sub figure approximately equals 7140 plus red 3629.84. The last line says A sub figure approximately equals 10,769.84 square meters.

 

try it

[ohm_question]146952[/ohm_question] [ohm_question]146953[/ohm_question]
 

Licenses & Attributions

CC licensed content, Original

  • Question ID 146953, 146952, 146949. Authored by: Lumen Learning. License: CC BY: Attribution.

CC licensed content, Specific attribution