How to Find Area and Perimeter of Composite Figures? (Examples) - BYJUS

Calculating Area, Perimeter of Composite Figures

Composite figures are figures made up of multiple shapes. We know to find the area and perimeter of simple shapes like triangles, circles, squares, etc. Here we will combine those formulas to find the area and perimeter of composite shapes. ...Read MoreRead Less

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Composite figures

Simple geometric shapes like triangles, squares, rectangles, semicircles, and other two-dimensional figures together make up composite figures. We can divide a composite figure or other irregular-shaped figures into simple, non-overlapping figures to find their area. We have to calculate the total area of the composite figure (Volume of a Composite Figure) by adding the areas of the simpler figures together.

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Estimating perimeter and area using a square grid

Square grids are used for various types of unit measurements and to compare sizes. On a square grid, we can draw various shapes and calculate the total area by counting the full and half squares. We can cut along a couple of lines that we’ve drawn. The shapes whose area have to be measured are placed in square grids. We must put the pieces together in a different shape and find the total area. Construct the various squares (1 x 1, 2 x 2, 3 x 3, 4 x 4) on the square grid, if the figure contains different shapes then draw the outlines on a square grid and count the unit squares to determine the area of each figure. Each square box has sides that measure 1 unit each. Observe the pattern to figure out that the rectangle’s area equals the length multiplied by the width.

 

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Let’s understand this topic using an example; let’s calculate the perimeter and area of the arrow shown below.

 

 

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Solution: A.

 

Step 1: Count how many grid squares the arrow covers. There are 16 of them.

 

Step 2: Count how many diagonal lengths there are around the arrow. There are four of them.

 

Step 3: Use 1.5 units for the length of the diagonals.

 

Length of 16 grid square lengths = 16 \(\times\) 1 = 16 units.

 

Length of 4 diagonal lengths = 4 \(\times\) 1.5 = 6 units.

 

So, the perimeter is about 16 + 6 = 22 Units.

 

B.

 

Step 1: Count how many squares are entirely within the figure. There are 14 of them.

 

Step 2: Count how many full squares there are in the diagram. There are 4 of them.

 

Area of 14 squares = 14 \(\times\) 1 = 14 Square units

 

Area of 4 half squares = 4 \(\times\) 0.5 = 2 Square units

 

So, the area is 14 + 2 = 16 Square units

Solved Area and Perimeter Composite Figures Examples

1)  Calculate the area and perimeter of the given figure.

 

 

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Triangle; base = 8ft, height = 10ft and semicircle; diameter = 12ft.

Solution:

 

A triangle and a semicircle together make up the figure.

 

The distance around the figure’s triangular part is 8 + 10 = 18 feet.

 

The circumference of a circle with a diameter of 12 feet is divided in half by the distance around the semicircle.

                   

\(\frac{C}{2}~=~\frac{\pi d}{2}\)          (Dividing the circumference by 2)

 

= \(\frac{3.14~\times ~12}{2}\)          (Substituting 3.14 for “\(\pi\)” and 12 for “d”)

                       

= 18.84              (Simplified)

 

So, the perimeter is 18 + 18.84 = 36.84 feet

 

The area of the triangle and the area of the semi circle must now be determined.

 

The area of the given triangle, A = \(\frac{1}{2}~bh\)

                                   

                                                     = \(\frac{1}{2}~(8)(10)\)

                                   

                                                     = 40 Square feet

 

 

The area of the given semicircle, A = \(\frac{\pi r^2}{2}\)

                                         

                                                         = \(\frac{3.14~\times~ 6^2}{2}\)   (The semicircle has a radius of \(\frac{12}{2}\) = 6 feet)

                                         

                                                         = 56.52 Square feet

 

So, the area is 40 + 56.52 = 96.52 square feet.

 

2) Calculate the area and perimeter of the given figure.

 

 

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Solution:

 

Four semicircles and a square make up the figure.

 

The circumference of a circle with a diameter of 5 inches is one-half the distance around the semicircle.

                 

\(\frac{C}{2}~=~\frac{\pi d}{2}\)          (Dividing the circumference by 2)

                         

= \(\frac{3.14~\times ~5}{2}\)           (Substituting 3.14 for “\(\pi\)” and 5 for “d”)

                   

= 7.85               (Simplified)

 

So, the perimeter is 4\(~\times~\) 3.14 = 12.56 inch. (since there are 4 semi circles so multiplying with 4)

 

The area of the square and the area of the semi circle must now be determined.

 

The area of the square, A = side\(~\times~\)side

                                          

                                          = 55

 

                                          = 25 Square inches

 

 

The area of the semicircle, A = \(\frac{\pi r^2}{2}\)

 

                                               = \(\frac{3.14~\times~ 2.5^2}{2}\) (The semicircle has a radius of 5= 2.5 inch)

 

                                               = 9.8125 Square inches

 

So, the area is 4 + (4\(~\times~\)9.8125) = 43.25 square inches

 

3) Calculate the area and perimeter of the given figure.

 

 

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Solution:

 

The rectangle and the semicircles make up the figure.

 

The figure’s rectangle is surrounded by a distance of

4 + 4 + 4 + 4 + 4 + 4 = 24 feet.

 

The distance around the semicircle is one- half the circumference of a circle with a diameter of 16 – (24) = 8 feet.

                 

\(\frac{C}{2}~=~\frac{\pi d}{2}\)      (Dividing the circumference by 2)

 

= \(\frac{3.14~\times~ 8}{2}\)       (Substituting 3.14 for “\(\pi\)” and 8 for “d”)

 

= 12.56          (Simplified)

 

So, the perimeter is about 24 + (2\(~\times~\) 12.56) = 49.12 feet. (since there are 2 semi circles we multiply the result with 2)

 

The area of the rectangle and the area of the semi circle must now be determined.

 

The area of the rectangle, A  = l\(~\times~\)w

 

                                               = 16\(~\times~\)4

 

                                               = 64 Square feet

 

The area of the semicircle, A = \(\frac{\pi r^2}{2}\)

 

                                               = \(\frac{3.14~\times~ 4^2}{2}\) (The semicircle has a radius of \(\frac{8}{2}\) = 3 feet)

 

                                               = 25.12 Square feet.

 

So, the area is 64 + (2\(~\times~\) 25.12) = 114.24 square feet.

 

4) The volleyball court’s center circle is painted blue and has a radius of 4 feet. The rest of the court has a brown stain. A 60 square feet area of the court can be stained with one gallon of wood stain. How many gallons of wood stain are required to cover the court’s brown areas?

 

Solution:

 

Step 1: Recognize the issue – you’ve been given the dimensions of a volleyball court.

 

When one gallon of wood stain covers 60 square feet, you must calculate the number of gallons of wood stain required to stain the brown portions of the court.

 

Step 2: Make a plan – Measure the rectangular court’s entire area. Then divide by 60 after subtracting the area of the center circle.

 

Step 3: Check and solve – Finding the area of the rectangle, A = l\(~\times~\)w

 

= 90\(~\times~\)60

 

= 5400 sq.ft

 

Finding the area of circle, A = \(\pi r^2\)

 

= 3.14\(~~\times~~4^2\)

 

= 50.24

 

Therefore, the area that is stained is about 5400 – 50.24 = 5450.24 square feet.

 

Because one gallon of wood stain covers 60 square feet, you will need 5450.24 ÷ 60 = 90 gallons of wood stain.

Frequently Asked Questions on Perimeter and Area of Composite Figures

Specific formulae can be used to find the areas of standard 3D solids. The area of a composite shape can be determined by breaking it down into individual standard solids first.

The best way to analyze composite figures is to break them down into smaller pieces with easily recognisable characteristics. Recalling the characteristics of simpler figures, such as squares and circles, is crucial and important.