How to Find the Area of the Shaded Region Using the Area Decomposition Method

Author:
Ray
Updated date:
May 29, 2021

Ray is a licensed engineer in the Philippines. He loves to write about mathematics and civil engineering.

How do you find the area of the shaded region? It has been a constant struggle for students taking up geometry classes on how to solve the shaded areas. The best way to find a shaded region area is to familiarize yourself with the area decomposition method. The Area decomposition method separates the basic shapes found in a given complex figure for fast and organized analysis. Two cases are present in obtaining the areas of the shaded regions - figures with holes and composite figures.

Case 1: Area of Composite Figures

A composite plane figure is a plane figure made up of different geometric figures whose areas can be determined. The total area of a composite plane figure is equal to the sum of the areas of its parts. If the given shape is a composite figure, subdivide it into different fundamental shapes or polygons of known area formulas. See the step-by-step procedure on how to compute the area of complex figures.

Determine the basic shapes represented in the problem. For composite shapes, subdivide the whole figure into forms. Once you identify the basic figures present, remember the area equations for each.
Calculate the areas of all shapes identified.
Find the area of the shaded region by adding all obtained area values of the different primary shapes to complete the whole figure. Use the formula shown:

Area of the shaded region = Area of Region 1 + Area of Region 2 + Area of nth Region

Case 2: Outer Area Minus Inner Area

For figures containing holes or cavities, consider the whole figure's area without the hole or cavity. The remaining area is obtained by subtracting the punch hole area from the entire figure. Use the step-by-step procedure shown below in solving for areas of shaded regions with holes.

Determine the basic shapes represented in the problem. Once you identify the basic figures present, remember the area equations for each figure.
Calculate the areas of both inner and outer shapes.
Find the area of the shaded region by subtracting the size area of the unshaded inner shape from the outer, larger shape. The area outside the inner figure is the part that indicates the area of interest. Use the formula shown:

Area of the shaded region = Area of the outer shape - Area of the unshaded inner shape

Example 1: Finding the Area of the Inscribed Four-Leaf Figure

If the area of the square shown in the figure below is 64 square inches. Find the area value of the inscribed four-leaf figure.

Solution

There are so many ways to solve this inscribed four-leaf figure problem. You can use simple geometrical analysis or a calculus method. However, this time, let us focus on the simple geometrical analysis. First and foremost, obtain the length of the square sides given its area of 64 square inches.

A_square = s²
A_square = 64 square inches
s² = 64
s = √64
s = 8

As you can observe from the photo shown, the curves from the four-leaf figure resemble that of a half-circle. Also, as we draw these half-circles running through the lines of the four-leaf, some areas are excluded. These areas are the unshaded regions we need to solve for the four-leaf area.

A_{unshaded region 1} = A_square - A_{semicircle #1} - A_{semicircle #2}
A_{unshaded region 1} = 8² - π(4)²/2 - π(4)²/2
A_{unshaded region 1} = 64 - 16π
A_{unshaded region 1} = 13.73 square inches

Example 2: Finding the Area of a Composite Figure Composed of a Triangle, Rectangle, and Semicircle

A composite figure consists of a triangle, rectangle, and semicircle. The triangle's altitude is three times the radius of the semicircle. The height of the rectangle is four times the radius of the semicircle. If the area of the composite figure is 201 square centimeters, find the radius of the semicircle.

Solution

For the given composite figure, it shows that all are shaded shapes. To solve this problem, directly add all areas of the triangle, rectangle, and semicircle. Let r be the radius of the semicircle. Given all regions are expressed in terms of r, solve for the value of r.

A_triangle = ½ x b x h
A_triangle = ½ (2r) (3r)
A_triangle = 3r²

A_rectangle = l x w
A_rectangle = (4r) (2r)
A_rectangle = 8r²

A_semicircle = π(r)²/2

Solve for the total area of the given complex shape by adding all area measurements obtained for triangle, rectangle, and semicircle.

Total Area = A_triangle + A_rectangle + A_semicircle
201 = 3r² + 8r² + (π/2)r²
201 = 11r² + (π/2)r²
r = 4 centimeters

Final Answer

The radius of the semicircle is 4 centimeters.

Example 3: Area of the Shaded Region in a Right Triangle

Solve for the area of the shaded region in the right triangle shown below.

Solution

It is a simple problem of finding the area of shaded regions on right triangles. Subtract the area value of the unshaded inner shape from the outer shape area to get the area measurement of the shaded region.

A_{outer shape} = ½ x b x h

A_{outer shape} = ½ (10) (8)

A_{outer shape} = 40 square inches

A_{inner shape} = ½ x b x h

A_{inner shape} = ½ (8) (6)

A_{inner shape} = 24 square inches

Finally, find the area of the shaded region.

A_{shaded region} = Area of the outer shape - Area of the unshaded inner shape
A_{shaded region} = 40 - 24
A_{shaded region} = 16 square inches

Final Answer

The area of the shaded region is equal to 16 square inches.

Example 4: Area of Unshaded Region in a Right Triangle

Given the dimensions of the right triangle shown, solve for the area of the unshaded region.

Solution

Treat the triangles as similar triangles. Then, subtract the area of the shaded region from the larger right triangle.

A_{bigger triangle} = ½ x b x h

A_{bigger triangle} = ½ (48 + 42) (74)

A_{bigger triangle} = 3330 square centimeters

A_{smaller triangle} = ½ x b x h

A_{smaller triangle}= ½ x (42) (48)

A_{smaller triangle}= 1008 square centimeters

Find the area of the unshaded region by subtracting the smaller triangle's area value from that of the larger triangle.

A_{unshaded region} = A_{bigger triangle} - A_{smaller triangle}
A_{unshaded region} = 3330 - 1008
A_{unshaded region} = 2322 square centimeters

Final Answer

The area of the unshaded region in the right triangle is 2322 square centimeters.

Example 5: Finding the Area of the Shaded Region of a Rectangle

Calculate the area of the shaded region of the rectangle shown below.

Solution

This problem is an example of case 2. Subtract the punch hole area, the inner rectangle, from the whole figure of dimensions 15 and 30 centimeters.

A_{larger rectangle} = l x w

A_{larger rectangle} = 30 x 15

A_{larger rectangle} = 450 square centimeters

A_{inner rectangle} = l x w

A_{inner rectangle} = 12 x 24

A_{inner rectangle} = 288 square centimeters

A_{shaded region} = A_{larger rectangle} - A_{inner rectangle}

A_{shaded region} = 450 - 288

A_{shaded region} = 162 square centimeters

Final Answer

The area of the shaded region in the rectangle is 162 square centimeters.

Example 6: Area of the Shaded Region of Square Diamond

Given that the larger square side is equal to 6 m, solve for the shaded area's value, the side length of the diamond square inscribed is 5 meters.

Solution

Find the area of the larger square using the formula s².

A_{larger square} = s²

A_{larger square} = 6 x 6

A_{larger square} = 36 square meters

Compute for the area value of the smaller square.

A_{smaller square} = s²
A_{smaller square} = 5 x 5
A_{smaller square} = 25 square meters

Find the area of the shaded region by subtracting the obtained area of the smaller square from the area measurement of the larger square shape.

A_{shaded region} = A_{bigger square} - A_{smaller square}
A_{shaded region} = 36 - 25
A_{shaded region} = 11 square meters

Final Answer

The area of the shaded region is equal to 11 square meters.

Example 7: Area of the Shaded Region in Between an Inscribed Triangle in a Circle

The diameter of the larger circle is 25 centimeters. If the triangle inside the circular shape is an equilateral triangle with a side length of 15 centimeters, what is the area of the shaded region?

Solution

Solve for the value of the larger circle given the diameter of 25 centimeters. Use the formula A = πD² / 4.

A_{bigger circle} = πD² / 4

A_{bigger circle} = π(25)² / 4

A_{bigger circle} = 625π/4 square centimeters

A_{bigger circle} = 490.87 square centimeters

Find the area value of the inner triangle using the formula shown below. There are many formulas to solve equilateral triangles' areas, such as heron's formula or the Pythagorean theorem. It is entirely up to you what to use.

A_{inner triangle} = √3/4 (a)²
A_{inner triangle} = √3/4 (15)²
A_{inner triangle} = 97.43 square centimeters

Finally, solve for the area measurement of the shaded region.

A_{shaded region} = A_{bigger circle} - A_{inner triangle}
A_{shaded region} = 490.87 - 97.43
A_{shaded region} = 393.44

Final Answer

The area of the shaded region is equal to 390.44 square cm.

Example 8: Calculating the Area of the Shaded Region in a Composite Shape

Calculate the shaded area of the complex figure with a circular hole in it, given that the diameter of the hole is 24 inches.

First, compute the area of the whole composite shape given the dimensions shown in the figure.

A_{composite figure} = (60 x 25) + (25 x 15)
A_{composite figure} = 1875 square inches

Then, compute the area of the inner circle, given that its diameter is 24 inches.

A_{inner circle} = πD² / 4
A_{inner circle} = π(24)² / 4
A_{inner circle} = 144π
A_{inner circle} = 452.39 square inches

Calculate the area of the shaded region by subtracting the area value of the inner circle from the area value of the composite shape.

A_{shaded region} = A_{composite figure} - A_{inner circle}
A_{shaded region} = 1875 - 452.39 square inches

Final Answer

The shaded area in the complex figure given is 452.39 square inches.

Example 9: Finding the Area of the Shaded Region

If triangle ABC is a right triangle, and BC is a semicircle, calculate the area of the shaded diagram below.

Solution

The total area is the sum of the areas of the semicircle BC and the right triangle ABC. Note that the length of the side and altitude of the right triangle are known. Solve for the area using the simple equation ½ (b) (h).

A_{right triangle} = ½ (b) (h)

A_{right triangle} = ½ (4) (3)

A_{right triangle} = 6 square units

Then, solve for the area of the semicircle BC. By observation, the diameter of semicircle BC is the hypotenuse of the right triangle ABC. Use the Pythagorean theorem to solve the hypotenuse length. Then proceed on getting the area of the semicircle. Let c be the hypotenuse.

c² = a² + b²

c = √a² + b²

c = √(3)² + (4)²

c = 5

A_semicircle = πD² / 8

A_semicircle = π(5)² / 8

A_semicircle = 25π / 8

A_semicircle = 9.82 square units

Finally, solve for the area of the shaded region by adding the area measurement of the right triangle and the area value of the semicircle.

A_{shaded region} = A_{right triangle} + A_semicircle
A_{shaded region} = 6 + 9.82
A_{shaded region} = 15.82 square units

Final Answer

The total area is equal to 15.82 square units.

Example 10: Finding the Area of a Complex Shape

If all angles are right angles in the figure below, find the total area of the complex shape using the area decomposition method. Take note that all angles are right angles.

Solution

Divide the given complex figure into six basic shapes. The total area is the sum of the rectangles, squares, and triangle areas. Given the values of areas of the basic shapes, add up all of them to compute the area value of the complex form. Just take note of the divisions made from the whole complex body.

A_{shaded region} = A₁ + A₂ + A₃ + A₄ + A₅ + A₆

A_{shaded region} = 16 + 4 + 4 + 4 + 2 + 2.5

A_{shaded region} = 32.5 square inches

Final Answer

The area of the complex shape is 32.5 square inches.

Example 11: Circle Inscribed in a Triangle

Find the area of the shaded region given that the radius of the inscribed circle is 16 cm and the side lengths of the triangle are 20 cm, 50 cm, and 75 cm.

Solution

Find the area of the outer triangle utilizing only the side lengths of the triangle. Use Heron's formula to solve the triangle's area value given the sides 20 cm, 50 cm, and 75 cm. But first, solve for the semi-perimeter of the triangle.

s = (a + b + c) / 2

s = (20 + 50 + 69) / 2

s = 69.5

A_triangle = √s(s-a)(s-b)(s-c)

A_triangle = √69.5 (69.5 - 20)(69.5 - 50)(69.5 - 69)

A_triangle = 183.15 square centimeters

You can also use the generated formula for the area of the circumscribed triangle given below.

A_triangle = ½ (r) (Perimeter of triangle)
A_triangle = ½ (2.64) (20 + 50 + 69)
A_triangle = 183.15 square centimeters

Then, solve for the area of the inner circle with radius r = 2.64 centimeters.

A_{inner circle} = πr²
A_{inner circle} =π(2.64)2
A_{inner circle} = 21.9 square centimeters

Calculate the area of the shaded region by subtracting the area measurement of the inscribed circle from the larger triangle.

A_{shaded region} = A_triangle - A_{inner circle}
A_{shaded region} = 183.15 - 21.9
A_{shaded region} = 161.25 square centimeters

Final Answer

The calculated area of the shaded region is 161.25 square centimeters.

Example 12: Area of the Shaded Region in a Circle

What is the area of the shaded region circle? Assume that the diameter of the larger circle is 30 centimeters and the radius of the punched circular hole is 6.5 centimeters.

Solution

Find the area of the shaded region by subtracting the area value of the smaller inner circle from the area value of the larger circle shape.

A_{inner circle} = πr²

A_{inner circle} = π(6.5)²

A_{inner circle} = 132.73 square centimeters

A_{outer circle} = πD² / 4

A_{outer circle} = π(30)² / 4

A_{outer circle} = 225π

A_{outer circle} = 706.86 square centimeters

A_{shaded region} = A_{outer circle} - A_{inner circle}

A_{shaded region} = 706.86 - 132.73

A_{shaded region} = 574.13 square centimeters

Final Answer

The area of the shaded region in the circle is 574.13 square centimeters.

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This content is accurate and true to the best of the author’s knowledge and is not meant to substitute for formal and individualized advice from a qualified professional.