Areas of Combinations of Plane Figures: Definition, Types & Examples

Areas of Combinations of Plane Figures: We come across many figures in our day-to-day life. A combination of plane figures forms various exciting designs like a square drawn with a semicircle on each side with the same diameter as the square’s side. What would be the area of this drawn figure? To calculate the area of such figures, we first identify the area of the plane figures in the combination separately and then sum up their area.

The area of the square and the area of the four semicircles should be added together. In this article, we will discuss how to find the area of such combinations of plane figures.

Introduction

In reality, you’ll come across a lot of situations where you’ll see a combination of figures. Flower beds, drain covers, table cover designs, and circular paths are just a few examples.

Area of Plane Figures

First, we’ll review the area of various two-dimensional shapes before discussing how to find the area of combinations of plane figures.

1. Area of Square \(=a^{2}\) square units, where \(a\) is the length of the side of the square
2. Area of Rectangle \(=l \times b\) square units, where \(l\) is the length of the rectangle, \(b\) is the breadth of the rectangle
3. Area of Circle \(=\pi r^{2}\) square units, where \(r\) is the radius of a circle.
4. Area of Triangle \(=\frac{1}{2} \times b \times h\) square units, where \(b\) is the base of the triangle and \(h\) is the height of the triangle
5. Area of Parallelogram \(=b \times h\) square units, where \(b\) is the base of the parallelogram, \(h\) is the vertical height of the parallelogram.
6. Area of Trapezium \(=\frac{1}{2}(a+b) \times h\) square units, where \(a\) and \(b\) are the length of the parallel sides and \(h\) is the height of trapezium.

How to Find the Areas of Combination of Plane Figures?

Adding Area – Two or More Regions: A compound shape made up of two or more non-overlapping shapes has a total area equal to the sum of their separate areas.

For example, let us find the area of the given figure.

We can see that the given figure is a combination of a rectangle and a semicircle. To find the area of the combined figure, we need to add the rectangle area and the area of the semicircle.

Area of the given figure \(=\) Area of rectangle \(+\) Area of the semicircle
\(=l \times b+\frac{1}{2} \times \pi r^{2}\)
\(=12 \times 5+\frac{1}{2} \times \frac{22}{7} \times\left(\frac{5}{2}\right)^{2}\)
\(=69.82 \,\mathrm{in}^{2}\)

Hence, the area of the given figure is \(69.82 \,\mathrm{in}^{2}\).

Subtracting Area – Two or More Regions: To calculate the area of composite figures made up of simple shapes that overlap, subtract the unshaded figure’s area from the overall area to get the shaded region’s area.

For example, let us find the area of the shaded region in the given figure.

We can see that the given figure is an overlap of a rectangle and a triangle. To find the area of the shaded figure, we need to subtract the triangle area from the area of the rectangle.

Area of the shaded figure \(=\) Area of the rectangle \(-\) Area of triangle
\(=l \times b-\frac{1}{2} \times b \times h\)
\(=22 \times 11-\frac{1}{2} \times 11 \times 5\)
\(=214.5 \,y d^{2}\)

Hence, the area of the shaded figure is \(214.5 \,y d^{2}\)

Examples of Area of Combinations of Plane Figures

1). The given figure is a combination of a semicircle and a rectangle. What is the area of the figure?

We can see that the given figure is a combination of a rectangle and a semicircle. To find the area of the combined figure, we need to add the rectangle area and the area of the semicircle.

Area of the combined figure \(=\) Area of rectangle \(+\) Area of the semicircle
\(=l \times b+\frac{1}{2} \times \pi r^{2}\)
\(=3 \times 6+\frac{1}{2} \times \frac{22}{7} \times(3)^{2}\)
\(=18+14.14=32.14 \,\mathrm{in}^{2}\)

Hence, the area of the combined figure is \(32.14 \,\mathrm{in}^{2}\).

2). Find the area of the combined figure. Remember to separate the figure and find the area of the parts, then combine the areas.

We can see that the given figure is a combination of a square and two semicircles. To find the area of the combined figure, we need to add the area of the square and the area of two semicircles.

Area of the combined figure \(=\) Area of square \(+\) Area of two semicircle
\(=a^{2}+2 \times \frac{1}{2} \times \pi r^{2}\)
\(=6 \times 6+2 \times \frac{1}{2} \times \frac{22}{7} \times(3)^{2}\)
\(=36+28.28=64.28 \,\mathrm{in}^{2}\)

Hence, the area of the combined figure is \(64.28 \,\mathrm{in}^{2}\).

Solved Examples – Areas of Combinations of Plane Figures

Q.1. Given a square of side \(14 \,\text {cm}\). Find the area of the shaded region given in the figure if the diameter of all the four semicircles is \(4 \,\text {cm}\).

Ans: Given, side of square \(A B C D=14 \mathrm{~cm}\)
Diameter of each semicircle \(=4 \mathrm{~cm}\)
The radius of each semicircle \(=\frac{4}{2}=2 \mathrm{~cm}\)
Area of each semicircle \(=\frac{1}{2} \pi r^{2}=\frac{1}{2} \times \pi \times 2^{2}\)
\(=\frac{1}{2} \times \pi \times 4=2 \pi \,\mathrm{cm}^{2}\)
Area of four semicircles \(=4 \times 2 \pi=8 \pi \,\mathrm{cm}^{2}\)
Area of the inner square \(=4 \times 4=16 \mathrm{~cm}^{2}\)
Area of unshaded region \(=8 \pi+16\)
Area of shaded region \(=\) Area of bigger square \(-\) Area of the unshaded region \(=14 \times 14-8 \pi-16\)
\(=196-8 \times 3.14-16\)
\(=196-25.12-16=154.88 \mathrm{~cm}^{2}\)
Hence, the required area is \(154.88 \mathrm{~cm}^{2}\)

Q.2. As seen in the figure, a doorway is decorated. There are four semicircles, the largest of which is \(112 \,\text {cm}\) in diameter. The other three semicircles are all the same size. The shape of the triangle is isosceles. Find the area of the shaded region.

Ans: Radius of the larger semicircle \(=\frac{112}{2}=56 \mathrm{~cm}\)
Diameter of the three small semicircles \(=\frac{112}{3} \mathrm{~cm}\)
The radius of the small semicircle \(=\frac{1}{2} \times \frac{112}{3}=\frac{56}{3} \mathrm{~cm}\)

Area of the larger semicircle \(=\frac{1}{2} \pi R^{2}=\frac{1}{2} \times \frac{22}{7} \times(56)^{2}=4928 \mathrm{~cm}^{2}\)
Area of the smaller semicircle \(=\frac{1}{2} \pi r^{2}=\frac{1}{2} \times \frac{22}{7} \times\left(\frac{56}{3}\right)^{2} \mathrm{~cm}^{2}\)
Area of the three smaller semicircles \(=3 \times \frac{1}{2} \pi r^{2}=3 \times \frac{1}{2} \times \frac{22}{7} \times\left(\frac{56}{3}\right)^{2}=1642.67 \mathrm{~cm}^{2}\)
In \(\triangle P Q R\),
\(\angle Q S P=\angle Q S R=90^{\circ}\)
Area of \(\triangle P Q R=\frac{1}{2} \times P R \times Q S\)
\(=\frac{1}{2} \times 112 \times 56=3136 \mathrm{~cm}^{2}\)
Area of the shaded region \(=\) Area of larger semicircle \(+3\, \times\) Area of smaller semicircle \(-\) Area of the triangle
\(=4928+1642.67-3136=3434.67 \mathrm{~cm}^{2}\)
Hence, the required area of the shaded region is \(3434.67 \mathrm{~cm}^{2}\).

Q.3. The cross-section of a tunnel consists of a semicircle shape surmounted on a rectangle whose shorter side is \(10 \,\text {m}\). If the perimeter of the cross-section is \(92 \,\text {m}\), find the width and height of the tunnel.

Ans: Let \(r\) be the radius of the semicircle.
The perimeter of the cross-section \(=A B+B C+E A+\) Circumference of the semicircle \(C D E\)
The length of the rectangle will be the diameter of the semicircle, \(2 r\).
\(\Rightarrow 92=2 r+10+10+\pi r\)
\(\Rightarrow 92=20+(2+\pi) r\)
\(\Rightarrow 92-20=(2+\pi) r\)
\(\Rightarrow 72=\left(2+\frac{22}{7}\right) r\)
\(\Rightarrow 72=\frac{36}{7} r\)
\(\Rightarrow r=\frac{72 \times 7}{36}=14 \mathrm{~m}\)
Therefore, the width of the tunnel \(A B=2 r=2 \times 14=28 \mathrm{~m}\)
Height of the tunnel \(=r+10=14+10=24 \mathrm{~m}\)

Q.4. Sia bought a table mat, as shown in the figure. It consists of \(6\) congruent semicircles. Find the area of the shaded region.

Ans: Given, length of the rectangle \(=42 \mathrm{~cm}\)
And semicircles are congruent.
So, the diameter of each semicircle \(=\frac{42}{3}=14 \mathrm{~cm}\)
The radius of each semicircle \(=\frac{14}{2}=7 \mathrm{~cm}\)
Width of rectangle \(=\) Radius of top semi-circle \(+\) Radius of bottom semi-circle
\(=7+7=14 \mathrm{~cm}\)
Area of the shaded region \(=\) Area of the rectangle \(-\) Area of \(6\) semicircles
\(=l \times w-6 \times \frac{1}{2} \pi r^{2}\)
\(=42 \times 14-6 \times \frac{1}{2} \times \frac{22}{7} \times 7^{2}\)
\(=588-462=126 \mathrm{~cm}^{2}\)

Q.5. The figure below shows a lawn consisting of a semicircle, a rectangle and an isosceles triangle. A pavement of \(3.5 \,\text {m}\) surrounds the semicircle. If the shaded region is covered with grass, find the area of the lawn covered with grass.

Ans: Given, the diameter of the semicircle covered with grass \(=14 \mathrm{~m}\)
So the radius \(=\frac{14}{2}=7 \mathrm{~m}\)
Also given, the length of pavement \(=3.5 \mathrm{~m}\)
Area of semicircle \(=\pi r^{2}=\pi(7)^{2}=\frac{22}{7} \times 7 \times 7=154 \mathrm{~m}^{2}\)
Therefore, the length of the rectangle \(=3.5+14+3.5=21 \mathrm{~m}\)
The breadth of rectangle \(=10 \mathrm{~m}\)
Area of rectangle \(= \text {length} \times \text {breadth} =21 \times 10=210 \mathrm{~m}^{2}\)
Height of isosceles triangle \(=40.5-10-7-3.5=20 \mathrm{~m}\)
The base of isosceles triangle \(=\) length of rectangle \(=21 \mathrm{~m}\)
Area of an isosceles triangle \(=\frac{1}{2} \times \text {base} \times \text {height}\)
\(=\frac{1}{2} \times 21 \times 20=210 \mathrm{~m}^{2}\)
Area of the shaded figure \(=\) Area of semicircle \(+\) Area of rectangle \(+\) Area of the isosceles triangle
\(=154+210+210=574 \mathrm{~m}^{2}\)
Hence, the area of the lawn covered with grass is \(574 \mathrm{~m}^{2}\).

Summary

In this article, we have discussed the combination of plane figures, the area of some plane figures, and how to find the area of the combination of plane figures and the solved examples.

FAQs – Areas of Combinations of Plane Figures

Q.1. What is an example of a plane figure?
Ans: A flat figure with closed lines in a single plane is known as a plane figure. The figure’s lines might be straight, curved, or a combination of both. Triangles, rectangles, squares, rhombuses, parallelograms, circles, ovals, hearts, pentagons, and hexagons are examples of planar figures.

Q.2. What is the perimeter and area of plane figures?
Ans: The length of a closed planar figure’s boundary is its perimeter. The area of a closed planar figure is the region enclosed by its boundary (surface).

Q.3. What is the definition of plane shape?
Ans: A plane shape is a closed, two-dimensional flat figure. The number of sides or corners of a plane shape differs from one to the next. A side is a straight line that forms part of a shape, while a corner is the intersection of two sides.

Q.4. Is rhombus a plane shape?
Ans: A rhombus is a quadrilateral with four equal sides and opposite sides parallel to each other.

Q.5. Are circles plane figures?
Ans: A circle is a planar figure bounded by a single curved line.