Area and Perimeter of Combinations of Plane Figures
Calculating Area of Regions formed by Combinations of 2D Shapes
In practical applications, we often encounter planar regions or figures that are not simple standard shapes like rectangles, triangles, or circles alone. Instead, they are composed of combinations of these basic geometric shapes, sometimes joined together, or sometimes with parts removed from a larger shape. To find the area of such combined figures or composite figures, we use the principle of breaking down the complex region into simpler, known geometric shapes whose area formulas are established and easy to apply.
Strategy for Finding the Area of Combined Shapes
The fundamental strategy for calculating the area of composite figures involves breaking the problem down into manageable steps:
-
Decomposition (or Addition/Subtraction Identification):
Analyze the given combined figure to determine how it can be viewed as a combination of basic shapes. This can involve:
- Decomposition: Dividing the complex figure into several smaller, non-overlapping basic geometric shapes (like rectangles, squares, triangles, circles, semicircles, sectors, trapeziums, etc.) that together make up the entire figure. Draw dividing lines if necessary to visualize these shapes.
- Addition/Subtraction: Identifying the figure as a larger basic shape from which one or more smaller basic shapes have been removed or cut out.
Often, a figure can be decomposed in multiple ways. Choose the method that seems simplest based on the dimensions provided.
-
Calculation of Individual Areas:
Once the figure is broken down into basic shapes, calculate the area of each individual basic shape using the appropriate area formula. Ensure all dimensions are in consistent units before performing calculations.
-
Combination of Areas:
Combine the areas calculated in the previous step to find the total area of the composite figure. The method of combination depends on how you decomposed or viewed the original figure:
- If the composite figure was formed by joining basic shapes, add the areas of all the constituent shapes.
- If the composite figure represents the area of a larger shape with a part removed, subtract the area of the removed part(s) from the area of the larger shape.
Basic Area Formulas for Reference
To perform the calculations in Step 2, you need to be familiar with the area formulas for common 2D shapes. Here is a consolidated table of frequently used formulas:
Shape | Figure | Formula for Area ($A$) | Variables Used |
---|---|---|---|
Rectangle |
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$\mathbf{A = l \times b}$ | $l$ = length, $b$ = breadth/width |
Square |
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$\mathbf{A = a^2}$ | $a$ = side length |
Triangle |
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$\mathbf{A = \frac{1}{2} \times b \times h}$ | $b$ = base, $h$ = height corresponding to base $b$ |
Triangle (using Heron's Formula) |
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$\mathbf{A = \sqrt{s(s-a)(s-b)(s-c)}}$ | $a, b, c$ = side lengths, $s = \frac{a+b+c}{2}$ (semi-perimeter) |
Equilateral Triangle |
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$\mathbf{A = \frac{\sqrt{3}}{4} a^2}$ | $a$ = side length |
Parallelogram |
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$\mathbf{A = b \times h}$ | $b$ = base, $h$ = height corresponding to base $b$ |
Rhombus / Kite |
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$\mathbf{A = \frac{1}{2} d_1 d_2}$ | $d_1, d_2$ = lengths of diagonals |
Trapezium |
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$\mathbf{A = \frac{1}{2} (a+b) h}$ | $a, b$ = lengths of parallel sides, $h$ = perpendicular height between parallel sides |
Circle |
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$\mathbf{A = \pi r^2}$ | $r$ = radius |
Semicircle |
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$\mathbf{A = \frac{1}{2} \pi r^2}$ | $r$ = radius |
Sector of a Circle |
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$\mathbf{A = \frac{\theta}{360^\circ} \times \pi r^2}$ (if $\theta$ in degrees) $\mathbf{A = \frac{1}{2} r^2 \theta}$ (if $\theta$ in radians) |
$r$ = radius, $\theta$ = sector angle at center |
Segment of a Circle |
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$\mathbf{A_{\text{segment}} = A_{\text{sector}} - A_{\triangle}}$ $\mathbf{A_{\text{segment}} = \left( \frac{\theta}{360^\circ} \pi r^2 \right) - \left( \frac{1}{2} r^2 \sin \theta \right)}$ (if $\theta$ in degrees) $\mathbf{A_{\text{segment}} = \frac{1}{2} r^2 (\theta - \sin\theta)}$ (if $\theta$ in radians) |
$r$ = radius, $\theta$ = angle subtended by the chord at the center |
Example Calculation
Let's illustrate the strategy with an example involving a combination of shapes.
Example 1. Find the area of the shaded region in the figure below, where ABCD is a square of side $14$ cm, and four identical circles are drawn such that each circle touches two other circles and two sides of the square.

Answer:
Given:
ABCD is a square with side length = $14$ cm.
Four identical circles are drawn inside the square, touching each other and the sides of the square as shown in the figure.
To Find:
Area of the shaded region (the region inside the square but outside the four circles).
Solution:
The shaded region is the area of the square minus the total area occupied by the four circles.
Area(Shaded Region) = Area(Square ABCD) - Area(Four Circles)
Step 1: Calculate the Area of the Square ABCD
The side length of the square is $a = 14$ cm.
Area(Square ABCD) $= a^2 = (14 \$ \text{cm})^2$
$= 14 \$ \text{cm} \times 14 \$ \text{cm}$
$= 196 \$ \text{cm}^2$
Step 2: Determine the Radius of Each Circle
Since the four identical circles are arranged within the square such that each touches two sides of the square and two other circles, their arrangement is symmetric. Consider the top side of the square (say, AB). Two circles touch this side. The distance along AB is covered by the diameters of these two circles. Similarly, along the side AD, the distance is covered by the diameters of two circles.
Let the radius of each identical circle be '$r$'. The diameter of each circle is $2r$.

The side length of the square is equal to the sum of the diameters of two circles arranged side-by-side.
Side length of square $= \text{Diameter} + \text{Diameter}$
"$14 \$ \text{cm} = 2r + 2r$"
[From the figure]
"$14 \$ \text{cm} = 4r$"
Solving for the radius $r$:
"$r = \frac{14}{4} \$ \text{cm} = \frac{7}{2} \$ \text{cm} = 3.5 \$ \text{cm}$"
Step 3: Calculate the Area of One Circle
Using the formula for the area of a circle, $A = \pi r^2$. We are given $\pi = \frac{22}{7}$. Using $r = \frac{7}{2}$ cm simplifies calculations with this value of $\pi$.
Area of one circle $= \pi r^2 = \frac{22}{7} \times \left(\frac{7}{2} \$ \text{cm}\right)^2$"
"$= \frac{22}{7} \times \frac{7}{2} \$ \text{cm} \times \frac{7}{2} \$ \text{cm}$"
"$= \frac{\cancel{22}^{11}}{\cancel{7}_1} \times \frac{\cancel{7}^1}{\cancel{2}_1} \times \frac{7}{2} \$ \text{cm}^2$"
[Cancelling 22 with 2, 7 with 7]
"$= 11 \times 1 \times \frac{7}{2} \$ \text{cm}^2 = \frac{77}{2} \$ \text{cm}^2$"
"$= 38.5 \$ \text{cm}^2$"
Step 4: Calculate the Total Area of the Four Circles
Since the four circles are identical, the total area occupied by them is 4 times the area of one circle.
Area of four circles $= 4 \times (\text{Area of one circle})$
"$= 4 \times 38.5 \$ \text{cm}^2$"
"$= 154.0 \$ \text{cm}^2$"
Step 5: Calculate the Area of the Shaded Region
Using the initial breakdown, the shaded area is the area of the square minus the combined area of the four circles.
Area(Shaded Region) = Area(Square ABCD) - Area(Four Circles)
"$= 196 \$ \text{cm}^2 - 154 \$ \text{cm}^2$"
Perform the subtraction:
$\begin{array}{cc} & 1 & 9 & 6 \\ - & 1 & 5 & 4 \\ \hline & & 4 & 2 \\ \hline \end{array}$"$\mathbf{= 42 \$\$ cm^2}$"
Therefore, the area of the shaded region is 42 square centimetres ($\text{cm}^2$).
Calculating Perimeter of Combined Plane Figures
The perimeter of a plane figure is defined as the total length of its boundary. When we deal with figures formed by combining two or more basic geometric shapes, the perimeter of the resulting figure is the length of the outer boundary of the combined shape. It is very important to understand that this is generally not the sum of the perimeters of the individual shapes that were combined, because internal boundaries created by joining the shapes are no longer part of the perimeter of the final figure.
Strategy for Finding the Perimeter of Combined Shapes
To calculate the perimeter of a combined or composite figure, follow these steps:
-
Identify the Outer Boundary:
Carefully examine the figure and visually trace the complete outer edge that encloses the entire shape. This outer boundary may consist of straight line segments, curved arcs, or a combination of both.
-
Break Down and Calculate Segment Lengths:
Identify each distinct segment (straight side or curved arc) that constitutes the outer boundary. Determine the length of each of these individual segments using the given dimensions and appropriate formulas. For straight sides, this is simply the length of the line segment. For curved parts, this will typically involve circumference or arc length formulas.
-
Sum the Lengths:
Add the lengths of all the identified outer boundary segments together. This sum is the total perimeter of the combined figure.
Crucial Distinction: Remember that any line segment or curve that is *internal* to the combined figure, even if it was originally a boundary of one of the component shapes, should not be included in the calculation of the perimeter of the combined figure. The perimeter is only the length of the boundary that separates the figure from the outside region.
Basic Formulas for Boundary Lengths
Calculating the lengths of the segments forming the boundary often requires using formulas for the perimeter or arc length of basic shapes:
Boundary Type | Figure/Concept | Formula for Length | Variables |
---|---|---|---|
Straight Line Segment | A line segment connecting two points. |
Length = Distance between endpoints. (Usually given or easily found from given dimensions). | |
Circumference of a Circle |
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$\mathbf{C = 2 \pi r}$ or $\mathbf{C = \pi d}$ | $r$ = radius, $d$ = diameter ($d=2r$) |
Arc Length of a Sector |
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If $\theta$ in degrees: $\mathbf{l = \frac{\theta}{360^\circ} \times 2 \pi r}$ If $\theta$ in radians: $\mathbf{l = r \theta}$ |
$r$ = radius, $\theta$ = sector angle at center |
Perimeter of a Semicircle (including diameter) |
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Perimeter = Arc Length + Diameter $\mathbf{P = \pi r + 2r = r(\pi + 2)}$ |
$r$ = radius |
Perimeter of a Quarter Circle (including two radii) |
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Perimeter = Arc Length + Radius + Radius $\mathbf{P = \frac{1}{4} (2\pi r) + 2r = \frac{\pi r}{2} + 2r = r(\frac{\pi}{2} + 2)}$ |
$r$ = radius |
Example Calculation
Example 1. Find the perimeter of the shaded region (a running track) shown below, which consists of a rectangle with two semicircular ends. The dimensions of the rectangular part are $100$ m by $42$ m.

Answer:
Given:
A running track shape formed by a rectangle combined with two semicircles attached to its shorter sides.
Dimensions of the rectangular part: Length ($L$) = $100$ m, Breadth ($B$) = $42$ m.
The semicircles are attached to the sides with length equal to the breadth of the rectangle.
To Find:
The perimeter of the entire running track figure (the outer boundary).
Solution:
The perimeter of the running track is the length of its outer boundary. Looking at the figure, the outer boundary consists of:
- Two straight line segments, which are the two longer sides of the rectangle (each of length $100$ m).
- Two semicircular arcs, which are the outer curves of the semicircular ends.
The shorter sides of the rectangle (each of length $42$ m) are internal boundaries in the combined figure and do not form part of the perimeter of the track.
Step 1: Identify the outer boundary segments and their lengths
The outer boundary segments are:
Length of one straight side $= 100 \$ \text{m}$
Length of the other straight side $= 100 \$ \text{m}$
The two semicircular ends have a diameter equal to the breadth of the rectangle, which is $42$ m.
Diameter of each semicircle, $d = 42 \$ \text{m}$
The radius of each semicircle is half of the diameter:
"$r = \frac{d}{2} = \frac{42}{2} \$ \text{m} = 21 \$ \text{m}$"
The length of the curved boundary of one semicircle is the arc length of a semicircle.
Arc length of one semicircle $= \frac{1}{2} \times \text{Circumference of full circle}$
"$= \frac{1}{2} \times 2 \pi r = \pi r$"
Using the value $\pi = \frac{22}{7}$ (a common approximation used for such problems):
"Arc length of one semicircle $= \frac{22}{7} \times 21 \$ \text{m}$"
[Substituting $\pi=22/7$, $r=21$]
Simplify by cancelling the common factor 7:
"$= \frac{22}{\cancel{7}_1} \times \cancel{21}^3 \$ \text{m} = 22 \times 3 \$ \text{m}$"
"$= 66 \$ \text{m}$"
There are two such semicircular ends, so the total length of the curved parts is:
Total length of curved parts $= 2 \times (\text{Arc length of one semicircle})$
"$= 2 \times 66 \$ \text{m} = 132 \$ \text{m}$"
Alternatively: The two semicircular arcs together form the full circumference of a circle with radius $r=21$ m. So, the total curved length is $2 \pi r = 2 \times \frac{22}{7} \times 21 = 2 \times 22 \times 3 = 132$ m.
Step 2: Sum the lengths of all outer boundary segments
The total perimeter of the running track figure is the sum of the lengths of the two straight parts and the total length of the two curved parts.
Total Perimeter $= (\text{Length of straight part 1}) + (\text{Length of straight part 2}) + (\text{Total length of curved parts})$
"$= 100 \$ \text{m} + 100 \$ \text{m} + 132 \$ \text{m}$"
[Substituting lengths]
"$= (100 + 100 + 132) \$ \text{m}$"
"$\mathbf{= 332 \$\$ m}$"
Therefore, the perimeter of the running track figure is 332 metres (m).
Notice that the internal line of length 42 m (the diameter of the semicircle/breadth of the rectangle) was NOT included in the perimeter calculation.
Solving Problems involving Areas and Perimeters of Combined Figures
This section focuses on applying the concepts of calculating areas and perimeters of individual basic plane figures (like squares, rectangles, triangles, circles, sectors, segments) to solve problems involving figures formed by combining these basic shapes. These combined figures, also called composite figures, are frequently encountered in real-world scenarios and geometry problems. Solving such problems requires a systematic approach involving visualization, decomposition, calculation, and careful combination of results.
Types of Problems
Problems involving combined figures can take various forms:
- Area of Shaded or Unshaded Regions: Finding the area of a specific part of a figure, which might be the area left after removing certain shapes or the area formed by the overlap of shapes.
- Cost Calculation: Determining the cost associated with covering an area (like flooring, painting, or plastering) or covering a length (like fencing, bordering, or running a wire) based on a given rate per unit area or unit length.
- Design and Layout Problems: Analyzing the geometric properties, areas, or perimeters of shapes used in architectural designs, engineering plans, decorative patterns, or land layouts.
- Practical Applications: Solving problems related to capacity (volume derived from area and height), material usage, or distance covered in contexts involving combined shapes.
Problem-Solving Steps
A structured approach is key to solving problems involving combined figures effectively:
-
Understand and Visualize:
Read the problem statement thoroughly. If a figure is provided, study it carefully. If not, try to draw a clear diagram based on the description. Clearly identify the specific region or boundary for which you need to calculate the area or perimeter.
-
Decompose or Identify Components:
Mentally (or by drawing auxiliary lines on the diagram) break down the combined figure into simpler, non-overlapping basic geometric shapes (e.g., squares, rectangles, triangles, circles, semicircles, sectors, segments, trapeziums). Alternatively, recognize the figure as a larger shape with parts removed.
-
Determine Necessary Dimensions:
Identify all the lengths, radii, heights, or angles required to calculate the areas or perimeters of the basic shapes identified in the previous step. These dimensions should be available directly from the problem statement or derivable using other given information and geometric properties (like Pythagoras theorem, properties of specific quadrilaterals, etc.). Ensure all dimensions are converted to consistent units if necessary.
-
Select Calculation Strategy:
- For Area: Decide whether you need to add the areas of the component shapes (if they form the region by joining) or subtract the areas of removed shapes from the area of a larger containing shape.
- For Perimeter: Trace the *outer* boundary of the entire combined figure. Identify each straight or curved segment that constitutes this outer boundary. Ignore any lines or curves that are internal to the final figure.
-
Calculate Individual Areas or Lengths:
Apply the appropriate area formula for each basic shape, or the length formula for each boundary segment identified. Perform these calculations systematically.
-
Combine Results:
Perform the final calculation by adding or subtracting the individual areas (for area problems) or summing the lengths of the outer boundary segments (for perimeter problems) according to the strategy chosen in Step 4.
-
State Final Answer with Units:
Write down the final answer clearly, ensuring that the correct units are attached (e.g., $\text{cm}^2$, $\text{m}^2$ for area; cm, m for perimeter). If the problem asks for cost, multiply the calculated area or perimeter by the given rate, ensuring unit consistency.
Example Problem
Example 1. In the figure, ABCD is a square of side $10$ cm. Semicircles are drawn with each side of the square as diameter. Find the area of the shaded region. (Use $\pi = 3.14$).

Answer:
Given:
ABCD is a square with side length $a = 10$ cm.
Four semicircles are drawn, taking each side of the square as a diameter.
Value of $\pi = 3.14$.
To Find:
Area of the shaded region (the regions within the square that are not covered by the overlap of the semicircles in the center).
Solution:
Let $A_{sh}$ be the area of the shaded region. Let the area of the central unshaded region (shaped like a flower with four petals) be $A_{un}$. The total area of the square is $A_{sq}$.
"$A_{sq} = A_{sh} + A_{un}$"
We can calculate the area of the square:
"$A_{sq} = (\text{side})^2 = (10 \$ \text{cm})^2 = 100 \$ \text{cm}^2$"
The shaded region is complicated to decompose directly into simple shapes. A better approach is to relate the areas of the semicircles to the areas of the shaded and unshaded regions.
Each semicircle has the side of the square as its diameter. So, the diameter of each semicircle is $10$ cm, and the radius is $r = \frac{10}{2} = 5$ cm.
The area of one semicircle is $\frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.14 \times (5 \$ \text{cm})^2 = \frac{1}{2} \times 3.14 \times 25 \$ \text{cm}^2$.
"$= \frac{1}{2} \times 78.5 \$ \text{cm}^2 = 39.25 \$ \text{cm}^2$"
There are four such semicircles. Let's consider adding the areas of the four semicircles.
Sum of Areas of 4 Semicircles $= 4 \times (\text{Area of one semicircle})$
"$= 4 \times 39.25 \$ \text{cm}^2 = 157.00 \$ \text{cm}^2$"
Now, let's analyze what regions are covered when we add the areas of the four semicircles.
The total area of the square consists of the shaded regions ($A_{sh}$) and the unshaded central region ($A_{un}$).
When we add the areas of the four semicircles, each point within the shaded region is covered by exactly one semicircle. Each point within the unshaded central region ($A_{un}$) is covered by exactly two semicircles (e.g., a point in the "petal" region I is covered by the semicircles on AB and AD).
So, the sum of the areas of the four semicircles is equal to the area of the shaded region counted once plus the area of the unshaded region counted twice.
Sum of Areas of 4 Semicircles $= A_{sh} + 2 \times A_{un}$
"$157 = A_{sh} + 2 A_{un}$"
...(i)
We also know the area of the square is the sum of the shaded and unshaded areas:
"$100 = A_{sh} + A_{un}$"
...(ii)
We have a system of two linear equations with two variables ($A_{sh}$ and $A_{un}$). We can solve for $A_{sh}$.
From (ii), $A_{un} = 100 - A_{sh}$. Substitute this into (i):
"$157 = A_{sh} + 2 (100 - A_{sh})$"
[Substitute $A_{un}$ from (ii) into (i)]
"$157 = A_{sh} + 200 - 2 A_{sh}$"
"$157 = 200 + (A_{sh} - 2 A_{sh})$"
"$157 = 200 - A_{sh}$"
Rearrange to solve for $A_{sh}$:
"$A_{sh} = 200 - 157$"
Perform the subtraction:
$\begin{array}{cc} & 2 & 0 & 0 \\ - & 1 & 5 & 7 \\ \hline & & 4 & 3 \\ \hline \end{array}$"$\mathbf{A_{sh} = 43 \$\$ cm^2}$"
Therefore, the area of the shaded region is 43 square centimetres ($\text{cm}^2$).
Alternate Approach (Calculating Unshaded Area First):
Let's find $A_{un}$ first from the system of equations:
"$A_{sh} + 2 A_{un} = 157$"
...(i)
"$A_{sh} + A_{un} = 100$"
...(ii)
Subtract equation (ii) from equation (i):
"$(A_{sh} + 2 A_{un}) - (A_{sh} + A_{un}) = 157 - 100$"
"$A_{sh} + 2 A_{un} - A_{sh} - A_{un} = 57$"
"$A_{un} = 57 \$ \text{cm}^2$"
Now substitute the value of $A_{un}$ into equation (ii) to find $A_{sh}$:
"$A_{sh} + A_{un} = 100$"
"$A_{sh} + 57 = 100$"
"$A_{sh} = 100 - 57$"
"$\mathbf{A_{sh} = 43 \$\$ cm^2}$"
Both approaches confirm that the area of the shaded region is 43 square centimetres ($\text{cm}^2$).