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| Perimeter and Area of a Circle | Perimeter and Area of Sector and Segment of a Circle | Perimeter and Area of Combinations of Plane Figures |
Chapter 12 Areas Related to Circles (Concepts)
Welcome to this fascinating chapter, Areas Related to Circles, where we extend our knowledge of mensuration beyond simple polygons and complete circles to explore the intricate calculations involving parts of circles and combinations of figures. Having mastered the basic formulas for the circumference and area of a full circle in previous studies, we now delve into the geometry of its constituent parts – sectors and segments – and learn how to precisely measure their boundaries and enclosed regions. This chapter bridges theoretical geometry with practical calculation, equipping you with the skills to analyze and quantify curved shapes encountered in design, architecture, engineering, and everyday life.
We begin by briefly revisiting the fundamental formulas that govern the entire circle. Recall that for a circle with radius $r$, its perimeter, known as the Circumference ($C$), is given by $C = 2\pi r$, and the measure of the space it encloses, its Area ($A$), is calculated as $A = \pi r^2$. These formulas, involving the ubiquitous mathematical constant $\pi$ (pi, approximately $3.14159...$), are the foundation upon which we build our understanding of circular parts. Our focus now shifts to dissecting the circle into meaningful components:
- Sector: Imagine slicing a pizza; each slice represents a sector. Formally, a sector is the region of the circle enclosed by two radii and the corresponding arc lying between the endpoints of those radii.
- Segment: Now imagine cutting the pizza slice straight across its crust; the piece containing the crust is a segment. Formally, a segment is the region enclosed by a chord and the corresponding arc.
For both sectors and segments, we distinguish between minor (smaller) and major (larger) parts, depending on the associated arc.
A key aspect of this chapter is developing formulas to calculate the arc length and area associated with a sector, defined by the angle $\theta$ (theta, measured in degrees) formed by the two radii at the center. We establish these proportional relationships:
- Length of an arc of a sector with angle $\theta$: This represents a fraction of the total circumference. The formula is: $$ \text{Arc Length} = \left( \frac{\theta}{360^\circ} \right) \times 2\pi r $$
- Area of a sector with angle $\theta$: This represents the same fraction of the total area of the circle. The formula is: $$ \text{Area of Sector} = \left( \frac{\theta}{360^\circ} \right) \times \pi r^2 $$
Calculating the Area of a Segment requires a slightly different approach. Since a segment is formed by a chord and an arc, its area is found by taking the area of the corresponding sector and subtracting the area of the triangle formed by the chord and the two radii connecting the chord's endpoints to the center. The calculation of this triangle's area might involve basic geometry (if it's equilateral or isosceles right-angled) or require trigonometric methods (like Area = $\frac{1}{2} r^2 \sin\theta$) depending on the central angle $\theta$.
The culmination of this chapter lies in applying these formulas to calculate the areas of complex shapes formed by combinations of plane figures. This involves analyzing diagrams featuring circles combined with squares, rectangles, triangles, or other circles, often requiring us to find the area of specifically shaded regions. Success in these problems hinges on two key skills:
- Visual Decomposition: The ability to break down the complex figure into simpler, recognizable geometric shapes whose areas we know how to calculate (circles, sectors, segments, triangles, rectangles, etc.).
- Strategic Calculation: Determining whether the desired area can be found by adding the areas of the simpler components or by subtracting the area of one component from another (e.g., finding the area of a path around a circular garden involves subtracting the area of the inner circle from the area of the outer circle).
Perimeter and Area of a Circle
In earlier geometry lessons, you were introduced to basic two-dimensional shapes, including the circle. You learned that a circle is defined by its centre and radius, and you were likely introduced to the concepts of its perimeter (circumference) and area. In this chapter, we revisit these concepts to solve problems involving areas related to circles, including parts of a circle like sectors and segments.
Basic Definitions
A circle is the set of all points in a plane that are at a fixed distance from a fixed point in the plane. The fixed point is called the centre of the circle, and the fixed distance is called the radius (denoted by $r$).
The diameter of a circle (denoted by $d$) is a line segment passing through the centre and connecting two points on the circle. The diameter is always twice the length of the radius: $d = 2r$.
Circumference (Perimeter) of a Circle
The circumference of a circle is the length of its boundary. It is the distance around the circle. The ratio of the circumference of any circle to its diameter is a constant value, which is represented by the Greek letter $\pi$ (pi).
$\frac{\text{Circumference}}{\text{Diameter}} = \pi$
Rearranging this definition gives the formula for the circumference in terms of the diameter:
$\text{Circumference} = \pi \times \text{Diameter} = \pi d$
Using the relationship $d = 2r$, the formula for the circumference in terms of the radius is:
$\text{Circumference} = \pi \times (2r) = 2\pi r$
The value of $\pi$ is an irrational number, meaning its decimal representation is non-terminating and non-repeating. Common approximations used for $\pi$ are $\frac{22}{7}$ and 3.14. Unless a specific value for $\pi$ is given in a problem, you can use $\frac{22}{7}$ or 3.14 as convenient.
The unit of circumference is a unit of length (e.g., centimetre (cm), metre (m), kilometre (km)).
Area of a Circle
The area of a circle is the measure of the two-dimensional region enclosed by the circle's boundary. It represents the amount of surface covered by the circle.
The formula for the area (A) of a circle with radius $r$ is:
$\text{Area} = \pi r^2$
Using the relationship $r = \frac{d}{2}$, the formula for the area in terms of the diameter is:
$\text{Area} = \pi \left(\frac{d}{2}\right)^2 = \pi \frac{d^2}{4}$
The unit of area is the square of the unit of length (e.g., square centimetre ($\text{cm}^2$), square metre ($\text{m}^2$), square kilometre ($\text{km}^2$)).
Derivation of the Area Formula
The formula for the area of a circle, $A = \pi r^2$, can be understood intuitively by imagining the circle being divided into a large number of equal sectors. If we arrange these sectors alternately, they form a shape that closely resembles a parallelogram or a rectangle.
Let's consider the properties of this resulting rectangle:
- The height of this rectangle is equal to the radius of the circle, $r$.
- The base of the rectangle is formed by the arcs of the sectors. Half of the sectors' arcs are on the top and half are on the bottom. Therefore, the length of the base is half the total circumference of the circle.
We can calculate the length of the base as follows:
$\text{Base} = \frac{1}{2} \times \text{Circumference}$
$\text{Base} = \frac{1}{2} \times (2\pi r) = \pi r$
Now, we find the area of the rectangle, which is equivalent to the area of the original circle:
$\text{Area of Circle} = \text{Area of Rectangle} = \text{Base} \times \text{Height}$
$\text{Area} = (\pi r) \times (r)$
$A = \pi r^2$
As we increase the number of sectors, the rearranged shape becomes closer and closer to a perfect rectangle, validating this derivation.
Area Between Two Concentric Circles (Annulus)
Two or more circles that share the same centre point are called concentric circles. The region between two concentric circles is known as an annulus or a circular ring.
To find the area of this ring, we calculate the area of the larger (outer) circle and subtract the area of the smaller (inner) circle.
Let:
- $R$ be the radius of the outer circle.
- $r$ be the radius of the inner circle.
The area of the outer circle is $\pi R^2$, and the area of the inner circle is $\pi r^2$.
The formula for the area of the annulus is:
$\text{Area of Annulus} = (\text{Area of Outer Circle}) - (\text{Area of Inner Circle})$
$A = \pi R^2 - \pi r^2$
We can simplify this formula by factoring out $\pi$:
$A = \pi (R^2 - r^2)$
Example 1. The radius of a circular field is 14 metres. Find its circumference and area. (Use $\pi = \frac{22}{7}$)
Answer:
To Find:
Circumference and Area of the circular field.
Given:
Radius of the circle $r = 14$ metres.
Use $\pi = \frac{22}{7}$.
Solution:
Using the formula for the circumference of a circle, $C = 2\pi r$:
$C = 2 \times \frac{22}{7} \times 14 \text{ m}$
Perform the multiplication. We can cancel out the common factor 7 from the denominator and 14 in the numerator:
$C = 2 \times 22 \times \frac{\cancel{14}^{2}}{\cancel{7}_{1}} \text{ m}$
$C = 2 \times 22 \times 2 \text{ m}$
$C = 88 \text{ m}$
... (1)
The circumference of the circular field is 88 metres.
Using the formula for the area of a circle, $A = \pi r^2$:
$A = \frac{22}{7} \times (14 \text{ m})^2$
Expand $(14 \text{ m})^2$ and rewrite the expression:
$A = \frac{22}{7} \times 14 \times 14 \text{ m}^2$
Cancel out the common factor 7:
$A = 22 \times \frac{\cancel{14}^{2}}{\cancel{7}_{1}} \times 14 \text{ m}^2$
$A = 22 \times 2 \times 14 \text{ m}^2$
Perform the multiplication:
$A = 44 \times 14 \text{ m}^2$
$A = 616 \text{ m}^2$
... (2)
Answer: The circumference of the circular field is 88 metres and its area is 616 square metres.
Example 2. If the circumference of a circle is 132 cm, find its radius and area. (Use $\pi = \frac{22}{7}$)
Answer:
To Find:
The radius and area of the circle.
Given:
Circumference of the circle $C = 132$ cm.
Use $\pi = \frac{22}{7}$.
Solution:
To find the area, we first need to determine the radius ($r$). We are given the circumference ($C$). The formula for circumference is $C = 2\pi r$. We can rearrange this formula to solve for the radius.
$C = 2\pi r$
Substitute the given circumference ($C = 132$) and the value of $\pi$ ($\frac{22}{7}$):
$132 = 2 \times \frac{22}{7} \times r$
Simplify the constant term on the right side:
$132 = \frac{44}{7} \times r$
Solve for $r$ by multiplying both sides of the equation by the reciprocal of $\frac{44}{7}$, which is $\frac{7}{44}$:
$r = 132 \times \frac{7}{44}$
Cancel out the common factor 44 (since $132 \div 44 = 3$):
$r = \cancel{132}^{3} \times \frac{7}{\cancel{44}_{1}}$
$r = 3 \times 7 = 21$ cm
... (1)
The radius of the circle is 21 cm.
Now, calculate the area of the circle using the formula $A = \pi r^2$ and the radius we just found:
Area $= \frac{22}{7} \times (21 \text{ cm})^2$
Expand $(21 \text{ cm})^2$ and rewrite the expression:
Area $= \frac{22}{7} \times 21 \times 21 \text{ cm}^2$
Cancel out the common factor 7:
Area $= 22 \times \frac{\cancel{21}^{3}}{\cancel{7}_{1}} \times 21 \text{ cm}^2$
Area $= 22 \times 3 \times 21 \text{ cm}^2$
Perform the multiplication:
Area $= 66 \times 21 \text{ cm}^2$
Area $= 1386 \text{ cm}^2$
... (2)
Answer: The radius of the circle is 21 cm and its area is 1386 cm$^2$.
Example 3. A circular park is surrounded by a path 7 m wide. The radius of the park is 28 m. Find the area of the path. (Use $\pi = \frac{22}{7}$)
Answer:
To Find:
The area of the path.
Given:
Radius of the inner circle (park), $r = 28$ m.
Width of the path = 7 m.
Solution:
The path surrounds the park, forming an annulus. The inner radius is the radius of the park itself.
$r = 28 \text{ m}$
The outer radius ($R$) is the radius of the park plus the width of the path.
$R = \text{Radius of park} + \text{Width of path}$
$R = 28 \text{ m} + 7 \text{ m} = 35 \text{ m}$
Now, we use the formula for the area of the annulus: $A = \pi (R^2 - r^2)$.
$A = \frac{22}{7} ( (35)^2 - (28)^2 )$
We can use the algebraic identity $a^2 - b^2 = (a-b)(a+b)$ to simplify the calculation.
$A = \frac{22}{7} ( (35 - 28)(35 + 28) )$
$A = \frac{22}{7} ( 7 \times 63 )$
Cancel the common factor 7:
$A = 22 \times \frac{\cancel{7}}{\cancel{7}} \times 63$
$A = 22 \times 63 \text{ m}^2$
$A = 1386 \text{ m}^2$
Answer: The area of the path is 1386 m$^2$.
Perimeter and Area of Sector and Segment of a Circle
A circle can be divided into various parts. The region enclosed by two radii and the arc between them is called a sector, much like a slice of pizza. The region enclosed by a chord and its corresponding arc is called a segment, similar to the crust portion of a pizza slice. Understanding how to calculate the perimeter and area of these parts is essential for solving problems involving circular regions.
Sector of a Circle
A sector of a circle is the region bounded by two radii and the arc connecting their endpoints on the circle. The angle formed by the two radii at the centre of the circle is called the angle of the sector or central angle, denoted by $\theta$.
Let $r$ be the radius of the circle and $\theta$ be the angle of the sector measured in degrees. A sector is essentially a fraction of the whole circle, and this fraction is determined by its central angle $\theta$ out of the total $360^\circ$.
- The sector with an angle less than $180^\circ$ is called the minor sector.
- The sector with an angle greater than $180^\circ$ (the rest of the circle) is called the major sector. The angle of the major sector is $360^\circ - \theta$.
1. Length of the Arc of a Sector
The arc of a sector is a piece of the circle's circumference. Its length is the same fraction of the total circumference as the sector's angle is of $360^\circ$.
$\text{Length of arc } (l) = \left(\frac{\theta}{360^\circ}\right) \times (\text{Circumference of circle})$
Since the circumference is $2\pi r$, the formula becomes:
$l = \frac{\theta}{360^\circ} \times 2\pi r$
... (1)
2. Area of a Sector
Similarly, the area of a sector is a fraction of the total area of the circle.
$\text{Area of sector} = \left(\frac{\theta}{360^\circ}\right) \times (\text{Area of circle})$
Since the area of the circle is $\pi r^2$, the formula is:
$\text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2$
... (2)
3. Perimeter of a Sector
The perimeter is the total length of the boundary of the sector. This includes two radii and the arc length.
$\text{Perimeter of sector} = r + r + l$
$\text{Perimeter of sector} = 2r + \frac{\theta}{360^\circ} \times 2\pi r$
... (3)
Segment of a Circle
A segment of a circle is the region enclosed by a chord and the arc corresponding to that chord. A chord divides a circle into two segments:
- Minor Segment: The region enclosed by the chord and the minor arc.
- Major Segment: The region enclosed by the chord and the major arc.
1. Area of a Segment
To find the area of a segment, we start with the area of the corresponding sector and subtract the area of the triangle formed by the two radii and the chord.
$\text{Area of segment} = \text{Area of sector} - \text{Area of triangle}$
Calculating the Area of the Triangle
The triangle formed (let's call it $\triangle \text{OAB}$) has two sides equal to the radius, $r$, and the angle between them is the central angle $\theta$.
- General Formula: The most general formula for the area of this triangle is derived from trigonometry:
$\text{Area}(\triangle \text{OAB}) = \frac{1}{2} r^2 \sin \theta$
... (4)
- Special Case ($\theta = 90^\circ$): The triangle is a right-angled triangle with base and height both equal to $r$.
$\text{Area}(\triangle \text{OAB}) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} r^2$
- Special Case ($\theta = 60^\circ$): The triangle is an equilateral triangle with all sides equal to $r$.
$\text{Area}(\triangle \text{OAB}) = \frac{\sqrt{3}}{4} (\text{side})^2 = \frac{\sqrt{3}}{4} r^2$
Final Formulas for Area of Segment
Area of Minor Segment:
$\text{Area} = \left(\frac{\theta}{360^\circ} \pi r^2\right) - \text{Area}(\triangle \text{OAB})$
... (5)
Area of Major Segment:
$\text{Area} = \text{Area of circle} - \text{Area of minor segment}$
... (6)
2. Perimeter of a Segment
The perimeter of a segment consists of the arc length and the length of the chord.
$\text{Perimeter} = \text{Length of arc} + \text{Length of chord}$
... (7)
The length of the arc is calculated using Formula (1). The length of the chord AB can be found using the formula $2r \sin(\theta/2)$.
Example 1. A sector of a circle with radius 7 cm has an angle of $90^\circ$. Find the length of the arc and the area of the sector. (Use $\pi = \frac{22}{7}$)
Answer:
To Find:
Length of the arc of the sector and the area of the sector.
Given:
Radius of the circle $r = 7$ cm.
Angle of the sector $\theta = 90^\circ$.
Solution:
Length of the arc:
Using Formula (1):
$\text{Length of arc} = \frac{90^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 7 \text{ cm}$
$= \frac{1}{4} \times 2 \times 22 \text{ cm} = \frac{44}{4} \text{ cm} = 11 \text{ cm}$
Area of the sector:
Using Formula (2):
$\text{Area of sector} = \frac{90^\circ}{360^\circ} \times \frac{22}{7} \times (7 \text{ cm})^2$
$= \frac{1}{4} \times \frac{22}{7} \times 49 \text{ cm}^2 = \frac{1}{4} \times 22 \times 7 \text{ cm}^2$
$= \frac{154}{4} \text{ cm}^2 = 38.5 \text{ cm}^2$
Answer: The length of the arc is 11 cm and the area of the sector is 38.5 cm$^2$.
Example 2. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding minor segment. (Use $\pi = 3.14$)
Answer:
To Find:
Area of the corresponding minor segment.
Given:
Radius $r = 10$ cm, Central angle $\theta = 90^\circ$.
Solution:
Area of minor segment = Area of minor sector $-$ Area of triangle.
Step 1: Calculate the area of the minor sector.
$\text{Area of minor sector} = \frac{90^\circ}{360^\circ} \times 3.14 \times (10)^2$
$= \frac{1}{4} \times 3.14 \times 100 = \frac{314}{4} = 78.5 \text{ cm}^2$
Step 2: Calculate the area of the triangle.
Since the angle is $90^\circ$, the triangle is a right-angled triangle with base and height equal to the radius (10 cm).
$\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 10 = 50 \text{ cm}^2$
Step 3: Calculate the area of the minor segment.
$\text{Area of minor segment} = 78.5 \text{ cm}^2 - 50 \text{ cm}^2 = 28.5 \text{ cm}^2$
Answer: The area of the corresponding minor segment is 28.5 cm$^2$.
Example 3. Find the perimeter of a sector of a circle with a radius of 14 cm and a central angle of $60^\circ$. (Use $\pi = \frac{22}{7}$)
Answer:
To Find:
Perimeter of the sector.
Given:
Radius $r = 14$ cm, Central angle $\theta = 60^\circ$.
Solution:
Perimeter of a sector is given by the formula $2r + l$, where $l$ is the arc length.
Step 1: Calculate the length of the arc ($l$).
$l = \frac{\theta}{360^\circ} \times 2\pi r = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 14$
$l = \frac{1}{6} \times 2 \times 22 \times \frac{\cancel{14}^2}{\cancel{7}_1} = \frac{1}{6} \times 88 = \frac{44}{3}$ cm
So, the arc length is approximately $14.67$ cm.
Step 2: Calculate the perimeter.
$\text{Perimeter} = 2r + l = 2(14) + \frac{44}{3}$
$\text{Perimeter} = 28 + \frac{44}{3} = \frac{28 \times 3 + 44}{3} = \frac{84 + 44}{3} = \frac{128}{3}$ cm
In decimal form, the perimeter is approximately $42.67$ cm.
Answer: The perimeter of the sector is $\frac{128}{3}$ cm or approximately 42.67 cm.
Example 4. In a circle of radius 21 cm, a chord subtends an angle of $60^\circ$ at the centre. Find the area of the minor segment and the major segment. (Use $\pi = \frac{22}{7}$ and $\sqrt{3} = 1.73$)
Answer:
To Find:
Area of the minor segment and major segment.
Given:
Radius $r = 21$ cm, Central angle $\theta = 60^\circ$.
Solution:
Step 1: Calculate the area of the minor sector.
$\text{Area of minor sector} = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times (21)^2$
$= \frac{1}{6} \times \frac{22}{7} \times 21 \times 21 = \frac{1}{6} \times 22 \times 3 \times 21 = 11 \times 21 = 231 \text{ cm}^2$
Step 2: Calculate the area of the triangle.
Since the angle is $60^\circ$ and the two sides are radii, the triangle is an equilateral triangle with side length 21 cm.
$\text{Area of triangle} = \frac{\sqrt{3}}{4} (\text{side})^2 = \frac{\sqrt{3}}{4} (21)^2$
$= \frac{1.73}{4} \times 441 = \frac{762.93}{4} \approx 190.73 \text{ cm}^2$
Step 3: Calculate the area of the minor segment.
$\text{Area of minor segment} = \text{Area of sector} - \text{Area of triangle}$
$= 231 - 190.73 = 40.27 \text{ cm}^2$
Step 4: Calculate the area of the major segment.
First, find the area of the entire circle.
$\text{Area of circle} = \pi r^2 = \frac{22}{7} \times (21)^2 = \frac{22}{7} \times 441 = 22 \times 63 = 1386 \text{ cm}^2$
Now, subtract the area of the minor segment.
$\text{Area of major segment} = \text{Area of circle} - \text{Area of minor segment}$
$= 1386 - 40.27 = 1345.73 \text{ cm}^2$
Answer: The area of the minor segment is approximately 40.27 cm$^2$, and the area of the major segment is approximately 1345.73 cm$^2$.
Perimeter and Area of Combinations of Plane Figures
In real-world applications, we often encounter shapes that are not simple circles, triangles, or squares, but are combinations of these basic plane figures. Examples include athletic tracks, window designs, or decorative patterns. To find the areas and perimeters of such composite shapes, we break them down into simpler, familiar figures.
The general strategy involves three main steps:
- Decomposition: Identify the basic shapes (like triangles, squares, circles, sectors) that make up the complex figure. This might involve seeing the figure as a sum of shapes or as a larger shape with smaller pieces removed.
- Calculation: Calculate the area and/or perimeter of each of the basic shapes using their standard formulas.
- Composition: Add or subtract the calculated values to find the total area or perimeter. For perimeter, it's crucial to only sum the lengths that form the outer boundary of the final shape.
Example 1. Find the area of the shaded region in the figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)
Answer:
To Find:
The area of the shaded region.
Given:
Equilateral triangle OAB with side length = 12 cm.
A circle with centre O and radius = 6 cm.
Solution:
Decomposition: The shaded region is a combination of two shapes: the equilateral triangle OAB and a major circular sector. The total area can be found by adding the area of the triangle and the area of the major sector of the circle.
The area of the minor sector (which overlaps with the triangle) should not be counted twice. A simple way to calculate the total area is:
Total Area = (Area of Equilateral Triangle) + (Area of Major Sector)
Step 1: Calculate the area of the equilateral triangle OAB.
The side length is $a = 12$ cm. The formula for the area of an equilateral triangle is $\frac{\sqrt{3}}{4} a^2$.
$\text{Area of } \triangle \text{OAB} = \frac{\sqrt{3}}{4} \times (12)^2 = \frac{\sqrt{3}}{4} \times 144 = 36\sqrt{3} \text{ cm}^2$
Using $\sqrt{3} = 1.73$:
$\text{Area of } \triangle \text{OAB} = 36 \times 1.73 = 62.28 \text{ cm}^2$
... (1)
Step 2: Calculate the area of the major sector of the circle.
Since $\triangle \text{OAB}$ is equilateral, the angle at vertex O is $\angle AOB = 60^\circ$. This angle defines the minor sector that lies inside the triangle.
The angle of the major sector is the rest of the circle: $\theta = 360^\circ - 60^\circ = 300^\circ$.
The radius is $r = 6$ cm. The area of the major sector is:
$\text{Area of major sector} = \frac{300^\circ}{360^\circ} \times \pi r^2 = \frac{5}{6} \times 3.14 \times (6)^2$
$= \frac{5}{6} \times 3.14 \times 36 = 5 \times 3.14 \times 6$
$= 30 \times 3.14 = 94.2 \text{ cm}^2$
... (2)
Step 3: Calculate the total area of the shaded region.
The total shaded area is the sum of the areas from Step 1 and Step 2.
$\text{Total Shaded Area} = \text{Area of } \triangle \text{OAB} + \text{Area of major sector}$
$= 62.28 \text{ cm}^2 + 94.2 \text{ cm}^2 = 156.48 \text{ cm}^2$
Answer: The area of the shaded region is 156.48 cm$^2$.
Example 2. A square handkerchief has a design made of nine circular designs each of radius 7 cm. Find the area of the remaining portion of the handkerchief.
Answer:
To Find:
The area of the remaining portion of the handkerchief.
Given:
Nine circular designs, each with radius $r = 7$ cm.
The circles are arranged in a 3x3 grid on a square handkerchief.
Solution:
Decomposition: The area of the remaining portion is the area of the entire square minus the total area of the nine circular designs.
Area of remaining portion = (Area of Square) - (Area of 9 Circles)
Step 1: Determine the side length of the square.
The nine circles are arranged in three rows and three columns. The diameter of one circle is $d = 2r = 2 \times 7 = 14$ cm.
The side of the square is equal to the length of three diameters placed side-by-side.
$\text{Side of square } (a) = 3 \times \text{diameter} = 3 \times 14 = 42$ cm
Step 2: Calculate the area of the square.
$\text{Area of square} = a^2 = (42)^2 = 1764 \text{ cm}^2$
... (1)
Step 3: Calculate the total area of the nine circles.
First, find the area of one circle using $\pi = \frac{22}{7}$.
$\text{Area of one circle} = \pi r^2 = \frac{22}{7} \times (7)^2 = \frac{22}{7} \times 49 = 22 \times 7 = 154 \text{ cm}^2$
Now, multiply by 9 to get the total area of the nine circles.
$\text{Area of 9 circles} = 9 \times 154 = 1386 \text{ cm}^2$
... (2)
Step 4: Calculate the area of the remaining portion.
Subtract the area of the circles from the area of the square.
$\text{Remaining Area} = \text{Area of Square} - \text{Area of 9 Circles}$
$= 1764 - 1386 = 378 \text{ cm}^2$
Answer: The area of the remaining portion of the handkerchief is 378 cm$^2$.
Example 3. The figure depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find the perimeter of the track along its inner edge.
Answer:
To Find:
The perimeter of the track along its inner edge.
Given:
Length of the inner straight segments = 106 m.
Distance between inner parallel segments = 60 m. This is the diameter of the inner semicircles.
Width of the track = 10 m (not needed for the inner perimeter).
Solution:
Decomposition: The inner perimeter (or the distance around the track along its inner edge) consists of two straight parts and two semicircular parts.
Inner Perimeter = (Length of Straight Part 1) + (Length of Straight Part 2) + (Circumference of Semicircle 1) + (Circumference of Semicircle 2)
Step 1: Calculate the total length of the straight parts.
There are two straight parts, each 106 m long.
$\text{Total straight length} = 106 \text{ m} + 106 \text{ m} = 212 \text{ m}$
... (1)
Step 2: Calculate the total length of the curved parts.
The two semicircular ends together form one full circle. The diameter of this inner circle is given as 60 m.
Therefore, the radius of the inner semicircles is $r = \frac{60}{2} = 30$ m.
The length of the two semicircular parts is equal to the circumference of a full circle with this radius.
$\text{Total curved length} = 2 \pi r = 2 \times \frac{22}{7} \times 30$
$= \frac{1320}{7} \text{ m}$
... (2)
Step 3: Calculate the total inner perimeter.
Add the lengths from Step 1 and Step 2.
$\text{Total Inner Perimeter} = 212 + \frac{1320}{7}$
$= \frac{212 \times 7 + 1320}{7} = \frac{1484 + 1320}{7} = \frac{2804}{7} \text{ m}$
Converting to a decimal, this is approximately 400.57 m.
Answer: The perimeter of the track along its inner edge is $\frac{2804}{7}$ m.