Samacheer Kalvi Class 8 Maths Solutions Chapter 2 Measurements Exercise 2.1

(i) Area of a circle	c. \( \pi r^2 \)
(ii) Circumference of a circle	d. \( 2\pi r \)
(iii) Area of the sector of a circle	e. \( \frac{\theta^\circ}{360^\circ} \times \pi r^2 \)
(iv) Circumference of a semicircle	b. \( (\pi + 2)r \)
(v) Area of a quadrant of a circle	a. \( \frac{1}{4} \pi r^2 \)

The correct matches are: (i) - c, (ii) - d, (iii) - e, (iv) - b, (v) - a. These formulas are essential for solving problems related to circles and their parts.
In simple words: Match each part of a circle or semicircle with its correct formula for area or circumference. For example, the area of a full circle is \( \pi r^2 \).

🎯 Exam Tip: Memorize these formulas correctly, as they are frequently used in geometry problems involving circles, sectors, and semicircles.

 

Question 3. Find the central angle of the shaded sectors (each circle is divided into equal sectors).
Answer: The central angle of each sector can be found by dividing \( 360^\circ \) by the number of equal parts (n) the circle is divided into.

In simple words: To find the angle of each slice in a circle, divide \( 360^\circ \) by how many equal slices there are. For example, if there are 2 slices, each angle is \( 180^\circ \).

🎯 Exam Tip: Remember that a full circle measures \( 360^\circ \). To find the central angle of equal sectors, always divide \( 360^\circ \) by the total number of sectors.

 

Question 4. For the sectors with given measures, find the length of the arc, area and perimeter. ( \( \pi = 3.14 \) )
(i) central angle \( 45^\circ \), \( r = 16 \) cm
Answer:
(i) Given: Central angle \( \theta = 45^\circ \), radius \( r = 16 \) cm, \( \pi = 3.14 \).
Length of the arc \( l = \frac{\theta^\circ}{360^\circ} \times 2\pi r \) units
\( l = \frac{45^\circ}{360^\circ} \times 2 \times 3.14 \times 16 \) cm
\( l = \frac{1}{8} \times 2 \times 3.14 \times 16 \) cm
\( l = 3.14 \times 4 \) cm
\( l = 12.56 \) cm. This calculation determines the curved length of the sector.
Area of the sector \( A = \frac{\theta^\circ}{360^\circ} \times \pi r^2 \) sq. units
\( A = \frac{45^\circ}{360^\circ} \times 3.14 \times 16 \times 16 \) cm\(^2\)
\( A = \frac{1}{8} \times 3.14 \times 256 \) cm\(^2\)
\( A = 3.14 \times 32 \) cm\(^2\)
\( A = 100.48 \) cm\(^2\)
Perimeter of the sector \( P = l + 2r \) units
\( P = 12.56 + 2(16) \) cm
\( P = 12.56 + 32 \) cm
\( P = 44.56 \) cm
(ii) Given: Central angle \( \theta = 120^\circ \), diameter \( d = 12.6 \) cm.
First, find the radius: \( r = \frac{d}{2} = \frac{12.6}{2} \) cm
\( r = 6.3 \) cm. The radius is always half of the diameter.
Length of the arc \( l = \frac{\theta^\circ}{360^\circ} \times 2\pi r \) units
\( l = \frac{120^\circ}{360^\circ} \times 2 \times 3.14 \times 6.3 \) cm
\( l = \frac{1}{3} \times 2 \times 3.14 \times 6.3 \) cm
\( l = 2 \times 3.14 \times 2.1 \) cm
\( l = 13.188 \) cm
\( l \approx 13.19 \) cm (rounded to two decimal places).
Area of the sector \( A = \frac{\theta^\circ}{360^\circ} \times \pi r^2 \) sq. units
\( A = \frac{120^\circ}{360^\circ} \times 3.14 \times 6.3 \times 6.3 \) cm\(^2\)
\( A = \frac{1}{3} \times 3.14 \times 39.69 \) cm\(^2\)
\( A = 3.14 \times 13.23 \) cm\(^2\)
\( A = 41.5422 \) cm\(^2\)
\( A \approx 41.54 \) cm\(^2\).
Perimeter of the sector \( P = l + 2r \) cm
\( P = 13.19 + 2(6.3) \) cm
\( P = 13.19 + 12.6 \) cm
\( P = 25.79 \) cm
In simple words: We used formulas to find the arc length (the curved edge), area (the space inside), and perimeter (total boundary length) for two different sectors. We first found the radius if only the diameter was given.

🎯 Exam Tip: Pay close attention to whether the question provides the radius or the diameter, and remember to convert diameter to radius (\( r = d/2 \)) before using the formulas. Use the correct value of \( \pi \) if specified.

 

Question 5. From the measures given below, find the area of the sectors.
(i) \( l = 48 \) m, \( r = 10 \) m
Answer:
(i) Given: Length of the arc \( l = 48 \) m, radius \( r = 10 \) m.
Area of the sector \( A = \frac{lr}{2} \) sq. units
\( A = \frac{48 \times 10}{2} \) m\(^2\)
\( A = 24 \times 10 \) m\(^2\)
\( A = 240 \) m\(^2\). This formula is especially useful when the central angle is not known.
(ii) Given: Length of the arc \( l = 12.5 \) cm, radius \( r = 6 \) cm.
Area of the sector \( A = \frac{lr}{2} \) sq. units
\( A = \frac{12.5 \times 6}{2} \)
\( A = 12.5 \times 3 \) cm\(^2\)
\( A = 37.5 \) cm\(^2\).
In simple words: When you know the length of the curved arc and the radius of a sector, you can find its area by multiplying them and dividing by 2.

🎯 Exam Tip: Remember the formula \( A = \frac{lr}{2} \) for the area of a sector. This is a common shortcut when the arc length \( l \) is provided directly.

 

Question 6. Find the central angle of each of the sectors whose measures are given below. ( \( \pi = \frac{22}{7} \) )
(i) area = 462 cm\(^2\), \( r = 21 \) cm
Answer:
(i) Given: Area of the sector \( A = 462 \) cm\(^2\), radius \( r = 21 \) cm, \( \pi = \frac{22}{7} \).
We know that the Area of a sector \( A = \frac{\theta^\circ}{360^\circ} \times \pi r^2 \).
So, \( 462 = \frac{\theta^\circ}{360^\circ} \times \frac{22}{7} \times 21 \times 21 \)
\( 462 = \frac{\theta^\circ}{360^\circ} \times 22 \times 3 \times 21 \)
\( 462 = \frac{\theta^\circ}{360^\circ} \times 1386 \)
Now, we solve for \( \theta^\circ \):
\( \theta^\circ = \frac{462 \times 360}{1386} \)
\( \theta^\circ = \frac{462 \times 360}{3 \times 462} \)
\( \theta^\circ = \frac{360}{3} \)
\( \theta^\circ = 120^\circ \). This shows how the angle depends on the area and radius.
(ii) Given: Radius of the sector \( r = 8.4 \) cm, length of the arc \( l = 4.4 \) cm.
We know that the Length of the arc \( l = \frac{\theta^\circ}{360^\circ} \times 2\pi r \).
So, \( 4.4 = \frac{\theta^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 8.4 \)
\( 4.4 = \frac{\theta^\circ}{360^\circ} \times 4 \times \frac{22}{7} \times 2.1 \) (Since \( 8.4/2.1 = 4 \))
\( 4.4 = \frac{\theta^\circ}{360^\circ} \times 44 \times 0.3 \) (Since \( 22 \times 2 \times 0.3 \times 2 = 44 \times 0.3 \times 2 \))
\( 4.4 = \frac{\theta^\circ}{360^\circ} \times 13.2 \)
Now, we solve for \( \theta^\circ \):
\( \theta^\circ = \frac{4.4 \times 360}{13.2} \)
\( \theta^\circ = \frac{44 \times 360}{132} \)
\( \theta^\circ = \frac{1 \times 360}{3} \)
\( \theta^\circ = 120^\circ \).
In simple words: We used the formulas for area and arc length of a sector, but this time we worked backward to find the central angle. We plugged in the given numbers and solved for \( \theta^\circ \).

🎯 Exam Tip: When finding the central angle, make sure to use the correct formula (area or arc length) based on the given information. Carefully simplify fractions to avoid errors.

 

Question 7. A circle of radius 120 m is divided into 8 equal sectors. Find the length of the arc of each of the sectors.
Answer: Given: Radius of the circle \( r = 120 \) m, Number of equal sectors \( n = 8 \).
First, find the central angle of each sector:
Central angle of each sector \( \theta^\circ = \frac{360^\circ}{n} \)
\( \theta^\circ = \frac{360^\circ}{8} \)
\( \theta^\circ = 45^\circ \). This is the angle each sector forms at the center.
Next, find the length of the arc of each sector:
Length of the arc \( l = \frac{\theta^\circ}{360^\circ} \times 2\pi r \) units
\( l = \frac{45^\circ}{360^\circ} \times 2\pi \times 120 \) m
\( l = \frac{1}{8} \times 2\pi \times 120 \) m
\( l = \frac{240\pi}{8} \) m
\( l = 30\pi \) m.
Another method to find the arc length when a circle is divided into equal sectors is: \( l = \frac{1}{n} \times 2\pi r \)
\( l = \frac{1}{8} \times 2\pi \times 120 \)
\( l = 30\pi \) m. Both methods yield the same result, showing flexibility in approach.
In simple words: We found the angle of one slice of the circle and then used that angle and the circle's radius to calculate the length of the curved edge of that slice.

🎯 Exam Tip: For problems involving equal sectors, you can use the simplified formula \( l = \frac{1}{n} \times 2\pi r \) to find the arc length, where \( n \) is the number of equal sectors. This often saves a step.

 

Question 8. A circle of radius 70 cm is divided into 5 equal sectors. Find the area of each of the sectors.
Answer: Given: Radius of the sector \( r = 70 \) cm, Number of equal sectors \( n = 5 \).
First, find the central angle of each sector:
Central angle of each sector \( \theta^\circ = \frac{360^\circ}{n} \)
\( \theta^\circ = \frac{360^\circ}{5} \)
\( \theta^\circ = 72^\circ \). Each segment of the circle covers \( 72^\circ \).
Next, find the area of each sector:
Area of the sector \( A = \frac{\theta^\circ}{360^\circ} \times \pi r^2 \) sq. units
\( A = \frac{72^\circ}{360^\circ} \times \pi \times 70 \times 70 \) cm\(^2\)
\( A = \frac{1}{5} \times \pi \times 4900 \) cm\(^2\)
\( A = 980\pi \) cm\(^2\). This calculation determines the space covered by one of the five segments.
In simple words: We figured out the angle for each of the 5 equal slices of the circle. Then, using that angle and the radius, we found how much space each slice takes up.

🎯 Exam Tip: When a circle is divided into equal sectors, you can directly calculate the area of one sector as \( A = \frac{1}{n} \times \pi r^2 \), where \( n \) is the number of equal sectors. This is a quicker way to solve such problems.

 

Question 9. Dhamu fixes a square tile of 30cm on the floor. The tile has a sector design on it as shown in the figure. Find the area of the sector. ( \( \pi = 3.14 \) ).
Answer: The figure shows a circular quadrant within a square tile.
Given: Side of the square = 30 cm.
For a circular quadrant within a square, the radius of the sector design is equal to the side of the square.
So, Radius of the sector design \( r = 30 \) cm.
A quadrant is a sector with a central angle of \( 90^\circ \), which is \( \frac{1}{4} \) of a full circle. This forms a right angle at the corner.
Area of the quadrant \( A = \frac{1}{4} \pi r^2 \) sq. units
\( A = \frac{1}{4} \times 3.14 \times 30 \times 30 \) cm\(^2\)
\( A = 3.14 \times 15 \times 15 \) cm\(^2\)
\( A = 3.14 \times 225 \) cm\(^2\)
\( A = 706.5 \) cm\(^2\) (approximately).
In simple words: The tile has a design that is one-fourth of a circle. The side of the square is the radius of this quarter circle. We used the formula for the area of a quadrant to find its size.

🎯 Exam Tip: Recognize that a "quadrant" is always a sector with a central angle of \( 90^\circ \), representing one-fourth of a circle. This simplifies the area calculation considerably.

 

Question 10. A circle is formed with 8 equal granite stones as shown in the figure each of radius 56 cm and whose central angle is 45°. Find the area of each of the granite stones. ( \( \pi = \frac{22}{7} \) )
Answer: Given: Number of equal sectors \( n = 8 \), Radius of the sector \( r = 56 \) cm, Central angle \( \theta = 45^\circ \), \( \pi = \frac{22}{7} \).
Area of each sector \( A = \frac{\theta^\circ}{360^\circ} \times \pi r^2 \) sq. units
\( A = \frac{45^\circ}{360^\circ} \times \frac{22}{7} \times 56 \times 56 \) cm\(^2\)
\( A = \frac{1}{8} \times \frac{22}{7} \times 56 \times 56 \) cm\(^2\)
\( A = \frac{1}{8} \times 22 \times 8 \times 56 \) cm\(^2\)
\( A = 22 \times 56 \) cm\(^2\)
\( A = 1232 \) cm\(^2\). This is the area of one granite stone, which is a sector of the circle.
Alternatively, since there are 8 equal sectors, we can use \( A = \frac{1}{n} \pi r^2 \):
\( A = \frac{1}{8} \times \frac{22}{7} \times 56 \times 56 \) cm\(^2\)
\( A = 1232 \) cm\(^2\) (approximately).
In simple words: We have a circle made of 8 equal pieces (like slices of a cake). We used the given angle and radius for one piece to find its area.

🎯 Exam Tip: When you are given both the central angle and the number of equal sectors, you can use either the \( \frac{\theta^\circ}{360^\circ} \times \pi r^2 \) formula or the \( \frac{1}{n} \times \pi r^2 \) formula, as they will give the same result for equal sectors. Choose the one you find easier to calculate.