Maharashtra Board Class 10 Maths Chapter 7 Mensuration Set 7.4 Solutions

Practice Set 7.4

Question 1. In the adjoining figure, A is the centre of the circle. \(\angle\text{ABC} = 45^\circ\) and \(\text{AC} = 7\sqrt{2}\text{ cm}\). Find the area of segment BXC. (\(\pi = 3.14\))
Answer: Given:
Radius of the circle (\(r\)) = \(\text{AC} = 7\sqrt{2}\text{ cm}\)
In \(\triangle\text{ABC}\),
\(\text{AB} = \text{AC}\) ...[Radii of the same circle]
\(\therefore \angle\text{ACB} = \angle\text{ABC} = 45^\circ\) ...[Isosceles triangle theorem]
Now, \(\angle\text{BAC} + \angle\text{ABC} + \angle\text{ACB} = 180^\circ\) ...[Sum of the measures of angles of a triangle]
\(\implies \angle\text{BAC} + 45^\circ + 45^\circ = 180^\circ\)
\(\implies \angle\text{BAC} + 90^\circ = 180^\circ\)
\(\implies \angle\text{BAC} = 90^\circ\)
\(\therefore\) Central angle (\(\theta\)) = \(90^\circ\)

Area of segment BXC = \(r^2 \left[ \frac{\pi\theta}{360} - \frac{\sin\theta}{2} \right]\)
\(\implies \text{Area of segment BXC} = (7\sqrt{2})^2 \left[ \frac{3.14 \times 90}{360} - \frac{\sin 90^\circ}{2} \right]\)
\(\implies \text{Area of segment BXC} = 98 \left[ \frac{3.14}{4} - \frac{1}{2} \right]\)
\(\implies \text{Area of segment BXC} = 98 \left[ 0.785 - 0.5 \right]\)
\(\implies \text{Area of segment BXC} = 98 \times 0.285\)
\(\implies \text{Area of segment BXC} = 27.93\text{ sq. cm.}\)
This calculated value represents the region enclosed between the chord and the corresponding arc. Therefore, the area of segment BXC is \(27.93\text{ sq. cm.}\)
In simple words: To find the area of the segment, we first find the central angle of the sector which is \(90^\circ\). Then, we subtract the area of the right-angled triangle from the area of the sector to get the remaining segment area of \(27.93\text{ sq. cm.}\)

🎯 Exam Tip: Always show the step-by-step calculation of the central angle \(\theta\) using the isosceles triangle property before substituting values into the segment area formula to secure full marks.

Question 1. In the adjoining figure, A is the centre of the circle. \(\angle ABC = 45^\circ\), \(AC = 7\sqrt{2}\) cm. Find the area of segment BXC. (\(\pi = 3.14\))
Answer:
In \(\Delta ABC\),
\(AC = AB\) ... [Radii of same circle]
\(\therefore \angle ABC = \angle ACB\) ... [Isosceles triangle theorem]
\(\dots \angle ABC = \angle ACB = 45^\circ\)
In \(\Delta ABC\),
\(\angle ABC + \angle ACB + \angle BAC = 180^\circ\) ... [Sum of the measures of angles of a triangle is \(180^\circ\)]
\(\therefore 45^\circ + 45^\circ + \angle BAC = 180^\circ\)
\(\therefore 90^\circ + \angle BAC = 180^\circ\)
\(\therefore \angle BAC = 90^\circ\)
Let \(\angle BAC = \theta = 90^\circ\)
Radius of the circle, \(r = AC = 7\sqrt{2}\) cm
Area of segment BXC = \(r^2 \left[ \frac{\pi\theta}{360} - \frac{\sin\theta}{2} \right]\)
\(= (7\sqrt{2})^2 \left[ \frac{3.14 \times 90}{360} - \frac{\sin 90}{2} \right]\)
\(= 49 \times 2 \left[ \frac{3.14}{4} - \frac{1}{2} \right]\)
\(= 98 \left[ \frac{3.14}{4} - \frac{2}{4} \right]\)
\(= 98 \left[ \frac{3.14 - 2}{4} \right]\)
\(= 98 \left[ \frac{1.14}{4} \right]\)
\(= 98 \times [0.285]\)
\(= 27.93\text{ cm}^2\)
\(\therefore\) The area of segment BXC is \(27.93\text{ cm}^2\). This calculation confirms that the segment's area is determined by subtracting the triangle's area from the sector's area.
In simple words: To find the area of the segment, we first find the central angle of the triangle, which is \(90^\circ\). Then, we use the segment area formula with the radius and this angle to calculate the final area.

🎯 Exam Tip: Remember to state the Isosceles Triangle Theorem when equating the base angles. Writing down the formula for the area of a segment clearly before substituting values helps secure step-wise marks.

Question 2. In the adjoining figure, O is the centre of the circle. m(arc PQR) = \(60^\circ\), OP = 10 cm. Find the area of the shaded region. (\(\pi = 3.14, \sqrt{3} = 1.73\))
Answer:
Given:
Radius of the circle, \(r = OP = 10\) cm
Measure of arc PQR = \(\theta = 60^\circ\)
Area of the shaded region (segment PQR) = \(r^2 \left[ \frac{\pi\theta}{360} - \frac{\sin\theta}{2} \right]\)
\(= 10^2 \left[ \frac{3.14 \times 60}{360} - \frac{\sin 60}{2} \right]\)
\(= 100 \left[ \frac{3.14}{6} - \frac{\sqrt{3}}{4} \right]\)
\(= 100 \left[ 0.5233 - \frac{1.73}{4} \right]\)
\(= 100 [0.5233 - 0.4325]\)
\(= 100 \times 0.0908\)
\(= 9.08\text{ cm}^2\)
\(\dots\) The area of the shaded region is \(9.08\text{ cm}^2\). This represents the region bounded by the chord and the corresponding arc.
In simple words: The shaded region is a segment of the circle. We find its area by using the formula that subtracts the area of the triangle from the area of the sector using the given radius of 10 cm and angle of \(60^\circ\).

🎯 Exam Tip: Be careful with decimal calculations when substituting \(\sqrt{3} = 1.73\). Keep at least four decimal places during intermediate steps to avoid rounding errors in your final answer.

Exercise 7.4

Question 2. In the adjoining figure, \( m(\text{arc } PQR) = 60^\circ \), radius \( (r) = OP = 10\text{ cm} \). Find the area of the shaded region. (\( \pi = 3.14, \sqrt{3} = 1.73 \))
Answer:
Given:
\( m(\text{arc } PQR) = 60^\circ \)
Radius \( (r) = OP = 10\text{ cm} \)

To find: Area of shaded region.

Solution:
\( \angle POR = m(\text{arc } PQR) \) ...[Measure of central angle]

\( \therefore \angle POR = \theta = 60^\circ \)

\( A(\text{segment } PQR) = r^2 \left[ \frac{\pi\theta}{360} - \frac{\sin\theta}{2} \right] \)
\( = 10^2 \left[ \frac{3.14 \times 60}{360} - \frac{\sin 60^\circ}{2} \right] \)
\( = 100 \left[ \frac{3.14}{6} - \frac{\sqrt{3}}{2} \times \frac{1}{2} \right] \)
\( = 100 \left[ \frac{3.14}{6} - \frac{1.73}{4} \right] \)
\( = 100 \left[ \frac{3.14 \times 2}{6 \times 2} - \frac{1.73 \times 3}{4 \times 3} \right] \)
\( = 100 \left[ \frac{6.28}{12} - \frac{5.19}{12} \right] \)
\( = 100 \left[ \frac{6.28 - 5.19}{12} \right] \)
\( = 100 \left[ \frac{1.09}{12} \right] \)
\( = 100 [0.0908] \)
\( = 9.08\text{ cm}^2 \)

This calculation shows that the area of the segment is indeed very small compared to the entire circle.

\( \therefore \) The area of the shaded region is \( 9.08\text{ cm}^2 \).
In simple words: To find the area of the shaded slice (segment) of the circle, we subtract the area of the triangle from the area of the sector. Using the formula with the given radius of 10 cm and angle of 60 degrees, we get approximately 9.08 square centimeters.

🎯 Exam Tip: Always write down the formula for the area of a segment clearly before substituting values to secure step-wise marks.

Question 3. In the adjoining figure, if A is the centre of the circle, \( \angle PAR = 30^\circ \), \( AP = 7.5 \), find the area of the segment PQR. (\( \pi = 3.14 \))
Answer:
Given:
Centre of the circle = A
Radius \( (r) = AP = 7.5 \)
Central angle \( (\theta) = \angle PAR = 30^\circ \)
\( \pi = 3.14 \)

To find: Area of segment PQR

Solution:
\( A(\text{segment } PQR) = r^2 \left[ \frac{\pi\theta}{360} - \frac{\sin\theta}{2} \right] \)
\( = (7.5)^2 \left[ \frac{3.14 \times 30}{360} - \frac{\sin 30^\circ}{2} \right] \)
\( = 56.25 \left[ \frac{3.14}{12} - \frac{1}{2 \times 2} \right] \)
\( = 56.25 \left[ 0.2617 - 0.25 \right] \)
\( = 56.25 \left[ 0.0117 \right] \)
\( = 0.658\text{ units}^2 \)

This small value is expected because a 30-degree central angle creates a very narrow segment.

\( \therefore \) The area of the segment PQR is \( 0.66\text{ units}^2 \).
In simple words: We find the area of the segment by using the formula with radius 7.5 and angle 30 degrees. Subtracting the triangle's area from the sector's area gives us a very small area of about 0.66 square units.

🎯 Exam Tip: When the unit of radius is not specified (like cm or m), write the final answer with "sq. units" or "\(\text{units}^2\)" to keep your answer mathematically complete.

Question 1. Given central angle \( (\theta) = \angle PAR = 30^\circ \) and radius \( (r) = AP = 7.5 \), find the area of segment PQR.
Diagram Details:

• Circle with center A
• Radius AP = 7.5
• Central angle \(\angle PAR = 30^\circ\)
• Shaded segment PQR

Answer:
Given:
Central angle \( (\theta) = \angle PAR = 30^\circ \)
Radius \( (r) = AP = 7.5 \)

To find: Area of segment PQR.

Solution:
Let \( \angle PAR = \theta = 30^\circ \)
Area of segment PQR = \( r^2 \left[ \frac{\pi\theta}{360} - \frac{\sin\theta}{2} \right] \)
\( \implies \) Area of segment PQR = \( (7.5)^2 \left[ \frac{3.14 \times 30}{360} - \frac{\sin 30^\circ}{2} \right] \)
\( \implies \) Area of segment PQR = \( 56.25 \left[ \frac{3.14}{12} - \frac{1/2}{2} \right] \)
\( \implies \) Area of segment PQR = \( 56.25 \left[ \frac{3.14}{12} - \frac{1}{4} \right] \)
\( \implies \) Area of segment PQR = \( 56.25 \left[ \frac{3.14}{12} - \frac{1 \times 3}{4 \times 3} \right] \)
\( \implies \) Area of segment PQR = \( 56.25 \left[ \frac{3.14}{12} - \frac{3}{12} \right] \)
\( \implies \) Area of segment PQR = \( 56.25 \left( \frac{3.14 - 3}{12} \right) \)
\( \implies \) Area of segment PQR = \( 56.25 \left( \frac{0.14}{12} \right) \)
\( \implies \) Area of segment PQR = \( \frac{7.875}{12} \)
\( \implies \) Area of segment PQR = \( 0.65625 \) sq. units.

Therefore, the area of segment PQR is \( 0.65625 \) sq. units. This precise calculation helps in determining the exact region bounded by the chord and the arc.
In simple words: To find the area of the small slice (segment) of the circle, we subtract the area of the triangle from the area of the sector. Using the formula with the given angle of 30 degrees and radius of 7.5, we get 0.65625 square units.

🎯 Exam Tip: Remember to convert the trigonometric value \(\sin 30^\circ = \frac{1}{2}\) correctly and keep your decimal calculations precise to avoid rounding errors.

Question 4. In the adjoining figure, if O is the centre of the circle, PQ is a chord, \( \angle POQ = 90^\circ \), area of shaded region is 114 cm\(^2\), find the radius of the circle, (\( \pi = 3.14 \))
Diagram Details:

• Circle with center O
• Chord PQ
• Central angle \(\angle POQ = 90^\circ\)
• Shaded segment PRQ

Answer:
Given:
Central angle \( (\theta) = \angle POQ = 90^\circ \)
Area of shaded region (segment PRQ) = \( 114 \text{ cm}^2 \)
\( \pi = 3.14 \)

To find: Radius of the circle \( (r) \).

Solution:
Area of segment PRQ = \( r^2 \left[ \frac{\pi\theta}{360} - \frac{\sin\theta}{2} \right] \)
\( \implies 114 = r^2 \left[ \frac{3.14 \times 90}{360} - \frac{\sin 90^\circ}{2} \right] \)
\( \implies 114 = r^2 \left[ \frac{3.14}{4} - \frac{1}{2} \right] \)
\( \implies 114 = r^2 \left[ 0.785 - 0.5 \right] \)
\( \implies 114 = r^2 [0.285] \)
\( \implies r^2 = \frac{114}{0.285} \)
\( \implies r^2 = 400 \)
\( \implies r = \sqrt{400} \)
\( \implies r = 20 \text{ cm} \)

Therefore, the radius of the circle is \( 20 \text{ cm} \). This shows how the area of a segment directly relates to the size of the circle's radius.
In simple words: We are given the area of the shaded segment as 114 square centimeters and a central angle of 90 degrees. By putting these values into our segment area formula, we can work backward to find that the radius of the circle is 20 cm.

🎯 Exam Tip: When the central angle is \(90^\circ\), \(\sin 90^\circ = 1\). Simplifying the fraction \(\frac{90}{360}\) to \(\frac{1}{4}\) early makes the decimal division much easier to solve.

Question 4. Find the radius of a circle if the area of the segment PRQ is \( 114\text{ cm}^2 \) and the central angle \( (\theta) \) is \( 90^\circ \). (Use \( \pi = 3.14 \))
Answer:
Given:
Central angle \( (\theta) = \angle\text{POQ} = 90^\circ \)
Area of segment PRQ = \( 114\text{ cm}^2 \)
To find: Radius \( (r) \)

Solution:
Area of segment PRQ = \( r^2 \left[ \frac{\pi\theta}{360} - \frac{\sin\theta}{2} \right] \)
\( \therefore 114 = r^2 \left[ \frac{3.14 \times 90}{360} - \frac{\sin 90^\circ}{2} \right] \)
\( \therefore 114 = r^2 \left[ \frac{3.14}{4} - \frac{1}{2} \right] \)
\( \dots [\because \sin 90^\circ = 1] \)
\( \dots 114 = r^2 \left[ \frac{3.14}{4} - \frac{1 \times 2}{2 \times 2} \right] \)
\( \therefore 114 = r^2 \left[ \frac{3.14}{4} - \frac{2}{4} \right] \)
\( \dots [\text{Making denominators equal}] \)
\( \therefore 114 = r^2 \left[ \frac{3.14 - 2}{4} \right] \)
\( \therefore 114 = r^2 \times \frac{1.14}{4} \)
\( \therefore r^2 = \frac{114 \times 4}{1.14} \)
\( \dots r^2 = \frac{11400 \times 4}{114} \)
\( \therefore r^2 = 100 \times 4 \)
\( \therefore r = 10 \times 2 \)
\( \dots [\text{Taking square root of both sides}] \)
\( \therefore r = 20\text{ cm} \)
The radius of the circle is \( 20\text{ cm} \).
In simple words: To find the radius, we use the formula for the area of a segment. By substituting the given area of 114 and angle of 90 degrees, we simplify the equation step-by-step to find that the radius is 20 cm.

🎯 Exam Tip: When simplifying fractions with decimals, multiply both the numerator and denominator by 100 to eliminate the decimal point. This makes division much easier and prevents calculation errors.

Question 5. A chord PQ of a circle with radius 15 cm subtends an angle of 60° with the centre of the circle. Find the area of the minor as well as the major segment. (\( \pi = 3.14 \), \( \sqrt{3} = 1.73 \))
Answer:
Given:
Radius \( (r) = 15\text{ cm} \)
Central angle \( (\theta) = 60^\circ \)

To find: Areas of major and minor segments.

Solution:
Let chord PQ subtend \( \angle\text{POQ} = 60^\circ \) at the centre.
\( \therefore \theta = 60^\circ \)

1. Area of minor segment:
Area of minor segment = \( r^2 \left[ \frac{\pi\theta}{360} - \frac{\sin\theta}{2} \right] \)
\( = 15^2 \left[ \frac{3.14 \times 60}{360} - \frac{\sin 60^\circ}{2} \right] \)
\( = 225 \left[ \frac{3.14}{6} - \frac{\sqrt{3}}{2 \times 2} \right] \)
\( = 225 \left[ 0.5233 - \frac{1.73}{4} \right] \)
\( = 225 \left[ 0.5233 - 0.4325 \right] \)
\( = 225 \times 0.0908 \)
\( = 20.43\text{ cm}^2 \) (approx)

2. Area of major segment:
Area of circle = \( \pi r^2 \)
\( = 3.14 \times 15^2 \)
\( = 3.14 \times 225 \)
\( = 706.5\text{ cm}^2 \)

Area of major segment = Area of circle - Area of minor segment
\( = 706.5 - 20.43 \)
\( = 686.07\text{ cm}^2 \) (approx)

Thus, the area of the minor segment is \( 20.43\text{ cm}^2 \) and the area of the major segment is \( 686.07\text{ cm}^2 \).
In simple words: First, we find the area of the minor segment using the formula with the radius and the 60-degree angle. Then, we subtract this minor segment's area from the total area of the circle to get the area of the larger (major) segment.

🎯 Exam Tip: Remember that the area of a major segment is always the total area of the circle minus the area of the minor segment. Keep your decimal calculations precise up to two decimal places to secure full marks.

Question 1. Find the area of the minor segment and major segment of a circle of radius 15 cm when the measure of the arc is \( 60^\circ \). (Use \( \pi = 3.14 \), \( \sqrt{3} = 1.73 \))
Answer:
Given:
Radius of the circle \( (r) = 15\text{ cm} \)
Central angle \( (\theta) = 60^\circ \)

Diagram Details:

• Circle with center O
• Radius OP = OQ = 15 cm
• Central angle \( \angle\text{POQ} = 60^\circ \)
• Shaded minor segment PSQ

\( \text{Area of minor segment} = r^2 \left[ \frac{\pi \theta}{360} - \frac{\sin \theta}{2} \right] \)
\( = 15^2 \left[ \frac{3.14 \times 60}{360} - \frac{\sin 60^\circ}{2} \right] \)
\( = 225 \left[ \frac{3.14}{6} - \frac{\sqrt{3}}{2} \times \frac{1}{2} \right] \)
\( = 225 \left[ \frac{3.14}{6} - \frac{1.73}{4} \right] \)
\( = 225 \left[ \frac{3.14 \times 2}{6 \times 2} - \frac{1.73 \times 3}{4 \times 3} \right] \)
\( = 225 \left[ \frac{6.28}{12} - \frac{5.19}{12} \right] \)
\( = 225 \left[ \frac{6.28 - 5.19}{12} \right] \)
\( = 225 \left[ \frac{1.09}{12} \right] \)
\( = 225 \times [0.0908] \)
\( = 20.43\text{ cm}^2 \)

\( \therefore \text{area of minor segment} = 20.43\text{ cm}^2 \)

\( \text{Area of circle} = \pi r^2 \)
\( = 3.14 \times 15 \times 15 \)
\( = 3.14 \times 225 \)
\( = 706.5\text{ cm}^2 \)

\( \text{Area of major segment} = \text{Area of circle} - \text{area of minor segment} \)
\( = 706.5 - 20.43 \)
\( = 686.07\text{ cm}^2 \)

\( \text{Area of major segment } 686.07\text{ cm}^2 \)
\( \therefore \text{The area of minor segment is } 20.43\text{ cm}^2 \text{ and the area of major segment is } 686.07\text{ cm}^2. \)
In simple words: To find the area of the minor segment, we use the formula that subtracts the triangular part from the sector. To find the major segment, we simply subtract this minor segment's area from the total area of the circle.

🎯 Exam Tip: Always write down the given values and the formula clearly before starting calculations to secure step-wise marks, and do not forget to write the final unit as \( \text{cm}^2 \).