Maharashtra Board Class 10 Maths Chapter 7 Mensuration Set 7.3 Solutions

Question 1. Radius of a circle is 10 cm. Measure of an arc of the circle is 54°. Find the area of the sector associated with the arc. (π = 3.14)
Answer: Given: Radius (r) = 10 cm, Measure of the arc (\( \theta \)) = 54° To find: Area of the sector. Solution: Area of sector = \( \frac{\theta}{360} \times \pi r^2 \) = \( \frac{54}{360} \times 3.14 \times (10)^2 \) = \( \frac{3}{20} \times 3.14 \times 100 \) = \( 3 \times 3.14 \times 5 \) = \( 15 \times 3.14 \) = \( 47.1 \text{ cm}^2 \) The area of the sector is 47.1 cm².
In simple words: To find the area of a sector, we use the formula involving the central angle and radius, substituting the given values to calculate the final area.

🎯 Exam Tip: Remember the formula for the area of a sector and ensure correct unit conversion for radius and area in your final answer.

Question 2. Measure of an arc of a circle is 80° and its radius is 18 cm. Find the length of the arc. (π = 3.14)
Answer: Given: Radius (r) = 18 cm, Measure of the arc (\( \theta \)) = 80° To find: Length of the arc. Solution: Length of arc = \( \frac{\theta}{360} \times 2\pi r \) = \( \frac{8}{360} \times 2 \times 3.14 \times 18 \) = \( 2 \times 2 \times 3.14 \times 18 \) = \( 25.12 \text{ cm} \) The length of the arc is 25.12 cm.
In simple words: The length of an arc is calculated by finding the fraction of the circle's circumference determined by the central angle.

🎯 Exam Tip: Pay close attention to the value of pi (π) provided in the question, and remember to use the correct formula for arc length.

Question 3. Radius of a sector of a circle is 3.5 cm and length of its arc is 2.2 cm. Find the area of the sector.
Answer: Solution: Given: Radius (r) = 3.5 cm, length of arc (l) = 2.2 cm To find: Area of the sector. Solution: Area of sector = \( \frac{l \times r}{2} \) = \( \frac{2.2 \times 3.5}{2} \) = \( 1.1 \times 3.5 = 3.85 \text{ cm}^2 \) The area of the sector is 3.85 cm².
In simple words: When the arc length and radius are known, the area of a sector can be found by multiplying them and dividing by two.

🎯 Exam Tip: There are two main formulas for the area of a sector; choose the one that best suits the given information (central angle and radius OR arc length and radius).

Question 4. Radius of a circle is 10 cm. Area of a sector of the circle is 100 cm². Find the area of its corresponding major sector, (π = 3.14)
Answer: Given: Radius (r) = 10 cm, area of minor sector =100 cm² To find: Area of maj or sector. Solution: Area of circle = \( \pi r^2 \) = \( 3.14 \times (10)^2 \) = \( 3.14 \times 100 = 314 \text{ cm}^2 \) Now, area of major sector = area of circle - area of minor sector = \( 314 - 100 \) = \( 214 \text{ cm}^2 \) The area of the corresponding major sector is 214 cm².
In simple words: To find the area of the major sector, subtract the area of the minor sector from the total area of the circle.

🎯 Exam Tip: Clearly distinguish between minor and major sectors. The sum of their areas always equals the total area of the circle.

Question 5. Area of a sector of a circle of radius 15 cm is 30 cm². Find the length of the arc of the sector.
Answer: Given: Radius (r) = 15 cm, area of sector = 30 cm² To find: Length of the arc (l). Solution: Area of sector = \( \frac{\text{length of the arc} \times \text{radius}}{2} \)
\( \implies 30 = \frac{l \times 15}{2} \)
\( \implies l = \frac{30 \times 2}{15} = 4 \text{ cm} \) The length of the arc is 4 cm.
In simple words: We can find the arc length by rearranging the sector area formula, using the given area and radius.

🎯 Exam Tip: Be careful with algebraic manipulation when rearranging formulas to solve for a specific unknown, such as arc length or radius.

Question 6. In the adjoining figure, radius of the circle is 7 cm and m (arc MBN) = 60°, find (i) Area of the circle. (ii) A(O-MBN). (iii) A(O-MCN).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त का चित्र है जिसका केंद्र O है। वृत्त पर M, B, N बिंदु स्थित हैं, जिससे एक त्रिज्यखंड O-MBN बनता है। M और N को मिलाने वाली जीवा MBN द्वारा केंद्र पर 60 डिग्री का कोण (∠MON) बनता है। वृत्त पर C बिंदु इस त्रिज्यखंड के बाहर स्थित है।
Answer: Given: radius (r) = 7 cm, m(arc MBN) = \( \theta \) = 60° Solution: (i) Area of circle = \( \pi r^2 \) = \( \frac{22}{7} \times (7)^2 \) = \( 22 \times 7 \) = \( 154 \text{ cm}^2 \) The area of the circle is 154 cm². (ii) Central angle (\( \theta \)) = \( \angle \text{MON} = 60^\circ \) Area of sector = \( \frac{\theta}{360} \times \pi r^2 \)
\( \implies \) A(O - MBN) = \( \frac{60}{360} \times \frac{22}{7} \times (7)^2 \) = \( \frac{1}{6} \times 22 \times 7 \) = \( 25.67 \text{ cm}^2 \) = \( 25.7 \text{ cm}^2 \) A(O-MBN) = 25.7 cm². (iii) Area of major sector = area of circle - area of minor sector
\( \implies \) A(O-MCN) = Area of circle - A(O-MBN) = \( 154 - 25.7 \)
\( \implies \) A(O-MCN) = \( 128.3 \text{ cm}^2 \)
In simple words: This problem involves calculating the area of the full circle, then a minor sector based on its central angle, and finally the major sector by subtracting the minor sector's area from the total circle's area.

🎯 Exam Tip: When dealing with sectors, ensure you correctly identify whether you're calculating a minor or major sector, and use the appropriate central angle or subtraction method.

Question 7. In the adjoining figure, radius of circle is 3.4 cm and perimeter of sector P-ABC is 12.8 cm. Find A(P-ABC).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्ताकार त्रिज्यखंड P-ABC का चित्र है, जिसमें P केंद्र है। PA और PC त्रिज्यखंड की त्रिज्याएँ हैं, जिनकी लंबाई 3.4 cm है। A से C तक एक वृत्ताकार चाप ABC है।
Answer: Solution: Given: Radius (r) = 3.4 cm, perimeter of sector 12.8 cm To find: A(P-ABC) Perimeter of sector = length of arc ABC + AP + CP
\( \implies 12.8 = l + 3.4 + 3.4 \)
\( \implies 12.8 = l + 6.8 \)
\( \implies l = 12.8 - 6.8 = 6 \text{ cm} \) A(P-ABC) = \( \frac{\text{length of the arc} \times \text{radius}}{2} \) = \( \frac{6 \times 3.4}{2} \) = \( 10.2 \text{ cm}^2 \) A(P-ABC) = 10.2 cm².
In simple words: First, calculate the arc length from the given perimeter and radii, then use the arc length and radius to find the sector's area.

🎯 Exam Tip: The perimeter of a sector includes two radii and the arc length; ensure all components are added correctly before solving for unknowns.

Question 8. In the adjoining figure, O is the centre of the sector. \( \angle \text{ROQ} = \angle \text{MON} = 60^\circ \). OR = 7 cm, and OM = 21 cm. Find the lengths of arc RXQ and \( (\pi = 2) \)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक चित्र है जिसमें एक ही केंद्र O से दो संकेंद्रित त्रिज्यखंड दिखाए गए हैं, छोटा त्रिज्यखंड O-RXQ है, जहाँ OR त्रिज्या है। बड़ा त्रिज्यखंड O-MYN है, जहाँ OM त्रिज्या है। दोनों त्रिज्यखंडों का केंद्रीय कोण 60° है।
Answer: Given: \( \angle \text{ROQ} = \angle \text{MON} = 60^\circ \), radius (r) = OR = 7 cm, radius (R) = OM = 21 cm To find: Lengths of arc RXQ and arc MYN. Solution: (i) Length of arc RXQ = \( \frac{\theta}{360} \times 2\pi r \) = \( \frac{60}{360} \times 2 \times \frac{22}{7} \times 7 \) = \( \frac{1}{6} \times 2 \times 22 \) = \( 7.33 \text{ cm} \) (ii) Length of arc MYN = \( \frac{\theta}{360} \times 2\pi R \) = \( \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 \) = \( \frac{1}{6} \times 2 \times 22 \times 3 \) = \( 22 \text{ cm} \) The lengths of arc RXQ and arc MYN are 7.33 cm and 22 cm respectively.
In simple words: For concentric sectors with the same angle but different radii, calculate the arc length for each separately using their respective radii.

🎯 Exam Tip: When dealing with concentric sectors, remember to use the correct radius for each arc length calculation, even if the central angle is the same.

Question 9. In the adjoining figure, if A(P-ABC) = 154 cm², radius of the circle is 14 cm, find (i) \( \angle \text{APC} \), (ii) l(arc ABC).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त का चित्र है जिसका केंद्र P है। वृत्त पर A, B, C बिंदु स्थित हैं, जिससे एक त्रिज्यखंड P-ABC बनता है। P से A और P से C वृत्त की त्रिज्याएँ हैं।
Answer: Given: A(P – ABC) = 154 cm², radius (r) = 14 cm Solution: (i) Let \( \angle \text{APC} = \theta \) A(P-ABC) = \( \frac{\theta}{360} \times \pi r^2 \)
\( \implies 154 = \frac{\theta}{360} \times \frac{22}{7} \times 14^2 \)
\( \implies \theta = \frac{154 \times 360 \times 7}{22 \times 14^2} \) = \( \frac{154 \times 360 \times 7}{22 \times 14 \times 14} \) = \( \frac{7 \times 360}{14 \times 2} = \frac{360}{2 \times 2} \) = \( 90^\circ \)
\( \implies \angle \text{APC} = 90^\circ \) (ii) l (arc ABC) = \( \frac{\theta}{360} \times 2\pi r \) = \( \frac{90}{360} \times 2 \times \frac{22}{7} \times 14 \) = \( \frac{1}{4} \times 2 \times 22 \times 2 \) = \( 22 \text{ cm} \) l (arc ABC) = 22 cm
In simple words: First, use the given sector area and radius to find the central angle, then use that angle to calculate the length of the arc.

🎯 Exam Tip: When working with the area of a sector, remember that the central angle is directly proportional to both the area and the arc length.

Question 10. Radius of a sector of a circle is 7 cm. If measure of arc of the sector is (i) 30° (ii) 210° (iii) three right angles, find the area of the sector in each case.
Answer: Given: Radius (r) = 7 cm To find: Area of the sector. Solution: (i) Measure of the arc (\( \theta \)) = 30° Area of sector = \( \frac{\theta \times \pi r^2}{360} \) = \( \frac{30}{360} \times \frac{22}{7} \times (7)^2 \) = \( \frac{1}{12} \times 22 \times 7 \) = \( 12.83 \text{ cm}^2 \) Area of the sector is 12.83 cm². (ii) Measure of the arc (\( \theta \)) = 210° Area of sector = \( \frac{\theta}{360} \times \pi r^2 \) = \( \frac{210}{360} \times \frac{22}{7} \times (7)^2 \) = \( \frac{7}{12} \times 22 \times 7 \) = \( 89.83 \text{ cm}^2 \) Area of the sector is 89.83 cm². (iii) Measure of the arc (\( \theta \)) = 3 right angle = \( 3 \times 90^\circ = 270^\circ \) Area of sector = \( \frac{\theta}{360} \times \pi r^2 \) = \( \frac{270}{360} \times \frac{22}{7} \times (7)^2 \) = \( \frac{3}{4} \times 22 \times 7 \) = \( 115.50 \text{ cm}^2 \) Area of the sector is 115.50 cm².
In simple words: The area of a sector directly depends on its central angle; larger angles result in larger areas, given a constant radius.

🎯 Exam Tip: Always convert "right angles" into degrees (90°) before using them in formulas for area or arc length calculations.

Question 11. The area of a minor sector of a circle is 3.85 cm² and the measure of its central angle is 36°. Find the radius of the circle.
Answer: Given: Area of minor sector = 3.85 cm², central angle (\( \theta \)) = 36° To find: Radius of the circle (r). Solution: Area of minor sector = \( \frac{\theta}{360} \times \pi r^2 \)
\( \implies 3.85 = \frac{36}{360} \times \frac{22}{7} \times r^2 \)
\( \implies 3.85 = \frac{1}{10} \times \frac{22}{7} \times r^2 \)
\( \implies r^2 = \frac{3.85 \times 10 \times 7}{22} \) = \( \frac{385}{100} \times \frac{10}{22} \times 7 \) = \( \frac{385}{10} \times \frac{7}{22} \) = \( \frac{77}{2} \times \frac{7}{22} \)
\( \implies r^2 = \frac{7 \times 7}{2 \times 2} \)
\( \implies r = \frac{7}{2} \) ...[Taking square root of both sides] = \( 3.5 \text{ cm} \) The radius of the circle is 3.5 cm.
In simple words: By rearranging the sector area formula, we can solve for the radius if the area and central angle are known.

🎯 Exam Tip: Algebraic rearrangement for \( r^2 \) must be meticulous; ensure all terms are correctly moved and simplified to avoid calculation errors.

Question 12. In the given figure, JPQRS is a rectangle. If PQ = 14 cm, QR = 21 cm, find the areas of the parts x, y and z.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक आयत PQRS का चित्र है। शीर्ष P से त्रिज्या PQ का एक वृत्त का चतुर्थांश बनाया गया है, जो क्षेत्र 'x' को दर्शाता है। शीर्ष R से भी त्रिज्या RA का एक वृत्त का चतुर्थांश बनाया गया है, जो क्षेत्र 'y' को दर्शाता है। आयत के शेष केंद्रीय भाग को 'z' के रूप में लेबल किया गया है। चित्र में अतिरिक्त बिंदु A (QR पर) और B (PS पर) दिखाए गए हैं जो चतुर्थांशों की सीमाओं को परिभाषित करते हैं।
Answer: Given: In rectangle PQRS, PQ = 14 cm, QR = 21 cm To find: Areas of the parts x, y and z. Solution: \( \angle \text{Q} = \angle \text{R} = \theta = 90^\circ \) ...[Angles of a rectangle] Area of part x = \( \frac{\theta}{360} \times \pi r^2 \) = \( \frac{90}{360} \times \frac{22}{7} \times \text{PQ}^2 \) = \( \frac{1}{4} \times \frac{22}{7} \times 14^2 \) = \( \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 \) = \( \frac{1}{2} \times 11 \times 2 \times 14 \) = \( 11 \times 14 \) = \( 154 \text{ cm}^2 \) For the sector (Q – PA), PQ = QA ...[Radii of the same circle] QA = 14 cm Now, QR = QA + AR ... [Q – A – R]
\( \implies 21 = 14 + \text{AR} \)
\( \implies \text{AR} = 7 \text{ cm} \) Area of part y = \( \frac{\theta}{360} \times \pi R^2 \) = \( \frac{90}{360} \times \frac{22}{7} \times (\text{AR})^2 \) = \( \frac{1}{4} \times \frac{22}{7} \times (7)^2 \) = \( \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 \) = \( \frac{1}{4} \times 22 \times 7 \) = \( \frac{154}{4} \) = \( 38.5 \text{ cm}^2 \) Area of rectangle = length \( \times \) breadth area of JPQRS = PQ \( \times \) QR = \( 14 \times 21 \) = \( 294 \text{ cm}^2 \) Area of part z = area of JPQRS - area of part x – area of part y = \( 294 - 154 - 38.5 \) = \( 101.5 \text{ cm}^2 \) The area of part x is 154 cm², the area of part y is 38,5 cm² and the area of part z is 101.5 cm².
In simple words: This problem requires calculating the areas of two quarter-circle sectors within a rectangle and then subtracting those from the total rectangle area to find the remaining central area.

🎯 Exam Tip: For composite figures, break them down into simpler geometric shapes. Remember that the angles of a rectangle are 90°, making sectors at its corners quadrants.

Question 13. \( \triangle \text{ALMN} \) is an equilat triangle. LM = 14 cm. As shown in figure, three sectors are drawn with vertices as centres and radius 7 cm. Find, (i) A (\( \triangle \text{LMN} \)). (ii) Area of any one of the sectors. (iii) Total area of all the three sectors. (iv) Area of the shaded region. (\( \sqrt{3} = 1.732 \))
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समबाहु त्रिभुज LMN का चित्र है जिसकी भुजा LM = 14 cm है। त्रिभुज के प्रत्येक शीर्ष (L, M, N) को केंद्र मानकर 7 cm त्रिज्या वाले तीन त्रिज्यखंड खींचे गए हैं। ये त्रिज्यखंड त्रिभुज के अंदर एक छायांकित क्षेत्र बनाते हैं।
Answer: Given: In equilateral triangle LMN, LM = 14 cm, radius of sectors (r) = 7 cm Solution: (i) \( \triangle \text{LMN} \) is an equilateral triangle. A(\( \triangle \text{LMN} \)) = \( \frac{\sqrt{3}}{4} \text{LM}^2 \) = \( \frac{\sqrt{3}}{4} \times 14^2 \) = \( 49 \times 1.732 \) = \( 84.868 \) = \( 84.87 \text{ cm}^2 \) (ii) Central angle (\( \theta \)) = 60° ...[Angle of an equilateral triangle] Area of sector = \( \frac{\theta}{360} \times \pi r^2 \) = \( \frac{60}{360} \times \frac{22}{7} \times 7^2 \) = \( \frac{1}{6} \times 22 \times 7 \) = \( \frac{11 \times 7}{3} \) = \( \frac{77}{3} \) = \( 25.67 \text{ cm}^2 \) Area of one sector = 25.67 cm². (iii) Total area of all three sectors = \( 3 \times \) Area of one sector = \( 3 \times 25.67 \) = \( 77.01 \text{ cm}^2 \) Total area of all three sectors = 77.01 cm². (iv) Area of shaded region = A(\( \triangle \text{LMN} \)) – total area of all three sectors = \( 84.87 - 77.01 \) = \( 7.86 \text{ cm}^2 \) Area of shaded region = 7.86 cm.
In simple words: First, find the area of the equilateral triangle, then calculate the area of one sector (since all three are identical), find the total area of the three sectors, and finally subtract this total from the triangle's area to get the shaded region.

🎯 Exam Tip: Remember that all angles in an equilateral triangle are 60°. Be careful with calculations involving square roots and rounding off intermediate values if needed for final answers.

Question 1. Complete the following table with the help of given figure. (Textbook pg. no. 149)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त का चित्र है जिसका केंद्र O है। वृत्त पर A, X, B, Y बिंदु स्थित हैं। चाप AXB का केंद्रीय कोण 100° है। चाप AYB, AXB का प्रमुख चाप है।
Answer: Solution:

In simple words: The measure of a minor arc is given by its central angle, while the major arc's measure is 360 degrees minus the minor arc's measure.

🎯 Exam Tip: Always remember that the sum of the minor arc and its corresponding major arc's measures is 360°.

Question 2. Observe the figures below. Radii of all circles are equal. Observe the areas of the shaded regions and complete the following table. (Textbook pg. no. 150)
ℹ️ चित्र व्याख्या (Diagram Explanation): चार वृत्ताकार त्रिज्यखंडों का एक सेट दिखाया गया है, सभी समान त्रिज्या 'r' के साथ। पहला त्रिज्यखंड एक पूरा वृत्त है (A1, 360°), दूसरा एक अर्धवृत्त है (A2, 180°), तीसरा एक चतुर्थांश है (A3, 90°), और चौथा एक 60° का त्रिज्यखंड है (A4)। नीचे एक सामान्य त्रिज्यखंड A भी दिखाया गया है, जिसका केंद्रीय कोण θ है।
Answer: Solution:

Thus, if measure of an arc of a circle is \( \theta \), then Area of sector (A) = \( \frac{\theta}{360} \times \) Area of circle
\( \implies \) Area of sector (A) = \( \frac{\theta}{360} \times \pi r^2 \)
\( \implies \frac{A}{\pi r^2} = \frac{\theta}{360} \) i.e. \( \frac{\text{Area of the sector}}{\text{Area of the circle}} = \frac{\theta}{360} \)
In simple words: The ratio of a sector's area to the total circle's area is equal to the ratio of its central angle to 360 degrees.

🎯 Exam Tip: Understanding the proportional relationship between the central angle, arc length, and sector area is fundamental for solving problems related to circles.

Question 3. In the following figures, radii of all circles are equal. Observe the length of arc in each figure and complete the table. (Textbook pg. no. 151)
ℹ️ चित्र व्याख्या (Diagram Explanation): चार वृत्ताकार चापों का एक सेट दिखाया गया है, सभी समान त्रिज्या 'r' के साथ। पहला चाप एक पूर्ण वृत्त की परिधि है (l1, 360°), दूसरा एक अर्धवृत्त का चाप है (l2, 180°), तीसरा एक चतुर्थांश का चाप है (l3, 90°), और चौथा एक 60° के त्रिज्यखंड का चाप है (l4)। नीचे एक सामान्य चाप 'l' भी दिखाया गया है, जिसका केंद्रीय कोण θ है।
Answer: Solution:

Thus, if the measure of an arc of a circle is \( \theta \), then Length of arc (l) = \( \frac{\theta}{360} \times \) circumference of circle
\( \implies \) Length of arc (l) = \( \frac{\theta}{360} \times 2\pi r \)
\( \implies \frac{l}{2\pi r} = \frac{\theta}{360} \) i.e., \( \frac{\text{Length of arc}}{\text{Circumference}} = \frac{\theta}{360} \)
In simple words: The ratio of an arc's length to the total circumference of the circle is equivalent to the ratio of its central angle to 360 degrees.

🎯 Exam Tip: Remember that the arc length formula is a direct proportion of the circle's circumference, scaled by the central angle over 360 degrees.