Maharashtra Board Class 10 Maths Chapter 7 Mensuration Set 7 Solutions

Question 1. Choose the correct alternative answer for each of the following questions.
(i) The ratio of circumference and area of a circle is 2: 7. Find its circumference.
(A) 14 π
(B) \( \frac{7}{\pi} \)
(C) 7π
(D) \( \frac{14}{\pi} \)
Answer:
\[ \frac{\text{Circumference}}{\text{Area of circle}} = \frac{2}{7} \]
\[ \frac{2\pi r}{\pi r^2} = \frac{2}{7} \]
\[ \frac{2}{r} = \frac{2}{7} \]
\( r = 7 \)
Circumference = \( 2\pi r = 2 \times \pi \times 7 = 14\pi \)
Answer: (A) 14π
In simple words: Given the ratio of circumference to area, we use the formulas for both to derive the radius. With the radius determined, the circumference is then straightforward to calculate.

🎯 Exam Tip: Remember the formulas for circumference (\(2\pi r\)) and area (\(\pi r^2\)) of a circle. Simplifying algebraic expressions involving \(r\) accurately is crucial for success.

Question 1.
(ii) If measure of an arc of a circle is 160° and its length is 44 cm, find the circumference of the circle.
(A) 66 cm
(B) 44 cm
(C) 160 cm
(D) 99 cm
Answer:
Length of arc = \( \frac{\theta}{360} \times 2\pi r \)
\( 44 = \frac{160}{360} \times 2\pi r \)
\( 2\pi r = \frac{44 \times 360}{160} = 99 \text{ cm} \)
Circumference = 99 cm
Answer: (D) 99 cm
In simple words: The length of an arc is a fraction of the circle's total circumference, determined by the central angle. By plugging in the given arc length and angle into the formula, we can solve for the full circumference.

🎯 Exam Tip: The formula for arc length, \( L = \frac{\theta}{360} \times 2\pi r \), is vital. Remember that \(2\pi r\) represents the full circumference, which can be directly solved for if arc length and angle are known.

Question 1.
(iii) Find the perimeter of a sector of a circle if its measure is 90° and radius is 7 cm.
(A) 44 cm
(B) 25 cm
(C) 36 cm
(D) 56 cm
Answer:
Perimeter of sector = Length of arc + 2r
\( = \frac{\theta}{360} \times 2\pi r + 2r \)
\( = \frac{90}{360} \times 2 \times \frac{22}{7} \times 7 + 2 \times 7 \)
\( = 11 + 14 \)
\( = 25 \text{ cm} \)
Answer: (B) 25 cm
In simple words: The perimeter of a sector is found by adding the length of its arc and the lengths of the two radii that form its boundaries.

🎯 Exam Tip: Distinguish between arc length and sector perimeter. The perimeter includes the two radii, while arc length is just the curved part. Use \( \pi = \frac{22}{7} \) for calculations involving multiples of 7.

Question 1.
(iv) Find the curved surface area of a cone of radius 7 cm and height 24 cm.
(A) 440 cm²
(B) 550 cm²
(C) 330 cm²
(D) 110 cm²
Answer:
Slant height of cone \( (l) = \sqrt{r^2 + h^2} \)
\( = \sqrt{7^2 + 24^2} \)
\( = \sqrt{49 + 576} \)
\( = \sqrt{625} = 25 \text{ cm} \)
Curved surface area of cone = \( \pi rl \)
\( = \frac{22}{7} \times 7 \times 25 \)
\( = 550 \text{ cm}^2 \)
Answer: (B) 550 cm²
In simple words: To find the curved surface area of a cone, we first calculate its slant height using the Pythagorean theorem with the radius and height. Then, we apply the formula for curved surface area involving the radius, slant height, and pi.

🎯 Exam Tip: Always calculate the slant height (\(l\)) first when dealing with a cone's surface area, using \( l = \sqrt{r^2 + h^2} \). The curved surface area formula is \( \pi rl \), not including the base.

Question 1.
(v) The curved surface area of a cylinder is 440 cm² and its radius is 5 cm. Find its height.
(A) \( \frac{44}{\pi} \) cm
(B) 22π cm
(C) 44π cm
(D) \( \frac{22}{\pi} \) cm
Answer:
Curved surface area of cylinder = \( 2\pi rh \)
\( 440 = 2 \times \pi \times 5 \times h \)
\( h = \frac{440}{10\pi} = \frac{44}{\pi} \text{ cm} \)
Answer: (A) 44/π cm
In simple words: Using the formula for the curved surface area of a cylinder and the given values for area and radius, we can algebraically rearrange and solve for the unknown height.

🎯 Exam Tip: Know the curved surface area formula for a cylinder: \( 2\pi rh \). When an area is given, use it to find an unknown dimension by isolating the variable in the formula.

Question 1.
(vi) A cone was melted and cast into a cylinder of the same radius as that of the base of the cone. If the height of the cylinder is 5 cm, find the height of the cone.
(A) 15 cm
(B) 10 cm
(C) 18 cm
(D) 5 cm
Answer:
Volume of cone = Volume of cylinder
\( \frac{1}{3} \pi r^2 h = \pi r^2 H \)
\( \frac{h}{3} = 5 \)
\( h = 15 \text{ cm} \)
Answer: (A) 15 cm
In simple words: When a solid is melted and recast into another shape, its volume remains constant. By equating the volumes of the cone and cylinder, and knowing their radii are the same, we can easily find the cone's height.

🎯 Exam Tip: For problems involving melting and recasting, the principle of conservation of volume is key. Set the initial volume equal to the final volume and solve for the unknown dimension.

Question 1.
(vii) Find the volume of a cube of side 0.01 cm.
(A) 1 cm
(B) 0.001 cm³
(C) 0.0001 cm³
(D) 0.000001 cm³
Answer:
Volume of cube = \( (\text{side})^3 \)
\( = (0.01)^3 = 0.000001 \text{ cm}^3 \)
Answer: (D) 0.000001 cm³
In simple words: The volume of a cube is calculated by cubing the length of its side. For a side of 0.01 cm, the volume is \( (0.01)^3 \).

🎯 Exam Tip: The formula for the volume of a cube is \( \text{side}^3 \). Pay close attention to decimal calculations, especially when cubing small numbers.

Question 1.
(viii) Find the side of a cube of volume 1 m³
(A) 1 cm
(B) 10 cm
(C) 100 cm
(D) 1000 cm
Answer:
Volume of cube = \( (\text{side})^3 \)
\( \therefore 1 = (\text{side})^3 \)
\( \therefore \text{Side} = 1 \text{ m} \)
\( = 100 \text{ cm} \)
Answer: (C) 100 cm
In simple words: Given the volume of a cube, its side length is found by taking the cube root of the volume. Remember to convert units if the answer is required in a different unit (e.g., meters to centimeters).

🎯 Exam Tip: Be mindful of unit conversions (e.g., 1 meter = 100 centimeters). Ensure the final answer is presented in the unit requested by the question.

Question 2. A washing tub in the shape of a frustum of a cone has height 21 cm. The radii of the circular top and bottom are 20 cm and 15 cm respectively. What is the capacity of the tub? \( (\pi = \frac{22}{7}) \)
Answer:
Given: For the frustum-shaped tub,
height (h) = 21 cm,
radii (r₁) = 20 cm, and (r₂) = 15 cm
To find: Capacity (volume) of the tub.
Solution:
Volume of frustum = \( \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 \times r_2) \)
Volume of frustum = \( \frac{1}{3} \times \frac{22}{7} \times 21 (20^2 + 15^2 + 20 \times 15) \)
\( = 22 (400 + 225 + 300) \)
\( = 22 \times 925 \)
\( = 20350 \text{ cm}^3 \)
\( = \frac{20350}{1000} \text{ litres} \)
...[1litre = 1000 cm³]
\( = 20.35 \text{ litres} \)
\( \therefore \) The capacity of the tub is 20.35 litres.
In simple words: The volume of a frustum is calculated using a specific formula involving its height and the radii of its two circular bases. After calculating the volume in cubic centimeters, it is converted to liters using the conversion factor (1 liter = 1000 cm³).

🎯 Exam Tip: The formula for the volume of a frustum is crucial. Pay close attention to squaring radii and multiplying them, as these are common areas for calculation errors. Remember to convert cm³ to liters if required.

Question 3. Some plastic balls of radius 1 cm were melted and cast into a tube. The thickness, length and outer radius of the tube were 2 cm, 90 cm and 30 cm respectively. How many balls were melted to make the tube?
Answer:
Given: For the cylindrical tube,
height (h) = 90 cm,
outer radius (R) = 30 cm,
thickness = 2 cm
For the plastic spherical ball,
radius (r₁) = 1 cm
To find: Number of balls melted.
Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): एक खोखले बेलनाकार ट्यूब का चित्र दिखाया गया है। ट्यूब की बाहरी त्रिज्या 30 सेमी है, दीवार की मोटाई 2 सेमी है और लंबाई (ऊंचाई) 90 सेमी है। यह एक पाइप जैसा दिखता है।
Inner radius of tube (r)
= outer radius – thickness of tube
\( = 30 - 2 \)
\( = 28 \text{ cm} \)
Volume of plastic required for the tube = Outer volume of tube – Inner volume of hollow tube
\( = \pi R^2 h – \pi r^2 h \)
\( = \pi h(R^2 – r^2) \)
\( = \pi \times 90 (30^2 – 28^2) \)
\( = \pi \times 90 (30 + 28) (30 – 28) \)
...[ \( \therefore a^2 – b^2 = (a + b)(a – b) \) ]
\( = 90 \times 58 \times 2 \pi \text{ cm}^3 \)
\( = 10440\pi \text{ cm}^3 \)
Volume of one plastic ball = \( \frac{4}{3} \pi r_1^3 \)
\( = \frac{4}{3} \pi \times 1^3 \)
\( = \frac{4}{3} \pi \text{ cm}^3 \)
Number of balls to be melted = \( \frac{\text{Volume of plastic required for the tube}}{\text{Volume of one plastic ball}} \)
\( = \frac{10440\pi}{\frac{4}{3}\pi} \)
\( = \frac{90 \times 58 \times 2 \times 3}{4} \)
\( = 7830 \)
\( \therefore \) 7830 plastic balls were melted to make the tube.
In simple words: To find the number of balls, first calculate the volume of plastic in the hollow tube by subtracting the inner volume from the outer volume. Then, calculate the volume of a single spherical ball. Finally, divide the total volume of plastic by the volume of one ball.

🎯 Exam Tip: Remember that melting and recasting implies conservation of volume. For hollow cylinders, use \( \pi h(R^2 - r^2) \) to find the material volume. Be careful with calculations involving fractions and \( \pi \).

Question 4. A metal parallelopiped of measures 16 cm × 11cm × 10cm was melted to make coins. How many coins were made if the thickness and diameter of each coin was 2 mm and 2 cm respectively?
Answer:
Given: For the parallelopiped,
length (l) = 16 cm, breadth (b) = 11 cm,
height (h) = 10 cm
For the cylindrical coin,
thickness (H) = 2 mm,
diameter (D) = 2 cm
To find: Number of coins made.
Solution:
Volume of parallelopiped = \( l \times b \times h \)
\( = 16 \times 11 \times 10 \)
\( = 1760 \text{ cm}^3 \)
Thickness of coin (H) = 2 mm
\( = 0.2 \text{ cm} \)
...[ \( \therefore 1 \text{ cm} = 10 \text{ mm} \) ]
Diameter of coin (D) = 2 cm
Radius of coin (R) = \( \frac{D}{2} = \frac{2}{2} = 1 \text{ cm} \)
Volume of one coin = \( \pi R^2 H \)
\( = \frac{22}{7} \times 1^2 \times 0.2 \)
\( = \frac{4.4}{7} \text{ cm}^3 \)
Number of coins that were made = \( \frac{\text{Volume of parallelopiped}}{\text{Volume of one coin}} \)
\( = \frac{1760}{\frac{4.4}{7}} \)
\( = \frac{1760 \times 7}{4.4} = \frac{1760 \times 7 \times 10}{44} \)
\( = 2800 \)
\( \therefore \) 2800 coins were made by melting the parallelopiped.
In simple words: To find the number of coins, first calculate the volume of the parallelopiped. Then, calculate the volume of a single cylindrical coin, remembering to convert units for consistency. Finally, divide the total volume of metal by the volume of one coin.

🎯 Exam Tip: Always ensure consistent units throughout the problem (e.g., all in cm). The conservation of volume principle applies here. The formula for a cylinder's volume \( \pi r^2 h \) is used for coins, where height is thickness.

Question 5. The diameter and length of a roller is 120 cm and 84 cm respectively. To level the ground, 200 rotations of the roller are required. Find the expenditure to level the ground at the rate of Rs. 10 per sq.m.
Answer:
Given: For the cylindrical roller,
diameter (d) = 120 cm,
length = height (h) = 84 cm
To find: Expenditure of levelling the ground.
Solution:
Diameter of roller (d) = 120 cm
Radius of roller (r) = \( \frac{d}{2} = \frac{120}{2} = 60 \text{ cm} \)
Curved surface area of roller = \( 2\pi rh \)
\( = 2 \times \frac{22}{7} \times 60 \times 84 \)
\( = 2 \times 22 \times 60 \times 12 \)
\( = 31680 \text{ cm}^2 \)
\( = \frac{31680}{100 \times 100} \text{ m}^2 \)
...[ \( \therefore 1 \text{ m} = 100 \text{ cm} \) ]
\( = 3.168 \text{ m}^2 \)
Now, area of ground levelled in one rotation = curved surface area of roller
\( = 3.168 \text{ m}^2 \)
\( \therefore \) Area of ground levelled in 200 rotations
\( = 3.168 \times 200 \)
\( = 633.6 \text{ m}^2 \)
Rate of levelling = Rs. 10 per m²
\( \therefore \) Expenditure of levelling the ground
\( = 633.6 \times 10 = 6336 \)
\( \therefore \) The expenditure of levelling the ground is Rs. 6336.
In simple words: First, calculate the curved surface area of the roller, which represents the area covered in one rotation. Convert this area to square meters. Then, multiply this area by the number of rotations to get the total area leveled. Finally, multiply the total area by the given rate to find the total expenditure.

🎯 Exam Tip: Remember that a roller's curved surface area is the area it covers in one rotation. Ensure all unit conversions (cm² to m²) are done correctly before calculating total area and expenditure. Pay attention to the \( \pi \) value if not specified.

Question 6. The diameter and thickness of a hollow metal sphere are 12 cm and 0.01 m respectively. The density of the metal is 8.88 gm per cm³. Find the outer surface area and mass of the sphere, [\( \pi = 3.14 \)]
Answer:
Given: For the hollow sphere,
diameter (D) = 12 cm, thickness = 0.01 m
density of the metal = 8.88 gm per cm³
To find:
(i) Outer surface area of the sphere
(ii) Mass of the sphere.

ℹ️ चित्र व्याख्या (Diagram Explanation): एक खोखले गोले का अनुप्रस्थ काट (cross-section) दिखाया गया है। गोले का बाहरी व्यास 12 सेमी है और धातु की मोटाई 0.01 मीटर है, जो दर्शाता है कि यह एक मोटी दीवार वाला खोखला गोला है।
Solution:
Diameter of the sphere (D) = 12 cm
\( \therefore \) Radius of sphere (R) = \( \frac{d}{2} = \frac{12}{2} = 6 \text{ cm} \)
\( \therefore \) Surface area of sphere = \( 4\pi R^2 \)
\( = 4 \times 3.14 \times 6^2 \)
\( = 452.16 \text{ cm}^2 \)
Thickness of sphere = 0.01 m
\( = 0.01 \times 100 \text{ cm} \)
...[ \( \therefore 1 \text{ m} = 100 \text{ cm} \) ]
\( = 1 \text{ cm} \)
\( \therefore \) Inner radius of the sphere (r)
= Outer radius – thickness of sphere
\( = 6 - 1 = 5 \text{ cm} \)
\( \therefore \) Volume of hollow sphere
= Volume of outer sphere – Volume of inner sphere
\( = \frac{4}{3} \pi R^3 – \frac{4}{3} \pi r^3 \)
\( = \frac{4}{3} \pi (R^3 – r^3) \)
\( = \frac{4}{3} \times 3.14 \times (6^3 – 5^3) \)
\( = \frac{4}{3} \times 3.14 \times (216 – 125) \)
\( = \frac{4}{3} \times 3.14 \times 91 \)
\( = \frac{1142.96}{3} \)
\( = 380.986 \)
\( = 380.99 \text{ cm}^3 \)
Now, density of metal = \( \frac{\text{Mass of sphere}}{\text{Volume of sphere}} \)
\( 8.88 = \frac{\text{Mass of sphere}}{380.99} \)
Mass of sphere = \( 8.88 \times 380.99 \)
\( = 3383.19 \text{ gm} \)
\( \therefore \) The outer surface area and the mass of the sphere are 452.16 cm² and 3383.19 gm respectively.
In simple words: First, calculate the outer surface area of the sphere. Then, determine the inner radius by subtracting the thickness from the outer radius. Calculate the volume of the hollow sphere by subtracting the inner volume from the outer volume. Finally, use the density formula (Mass = Density × Volume) to find the mass of the metal.

🎯 Exam Tip: Remember to convert all units to be consistent (e.g., meters to centimeters) before calculations. For hollow shapes, material volume is the difference between outer and inner volumes. Mass calculations require density and volume.

Question 7. A cylindrical bucket of diameter 28 cm and height 20 cm was full of sand. When the sand in the bucket was poured on the ground, the sand got converted into a shape of a cone. If the height of the cone was 14 cm, what was the base area of the cone?
Answer:
Given: For the cylindrical bucket,
diameter (d) = 28 cm, height (h) = 20 cm
For the conical heap of sand,
height (H) = 14 cm
To find: Base area of the cone (\( \pi R^2 \)).
Solution:
Diameter of the bucket (d) = 28 cm
\( \therefore \) Radius of bucket (r) = \( \frac{d}{2} = \frac{28}{2} = 14 \text{ cm} \)
Volume of bucket = \( \pi r^2 h \)
\( = \frac{22}{7} \times 14^2 \times 20 \)
\( = 22 \times 14 \times 2 \times 20 \)
\( = 12320 \text{ cm}^3 \)
Volume of conical heap = \( \frac{1}{3} \pi R^2 H \)
\( = \frac{1}{3} \times \pi R^2 \times 14 \)
\( = \frac{14}{3} \pi R^2 \text{ cm}^2 \)
But, volume of bucket = volume of conical heap
\( \therefore 12320 = \frac{14}{3} \pi R^2 \)
\( \therefore \pi R^2 = \frac{12320 \times 3}{14} \)
\( = 2640 \text{ cm}^2 \)
The base area of the cone is 2640 cm².
In simple words: When sand is poured from a cylindrical bucket to form a cone, their volumes are equal. Calculate the volume of sand in the cylinder. Then, equate this to the volume of the cone, using the given cone height to solve for its base area.

🎯 Exam Tip: Remember the principle of conservation of volume when reshaping solids. The key is to correctly apply the volume formulas for both the cylinder (\( \pi r^2 h \)) and the cone (\( \frac{1}{3} \pi R^2 H \)) and solve for the required unknown.

Question 8. The radius of a metallic sphere is 9 cm. It was melted to make a wire of diameter 4 mm. Find the length of the wire.
Answer:
Given: For metallic sphere,
radius (R) = 9 cm
For the cylindrical wire,
diameter (d) = 4 mm
To find: Length of wire (h).
Solution:
Volume of sphere = \( \frac{4}{3} \pi R^3 \)
\( = \frac{4}{3} \times \pi \times 9^3 \)
\( = 972\pi \text{ cm}^3 \)
Diameter of wire (d) = 4 mm
\( = \frac{4}{10} \text{ cm} \)
...[ \( \therefore 1 \text{ cm} = 10 \text{ mm} \) ]
\( = 0.4 \text{ cm} \)
Radius of wire (r) = \( \frac{d}{2} = \frac{0.4}{2} = 0.2 \text{ cm} \)
Volume of wire = \( \pi r^2 h \)
\( = \pi (0.2)^2 h = 0.04\pi h \text{ cm}^3 \)
But, volume of wire = volume of sphere
\( 0.04 \pi h = 972\pi \)
\( h = \frac{972}{0.04} \)
\( = \frac{97200}{4} \)
\( = 24300 \text{ cm} \)
\( = \frac{24300}{100} \text{ m} \)
...[ \( \therefore 1 \text{ m} = 100 \text{ cm} \) ]
\( h = 243 \text{ m} \)
\( \therefore \) The length of the wire is 243 m.
In simple words: Melting a sphere to form a wire means their volumes are equal. First, calculate the volume of the sphere. Then, convert the wire's diameter to centimeters and calculate its radius. Equate the sphere's volume to the wire's cylindrical volume and solve for the wire's length, finally converting it to meters.

🎯 Exam Tip: Ensure all dimensions are in consistent units (e.g., cm) before calculating volumes. The conservation of volume is key. Remember the volume formulas for a sphere (\( \frac{4}{3} \pi r^3 \)) and a cylinder (\( \pi r^2 h \)).

Question 9. The area of a sector of a circle of 6 cm radius is 157t sq.cm. Find the measure of the arc and length of the arc corresponding to the sector.
Answer:
Given: Radius (r) = 6 cm,
area of sector = 15π cm²
To find:
(i) Measure of the arc (θ),
(ii) Length of the arc (l)
Solution:
Area of sector = \( \frac{\theta}{360} \times \pi r^2 \)
\( 15\pi = \frac{\theta}{360} \times \pi \times 6^2 \)
\( 15\pi = \frac{\theta}{360} \times \pi \times 36 \)
\( 15 = \frac{\theta}{10} \)
\( \theta = 150^\circ \)
Also, area of sector = \( \frac{\text{length of the arc} \times \text{radius}}{2} \)
\( 15\pi = \frac{l \times 6}{2} \)
\( l = \frac{15\pi \times 2}{6} = 5\pi \text{ cm} \)
\( \therefore \) The measure of the arc and the length of the arc are 150° and 5π cm respectively.
In simple words: We can find the measure of the arc (central angle) by using the area of the sector formula. Once the angle is known, we can find the length of the arc using its formula, or alternatively, use the relationship between the sector's area, arc length, and radius.

🎯 Exam Tip: Be aware of two key formulas for sector area: \( \frac{\theta}{360} \times \pi r^2 \) and \( \frac{1}{2} \times l \times r \). These allow you to find either the angle (\(\theta\)) or the arc length (\(l\)) if other values are known.

Question 10. In the adjoining figure, seg AB is a chord of a circle with centre P. If PA = 8 cm and distance of chord AB from the centre P is 4 cm, find the area of the shaded portion.
Answer:
\( (\pi = 3.14, \sqrt{3} = 1.73) \)
Given: Radius (r) = PA = 8 cm,
PC = 4 cm
To find: Area of shaded region.
Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त जिसका केंद्र P है, में एक जीवा AB दिखाई गई है। P से A तक की त्रिज्या 8 सेमी है। केंद्र P से जीवा AB तक की लंबवत दूरी 4 सेमी है, जो C पर जीवा को काटती है। P, A, B से मिलकर एक त्रिभुज बनता है, और P, A, C से मिलकर एक समकोण त्रिभुज बनता है। इसमें एक सेक्टर P-ADB और एक त्रिभुज PAB है, और छायांकित भाग सेक्टर में से त्रिभुज को घटाकर प्राप्त किया जाता है।
Let \( \angle\text{APC} = \theta_1 \)
In \( \triangle\text{ACP} \), \( \angle\text{ACP} = 90^\circ \)
\( \cos \theta_1 = \frac{\text{PC}}{\text{AP}} = \frac{4}{8} = \frac{1}{2} \)
But, \( \cos 60^\circ = \frac{1}{2} \)
\( \therefore \theta_1 = 60^\circ \)
Similarly, we can show that, \( \angle\text{BPC} = 60^\circ \)
\( \angle\text{APB} = \angle\text{APC} + \angle\text{BPC} \)
...[Angle sum property]
\( \therefore \theta = 60^\circ + 60^\circ = 120^\circ \)
A(P-ADB) = \( \frac{\theta}{360} \times \pi r^2 \)
\( = \frac{120}{360} \times 3.14 \times 8^2 \)
\( = \frac{1}{3} \times 3.14 \times 64 \)
\( = 66.98 \text{ cm}^2 \)
In \( \triangle\text{APC} \),
\( \sin \theta_1 = \frac{\text{AC}}{\text{AP}} \)
\( \sin 60^\circ = \frac{\text{AC}}{8} \)
\( \frac{\sqrt{3}}{2} = \frac{\text{AC}}{8} \)
\( \text{AC} = 4\sqrt{3} \text{ cm} \)
Now, AB = 2 AC
Perpendicular drawn from the centre of the circle to the chord bisects the chord
\( = 2 \times 4\sqrt{3} \)
\( = 8\sqrt{3} \text{ cm} \)
A(\( \triangle\text{APB} \)) = \( \frac{1}{2} \times \text{AB} \times \text{PC} \)
\( = \frac{1}{2} \times 8\sqrt{3} \times 4 \)
\( = 16\sqrt{3} \)
\( = 16 \times 1.73 \)
\( = 27.68 \text{ cm}^2 \)
Area of shaded region = A(P-ADB) – A(\( \triangle\text{APB} \))
\( = 66.98 - 27.68 \)
\( = 39.30 \text{ cm}^2 \)
\( \therefore \) The area of the shaded region is 39.30 cm².
In simple words: To find the shaded area, we calculate the area of the sector formed by the chord and then subtract the area of the triangle formed by the chord and the two radii. Trigonometry is used to find angles and lengths within the triangle.

🎯 Exam Tip: For shaded region problems, break down the area into simpler geometric shapes. Often, it involves finding the area of a sector and subtracting the area of a triangle. Master trigonometric ratios for right-angled triangles to find unknown lengths and angles.

Question 11. In the adjoining figure, square ABCD is inscribed in the sector A-PCQ. The radius of sector C-BXD is 20 cm. Complete the following activity to find the area of shaded region.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक बड़े वृत्त के सेक्टर A-PCQ को दर्शाता है जिसमें एक वर्ग ABCD अंतर्निहित है। वर्ग का एक शीर्ष C है, और दूसरा सेक्टर C-BXD है जिसकी त्रिज्या 20 सेमी है। चित्र में कई छायांकित क्षेत्र हैं, और मुख्य लक्ष्य इन छायांकित क्षेत्रों का कुल क्षेत्रफल ज्ञात करना है।
Solution:
Side of square ABCD
= radius of sector C-BXD = [20] cm
Area of square = \( (\text{side})^2 = 20^2 = 400 \text{ cm}^2 \) ....(i)
Area of shaded region inside the square = Area of square ABCD – Area of sector C-BXD
= Area of square ABCD \( - \) Area of sector C-BXD
\( = 400 - \frac{\theta}{360} \times \pi r^2 \)
\( = 400 - \frac{90}{360} \times 3.14 \times 20^2 \)
\( = 400 - 314 \)
\( = 86 \text{ cm}^2 \)
Radius of bigger sector
= Length of diagonal of square ABCD
\( = \sqrt{2} \times \text{side} \)
\( = 20 \sqrt{2} \text{ cm} \)
Area of the shaded regions outside the square
= Area of sector A-PCQ – Area of square ABCD
= A(A – PCQ) – A(ABCD)
\( = \left( \frac{\theta}{360} \times \pi r^2 \right) - \text{AB}^2 \)
\( = \left( \frac{90}{360} \times 3.14 \times (20\sqrt{2})^2 \right) - (20)^2 \)
\( = \left( \frac{1}{4} \times 3.14 \times 800 \right) - 400 \)
\( = (3.14 \times 200) - 400 \)
\( = 628 - 400 \)
\( = 228 \text{ cm}^2 \)
Total area of the shaded region = \( 86 + 228 \)
\( = 314 \text{ cm}^2 \)

Alternate method:
A(C-BXD) = \( \frac{\theta}{360} \times \pi r^2 \)
\( = \frac{90}{360} \times 3.14 \times 20 \times 20 \)
\( = 3.14 \times 5 \times 20 \)
\( = 314 \text{ cm}^2 \)
ABCD is a square. ... [Given]
Side of ABCD = radius of sector (C-BXD)
\( = 20 \text{ cm} \)
Radius of sector (A-PCQ) = Diagonal
\( = \sqrt{2} \times \text{side} \)
\( = \sqrt{2} \times 20 \)
\( = 20 \sqrt{2} \text{ cm} \)
A(A-PCQ) = \( \frac{\theta}{360} \times \pi r^2 \)
\( = \frac{90}{360} \times 3.14 \times (20\sqrt{2})^2 \)
\( = \frac{1}{4} \times 3.14 \times 20 \times 20 \times 2 \)
\( = 628 \text{ cm}^2 \)
Now, Area of shaded region
= A(A-PCQ) – A(C-BXD)
\( = 628 - 314 \)
\( = 314 \text{ cm}^2 \)
\( \therefore \) The area of the shaded region is 314 cm².
In simple words: The total shaded area is found by adding two parts: the shaded region inside the square (square area minus the small sector) and the shaded region outside the square (larger sector area minus the square area). The alternate method calculates the two sector areas and subtracts them.

🎯 Exam Tip: For complex shaded region problems, break the figure into known geometric shapes (squares, sectors, triangles). Pay close attention to the radii and angles for each sector. The diagonal of a square with side 's' is \( s\sqrt{2} \).

Alternate Method:

A(C-BXD) = \( \frac{\theta}{360} \times \pi r^2 \)
= \( \frac{90}{360} \times 3.14 \times 20 \times 20 \)
= \( 3.14 \times 5 \times 20 \)
= \( 314 \text{ cm}^2 \)

□ABCD is a square. ... [Given]
Side of □ABCD = radius of sector (C-BXD)
= 20 cm
Radius of sector (A-PCQ) = Diagonal
= \( \sqrt{2} \times \) side
= \( \sqrt{2} \times 20 \)
= \( 20 \sqrt{2} \) cm

A(A-PCQ) = \( \frac{\theta}{360} \times \pi r^2 \)
= \( \frac{90}{360} \times 3.14 \times (20\sqrt{2})^2 \)
= \( \frac{1}{4} \times 3.14 \times 20 \times 20 \times 2 \)
= \( 628 \text{ cm}^2 \)

Now, Area of shaded region
= A(A-PCQ) - A(C-BXD)
= 628 - 314
= 314 cm²
∴ The area of the shaded region is 314 cm².
In simple words: This alternate method calculates the area of the larger sector A-PCQ and subtracts the area of the smaller sector C-BXD directly to find the total shaded region. Both methods give the same result of 314 cm².

🎯 Exam Tip: Multiple approaches can lead to the correct answer. Understanding how to interpret the shaded region is key—either by summing individual shaded parts or by finding the difference between larger geometric figures.

Question 12. In the adjoining figure, two circles with centres O and P are touching internally at point A. If BQ = 9, DE = 5, complete the following activity to find the radii of the circles.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो वृत्तों को दर्शाता है जो बिंदु A पर आंतरिक रूप से स्पर्श कर रहे हैं। बड़ा वृत्त O केंद्र वाला है और छोटा वृत्त P केंद्र वाला है। वृत्त के अंदर कुछ बिंदु (B, Q, D, E, F) और रेखाखंड दिए गए हैं, जिनमें BQ=9 और DE=5 दिया गया है।
Let the radius of the bigger circle be R and that of smaller circle be r.
OA, OB, OC and OD are the radii of the bigger circle.
∴ OA = OB = OC = OD = R
PQ = PA = r
OQ + BQ = OB ... [B - Q - O]
OQ = OB - BQ = R - 9
OE + DE = OD ....[D - E - O]
OE = OD - DE = [R - 5]
As the chords QA and EF of the circle with centre P intersect in the interior of the circle, so by the property of internal division of two chords of a circle,
OQ × OA = OE × OF
∴ (R - 9) × R = (R - 5) × (R - 5) ...[∵ OE = OF]
∴ R² - 9R = R² - 10R + 25
∴ -9R + 10R = 25
∴ R = [25units]
AQ = AB - BQ = 2r ....[B-Q-A]
∴ 2r = 50 - 9 = 41
∴ r = \( \frac{41}{2} \) = 20.5 units
In simple words: This problem uses the properties of circles, specifically the radii, internal tangency, and the property of chords intersecting inside a circle, to set up equations. By relating the given segment lengths to the radii of the two circles, we can solve for R (radius of the bigger circle) and r (radius of the smaller circle).

🎯 Exam Tip: For geometry problems involving intersecting chords or tangents, always recall the relevant theorems and properties. Clearly label radii and known segments to formulate equations correctly.