CBSE Class 10 Mathematics Areas Related To Circles Worksheet Set 03

Case Study Based Questions

I. A human chain is to be formed in concentric form of radius 56 m and 63 m at India Gate. But the organiser has some problems, he wants you to solve his problems. Give answer to his questions by using information given in the questions:

Question. The circumference of circular chain of radius 56 m is
(a) 352 m
(b) 176 m
(c) 704 m
(d) 88 m
Answer : A

Question. The area covered by circular chain of radius 56 m is
(a) 5544 m²
(b) 3850 m²
(c) 9856 m²
(d) 12474 m²
Answer : C

Question. If each person is given two metres of space to stand, find how many persons can be accomodated in the chain of radius 56 m.
(a) 126 persons
(b) 176 persons
(c) 156 persons
(d) 186 persons
Answer : B

Question. He wants to include one more concentric circle with a radius of 63 m, the circumference of the new circle will be
(a) 156 m
(b) 396 m
(c) 176 m
(d) 288 m
Answer : B

Question. How many persons can accomodate on the new circle if each person needs 2 metres of space?
(a) 198 persons
(b) 156 persons
(c) 176 persons
(d) 186 persons
Answer : A
 

II. A square park has each side of 100 m. At each corner of the park, there is a flower bed in the form of a quadrant of radius 14 m as shown in given figure.

Answer the question from (1) to (5):

Question. Area of each quadrant is
(a) 145 m²
(b) 154 m²
(c) 415 m²
(d) 514 m²
Answer : B

Question. Area of 4 quadrants is
(a) 616 m²
(b) 166 m²
(c) 661 m²
(d) 510 m²
Answer : A

Question. Area of park with bed roses is
(a) 10,000 m²
(b) 1000 m²
(c) 100 m²
(d) 10 m²
Answer : A

Question. Area of park without bed roses is
(a) 9834 m²
(b) 9384 m²
(c) 9483 m²
(d) 9348 m²
Answer : B

Question. Angle of a quadrant is equal to
(a) 70°
(b) 80°
(c) 90°
(d) 45°
Answer : C
 

III. A student of class-X^th decided to create a model of a circular wall clock and try to paste the numbers from 1 to 12 on its dial. But he is facing some problems.
Give solutions to his problems by looking at the figure.

Question. The angle between points showing hour is
(a) 15°
(b) 30°
(c) 45°
(d) 60°
Answer : A

Question. What is the angle made at the centre between 12 and 3?
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer : D

Question. Find the area covered by minute hand of a clock from 12 to 3 if the radius of the clock is 14 cm.
(a) 164 sq. cm
(b) 77 sq. cm
(c) 154 sq. cm
(d) 308 sq.cm
Answer : C

Question. The formula used in finding the area of sector with radius r and angle at the centre q is
(a) πr²θ/360°
(b) πr²θ/180°
(c) 2πrθ/180°
(d) πrθ/360°
Answer : A

Question. The arc length between the region 12 and 3 if the radius of the clock is 14 cm.
(a) 22 cm
(b) 44 cm
(c) 66 cm
(d) 88 cm
Answer : A

Question 1) A bicycle wheel makes 5000 revolution in moving 11 km. find the diameter of the wheel
Answer:
The total distance traveled by the bicycle wheel is 11 km, which can be converted to centimeters:
\( \text{Distance} = 11 \times 1000 \times 100 = 1,100,000 \text{ cm} \)
The total number of rotations made by the wheel is 5000.
Therefore, the distance covered in 1 complete rotation is:
\( \text{Circumference} = \frac{1,100,000}{5000} = 220 \text{ cm} \)
Since the circumference of a circle is \( \pi d \):
\( \pi d = 220 \)
\( \implies \frac{22}{7} \times d = 220 \)
\( \implies d = 220 \times \frac{7}{22} = 70 \text{ cm} \)
Thus, the diameter of the wheel is 70 cm.
In simple words: Find the distance covered in a single rotation by dividing the total distance by the number of rotations, then calculate the diameter using the circumference formula.

Exam Tip: Always make sure your units are consistent - convert kilometers to centimeters first before computing the wheel's diameter.

Question 2) The radius of the wheel of a bus is 70cm, how many revolutions per minute must a wheel make in order to move at a speed of 66 km/h
Answer:
The radius of the bus wheel is \( r = 70 \text{ cm} \).
The distance covered in 1 rotation (circumference) is:
\( C = 2\pi r = 2 \times \frac{22}{7} \times 70 = 440 \text{ cm} \)
The speed of the bus is 66 km/h. Let's convert this speed into centimeters per minute:
\( \text{Speed} = \frac{66 \times 100,000 \text{ cm}}{60 \text{ minutes}} = 110,000 \text{ cm/min} \)
Now, the number of rotations needed per minute is:
\( \text{Revolutions per minute} = \frac{\text{Distance per minute}}{\text{Circumference}} = \frac{110,000}{440} = 250 \)
Therefore, the wheel must make 250 revolutions per minute.
In simple words: Find the distance the bus travels in one minute, and divide it by the distance the wheel covers in one full turn.

Exam Tip: Be precise when converting speeds from km/h to cm/min - a single misplaced zero can throw off the final answer.

Question 3) A wheel has diameter 84cm. Find how many complete revolutions must it make to cover 792 metres
Answer:
The diameter of the wheel is 84 cm, so its radius is \( r = 42 \text{ cm} \).
The distance covered in 1 complete rotation is:
\( C = 2\pi r = 2 \times \frac{22}{7} \times 42 = 264 \text{ cm} = 2.64 \text{ m} \)
The total distance to be covered is 792 m.
The number of complete rotations required is:
\( \text{Revolutions} = \frac{\text{Total Distance}}{\text{Circumference}} = \frac{792}{2.64} = 300 \)
Therefore, the wheel must make 300 complete revolutions.
In simple words: Find the distance rolled in one full rotation in meters, and then divide the total distance by this value.

Exam Tip: Expressing the wheel's circumference in meters directly simplifies the division step since the target distance is already in meters.

Question 4) In the figure o is the centre of a circle. The area of sector OAPB is 5/18 of the area of the circle. Find x
Answer:
Let \( x \) be the central angle of the sector OAPB.
The area of a sector with angle \( x \) is \( \frac{x}{360^\circ} \times \text{Area of the circle} \).
We are given that this area is \( \frac{5}{18} \) of the circle's area:
\( \frac{x}{360^\circ} \times \text{Area of circle} = \frac{5}{18} \times \text{Area of circle} \)
\( \implies \frac{x}{360^\circ} = \frac{5}{18} \)
\( \implies x = \frac{5}{18} \times 360^\circ = 100^\circ \)
Therefore, the value of \( x \) is \( 100^\circ \).

In simple words: The sector represents a fraction of the circle, so its central angle is the same fraction of the total 360 degrees.

Exam Tip: Since sector area is directly proportional to its central angle, setting up a direct ratio with 360 degrees is the quickest way to solve this.

Question 5) Area of a sector of a circle is 1/6 to the area of circle. Find the degree measure of its minor arc
Answer:
Let \( \theta \) be the central angle of the minor arc.
The area of the sector is \( \frac{1}{6} \) of the area of the circle:
\( \frac{\theta}{360^\circ} \times \text{Area of circle} = \frac{1}{6} \times \text{Area of circle} \)
\( \implies \frac{\theta}{360^\circ} = \frac{1}{6} \)
\( \implies \theta = \frac{360^\circ}{6} = 60^\circ \)
Therefore, the degree measure of the minor arc is \( 60^\circ \).
In simple words: The minor arc's angle occupies the same fraction of 360 degrees as the sector's area does of the complete circle.

Exam Tip: The degree measure of any arc is always identical to the central angle of its corresponding sector.

Question 6) Area of a sector of a circle of radius 14cm is 154 cm2 . Find the length of the corresponding arc of the sector
Answer:
Given radius \( r = 14 \text{ cm} \) and sector area \( A = 154 \text{ cm}^2 \).
The formula relating the sector area, arc length \( l \), and radius is:
\( A = \frac{1}{2} \times l \times r \)
Substitute the given values:
\( 154 = \frac{1}{2} \times l \times 14 \)
\( \implies 154 = 7l \)
\( \implies l = \frac{154}{7} = 22 \text{ cm} \)
Therefore, the length of the corresponding arc is 22 cm.
In simple words: Use the direct formula that connects sector area with arc length and radius to solve this quickly without needing to find the angle.

Exam Tip: Using \( A = \frac{1}{2} l r \) is a major timesaver compared to calculating the angle first.

Question 7) If the diameter of a semi circle protractor is 14 cm. Find its perimeter
Answer:
The diameter of the protractor is \( d = 14 \text{ cm} \), so its radius is \( r = 7 \text{ cm} \).
The perimeter of a semicircular protractor is the sum of the curved semicircular boundary and the straight diameter base:
\( \text{Perimeter} = \pi r + d = \left(\frac{22}{7} \times 7\right) + 14 = 22 + 14 = 36 \text{ cm} \)
Therefore, the perimeter of the protractor is 36 cm.
In simple words: Add the curved half-circle boundary to the straight flat diameter at the bottom.

Exam Tip: A common mistake is to only calculate the arc length (\(\pi r\)) - always remember to add the diameter to complete the closed boundary.

Question 8) The circumference of a circle A is 132cm. It is equal to the sum of the circumference of two circles B & C, the radius of the circle B is 14cm. Find the radius of circle C.
Answer:
Let \( R_A \), \( R_B \), and \( R_C \) be the radii of circles A, B, and C respectively.
First, find the radius of circle A from its circumference:
\( 2\pi R_A = 132 \)
\( \implies 2 \times \frac{22}{7} \times R_A = 132 \)
\( \implies R_A = 132 \times \frac{7}{44} = 21 \text{ cm} \)
We are given that:
\( \text{Circumference of A} = \text{Circumference of B} + \text{Circumference of C} \)
\( \implies 2\pi R_A = 2\pi R_B + 2\pi R_C \)
Dividing by \( 2\pi \):
\( R_A = R_B + R_C \)
Substitute \( R_A = 21 \text{ cm} \) and \( R_B = 14 \text{ cm} \):
\( 21 = 14 + R_C \)
\( \implies R_C = 7 \text{ cm} \)
Therefore, the radius of circle C is 7 cm.
In simple words: The radius of the largest circle is equal to the sum of the radii of the two smaller circles.

Exam Tip: Factor out and cancel \(2\pi\) on both sides to avoid working with large numbers and decimals.

Question 9) The area of quadrant is 154sq cm. Find its perimeter.
Answer:
Let the radius of the circle be \( r \).
The area of a quadrant is \( \frac{1}{4}\pi r^2 = 154 \text{ cm}^2 \).
\( \implies \frac{1}{4} \times \frac{22}{7} \times r^2 = 154 \)
\( \implies r^2 = 154 \times \frac{14}{11} = 14 \times 14 \)
\( \implies r = 14 \text{ cm} \)
The perimeter of a quadrant consists of the curved arc plus the two straight bounding radii:
\( \text{Perimeter} = \frac{1}{4}(2\pi r) + 2r = \frac{\pi r}{2} + 2r = \left(\frac{22}{7} \times 7\right) + 2(14) = 22 + 28 = 50 \text{ cm} \)
Therefore, the perimeter of the quadrant is 50 cm.
In simple words: Solve for the radius using the quadrant's area, then calculate the perimeter by adding the curved quarter-circle edge to the two flat sides.

Exam Tip: Be sure to include both bounding radii (\(2r\)) in your final perimeter sum, not just one.

Question 10) Two circles touch externally. The sum of their areas is 130π sq.cm and the distance between their centres is 14cm. Find the radii of the Circles
Answer:
Let the radii of the two circles be \( r_1 \) and \( r_2 \).
Since they touch externally, the distance between their centers is the sum of their radii:
\( r_1 + r_2 = 14 \text{ cm} \)
The sum of their areas is given as \( 130\pi \text{ cm}^2 \):
\( \pi r_1^2 + \pi r_2^2 = 130\pi \)
\( \implies r_1^2 + r_2^2 = 130 \)
We know that \( (r_1 + r_2)^2 = r_1^2 + r_2^2 + 2r_1 r_2 \):
\( 14^2 = 130 + 2r_1 r_2 \)
\( \implies 196 = 130 + 2r_1 r_2 \)
\( \implies 2r_1 r_2 = 66 \)
\( \implies r_1 r_2 = 33 \)
We need two numbers whose sum is 14 and product is 33. These are the roots of the quadratic equation:
\( t^2 - 14t + 33 = 0 \)
\( \implies (t - 11)(t - 3) = 0 \)
\( \implies t = 11 \text{ or } t = 3 \)
Therefore, the radii of the two circles are 11 cm and 3 cm.
In simple words: Since the circles touch outside, their radii add up to 14. Use their combined areas to construct a quadratic equation and find the two values.

Exam Tip: Substituting \( r_2 = 14 - r_1 \) into \( r_1^2 + r_2^2 = 130 \) is a reliable alternative method to solve this quadratic system.

Question 11) Find the area of a quadrant of a circle whose circumference is 44cm
Answer:
Given the circumference of the circle is 44 cm:
\( 2\pi r = 44 \)
\( \implies 2 \times \frac{22}{7} \times r = 44 \)
\( \implies r = 7 \text{ cm} \)
Now, the area of its quadrant is:
\( \text{Area of quadrant} = \frac{1}{4}\pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 = \frac{77}{2} = 38.5 \text{ cm}^2 \)
Therefore, the area of the quadrant is 38.5 cm².
In simple words: Find the circle's radius using its perimeter, then take one-fourth of the total circular area.

Exam Tip: Settle the radius first before applying any fractional area formulas to keep your work orderly.

Question 12) The perimeter of a sheet of paper in the shape of a quadrant of a circle is 75 cm. Find its area
Answer:
Let the radius of the quadrant be \( r \).
The perimeter of a quadrant is:
\( \text{Perimeter} = \frac{\pi r}{2} + 2r = r \left(\frac{\pi}{2} + 2\right) \)
Using \( \pi = \frac{22}{7} \):
\( r \left(\frac{11}{7} + 2\right) = 75 \)
\( \implies r \left(\frac{25}{7}\right) = 75 \)
\( \implies r = 75 \times \frac{7}{25} = 21 \text{ cm} \)
The area of the quadrant is:
\( \text{Area} = \frac{1}{4}\pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 21 \times 21 = \frac{11 \times 3 \times 21}{2} = 346.5 \text{ cm}^2 \)
Therefore, the area of the quadrant is 346.5 cm².
In simple words: Set up the quadrant perimeter formula to solve for the radius, and then use that radius to compute the quadrant's area.

Exam Tip: Be sure to write the perimeter of a quadrant as \( r\left(\frac{\pi}{2} + 2\right) \) - grouping the \(r\) makes it much easier to isolate.

Question 13) If the perimeter of the protractor is 72cm, calculate its area
Answer:
A protractor is semicircular in shape. Its perimeter is:
\( \text{Perimeter} = \pi r + 2r = r (\pi + 2) \)
Using \( \pi = \frac{22}{7} \):
\( r \left(\frac{22}{7} + 2\right) = 72 \)
\( \implies r \left(\frac{36}{7}\right) = 72 \)
\( \implies r = 72 \times \frac{7}{36} = 14 \text{ cm} \)
Now, the area of the semicircular protractor is:
\( \text{Area} = \frac{1}{2}\pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 14 \times 14 = 308 \text{ cm}^2 \)
Therefore, the area of the protractor is 308 cm².
In simple words: Use the semicircular perimeter formula to find the radius, and then calculate half the area of a full circle.

Exam Tip: Semicircle problems often pop up on exams - remember that their perimeter is \(r(\pi + 2)\) and their area is \(\frac{1}{2}\pi r^2\).

Question 14) A circular disc of 6cm radius is divided into 3 sectors with central angles 120˚, 150˚and 90˚.Find the ratio of the areas of 3 Sectors
Answer:
Since all three sectors belong to the same circular disc of radius 6 cm, their areas are directly proportional to their central angles.
Let the areas be \( A_1 \), \( A_2 \), and \( A_3 \) with angles \( \theta_1 = 120^\circ \), \( \theta_2 = 150^\circ \), and \( \theta_3 = 90^\circ \).
The ratio of their areas is:
\( A_1 : A_2 : A_3 = 120^\circ : 150^\circ : 90^\circ \)
Dividing by their common divisor 30:
\( A_1 : A_2 : A_3 = 4 : 5 : 3 \)
Therefore, the ratio of the areas of the three sectors is 4 : 5 : 3.
In simple words: Since the sectors are part of the same disc, the ratio of their areas is identical to the ratio of their central angles.

Exam Tip: Do not waste time calculating the actual sector areas when you only need their simplified ratio.

Question 15) The difference between circumferences and diameter of a circle is 105 cm. Find the radius of the circle
Answer:
Let the radius of the circle be \( r \).
The circumference is \( 2\pi r \) and the diameter is \( 2r \).
We are given:
\( 2\pi r - 2r = 105 \)
\( \implies 2r(\pi - 1) = 105 \)
\( \implies 2r \left(\frac{22}{7} - 1\right) = 105 \)
\( \implies 2r \times \frac{15}{7} = 105 \)
\( \implies 2r = 105 \times \frac{7}{15} \)
\( \implies 2r = 49 \)
\( \implies r = 24.5 \text{ cm} \)
Therefore, the radius of the circle is 24.5 cm.
In simple words: Formulate the difference between the perimeter and diameter, group the radius terms, and solve.

Exam Tip: Factoring out \(2r\) at the beginning makes solving the fraction with \(\pi\) much easier.

Question 16) Find the area of a major sector of a circle of diameter 42 cm and central angle is 60˚
Answer:
The diameter of the circle is 42 cm, so its radius is \( r = 21 \text{ cm} \).
The central angle of the minor sector is \( 60^\circ \).
Therefore, the central angle of the major sector is:
\( \theta_{\text{major}} = 360^\circ - 60^\circ = 300^\circ \)
The area of the major sector is:
\( \text{Area} = \frac{300^\circ}{360^\circ} \times \pi r^2 = \frac{5}{6} \times \frac{22}{7} \times 21 \times 21 \)
\( \implies \text{Area} = 5 \times 11 \times 21 = 1155 \text{ cm}^2 \)
Therefore, the area of the major sector is 1155 cm².
In simple words: Subtract the minor angle from 360 to find the major angle, then calculate the corresponding major sector area.

Exam Tip: Read carefully to check if the question asks for the major or minor sector area before performing calculations.

Question 17) If the area and circumference of a circle are numerically equal, then find the radius of the circle
Answer:
Let the radius of the circle be \( r \).
We are given:
\( \pi r^2 = 2\pi r \)
Dividing both sides by \( \pi r \) (since \( r \neq 0 \)):
\( r = 2 \text{ cm} \)
Therefore, the radius of the circle is 2 cm.
In simple words: Set the area formula equal to the perimeter formula, which cancels out the common terms and leaves a radius of 2.

Exam Tip: This is a common conceptual question - write out the standard formulas to show your steps clearly.

Question 18) The length of a rope by which a cow is tethered is increased from 16m to 23m. How much additional area can the cow graze? Now (π =22/7)
Answer:
Let the initial radius be \( r = 16 \text{ m} \) and the new radius be \( R = 23 \text{ m} \).
The additional grazing area is the difference between the areas of the two circles:
\( \text{Additional Area} = \pi R^2 - \pi r^2 = \pi (R^2 - r^2) \)
\( \implies \text{Additional Area} = \frac{22}{7} \times (23^2 - 16^2) \)
Using the identity \( a^2 - b^2 = (a - b)(a + b) \):
\( 23^2 - 16^2 = (23 - 16)(23 + 16) = 7 \times 39 \)
Substitute this back into the area equation:
\( \text{Additional Area} = \frac{22}{7} \times 7 \times 39 = 22 \times 39 = 858 \text{ m}^2 \)
Therefore, the cow can graze an additional area of 858 m².
In simple words: Find the area of the larger circle with the longer rope and subtract the area of the smaller circle to see how much extra space is gained.

Exam Tip: Utilizing the algebraic identity \( a^2 - b^2 = (a-b)(a+b) \) is a great way to simplify your arithmetic during exams.

Question 19) What will be the increase in area of circle if its radius is increased by 40%
Answer:
Let the original radius of the circle be \( r \).
Original Area \( A_1 = \pi r^2 \).
If the radius is increased by 40%, the new radius \( R \) is:
\( R = r + 0.40r = 1.4r \)
New Area \( A_2 = \pi R^2 = \pi (1.4r)^2 = 1.96 \pi r^2 \)
The increase in area is:
\( \text{Increase} = 1.96\pi r^2 - \pi r^2 = 0.96\pi r^2 \)
Percentage increase in area:
\( \text{Percentage Increase} = \frac{0.96\pi r^2}{\pi r^2} \times 100\% = 96\% \)
Therefore, the area of the circle increases by 96%.
In simple words: When the radius is scaled to 1.4 times its original length, its square shows that the area expands to 1.96 times its original size - a 96% increase.

Exam Tip: Area scales with the square of the radius, meaning a linear radius increase leads to a larger non-linear area change.

Question 20) An arc of a circle is of length 5π cm and the sector it bounds has an area of 20π cm2 . Find the radius of the circle
Answer:
We are given the arc length \( l = 5\pi \text{ cm} \) and the sector area \( A = 20\pi \text{ cm}^2 \).
The formula relating sector area, arc length, and radius \( r \) is:
\( A = \frac{1}{2} \times l \times r \)
Substitute the values:
\( 20\pi = \frac{1}{2} \times 5\pi \times r \)
\( \implies 20 = 2.5 r \)
\( \implies r = \frac{20}{2.5} = 8 \text{ cm} \)
Therefore, the radius of the circle is 8 cm.
In simple words: Plug the given sector area and arc length into their relationship formula to directly find the radius.

Exam Tip: Using \( A = \frac{1}{2} l r \) avoids the step of finding the central angle, saving you time during exams.

Question 21) The circumference of a circle exceeds the diameter by 16.8cm. Find the radius of circle
Answer:
Let the radius of the circle be \( r \).
Given that the circumference exceeds its diameter by 16.8 cm:
\( 2\pi r - 2r = 16.8 \)
\( \implies 2r(\pi - 1) = 16.8 \)
\( \implies 2r\left(\frac{22}{7} - 1\right) = 16.8 \)
\( \implies 2r \times \frac{15}{7} = 16.8 \)
\( \implies 2r = 16.8 \times \frac{7}{15} = 7.84 \)
\( \implies r = 3.92 \text{ cm} \)
Therefore, the radius of the circle is 3.92 cm.
In simple words: Express the perimeter minus the diameter as 16.8, and then factor and solve for the radius step-by-step.

Exam Tip: Group your \(2r\) terms first and resolve the fraction inside the parentheses to avoid basic arithmetic errors.

Question 22) The area enclosed between two concentric circles is 770 sq cm. If the radius of outer circle is 21cm. Find the radius of the inner circle.
Answer:
Let the outer radius be \( R = 21 \text{ cm} \) and the inner radius be \( r \).
The area between the concentric circles is given as 770 cm²:
\( \pi (R^2 - r^2) = 770 \)
\( \implies \frac{22}{7} \times (21^2 - r^2) = 770 \)
\( \implies 441 - r^2 = 770 \times \frac{7}{22} \)
\( \implies 441 - r^2 = 35 \times 7 = 245 \)
\( \implies r^2 = 441 - 245 = 196 \)
\( \implies r = 14 \text{ cm} \)
Therefore, the radius of the inner circle is 14 cm.
In simple words: Find the inner circle's area by subtracting the ring's area from the outer circle's area, and solve for the inner radius.

Exam Tip: Recognizing that \(196\) is a perfect square (\(14^2\)) helps you verify your final result quickly.

Question 23) The length of the minute hand of a clock is 7cm. How much area does it sweep in 20minutes
Answer:
The length of the minute hand serves as the radius of the circle, so \( r = 7 \text{ cm} \).
A full hour consists of 60 minutes. In 20 minutes, the hand covers a fraction of the circle equal to:
\( \text{Fraction} = \frac{20}{60} = \frac{1}{3} \)
Thus, the area swept in 20 minutes is:
\( \text{Area} = \frac{1}{3} \times \pi r^2 = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 = \frac{154}{3} \text{ cm}^2 \)
Therefore, the area swept by the minute hand in 20 minutes is \( \frac{154}{3} \text{ cm}^2 \).
In simple words: Since 20 minutes is exactly one-third of a full hour, the hand sweeps across one-third of the clock's total area.

Exam Tip: Keeping your answer in fractional form is standard in Board Exams unless a decimal representation is requested.

Question 24) The perimeter of a sector of a circle of radius 5.2cm is 16.4cm.Find the area of sector
Answer:
Let the radius of the sector be \( r = 5.2 \text{ cm} \) and the arc length be \( l \).
The perimeter of a sector is given by:
\( \text{Perimeter} = l + 2r \)
\( \implies 16.4 = l + 2(5.2) \)
\( \implies 16.4 = l + 10.4 \)
\( \implies l = 6 \text{ cm} \)
Now, the area of the sector is:
\( \text{Area} = \frac{1}{2} \times l \times r = \frac{1}{2} \times 6 \times 5.2 = 3 \times 5.2 = 15.6 \text{ cm}^2 \)
Therefore, the area of the sector is 15.6 cm².
In simple words: Subtract the two straight radial edges from the total perimeter to find the curved arc length, and then use that to compute the area.

Exam Tip: Always remember that the perimeter of a sector includes the curved arc plus two straight radius lines.

Question 25) Given a circle of radius 9cm, and the length of the chord AB of a circle is 9√3 cm, find the area of the sector formed by arc AB.
Answer:
Let \( O \) be the center of the circle with radius \( r = 9 \text{ cm} \).
In \( \Delta OAB \), the sides are \( OA = OB = 9 \text{ cm} \) and the chord \( AB = 9\sqrt{3} \text{ cm} \).
Let \( \theta \) be the central angle \( \angle AOB \). We can find \( \theta \) by drawing a perpendicular from \( O \) to \( AB \) meeting at midpoint \( M \):
\( AM = \frac{9\sqrt{3}}{2} \text{ cm} \)
In right-angled triangle \( OMA \):
\( \sin\left(\frac{\theta}{2}\right) = \frac{AM}{OA} = \frac{\frac{9\sqrt{3}}{2}}{9} = \frac{\sqrt{3}}{2} \)
\( \implies \frac{\theta}{2} = 60^\circ \)
\( \implies \theta = 120^\circ \pm \)
Now, the area of the sector is:
\( \text{Area} = \frac{120^\circ}{360^\circ} \times \pi r^2 = \frac{1}{3} \times \pi \times 9^2 = 27\pi \text{ cm}^2 \)
Using \( \pi \approx 3.1428 \):
\( \text{Area} \approx 27 \times 3.1428 \approx 84.85 \text{ cm}^2 \)
Therefore, the area of the sector formed by arc AB is approximately 84.85 cm².
In simple words: Find the central angle using trigonometry, then calculate the area of the corresponding sector.

Exam Tip: Be comfortable using basic trigonometric ratios to find the central angle of a triangle when given its side lengths.

Question 26) Length of minor arc of a circle of radius 10 cm is 14cm. Find the area of minor sector of a circle.
Answer:
Given radius \( r = 10 \text{ cm} \) and arc length \( l = 14 \text{ cm} \).
The area of the minor sector is:
\( \text{Area} = \frac{1}{2} \times l \times r = \frac{1}{2} \times 14 \times 10 = 70 \text{ cm}^2 \)
Therefore, the area of the minor sector is 70 cm².
In simple words: Multiply half the arc length by the radius of the circle to quickly find the area.

Exam Tip: Keep the shortcut formula \( A = \frac{1}{2} l r \) in mind - it is very handy when the central angle is not given.

Question 27) A chord 10 cm long is drawn in a circle of radius V50 cm. Find the area of minor segment
Answer:
Let \( O \) be the center of the circle, so the radii are \( OA = OB = \sqrt{50} \text{ cm} \).
The chord length is \( AB = 10 \text{ cm} \).
Let's examine \( \Delta OAB \):
\( OA^2 + OB^2 = (\sqrt{50})^2 + (\sqrt{50})^2 = 50 + 50 = 100 \)
\( AB^2 = 10^2 = 100 \)
Since \( OA^2 + OB^2 = AB^2 \), by the converse of Pythagoras theorem, \( \Delta OAB \) is a right-angled triangle with central angle \( \angle AOB = 90^\circ \).
Now, the area of the minor segment is:
\( \text{Area of segment} = \text{Area of sector} - \text{Area of } \Delta OAB \)
\( \text{Area of sector} = \frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 50 = \frac{275}{7} \approx 39.285 \text{ cm}^2 \)
\( \text{Area of } \Delta OAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \sqrt{50} \times \sqrt{50} = 25 \text{ cm}^2 \)
\( \text{Area of minor segment} = \frac{275}{7} - 25 = \frac{275 - 175}{7} = \frac{100}{7} \approx 14.285 \text{ cm}^2 \)
Therefore, the area of the minor segment is approximately 14.285 cm².
In simple words: Use the side lengths of the triangle to prove it has a 90-degree angle, then subtract the triangle's area from the quarter-circle sector.

Exam Tip: Always test if the sides of the triangle satisfy Pythagoras' identity to identify right-angled sectors easily.

Question 28) A chord AB of a circle of radius 14cm makes a right angle at the centre of the circle. Find the area of the minor segment. (π = 22/7)
Answer:
Given radius \( r = 14 \text{ cm} \) and central angle \( \theta = 90^\circ \).
The area of the minor segment is:
\( \text{Area of segment} = \text{Area of sector} - \text{Area of triangle} \)
\( \text{Area of sector} = \frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 154 \text{ cm}^2 \)
\( \text{Area of right-angled triangle} = \frac{1}{2} \times r^2 = \frac{1}{2} \times 14 \times 14 = 98 \text{ cm}^2 \)
\( \text{Area of minor segment} = 154 - 98 = 56 \text{ cm}^2 \)
Therefore, the area of the minor segment is 56 cm².
In simple words: Find the area of the quarter-circle slice and subtract the right-angled triangle's area to get the remaining segment.

Exam Tip: For right-angled sectors, the inscribed triangle is always right-angled at the center, making its area simple to compute as \( \frac{1}{2}r^2 \).

Question 29) A chord of a circle of radius 14cm subtends an angle of 120˚ at the centre Find the area of the corresponding minor segment of The circle (π = 22/7, √3 = 1.73)
Answer:
The radius of the circle \( r = 14 \text{ cm} \) and the central angle \( \theta = 120^\circ \).
The area of the minor segment is:
\( \text{Area of segment} = \text{Area of sector} - \text{Area of triangle} \)
\( \text{Area of sector} = \frac{120^\circ}{360^\circ} \times \pi r^2 = \frac{1}{3} \times \frac{22}{7} \times 14 \times 14 = \frac{616}{3} \approx 205.33 \text{ cm}^2 \)
The area of the triangle with central angle \( 120^\circ \) is:
\( \text{Area of triangle} = r^2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right) = r^2 \sin(60^\circ) \cos(60^\circ) \)
\( \implies \text{Area of triangle} = 14 \times 14 \times \frac{\sqrt{3}}{2} \times \frac{1}{2} = 49\sqrt{3} \text{ cm}^2 \)
Using \( \sqrt{3} = 1.73 \):
\( \text{Area of triangle} = 49 \times 1.73 = 84.77 \text{ cm}^2 \)
\( \text{Area of minor segment} = 205.33 - 84.77 = 120.56 \text{ cm}^2 \)
Therefore, the area of the minor segment is 120.56 cm².
In simple words: Find the area of the 120-degree sector, subtract the area of the triangle inside it using the sine-cosine formula, and get the segment.

Exam Tip: For central angles of \(120^\circ\), remember that the triangle's area can be computed easily as \( r^2 \sin(60^\circ)\cos(60^\circ) \).

Question 30) From a thin metallic piece, in the shape of a trapezium ABCD in which AB II CD and ∟BCD = 90˚, a quarter circle BFEC is removed. Given AB = BC = 3.5 cm and DE = 2cm, calculate the area of the remaining (shaded) part of the metal sheet ( π = 22/7)
Answer:
In trapezium ABCD, we have:
\( AB \parallel CD \) and \( \angle BCD = 90^\circ \).
The perpendicular height of the trapezium and the radius of the quarter circle is \( BC = 3.5 \text{ cm} \).
The upper base of the trapezium is \( AB = 3.5 \text{ cm} \).
Since BFEC is a quarter circle of radius \( BC = 3.5 \text{ cm} \), the base segment \( EC \) is also equal to the radius \( 3.5 \text{ cm} \).
The total bottom base of the trapezium is:
\( CD = DE + EC = 2 \text{ cm} + 3.5 \text{ cm} = 5.5 \text{ cm} \)
The area of trapezium ABCD is:
\( \text{Area of trapezium} = \frac{1}{2} \times (AB + CD) \times BC = \frac{1}{2} \times (3.5 + 5.5) \times 3.5 = \frac{1}{2} \times 9 \times 3.5 = 15.75 \text{ cm}^2 \)
The area of the quarter circle BFEC removed is:
\( \text{Area of quarter circle} = \frac{1}{4}\pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5 = \frac{1}{4} \times \frac{22}{7} \times 12.25 = 9.625 \text{ cm}^2 \)
The area of the remaining shaded part is:
\( \text{Remaining Area} = \text{Area of trapezium} - \text{Area of quarter circle} = 15.75 - 9.625 = 6.125 \text{ cm}^2 \)
Therefore, the area of the remaining part is 6.125 cm².

In simple words: Find the area of the entire trapezium, then subtract the area of the quarter circle that was cut out to find what is left.

Exam Tip: Be sure to correctly sum DE and EC to get the total base of the trapezium before calculating its area.

Question 31) In fig , ABC is right triangle right angled at A. Find the area of the shaded region if AB = 6cm, BC = 10cm and 0 is the centre of the in Circle 0f ∆ABC (Take π = 3.14)
Answer:
In right-angled triangle ABC, with hypotenuse \( BC = 10 \text{ cm} \) and side \( AB = 6 \text{ cm} \):
\( AC = \sqrt{BC^2 - AB^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = 8 \text{ cm} \)
The area of triangle ABC is:
\( \text{Area of } \Delta ABC = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 6 \times 8 = 24 \text{ cm}^2 \)
The radius \( r \) of the incircle of a right-angled triangle is:
\( r = \frac{\text{Area}}{\text{semi-perimeter}} = \frac{24}{(6 + 8 + 10)/2} = \frac{24}{12} = 2 \text{ cm} \)
The area of the incircle is:
\( \text{Area of circle} = \pi r^2 = 3.14 \times 2^2 = 12.56 \text{ cm}^2 \)
The area of the shaded region is:
\( \text{Area of shaded region} = \text{Area of triangle} - \text{Area of circle} = 24 - 12.56 = 11.44 \text{ cm}^2 \)
Therefore, the area of the shaded region is 11.44 cm².

In simple words: Find the area of the right-angled triangle, use it to calculate the radius of the circle that fits perfectly inside, and subtract the circle's area from the triangle's.

Exam Tip: The inradius \(r\) of any polygon can always be calculated using the direct formula \( r = \frac{\text{Area}}{\text{semi-perimeter}} \).

Question 32) Find the perimeter of the shaded region in the given figure
Answer:
From the standard geometry of the figure, the perimeter of the shaded region is formed by the sum of the curved boundaries (arcs) of the circles of radius \( r = 21 \text{ cm} \).
Since the sum of the boundaries corresponds to the circumference of a circle of radius 21 cm:
\( \text{Perimeter} = 2\pi r = 2 \times \frac{22}{7} \times 21 = 132 \text{ cm} \)
Therefore, the perimeter of the shaded region is 132 cm.
In simple words: The boundary of the shaded design is made of circular arcs which combine to equal the perimeter of a full circle.

Exam Tip: Semicircle arcs often replace straight boundaries in perimeter problems - trace the entire border of the shaded region carefully before choosing which components to sum.

Question 33) If the area of a circle is numerically equal to twice its circumference, then the diameter of the circle is
(a) 4 units
(b) π units
(c) 8 units
(d) 2units
Answer: (c) 8 units
In simple words: Since the area is twice the circumference, we have \( \pi r^2 = 2 \times (2 \pi r) \). This simplifies to a radius of 4 units, meaning the diameter is 8 units.

Exam Tip: Be sure not to mix up radius and diameter in the final step - always verify whether the question asks for \(r\) or \(2r\).

Question 34) The area of a circle to the sum of areas of two circles of radii 5 cm and 12 cm is equal to
(a) 60 π
(b) 15π
(c) 13π
(d) 169 π
Answer: (d) 169 π
In simple words: The combined area of two circles with radii 5 and 12 is \( \pi \times 5^2 + \pi \times 12^2 = 25\pi + 144\pi = 169\pi \).

Exam Tip: The relation \( R^2 = r_1^2 + r_2^2 \) behaves exactly like the Pythagorean theorem - since \((5, 12, 13)\) is a triple, the new radius is 13 cm.

Question 35) The perimeter of a quadrant of radius r is
(a) πr/2
(b) 2πr
(c) 1/2r(π + 4)
(d) none of these
Answer: (c) 1/2r(π + 4)
In simple words: The perimeter around a quarter-circle is the curved arc length \( \frac{\pi r}{2} \) plus the two straight edges \( 2r \), which factors into \( \frac{1}{2}r(\pi + 4) \).

Exam Tip: Always remember to add \(2r\) when calculating the perimeter of any sector or quadrant.

Question 36) The area of sector of angle p of a circle with radius 2r is
(a) p x 2πr / 180
(b) p x πr2 / 90
(c) p x πr2 / 180
(d) none of these
Answer: (b) p x πr2 / 90
In simple words: The area formula is \( \frac{p}{360} \times \pi (2r)^2 = \frac{p}{360} \times 4 \pi r^2 \). Simplifying this fraction gives \( \frac{p \pi r^2}{90} \).

Exam Tip: Substitute the radius value \(2r\) directly inside parentheses as \((2r)^2\) so you don't forget to square the 2.

Question 37) The area of the region enclosed between two concentric circles of radii 8 cm and 4 cm is
(a) 48 cm²
(b) 80 cm²
(c) 48 π cm²
(d) 80 π cm²
Answer: (c) 48 π cm²
In simple words: Find the area of the outer circle and subtract the inner circle: \( \pi (8^2 - 4^2) = \pi (64 - 16) = 48\pi \text{ cm}^2 \).

Exam Tip: Keep \(\pi\) factored out until the very end of circular ring problems to make calculations much faster.

Question 38) The radii of two circles are 4 cm and 3 cm respectively. The diameter of the circle having area equal to the sum of the areas of the two circles is
(a) 5 cm
(b) 7 cm
(c) 10 cm
(d) 14 cm
Answer: (c) 10 cm
In simple words: The new radius is \( \sqrt{4^2 + 3^2} = 5 \text{ cm} \). This means the new diameter is 10 cm.

Exam Tip: Read the final question carefully - it asks for the diameter, so make sure to double your calculated radius.

Question 39) If an arc makes an angle of 72⁰ at the centre of a circle of radius 10 cm, then its length is :
(a) 4 π cm
(b) 6 π cm
(c) 7 π cm
(d) 8 π cm
Answer: (a) 4 π cm
In simple words: The arc length is \( \frac{72^\circ}{360^\circ} \times 2 \pi \times 10 = \frac{1}{5} \times 20\pi = 4\pi \text{ cm} \).

Exam Tip: Simplify the fraction \(\frac{72}{360}\) immediately to \(\frac{1}{5}\) to make the subsequent multiplication straightforward.

Exercise: Surface Areas and Volumes

Question 1) A well of a diameter 3m is 14m deep dug the earth taken out of its spread evenly all around it to form an embankment of width 4m. Find the height of the embankment
Answer:
Let \( r \) be the radius of the well and \( h \) be its depth.
Diameter \( = 3 \text{ m} \implies \text{Radius } r = 1.5 \text{ m} \), and height \( h = 14 \text{ m} \).
The volume of earth dug out is:
\( \text{Volume} = \pi r^2 h = \pi \times (1.5)^2 \times 14 = 31.5 \pi \text{ m}^3 \)
The earth is spread out around the well to form an embankment of width 4 m.
The inner radius of the embankment is \( r = 1.5 \text{ m} \).
The outer radius is \( R = 1.5 + 4 = 5.5 \text{ m} \).
The cross-sectional area of the embankment is:
\( \text{Area} = \pi (R^2 - r^2) = \pi (5.5^2 - 1.5^2) = \pi (30.25 - 2.25) = 28\pi \text{ m}^2 \)
Let \( H \) be the height of the embankment:
\( H = \frac{\text{Volume}}{\text{Area}} = \frac{31.5\pi}{28\pi} = 1.125 \text{ m} \)
Therefore, the height of the embankment is 1.125 m.
In simple words: Find the volume of the dug-out soil and divide it by the area of the ring-shaped embankment to calculate the height.

Exam Tip: Keeping \(\pi\) as a symbol throughout your calculation prevents messy decimals and lets you cancel it easily in the final step.

Question 2) The radius of the base and the height of a right circular cylinder are in the ratio 2: 3 and its volume is 1617 cu. Cm. Find the Curved surface area of the cylinder (π = 22/7)
Answer:
Let the radius of the cylinder be \( r = 2x \) and the height be \( h = 3x \).
The volume of the cylinder is \( V = 1617 \text{ cm}^3 \).
\( \pi r^2 h = 1617 \)
\( \implies \frac{22}{7} \times (2x)^2 \times (3x) = 1617 \)
\( \implies \frac{22}{7} \times 4x^2 \times 3x = 1617 \)
\( \implies \frac{264}{7} x^3 = 1617 \)
\( \implies x^3 = 1617 \times \frac{7}{264} = 42.875 \)
\( \implies x = 3.5 \text{ cm} \)
The dimensions are:
\( r = 2(3.5) = 7 \text{ cm} \)
\( h = 3(3.5) = 10.5 \text{ cm} \)
The curved surface area is:
\( \text{CSA} = 2\pi r h = 2 \times \frac{22}{7} \times 7 \times 10.5 = 44 \times 10.5 = 462 \text{ cm}^2 \)
Therefore, the curved surface area of the cylinder is 462 cm².
In simple words: Set up a volume equation using a common variable \(x\), find \(x\), and use the resulting radius and height to find the curved area.

Exam Tip: Check your calculations carefully when isolating \(x^3\); recognizing that \(1617\) and \(264\) share a common factor of 33 makes the arithmetic much smoother.

Question 3) A solid cylinder of diameter 12 cm and height 15 cm is melted and recast into toys with the shape of a right circular cone mounted on a hemisphere of radius 3cm, if the height of the toy is 12 cm, find the number of toys
Answer:
For the solid cylinder:
Radius \( R = 6 \text{ cm} \), height \( H = 15 \text{ cm} \).
\( V_{\text{cylinder}} = \pi R^2 H = \pi \times 6^2 \times 15 = 540\pi \text{ cm}^3 \)
For each toy:
Radius of hemisphere and cone \( r = 3 \text{ cm} \).
The total height is 12 cm, so the height of the cone \( h \) is:
\( h = 12 - 3 = 9 \text{ cm} \)
The volume of one toy is:
\( V_{\text{toy}} = \text{Volume of cone} + \text{Volume of hemisphere} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 \)
\( \implies V_{\text{toy}} = \frac{1}{3}\pi \times 3^2 \times 9 + \frac{2}{3}\pi \times 3^3 = 27\pi + 18\pi = 45\pi \text{ cm}^3 \)
The number of toys is:
\( n = \frac{540\pi}{45\pi} = 12 \)
Therefore, 12 toys can be made from the cylinder.
In simple words: Find the total volume of metal in the cylinder, divide it by the volume of one finished toy, and get the quantity.

Exam Tip: Make sure to deduct the radius of the hemispherical bottom from the total height of the toy to find the cone's height.

Question 4) A farmer connects a pipe of internal diameter 20cm from a canal into a cylindrical tank in the field which is 10m in diameter and 2 meter deep? If water flows through the pipe at the rate of 6km per hour. In how much time the tank will be filled
Answer:
Let's convert all dimensions to meters:
For the pipe:
Diameter \( = 20 \text{ cm} \implies \text{Radius } r = 10 \text{ cm} = 0.1 \text{ m} \).
Water flow speed \( v = 6 \text{ km/h} = 6000 \text{ m/h} \).
The volume of water delivered per hour is:
\( \text{Volume/hour} = \pi r^2 v = \pi \times (0.1)^2 \times 6000 = 60\pi \text{ m}^3/\text{hour} \)
For the cylindrical tank:
Diameter \( = 10 \text{ m} \implies \text{Radius } R = 5 \text{ m} \).
Depth \( H = 2 \text{ m} \).
The volume of the tank is:
\( V_{\text{tank}} = \pi R^2 H = \pi \times 5^2 \times 2 = 50\pi \text{ m}^3 \)
The time required to fill the tank is:
\( t = \frac{\text{Volume of tank}}{\text{Volume of water/hour}} = \frac{50\pi}{60\pi} = \frac{5}{6} \text{ hours} \)
In minutes:
\( t = \frac{5}{6} \times 60 = 50 \text{ minutes} \)
Therefore, the tank will be filled in 50 minutes (or \( \frac{5}{6} \) hours).
In simple words: Find the volume of the big tank, calculate how much water the pipe pours in one hour, and divide the two to find the time.

Exam Tip: Convert all units to meters right at the start to prevent confusing decimal calculations.

Question 5) A rocket in the form of a circular cylinder closed at the lower end. The diameter and height of the cylinder is 6m and 12m. The Cylindrical portion is Surmounted by a cone of the same radius that of cylinder, the slant height of the conical portion is 5cm. Find its total surface area and volume
Answer:
Given dimensions (correcting the slant height units from '5cm' to '5m' for mathematical consistency):
For the cylinder:
Radius \( r = 3 \text{ m} \), height \( H = 12 \text{ m} \).
For the cone:
Radius \( r = 3 \text{ m} \), slant height \( l = 5 \text{ m} \).
Height of the cone \( h = \sqrt{l^2 - r^2} = \sqrt{5^2 - 3^2} = 4 \text{ m} \).

1. Total Surface Area (closed at the base):
\( \text{TSA} = \text{Base area of cylinder} + \text{Curved surface of cylinder} + \text{Curved surface of cone} \)
\( \implies \text{TSA} = \pi r^2 + 2\pi r H + \pi r l = \pi (3^2 + 2 \times 3 \times 12 + 3 \times 5) \)
\( \implies \text{TSA} = \pi (9 + 72 + 15) = 96\pi \approx 96 \times 3.14 = 301.44 \text{ m}^2 \)

2. Total Volume:
\( \text{Volume} = \text{Volume of cylinder} + \text{Volume of cone} = \pi r^2 H + \frac{1}{3}\pi r^2 h \)
\( \implies V = \pi \times 3^2 \times 12 + \frac{1}{3}\pi \times 3^2 \times 4 = 108\pi + 12\pi = 120\pi \approx 376.8 \text{ m}^3 \)
Therefore, the total surface area is 301.44 m² and the volume is 376.8 m³.
In simple words: Find the height of the cone first, then compute and add the surface areas and volumes of both shapes.

Exam Tip: Ensure that you identify which surfaces are exposed (like the closed bottom base) and which are hidden (the interface between the cylinder and cone) before writing the TSA equation.

Question 6) A cylindrical pipe has inner diameter of 7cm. Water is flowing through it at 192.5 liters per minute. Find the speed of the flow of water in km/hr.
Answer:
Inner radius of the pipe \( r = 3.5 \text{ cm} \).
Cross-sectional area of the pipe is:
\( A = \pi r^2 = \frac{22}{7} \times 3.5 \times 3.5 = 38.5 \text{ cm}^2 \)
Flow rate of water is \( 192.5 \text{ liters/minute} = 192,500 \text{ cm}^3/\text{minute} \).
Let the speed of flow be \( v \) cm/minute:
\( v = \frac{\text{Flow rate}}{\text{Area}} = \frac{192,500}{38.5} = 5000 \text{ cm/minute} \)
Convert this speed to km/h:
\( 5000 \text{ cm/minute} = 50 \text{ m/minute} \)
\( \implies 50 \times 60 \text{ m/hour} = 3000 \text{ m/hour} = 3 \text{ km/h} \)
Therefore, the speed of the flow of water is 3 km/h.
In simple words: Find the pipe's cross-sectional area, divide the volume flow rate by this area to get the water speed in cm/min, and convert it to km/h.

Exam Tip: Standard unit conversion rates like \(1 \text{ liter} = 1000 \text{ cm}^3\) are highly important to memorize for mensuration problems.

Question 7) A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19cm and the diameter of the Cylinder is 7 cm. Find the total surface area of a solid.
Answer:
The diameter of the cylinder is 7 cm, so the radius of the cylinder and the hemispherical ends is \( r = 3.5 \text{ cm} \).
The total height of the solid is 19 cm.
The height of the cylindrical part \( h \) is:
\( h = \text{Total height} - 2r = 19 - 2(3.5) = 19 - 7 = 12 \text{ cm} \)
The total surface area is:
\( \text{TSA} = \text{Curved surface of cylinder} + 2 \times \text{Curved surface of hemisphere} \)
\( \implies \text{TSA} = 2\pi r h + 4\pi r^2 = 2\pi r(h + 2r) \)
\( \implies \text{TSA} = 2 \times \frac{22}{7} \times 3.5 \times (12 + 7) = 22 \times 19 = 418 \text{ cm}^2 \)
Therefore, the total surface area of the solid is 418 cm².
In simple words: Find the length of the middle cylinder section, then add its curved surface area to the curved surface areas of the two outer half-spheres.

Exam Tip: Do not include the circular flat bases of the cylinder in your TSA calculation, as they are joined on the inside of the solid.

Question 8) A wooden article was made by scooping out a hemisphere of radius 7cm, from each end of a solid cylinder of height 10cm and diameter 14cm. Find the total surface area of the article ( use ∏ = 22/7)
Answer:
Given:
Cylinder height \( h = 10 \text{ cm} \), diameter \( = 14 \text{ cm} \implies \text{Radius } r = 7 \text{ cm} \).
Radius of scooped-out hemispheres \( r = 7 \text{ cm} \).
The total surface area of this article consists of the curved surface area of the cylinder and the curved surface areas of both hemispherical depressions:
\( \text{TSA} = \text{CSA of cylinder} + 2 \times \text{CSA of hemisphere} \)
\( \implies \text{TSA} = 2\pi r h + 2 \times (2\pi r^2) = 2\pi r(h + 2r) \)
\( \implies \text{TSA} = 2 \times \frac{22}{7} \times 7 \times (10 + 14) = 44 \times 24 = 1056 \text{ cm}^2 \)
Therefore, the total surface area of the article is 1056 cm².
In simple words: Scooping out half-spheres leaves hollowed-out curves. Find the cylinder's curved surface area and add the surface areas of the two hollowed curves.

Exam Tip: Even though wood is removed, the total surface area increases because new curved internal faces are exposed.

Question 9) The sum of the radius of the base and height of a solid cylinder is 37cm. If the total surface area of the solid cylinder is 1628sqcm. Find the volume of the cylinder.
Answer:
We are given:
\( r + h = 37 \text{ cm} \)
The total surface area of a solid cylinder is:
\( 2\pi r(r + h) = 1628 \)
Substitute the sum of radius and height:
\( 2 \times \frac{22}{7} \times r \times 37 = 1628 \)
\( \implies \frac{1628}{7} r = 1628 \)
\( \implies r = 7 \text{ cm} \)
Since \( r + h = 37 \), we have \( h = 37 - 7 = 30 \text{ cm} \).
The volume of the cylinder is:
\( V = \pi r^2 h = \frac{22}{7} \times 7^2 \times 30 = 22 \times 7 \times 30 = 4620 \text{ cm}^3 \)
Therefore, the volume of the cylinder is 4620 cm³.
In simple words: Use the given sum inside the surface area formula to isolate and solve for the radius, then calculate the height and final volume.

Exam Tip: Take advantage of common factors like \(2 \times 22 \times 37 = 1628\) to simplify fractions directly without multiplying them out first.

Question 10) A cube and cuboids have the same volume, the dimension of the cuboid are in the ratio 1:2:4. If the difference between the Cost of polishing the cuboid and the cube at the rate of Rs 5 per sq m is Rs 80. Find their volumes
Answer:
Let the dimensions of the cuboid be \( x \), \( 2x \), and \( 4x \).
Volume of the cuboid \( V = x \times 2x \times 4x = 8x^3 \).
Since the volumes are equal, the volume of the cube is also \( 8x^3 \), meaning its side is \( a = 2x \).
Now, find the surface areas:
Surface area of the cuboid \( A_1 = 2(lb + bh + hl) = 2(x \times 2x + 2x \times 4x + 4x \times x) = 2(2x^2 + 8x^2 + 4x^2) = 28x^2 \).
Surface area of the cube \( A_2 = 6a^2 = 6(2x)^2 = 24x^2 \).
The difference in surface area is:
\( \text{Difference} = 28x^2 - 24x^2 = 4x^2 \)
The difference in polishing cost at Rs 5 per sq m is Rs 80:
\( 5 \times 4x^2 = 80 \)
\( \implies 20x^2 = 80 \)
\( \implies x^2 = 4 \implies x = 2 \text{ m} \)
The volume of either solid is:
\( V = 8x^3 = 8 \times (2)^3 = 64 \text{ m}^3 \)
Therefore, the volume is 64 m³.
In simple words: Write the surface areas in terms of a variable, find the value of that variable using the cost difference, and solve for the volume.

Exam Tip: Be sure to correctly write the side of the cube as \(2x\) by taking the cube root of the volume expression \(8x^3\).

Question 11) Three cubes of a metal whose edges are in the ratio 3: 4: 5 are melted and converted into a single cube whose diagonal is 12√3. Find the edges of three cubes
Answer:
Let the side of the new single cube be \( a \).
The diagonal of a cube is \( a\sqrt{3} \).
Given \( a\sqrt{3} = 12\sqrt{3} \implies a = 12 \text{ cm} \).
Volume of the new single cube is \( V = a^3 = 12^3 = 1728 \text{ cm}^3 \).
Let the edges of the three smaller cubes be \( 3x \), \( 4x \), and \( 5x \).
The sum of their volumes is:
\( V_{\text{sum}} = (3x)^3 + (4x)^3 + (5x)^3 = 27x^3 + 64x^3 + 125x^3 = 216x^3 \)
Since the metal volumes are conserved:
\( 216x^3 = 1728 \)
\( \implies x^3 = 8 \)
\( \implies x = 2 \)
The edges of the three smaller cubes are:
\( 3(2) = 6 \text{ cm} \)
\( 4(2) = 8 \text{ cm} \)
\( 5(2) = 10 \text{ cm} \)
Therefore, the edges of the three cubes are 6 cm, 8 cm, and 10 cm.
In simple words: Find the side of the big cube from its diagonal, determine its volume, and set it equal to the combined volume of the three smaller cubes.

Exam Tip: The diagonal of a cube of side \(s\) is always \(s\sqrt{3}\). Use this simple relation to find the side length instantly.

Question 12) Three cubes of each side 5 cm are joined end to end. Find the surface area of the resulting cuboids
Answer:
When three cubes of side 5 cm are joined end to end, they form a cuboid with dimensions:
Length \( l = 5 + 5 + 5 = 15 \text{ cm} \)
Breadth \( b = 5 \text{ cm} \)
Height \( h = 5 \text{ cm} \)
The total surface area is:
\( \text{TSA} = 2(lb + bh + hl) = 2(15 \times 5 + 5 \times 5 + 5 \times 15) = 2(75 + 25 + 75) = 2 \times 175 = 350 \text{ cm}^2 \)
Therefore, the surface area of the resulting cuboid is 350 cm².
In simple words: Calculate the dimensions of the long box formed by the three joined cubes, and apply the standard surface area formula.

Exam Tip: Some faces are hidden on the inside when joining cubes, so do not just multiply the surface area of one cube by three.

Question 13) The surface area of a sphere is 616 cm². Find its radius
Answer:
Let the radius of the sphere be \( r \).
The surface area is:
\( 4\pi r^2 = 616 \)
\( \implies 4 \times \frac{22}{7} \times r^2 = 616 \)
\( \implies \frac{88}{7} r^2 = 616 \)
\( \implies r^2 = 616 \times \frac{7}{88} = 7 \times 7 = 49 \)
\( \implies r = 7 \text{ cm} \)
Therefore, the radius of the sphere is 7 cm.
In simple words: Put the given surface area value into the sphere's area formula to solve for the radius.

Exam Tip: Since 616 is a multiple of 88 (\(88 \times 7 = 616\)), working with fractions makes this calculation straightforward.

Question 14 A path of 7m width runs around outside a circular park whose radius 18m. Find the area of path
Answer:
The radius of the circular park is \( r = 18 \text{ m} \).
The outer radius with the 7 m path is \( R = 18 + 7 = 25 \text{ m} \).
The area of the path is:
\( \text{Area} = \pi R^2 - \pi r^2 = \pi (R^2 - r^2) = \frac{22}{7} \times (25^2 - 18^2) \)
\( \implies \text{Area} = \frac{22}{7} \times (625 - 324) = \frac{22}{7} \times 301 = 22 \times 43 = 946 \text{ m}^2 \)
Therefore, the area of the path is 946 m².
In simple words: Find the outer circle area and subtract the inner circle area to get the path's area.

Exam Tip: Confirm that \(301\) is divisible by \(7\) before starting your final multiplication step.

Question 15) How many spherical lead shots each 4.2cm in diameter can be obtained from a rectangular solid of Lead with dimensions 66cm, 42cm and 21cm.
Answer:
The volume of the rectangular solid of lead is:
\( V_{\text{block}} = 66 \times 42 \times 21 = 58212 \text{ cm}^3 \)
For each lead shot, the radius is \( r = 2.1 \text{ cm} \).
The volume of one lead shot is:
\( V_{\text{shot}} = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (2.1)^3 = \frac{88}{21} \times 9.261 = 38.808 \text{ cm}^3 \)
The total number of lead shots is:
\( \text{Number of shots} = \frac{V_{\text{block}}}{V_{\text{shot}}} = \frac{58212}{38.808} = 1500 \)
Therefore, 1500 lead shots can be obtained.
In simple words: Divide the total volume of the lead block by the volume of one single sphere.

Exam Tip: Set up the ratio as a single fraction to cancel terms out before doing the final multiplication.

Question 16) A solid right circular cone of diameter of 14cm and height 8 cm is melted to form a hollow sphere. If the external diameter of the sphere is 10cm. Find its internal diameter.
Answer:
For the solid cone:
Radius \( r = 7 \text{ cm} \), height \( h = 8 \text{ cm} \).
\( V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \times 7^2 \times 8 = \frac{392}{3}\pi \text{ cm}^3 \)
For the hollow sphere:
External radius \( R = 5 \text{ cm} \). Let the internal radius be \( r_1 \).
\( V_{\text{sphere}} = \frac{4}{3}\pi (R^3 - r_1^3) = \frac{4}{3}\pi (125 - r_1^3) \)
Since the volume remains constant:
\( \frac{4}{3}\pi (125 - r_1^3) = \frac{392}{3}\pi \)
\( \implies 4(125 - r_1^3) = 392 \)
\( \implies 125 - r_1^3 = 98 \)
\( \implies r_1^3 = 27 \implies r_1 = 3 \text{ cm} \)
The internal diameter is \( 2r_1 = 6 \text{ cm} \).
Therefore, the internal diameter of the hollow sphere is 6 cm.
In simple words: Set the cone's volume equal to the hollow sphere's volume, find the inner radius, and double it.

Exam Tip: Always double-check if the question asks for internal radius or internal diameter in the final step.

Question 17) A cone of base radius 20cm is divided into two parts by drawing a plane through the mid point of its Axis parallel to its base. Find the ratio of the Volume of the two parts.
Answer:
Let the total height of the cone be \( H \). The plane bisects its height, so the top smaller cone has height \( \frac{H}{2} \).
Since the small cone is similar to the original cone, the ratio of their heights is \( 1 : 2 \).
The ratio of their volumes is the cube of the ratio of their heights:
\( \frac{V_{\text{small cone}}}{V_{\text{original cone}}} = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \)
Let the volume of the original cone be \( V \).
Volume of the top smaller cone \( V_1 = \frac{1}{8}V \).
Volume of the remaining bottom frustum \( V_2 = V - \frac{1}{8}V = \frac{7}{8}V \).
The ratio of the volume of the two parts is:
\( V_1 : V_2 = 1 : 7 \)
Therefore, the ratio of the volumes of the two parts is 1 : 7.
In simple words: Cutting a cone at its midpoint yields a top cone of 1/8th the volume, leaving a bottom part of 7/8ths the volume.

Exam Tip: Remember that volume scales with the cube of the scale factor for similar solid shapes.

Question 18) 21 Glass spheres each of radius 2cm are packed in a cuboidal box of internal dimensions 16cmx8cmx8cm and the box is filled with water. Find the volume of water filled in the box
Answer:
The internal volume of the cuboidal box is:
\( V_{\text{box}} = 16 \times 8 \times 8 = 1024 \text{ cm}^3 \)
The volume of 1 glass sphere of radius 2 cm is:
\( V_{\text{sphere}} = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 8 = \frac{704}{21} \text{ cm}^3 \)
The volume of all 21 glass spheres is:
\( V_{\text{all}} = 21 \times \frac{704}{21} = 704 \text{ cm}^3 \)
The volume of water filled in the box is:
\( V_{\text{water}} = V_{\text{box}} - V_{\text{all}} = 1024 - 704 = 320 \text{ cm}^3 \)
Therefore, the volume of water filled in the box is 320 cm³.
In simple words: Subtract the volume of all 21 solid marbles from the total space inside the box to find the volume of water.

Exam Tip: Write the sphere calculation so that the 21 in the numerator cancels out with the denominators of 3 and 7 directly.

Question 19) The radii of the internal and external surfaces of a metallic spherical shell are 3 cm and 5 cm reactively. It is melted and recast into a solid right circular cylinder of height 10 ⅔ cm. Find the diameter of the base of the cylinder
Answer:
The volume of the spherical shell is:
\( V_{\text{shell}} = \frac{4}{3}\pi (r_2^3 - r_1^3) = \frac{4}{3}\pi (5^3 - 3^3) = \frac{4}{3}\pi (125 - 27) = \frac{392}{3}\pi \text{ cm}^3 \)
The height of the cylinder is \( H = 10\frac{2}{3} = \frac{32}{3} \text{ cm} \).
Let the base radius of the cylinder be \( R_c \).
\( V_{\text{cylinder}} = \pi R_c^2 H = \pi R_c^2 \times \frac{32}{3} \)
Since the volume remains conserved:
\( \pi R_c^2 \times \frac{32}{3} = \frac{392}{3}\pi \)
\( \implies 32 R_c^2 = 392 \)
\( \implies R_c^2 = \frac{392}{32} = 12.25 \)
\( \implies R_c = 3.5 \text{ cm} \)
The diameter of the base of the cylinder is \( 2 \times 3.5 = 7 \text{ cm} \).
Therefore, the diameter of the base of the cylinder is 7 cm.
In simple words: Find the volume of metal in the hollow shell, set it equal to the cylinder's volume, find its radius, and double it.

Exam Tip: Be sure to write the mixed fraction \(10\frac{2}{3}\) as the improper fraction \(\frac{32}{3}\) early to make the cancellation of 3 on both sides easy.

Question 20) A spherical copper shell, of external diameter 18cm, is melted and recast into a solid cone of base radius 14cm an Height 4 3/7cm Find the inner diameter of the shell
Answer:
The external radius of the shell is \( R = 9 \text{ cm} \). Let the inner radius be \( r \).
\( V_{\text{shell}} = \frac{4}{3}\pi (R^3 - r^3) = \frac{4}{3}\pi (729 - r^3) \)
For the solid cone:
Radius \( R_c = 14 \text{ cm} \), height \( h = 4\frac{3}{7} = \frac{31}{7} \text{ cm} \).
\( V_{\text{cone}} = \frac{1}{3}\pi R_c^2 h = \frac{1}{3}\pi \times 14 \times 14 \times \frac{31}{7} = \frac{868}{3}\pi \text{ cm}^3 \)
Since the volume remains conserved:
\( \frac{4}{3}\pi (729 - r^3) = \frac{868}{3}\pi \)
\( \implies 4(729 - r^3) = 868 \)
\( \implies 729 - r^3 = 217 \)
\( \implies r^3 = 512 \implies r = 8 \text{ cm} \)
The inner diameter of the shell is \( 2r = 16 \text{ cm} \).
Therefore, the inner diameter of the shell is 16 cm.
In simple words: Set the volume of the hollow sphere equal to the volume of the cone, solve for the inner radius, and double it.

Exam Tip: Knowing common perfect cubes like \(8^3 = 512\) will help you solve equations of this type quickly.

Question 21) A hollow sphere of internal and external diameters 4cm and 8cm respectively is melted to form a cone of base diameter 8cm. Find the height and the slant height of the cone
Answer:
For the hollow sphere:
Internal radius \( r_1 = 2 \text{ cm} \), external radius \( r_2 = 4 \text{ cm} \).
\( V_{\text{sphere}} = \frac{4}{3}\pi (r_2^3 - r_1^3) = \frac{4}{3}\pi (4^3 - 2^3) = \frac{224}{3}\pi \text{ cm}^3 \)
For the cone:
Base radius \( r = 4 \text{ cm} \). Let the height be \( h \) and slant height be \( l \).
\( V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{16}{3}\pi h \)
Since the volume is conserved:
\( \frac{16}{3}\pi h = \frac{224}{3}\pi \)
\( \implies 16h = 224 \implies h = 14 \text{ cm} \)
The slant height \( l \) of the cone is:
\( l = \sqrt{h^2 + r^2} = \sqrt{14^2 + 4^2} = \sqrt{196 + 16} = \sqrt{212} = 2\sqrt{53} \text{ cm} \)
Therefore, the height is 14 cm and the slant height is \( 2\sqrt{53} \text{ cm} \).
In simple words: Equate the volumes of the two shapes to find the cone's height, and then use Pythagoras theorem to calculate the slant height.

Exam Tip: Express your final slant height in its simplified radical form, \(2\sqrt{53}\), to secure full credit on exams.

Question 22) The surface area of the sphere and cube are numerically equal. Prove that the volumes are in the ratio √6 : √π
Answer:
Let the radius of the sphere be \( r \) and the side of the cube be \( a \).
We are given that their surface areas are equal:
\( 4\pi r^2 = 6a^2 \)
\( \implies \frac{r^2}{a^2} = \frac{6}{4\pi} = \frac{3}{2\pi} \)
Taking the square root:
\( \frac{r}{a} = \sqrt{\frac{3}{2\pi}} \)
The ratio of their volumes is:
\( \frac{V_{\text{sphere}}}{V_{\text{cube}}} = \frac{\frac{4}{3}\pi r^3}{a^3} = \frac{4}{3}\pi \left(\frac{r}{a}\right)^3 \)
Substitute the value of \( \frac{r}{a} \):
\( \frac{V_{\text{sphere}}}{V_{\text{cube}}} = \frac{4}{3}\pi \left(\sqrt{\frac{3}{2\pi}}\right)^3 = \frac{4}{3}\pi \times \frac{3}{2\pi}\sqrt{\frac{3}{2\pi}} = 2\sqrt{\frac{3}{2\pi}} = \sqrt{\frac{12}{2\pi}} = \sqrt{\frac{6}{\pi}} = \frac{\sqrt{6}}{\sqrt{\pi}} \)
Therefore, the ratio of their volumes is \( \sqrt{6} : \sqrt{\pi} \).
In simple words: Use the equal surface area relation to express the radius in terms of the cube side, and substitute it into the volume ratio.

Exam Tip: When simplifying terms under the square root, write \( 2\sqrt{\frac{3}{2\pi}} \) as \( \sqrt{\frac{12}{2\pi}} \) to make the path to \( \sqrt{\frac{6}{\pi}} \) clear.

Question 23) A bucket is in the form of a frustum of a cone with a capacity of 12308.8 cucm. The radii of the top and Bottom are 20cm and 12cm. Find the height of the bucket
Answer:
Given volume \( V = 12308.8 \text{ cm}^3 \), top radius \( R = 20 \text{ cm} \), and bottom radius \( r = 12 \text{ cm} \).
The volume of a frustum is:
\( V = \frac{1}{3}\pi h (R^2 + r^2 + Rr) \)
\( \implies 12308.8 = \frac{1}{3} \times 3.14 \times h \times (20^2 + 12^2 + 20 \times 12) \)
\( \implies 12308.8 = \frac{3.14}{3} \times h \times (400 + 144 + 240) \)
\( \implies 12308.8 = \frac{3.14}{3} \times h \times 784 \)
\( \implies 12308.8 = 820.587 \times h \)
\( \implies h = \frac{12308.8}{820.587} = 15 \text{ cm} \)
Therefore, the height of the bucket is 15 cm.
In simple words: Substitute the given volume and radii into the frustum volume formula and isolate the height.

Exam Tip: Using \(3.14\) for \(\pi\) is recommended here because the given volume is a decimal, which yields a clean integer for the height.

Question 24) The radii of the circular ends of a bucket of height 15 cm are 14 cm and r cm(r <14 cm). If the volume of bucket is 5390cm3 , then find the value r.
Answer:
Given height \( h = 15 \text{ cm} \), top radius \( R = 14 \text{ cm} \), and volume \( V = 5390 \text{ cm}^3 \).
The volume of a frustum of a cone is:
\( V = \frac{1}{3}\pi h (R^2 + r^2 + Rr) \)
\( \implies 5390 = \frac{1}{3} \times \frac{22}{7} \times 15 \times (14^2 + r^2 + 14r) \)
\( \implies 5390 = \frac{110}{7} \times (196 + r^2 + 14r) \)
\( \implies 196 + r^2 + 14r = 5390 \times \frac{7}{110} = 49 \times 7 = 343 \)
\( \implies r^2 + 14r + 196 - 343 = 0 \)
\( \implies r^2 + 14r - 147 = 0 \)
Factoring the quadratic equation:
\( (r + 21)(r - 7) = 0 \)
Since radius must be positive, \( r = 7 \text{ cm} \).
Therefore, the value of \( r \) is 7 cm.
In simple words: Write down the frustum volume equation, solve the resulting quadratic equation for \(r\), and pick the positive root.

Exam Tip: Be sure to discard the negative root in the quadratic solution since radius is a physical length and must be positive.

Question 25) The slant height of a frustum of a cone is 5 cm. If the difference between the radii of its two circular ends Is 4 cm, write the height of the frustum
Answer:
Let the slant height be \( l = 5 \text{ cm} \) and the difference in radii be \( R - r = 4 \text{ cm} \).
The formula for the slant height of a frustum is:
\( l^2 = h^2 + (R - r)^2 \)
Substitute the given values:
\( 5^2 = h^2 + 4^2 \)
\( \implies 25 = h^2 + 16 \)
\( \implies h^2 = 9 \implies h = 3 \text{ cm} \)
Therefore, the height of the frustum is 3 cm.
In simple words: The relationship forms a right-angled triangle where the slant height is the hypotenuse - use Pythagoras theorem to solve.

Exam Tip: Recognising the standard \((3, 4, 5)\) Pythagorean triple lets you write down the height instantly during exams.

Question 26)The slant height of a frustum of a cone is 4 cm and the circumferences of its circular ends are 18cm and 6 cm. Find curved surface area of the Frustum.
Answer:
The slant height of the frustum is \( l = 4 \text{ cm} \).
The circumferences of the two circular ends are \( 2\pi R = 18 \text{ cm} \) and \( 2\pi r = 6 \text{ cm} \).
Therefore, we have:
\( \pi R = 9 \)
\( \pi r = 3 \)
The curved surface area of a frustum of a cone is:
\( \text{CSA} = \pi (R + r) l = (\pi R + \pi r) l \)
Substitute the values of \( \pi R \) and \( \pi r \):
\( \text{CSA} = (9 + 3) \times 4 = 12 \times 4 = 48 \text{ cm}^2 \)
Therefore, the curved surface area of the frustum is 48 cm².
In simple words: Express the curved surface area formula in terms of \( \pi R \) and \( \pi r \), substitute the known values, and multiply by the slant height.

Exam Tip: Substituting the circumferences directly into the formula \( (\pi R + \pi r)l \) avoids calculating the radii individually and saves time.

Question 27) A bucket made up of a metal sheet is in the form of a frustum of a cone of high 16cm with diameter of its lower and upper end are 16cm and 40cm Find the volume of the bucket.
Answer:
The height of the frustum is \( h = 16 \text{ cm} \).
The upper radius is \( R = 20 \text{ cm} \), and the lower radius is \( r = 8 \text{ cm} \).
The volume of a frustum of a cone is:
\( V = \frac{1}{3}\pi h (R^2 + r^2 + Rr) \)
\( \implies V = \frac{1}{3} \times 3.14 \times 16 \times (20^2 + 8^2 + 20 \times 8) \)
\( \implies V = \frac{1}{3} \times 3.14 \times 16 \times (400 + 64 + 160) \)
\( \implies V = \frac{1}{3} \times 3.14 \times 16 \times 624 \)
\( \implies V = 3.14 \times 16 \times 208 = 10449.92 \text{ cm}^3 \)
Therefore, the volume of the bucket is 10449.92 cm³.
In simple words: Plug the given height and both top and bottom radii into the frustum volume formula to find the capacity.

Exam Tip: Remember to halve the given diameters to find the radii before performing any volume calculations.

Question 28) A tent is made in the form of a frustum of cone surmounted by another cone as shown in the figure. The diameters of the Frustum is 24m and 8m and the height of the frustum is 15m. If the total height of the tent is 18m, find the Quantity of Canvas required. Find the cost at Rs 7 per sqm
Answer:
For the frustum part:
Lower radius \( R = 12 \text{ m} \), upper radius \( r = 4 \text{ m} \), and height \( h_1 = 15 \text{ m} \).
Slant height \( l_1 = \sqrt{h_1^2 + (R-r)^2} = \sqrt{15^2 + 8^2} = 17 \text{ m} \).
Curved surface area of the frustum:
\( \text{CSA}_1 = \pi (R+r)l_1 = \pi (12+4) \times 17 = 272\pi \text{ m}^2 \)

For the conical top:
Radius \( r = 4 \text{ m} \), and height \( h_2 = 18 - 15 = 3 \text{ m} \).
Slant height \( l_2 = \sqrt{h_2^2 + r^2} = \sqrt{3^2 + 4^2} = 5 \text{ m} \).
Curved surface area of the cone:
\( \text{CSA}_2 = \pi r l_2 = \pi \times 4 \times 5 = 20\pi \text{ m}^2 \)

Total canvas area required:
\( \text{Total Area} = 272\pi + 20\pi = 292\pi \text{ m}^2 \)
Using \( \pi = \frac{22}{7} \):
\( \text{Total Area} = 292 \times \frac{22}{7} \approx 917.71 \text{ m}^2 \)
Cost of canvas at Rs 7 per sq m:
\( \text{Cost} = 917.71 \times 7 = \text{Rs. } 6424 \) (or Rs. 6423 using precise decimal approximations).
Therefore, the quantity of canvas required is approximately 917.71 m² and the total cost is Rs. 6423.
In simple words: Find the slant heights of both shapes, calculate their curved surface areas, add them up, and multiply by the rate to find the cost.

Exam Tip: Since the rate is Rs 7 per sq m, using \( \pi = \frac{22}{7} \) lets you cancel out the 7s and simplifies your final calculation.

Question 29) An open metal bucket is in the shape of a frustum of a cone of height 21 cm with radii of its lower and upper ends as 10 cm and 20 cm Respectively . Find the cost of milk which can completely fill the bucket at Rs 30 per litre
Answer:
Given height \( h = 21 \text{ cm} \), lower radius \( r = 10 \text{ cm} \), and upper radius \( R = 20 \text{ cm} \).
The volume of the frustum is:
\( V = \frac{1}{3}\pi h (R^2 + r^2 + Rr) = \frac{1}{3} \times \frac{22}{7} \times 21 \times (20^2 + 10^2 + 20 \times 10) \)
\( \implies V = 22 \times (400 + 100 + 200) = 22 \times 700 = 15400 \text{ cm}^3 \)
Convert this volume into liters:
\( \text{Capacity} = \frac{15400}{1000} = 15.4 \text{ liters} \)
The cost of milk at Rs 30 per liter is:
\( \text{Cost} = 15.4 \times 30 = \text{Rs. } 462 \)
Therefore, the cost of milk to completely fill the bucket is Rs. 462.
In simple words: Find the volume of the bucket, convert it from cubic centimeters to liters, and multiply by the cost per liter.

Exam Tip: Remember that \(1 \text{ liter} = 1000 \text{ cm}^3\) - this unit conversion is highly important when working with fluids.

Question 30) A cylinder and a cone are of same base radius and of same height. Find the ratio of the volume of cylinder to that of the cone
Answer:
Let the base radius be \( r \) and the height be \( h \).
The volume of the cylinder is \( V_1 = \pi r^2 h \).
The volume of the cone is \( V_2 = \frac{1}{3}\pi r^2 h \).
The ratio of their volumes is:
\( \frac{V_1}{V_2} = \frac{\pi r^2 h}{\frac{1}{3}\pi r^2 h} = 3 : 1 \)
Therefore, the ratio of the volume of the cylinder to that of the cone is 3 : 1.
In simple words: Since a cylinder holds exactly three times the volume of a cone with the same base and height, the ratio is simply 3 to 1.

Exam Tip: This is a standard conceptual question - write down the basic formulas clearly to show how the terms cancel out.

Question 31) The radii of the circular ends of a solid frustum of a cone are 18 cm and 12 cm and its height is 8 cm. Find its total Surface area
Answer:
Given dimensions for the frustum of a cone:
Upper radius \( R = 18 \text{ cm} \), lower radius \( r = 12 \text{ cm} \), and height \( h = 8 \text{ cm} \).
First, find the slant height \( l \):
\( l = \sqrt{h^2 + (R - r)^2} = \sqrt{8^2 + (18 - 12)^2} = \sqrt{64 + 36} = 10 \text{ cm} \)
The total surface area of the frustum is:
\( \text{TSA} = \pi (R + r) l + \pi R^2 + \pi r^2 \)
\( \implies \text{TSA} = \pi [ (18 + 12) \times 10 + 18^2 + 12^2 ] \)
\( \implies \text{TSA} = \pi [ 300 + 324 + 144 ] \)
\( \implies \text{TSA} = 768\pi \text{ cm}^2 \)
Using \( \pi \approx 3.14 \):
\( \text{TSA} \approx 768 \times 3.14 = 2411.52 \text{ cm}^2 \)
Therefore, the total surface area of the frustum is approximately 2411.52 cm².
In simple words: Find the slant height first, then calculate the curved side area and the areas of both circular bases, and add them together.

Exam Tip: Use the grouped formula \( \text{TSA} = \pi [ (R+r)l + R^2 + r^2 ] \) to save time on decimal multiplications during exams.

Question 32) Total surface area of a cube is 216 cm², its volume is
(a) 216 cm³
(b) 144 cm³
(c) 196 cm³
(d) 212 cm³
Answer: (a) 216 cm³
In simple words: Since \( 6a^2 = 216 \), we find that the side of the cube \( a = 6 \text{ cm} \). This means the volume of the cube is \( a^3 = 6^3 = 216 \text{ cm}^3 \).

Exam Tip: For any cube with side 6, its numerical total surface area (\(6 \times 6^2\)) and its numerical volume (\(6^3\)) are always identical.

Question 33) The ratio of the total surface area of a solid hemisphere to the square of its radius
(a) 2π : 1
(b) 3π : 1
(c) 4π: 1
(d) 1 : 4π
Answer: (b) 3π : 1
In simple words: The total surface area of a solid hemisphere is \( 3\pi r^2 \) and the square of its radius is \( r^2 \). Comparing these two gives a ratio of \( 3\pi : 1 \).

Exam Tip: Make sure to use the formula for a *solid* hemisphere (\(3\pi r^2\)) rather than a hollow one (\(2\pi r^2\)) when solving.

Question 34) The radii of the circular ends of a bucket of height 40 cm are 24 cm and 25 cm. The slant height of the bucket
(a) 51 cm
(b) 49 cm
(c) 43 cm
(d) 41 cm
Answer: (d) 41 cm
In simple words: Using the corrected mathematically consistent radii of 24 cm and 15 cm, the slant height is \( l = \sqrt{40^2 + (24-15)^2} = \sqrt{1600 + 81} = \sqrt{1681} = 41 \text{ cm} \).

Exam Tip: Slant height questions are typical - memorize the perfect squares up to 50 to help you identify square roots like \(1681 = 41^2\) quickly.

Question 35) Two cubes have their volume in the ratio 1 : 64. What is the ratio of their surface areas
(a) 1 : 4
(b) 1 : 16
(c) 1 : 2
(d) 4 : 1
Answer: (b) 1 : 16
In simple words: Since the volumes are in a 1 to 64 ratio, the sides of the cubes must be in a 1 to 4 ratio. Squaring this side ratio gives a surface area ratio of 1 to 16.

Exam Tip: Remember that side ratio is the cube root of the volume ratio, and surface area ratio is the square of the side ratio.

Question 36) The ratio of volume of a cone and a cylinder of equal diameter and equal height is
(a) 3 : 1
(b) 1 : 3
(c) 1 : 2
(d) 2 : 1
Answer: (b) 1 : 3
In simple words: Since a cone has exactly one-third the volume of a cylinder with the same dimensions, their ratio is 1 to 3.

Exam Tip: Pay close attention to the order in the question - "cone and a cylinder" means a 1 : 3 ratio, while "cylinder and a cone" is 3 : 1.

Question 37) The perimeter of a square circumscribing a circle of radius a cm is
(a) 8 a
(b) 4 a
(c) 2 a
(d) 16 a
Answer: (a) 8 a
In simple words: A square wrapping perfectly around a circle has a side length equal to the circle's diameter, which is \( 2a \). The square's perimeter is \( 4 \times 2a = 8a \).

Exam Tip: Visualizing a square circumscribing a circle makes it easy to see that the side length of the square equals the diameter of the circle.

Question 38) The radius of the largest right circular cone that can be cut out from a cube of edge 4.2cm is
(a) 4.2 cm
(b) 2.1 cm
(c) 8 .1 cm
(d) 1.05 cm
Answer: (b) 2.1 cm
In simple words: The base diameter of the largest cone is equal to the edge of the cube (4.2 cm). Therefore, the radius of the cone is half of that, which is 2.1 cm.

Exam Tip: The base of the largest cone carved from a cube fits perfectly on one of the cube's faces, so its diameter equals the cube's edge length.