Samacheer Kalvi Class 10 Maths Solutions Chapter 7 Mensuration Exercise 7.1

Question 1. The radius and height of a cylinder are in the ratio 5 : 7 and its curved surface area is 5500 sq.cm. Find its radius and height.
Answer: Let the radius be \( 5x \) and the height be \( 7x \). The Curved Surface Area (C.S.A) of the cylinder is given as \( 5500 \) sq.cm. The formula for the C.S.A of a cylinder is \( 2\pi rh \).
Substituting the given values into the formula:
\( 2 \times \frac{22}{7} \times 5x \times 7x = 5500 \)
\( 2 \times 22 \times 5 \times x^2 = 5500 \)
\( x^2 = \frac{5500}{2 \times 22 \times 5} \)
\( x^2 = 25 \)
\( x = 5 \) cm
Now, we can find the actual radius and height:
Radius of the cylinder \( = 5 \times 5 = 25 \) cm
Height of the cylinder \( = 7 \times 5 = 35 \) cm. Always define variables clearly when working with ratios to simplify the problem.
In simple words: We first set the radius as 5 times a variable 'x' and the height as 7 times 'x'. Using the formula for curved surface area and the given area, we find the value of 'x'. Then, we multiply 'x' by 5 to get the radius and by 7 to get the height.

🎯 Exam Tip: When given ratios, express dimensions in terms of a common variable (like \( 5x \) and \( 7x \)) to make algebraic calculations easier and ensure proportionality.

Question 2. A solid iron cylinder has total surface area of 1848 sq.m. Its curved surface area is five-sixth of its total surface area. Find the radius and height of the iron cylinder.
Answer: We are given that the Total Surface Area (T.S.A) of the cylinder is \( 1848 \) sq.m. The formula for T.S.A is \( 2\pi r(h+r) \).
The Curved Surface Area (C.S.A) is \( \frac{5}{6} \) of its total surface area.
C.S.A \( = \frac{5}{6} \times 1848 = 5 \times 308 = 1540 \) sq.m.
The formula for C.S.A is \( 2\pi rh \). So, \( 2\pi rh = 1540 \) sq.m.
We know that T.S.A \( = 2\pi rh + 2\pi r^2 \).
Substitute the value of \( 2\pi rh \) into the T.S.A equation:
\( 1540 + 2\pi r^2 = 1848 \)
\( 2\pi r^2 = 1848 - 1540 \)
\( 2\pi r^2 = 308 \)
\( 2 \times \frac{22}{7} \times r^2 = 308 \)
\( r^2 = \frac{308 \times 7}{2 \times 22} = \frac{308 \times 7}{44} = 7 \times 7 = 49 \)
\( r = \sqrt{49} = 7 \) m. This calculation helps determine the base dimension.
Now, use \( 2\pi rh = 1540 \) to find the height \( h \):
\( 2 \times \frac{22}{7} \times 7 \times h = 1540 \)
\( 2 \times 22 \times h = 1540 \)
\( 44h = 1540 \)
\( h = \frac{1540}{44} = 35 \) m.
In simple words: We are given the total surface area and how the curved surface area relates to it. We use these facts to find the curved surface area. Then, by using the total surface area formula and the curved surface area, we can first find the radius. Once we have the radius, we use the curved surface area formula again to find the height.

🎯 Exam Tip: Remember the breakdown of Total Surface Area for a cylinder: T.S.A = Curved Surface Area + Area of two circular bases. This relationship is often key to solving such problems.

Question 3. The external radius and the length of a hollow wooden log are 16 cm and 13 cm respectively. If its thickness is 4 cm then find its T.S.A.
Answer: We are given the following information for the hollow wooden log:
External radius \( (R) = 16 \) cm
Thickness \( = 4 \) cm
Internal radius \( (r) = R - \text{thickness} = 16 - 4 = 12 \) cm
Length of the wooden log \( (h) = 13 \) cm
The formula for the Total Surface Area (T.S.A) of a hollow cylinder is \( 2\pi (R + r) (R - r + h) \). This formula accounts for both inner and outer curved surfaces and the two annular rings.
Substituting the values into the formula:
T.S.A \( = 2 \times \frac{22}{7} \times (16 + 12) (16 - 12 + 13) \) sq.cm
T.S.A \( = 2 \times \frac{22}{7} \times (28) (4 + 13) \) sq.cm
T.S.A \( = 2 \times \frac{22}{7} \times 28 \times 17 \) sq.cm
T.S.A \( = 2 \times 22 \times 4 \times 17 \) sq.cm (since \( \frac{28}{7} = 4 \))
T.S.A \( = 44 \times 68 \) sq.cm
T.S.A \( = 2992 \) sq.cm. Always calculate the internal radius carefully when thickness is provided.
In simple words: First, we find the inner radius by taking away the thickness from the outer radius. Then, we use the special formula for the total surface area of a hollow cylinder, plugging in the outer radius, inner radius, and length of the log to get the final area.

🎯 Exam Tip: For hollow objects, remember that the total surface area often involves external, internal, and annular (ring-shaped) areas. Calculate the internal radius accurately.

Question 4. A right angled triangle PQR where \( \angle Q = 90^\circ \) is rotated about QR and PQ. If QR = 16 cm and PR = 20 cm, compare the curved surface areas of the right circular cones so formed by the triangle.
Answer: First, we need to find the length of the side PQ using the Pythagorean theorem, since \( \triangle PQR \) is a right-angled triangle at Q.
\( QP^2 = PR^2 - QR^2 \)
\( QP^2 = 20^2 - 16^2 \)
\( QP^2 = 400 - 256 = 144 \)
\( QP = \sqrt{144} = 12 \) cm.
**Case 1: When the triangle is rotated about QR.**
In this case, the height of the cone \( h = QR = 16 \) cm.
The radius of the cone \( r = PQ = 12 \) cm.
The slant height of the cone \( l = PR = 20 \) cm.
Curved Surface Area \( (C.S.A_1) = \pi rl = \pi \times 12 \times 20 = 240\pi \) sq.cm.
**Case 2: When the triangle is rotated about PQ.**
In this case, the height of the cone \( h = PQ = 12 \) cm.
The radius of the cone \( r = QR = 16 \) cm.
The slant height of the cone \( l = PR = 20 \) cm.
Curved Surface Area \( (C.S.A_2) = \pi rl = \pi \times 16 \times 20 = 320\pi \) sq.cm.
Comparing the two curved surface areas: \( 320\pi \) sq.cm \( > 240\pi \) sq.cm.
Therefore, the C.S.A. of the cone formed when rotated about PQ is larger. Understanding which side becomes the radius and which becomes the height is critical.
In simple words: We first find the missing side of the triangle. Then, we look at two scenarios: one where the triangle spins around side QR, and another where it spins around side PQ. For each case, we figure out the cone's radius and slant height, calculate its curved surface area, and then compare which area is bigger.

🎯 Exam Tip: When a right-angled triangle is rotated to form a cone, the side about which it is rotated becomes the height, and the other perpendicular side becomes the radius of the cone.

Question 5. 4 persons live in a conical tent whose slant height is 19 cm. If each person requires 22 cm² of the floor area, then find the height of the tent.
Answer: The slant height of the cone \( (l) = 19 \) cm.
Each person requires \( 22 \) sq.cm of floor area.
For 4 persons, the total floor area required \( = 4 \times 22 = 88 \) sq.cm.
The floor of a conical tent is a circle, so its area is given by \( \pi r^2 \).
Thus, \( \pi r^2 = 88 \)
\( \frac{22}{7} \times r^2 = 88 \)
\( r^2 = \frac{88 \times 7}{22} \)
\( r^2 = 4 \times 7 = 28 \) sq.cm. The floor area helps us determine the base radius.
To find the height \( (h) \) of the tent, we use the Pythagorean relationship for a cone: \( l^2 = h^2 + r^2 \).
\( h^2 = l^2 - r^2 \)
\( h^2 = 19^2 - 28 \)
\( h^2 = 361 - 28 \)
\( h^2 = 333 \)
\( h = \sqrt{333} \)
\( h \approx 18.248 \) cm
Rounding to two decimal places, the height of the tent is approximately \( 18.25 \) cm.
In simple words: We first calculate the total floor space needed by all four people. Since the tent's floor is a circle, we use this area to find the radius of the tent's base. Finally, using the slant height and the base radius, we find the tent's actual height with a special triangle rule.

🎯 Exam Tip: The floor area of a conical tent refers to its circular base area \( (\pi r^2) \). Remember to apply the Pythagorean theorem \( (l^2 = h^2 + r^2) \) to find the height, radius, or slant height when two are known.

Question 6. A girl wishes to prepare birthday caps in the form of right circular cones for her birthday party using a sheet of paper whose area is 5720 cm², how many caps can be made with radius 5 cm and height 12 cm.
Answer: For one birthday cap (which is a right circular cone):
Radius \( (r) = 5 \) cm
Height \( (h) = 12 \) cm
First, calculate the slant height \( (l) \) of one cap using the formula \( l = \sqrt{h^2 + r^2} \).
\( l = \sqrt{12^2 + 5^2} \)
\( l = \sqrt{144 + 25} \)
\( l = \sqrt{169} \)
\( l = 13 \) cm.
Next, calculate the Curved Surface Area (C.S.A) of one cap, as this is the amount of paper needed for one cap. The formula is \( \pi rl \).
C.S.A. of a cap \( = \frac{22}{7} \times 5 \times 13 \) sq.cm
C.S.A. of a cap \( = \frac{1430}{7} \) sq.cm. This is the paper required for one cap.
Total area of the sheet of paper available \( = 5720 \) sq.cm.
To find the number of caps that can be made, divide the total area of the sheet by the C.S.A. of one cap:
Number of caps \( = \frac{\text{Total area of the paper}}{\text{Area of paper for one cap}} \)
Number of caps \( = \frac{5720}{\frac{1430}{7}} \)
Number of caps \( = 5720 \times \frac{7}{1430} \)
Number of caps \( = 4 \times 7 = 28 \) caps.
In simple words: We first find the slant height of a single birthday cap. Then, we calculate the curved surface area of one cap, which tells us how much paper each cap needs. Finally, we divide the total paper area by the area needed for one cap to find out how many caps can be made.

🎯 Exam Tip: For problems involving making items like caps or tents, focus on the curved surface area, as the base is usually open. Always use the Pythagorean theorem to find the slant height if not directly given.

Question 7. The ratio of the radii of two right circular cones of the same height is 1 : 3. Find the ratio of their curved surface areas when the height of each cone is 3 times the radius of the smaller cone.
Answer: Let the radius of the first (smaller) cone be \( r_1 = x \).
Since the ratio of the radii is 1:3, the radius of the second (larger) cone will be \( r_2 = 3x \).
The height of each cone \( (h) \) is 3 times the radius of the smaller cone, so \( h = 3x \).
**For the first cone:**
Radius \( r_1 = x \)
Height \( h_1 = 3x \)
Slant height \( l_1 = \sqrt{h_1^2 + r_1^2} = \sqrt{(3x)^2 + x^2} = \sqrt{9x^2 + x^2} = \sqrt{10x^2} = x\sqrt{10} \).
Curved Surface Area \( C.S.A_1 = \pi r_1 l_1 = \pi \times x \times x\sqrt{10} = \pi x^2\sqrt{10} \).
**For the second cone:**
Radius \( r_2 = 3x \)
Height \( h_2 = 3x \)
Slant height \( l_2 = \sqrt{h_2^2 + r_2^2} = \sqrt{(3x)^2 + (3x)^2} = \sqrt{9x^2 + 9x^2} = \sqrt{18x^2} = \sqrt{9 \times 2 \times x^2} = 3x\sqrt{2} \).
Curved Surface Area \( C.S.A_2 = \pi r_2 l_2 = \pi \times 3x \times 3x\sqrt{2} = 9\pi x^2\sqrt{2} \).
Now, find the ratio of their curved surface areas:
\( C.S.A_1 : C.S.A_2 = \pi x^2\sqrt{10} : 9\pi x^2\sqrt{2} \)
Divide both sides by \( \pi x^2 \) (assuming \( x \neq 0 \)):
\( = \sqrt{10} : 9\sqrt{2} \)
We can write \( \sqrt{10} \) as \( \sqrt{2 \times 5} = \sqrt{2} \times \sqrt{5} \):
\( = \sqrt{2}\sqrt{5} : 9\sqrt{2} \)
Divide both sides by \( \sqrt{2} \):
\( = \sqrt{5} : 9 \). Using a common variable like 'x' helps in maintaining the proportional relationships throughout the calculations.
In simple words: We set up the radii of the two cones using a variable 'x' and define their heights based on the smaller cone's radius. Then, for each cone, we calculate its slant height and curved surface area. Finally, we compare these two curved surface areas to find their simplified ratio.

🎯 Exam Tip: When dealing with ratios of geometric figures, represent dimensions with a common variable (e.g., \( x \)) to maintain proportionality. Carefully apply the slant height formula \( l = \sqrt{h^2+r^2} \).

Question 8. The radius of a sphere increases by 25%. Find the percentage increase in its surface area.
Answer: Let the original radius of the sphere be \( r \).
Original surface area of the sphere \( A_1 = 4\pi r^2 \).
If the radius increases by 25%, the new radius \( r' \) will be:
\( r' = r + \frac{25}{100}r = r + \frac{1}{4}r = \frac{4r+r}{4} = \frac{5r}{4} \).
The new surface area of the sphere \( A_2 \) will be:
\( A_2 = 4\pi (r')^2 = 4\pi \left(\frac{5r}{4}\right)^2 \)
\( A_2 = 4\pi \times \frac{25r^2}{16} = \frac{25\pi r^2}{4} \).
The difference in surface area (increase) \( = A_2 - A_1 \):
Difference \( = \frac{25\pi r^2}{4} - 4\pi r^2 \)
Difference \( = \pi r^2 \left(\frac{25}{4} - 4\right) \)
Difference \( = \pi r^2 \left(\frac{25-16}{4}\right) \)
Difference \( = \frac{9\pi r^2}{4} \).
Now, calculate the percentage increase in surface area:
Percentage increase \( = \frac{\text{Difference in surface area}}{\text{Original surface area}} \times 100\% \)
Percentage increase \( = \frac{\frac{9\pi r^2}{4}}{4\pi r^2} \times 100\% \)
Percentage increase \( = \frac{9\pi r^2}{4 \times 4\pi r^2} \times 100\% \)
Percentage increase \( = \frac{9}{16} \times 100\% \)
Percentage increase \( = 0.5625 \times 100\% = 56.25\% \). A small percentage change in radius can lead to a larger percentage change in surface area.
In simple words: We first write down the original surface area using the radius 'r'. Then, we find the new radius after it increases by 25%. We use this new radius to calculate the new surface area. We subtract the original area from the new area to find how much it increased, and then express this increase as a percentage of the original area.

🎯 Exam Tip: When a dimension (like radius) changes by a percentage, the area changes by a larger percentage because the dimension is squared in the area formula. Always convert percentage changes into a fractional or decimal multiplier for calculations.

Question 9. The internal and external diameters of a hollow hemispherical vessel are 20 cm and 28 cm respectively. Find the cost to paint the vessel all over at Rs. 0.14 per cm².
Answer: We are given the following:
Internal diameter \( = 20 \) cm, so internal radius \( (r) = \frac{20}{2} = 10 \) cm.
External diameter \( = 28 \) cm, so external radius \( (R) = \frac{28}{2} = 14 \) cm.
The Total Surface Area (T.S.A) of a hollow hemisphere, painted all over (which means inner surface, outer surface, and the top ring), is given by the formula \( \pi (3R^2 + r^2) \). This formula covers all surfaces to be painted.
T.S.A \( = \frac{22}{7} [3 \times (14)^2 + (10)^2] \) sq.cm
T.S.A \( = \frac{22}{7} [3 \times 196 + 100] \) sq.cm
T.S.A \( = \frac{22}{7} [588 + 100] \) sq.cm
T.S.A \( = \frac{22}{7} \times 688 \) sq.cm.
Cost of painting per sq.cm \( = \) Rs. \( 0.14 \).
Total cost of painting the vessel \( = \text{T.S.A} \times \text{Cost per sq.cm} \)
Total cost \( = \frac{22}{7} \times 688 \times 0.14 \) Rs.
Total cost \( = \frac{22}{7} \times 688 \times \frac{14}{100} \) Rs.
Total cost \( = 22 \times 688 \times \frac{2}{100} \) Rs. (since \( \frac{14}{7} = 2 \))
Total cost \( = \frac{30272}{100} \) Rs.
Total cost \( = 302.72 \) Rs.
In simple words: First, we find the inner and outer radii from the given diameters. Then, we use a special formula for a hollow hemisphere to calculate the total area that needs painting. Finally, we multiply this total area by the cost to paint each square centimeter to get the final total cost.

🎯 Exam Tip: For hollow hemispherical vessels, the painting involves three surfaces: the outer curved surface, the inner curved surface, and the area of the ring at the top. Ensure you use the correct combined formula \( \pi (3R^2 + r^2) \).

Question 10. The frustum shaped outer portion of the table lamp has to be painted including the top part. Find the total cost of painting the lamp if the cost of painting 1 sq.cm is Rs. 2.
Answer: From the given diagram of the frustum-shaped lamp:
Height \( (h) = 8 \) m
Larger radius \( (R) = 12 \) m
Smaller radius \( (r) = 6 \) m
First, calculate the slant height \( (l) \) of the frustum using the formula \( l = \sqrt{h^2 + (R-r)^2} \).
\( l = \sqrt{8^2 + (12-6)^2} \)
\( l = \sqrt{8^2 + 6^2} \)
\( l = \sqrt{64 + 36} \)
\( l = \sqrt{100} = 10 \) m.
The total area to be painted includes the Curved Surface Area (C.S.A) of the frustum and the area of its top circular part.
Total Area \( = \text{C.S.A. of frustum} + \text{Area of top circle} \)
Total Area \( = \pi (R+r)l + \pi r^2 \)
Total Area \( = \pi [(R+r)l + r^2] \)
Substituting the values:
Total Area \( = \frac{22}{7} [(12+6) \times 10 + 6^2] \) sq.m
Total Area \( = \frac{22}{7} [18 \times 10 + 36] \) sq.m
Total Area \( = \frac{22}{7} [180 + 36] \) sq.m
Total Area \( = \frac{22}{7} \times 216 \) sq.m
Total Area \( = \frac{4752}{7} \approx 678.857 \) sq.m.
Cost of painting per sq.cm is given as Rs. \( 2 \). Assuming the units are consistent and it means Rs. 2 per sq.m.
Total cost of painting \( = \text{Total Area} \times \text{Cost per unit area} \)
Total cost \( = \frac{4752}{7} \times 2 \) Rs.
Total cost \( = \frac{9504}{7} \approx 1357.714 \) Rs.
Rounding to two decimal places, the total cost of painting is Rs. \( 1357.72 \). Always ensure all relevant surfaces are included when calculating total area.
In simple words: First, we find the slant height of the frustum (the lamp's shape). Then, we calculate the total area to be painted, which includes the curved side and the top circular part. Finally, we multiply this total painted area by the cost per square meter to find the total painting expense.

🎯 Exam Tip: For a frustum-shaped object, make sure to include the area of the top base if it's part of the painting requirement. The slant height formula \( l = \sqrt{h^2 + (R-r)^2} \) is critical for finding the curved surface area.