Class 10 Maths Mensuration Solutions | Ex 7.1 Full Solved Questions (Cylinder, Cone, Frustum, Sphere)

Class 10 Maths: Mensuration Exercise 7.1 Solutions

Exercise 7.1

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.

Given:

• For a cylinder, \(r : h = 5 : 7\).
• Curved Surface Area (C.S.A.) of cylinder = \(5500 \text{ sq. cm.}\)

To find: Radius (r) and Height (h) of cylinder.

Let the common multiple be \(x\).

Therefore, Radius \(r = 5x\) and Height \(h = 7x\).

Formula for C.S.A. of cylinder: \(C.S.A. = 2 \pi r h\).

Substitute the given values:

$$5500 = 2 \times \frac{22}{7} \times (5x) \times (7x)$$ $$5500 = 2 \times 22 \times 5x \times x$$ $$5500 = 220x^2$$ $$x^2 = \frac{5500}{220}$$ $$x^2 = 25$$ $$x = 5$$

Now, calculate radius and height:

• Radius \(r = 5x = 5 \times 5 = 25 \text{ cm}\)
• Height \(h = 7x = 7 \times 5 = 35 \text{ cm}\)

The radius of the cylinder is 25 cm and the height is 35 cm.

2. A solid iron cylinder has total surface area of 1848 sq.cm. Its curved surface area is five – sixth of its total surface area. Find the radius and height of the iron cylinder.

Given:

• Total Surface Area (T.S.A.) of cylinder = \(1848 \text{ sq. cm.}\)
• Curved Surface Area (C.S.A.) = \(\frac{5}{6}\) of T.S.A.

To find: Radius (r) and Height (h).

Calculate C.S.A.:

$$C.S.A. = \frac{5}{6} \times 1848$$ $$C.S.A. = 5 \times 308$$ $$C.S.A. = 1540 \text{ sq. cm.}$$

Formula for T.S.A. of a solid cylinder:

$$T.S.A. = C.S.A. + 2 \pi r^2$$

Substitute the values to find \(2 \pi r^2\):

$$1848 = 1540 + 2 \pi r^2$$ $$2 \pi r^2 = 1848 - 1540$$ $$2 \pi r^2 = 308$$

Now find \(r\):

$$2 \times \frac{22}{7} \times r^2 = 308$$ $$r^2 = \frac{308 \times 7}{2 \times 22}$$ $$r^2 = \frac{308 \times 7}{44}$$ $$r^2 = 7 \times 7$$ $$r^2 = 49$$ $$r = 7 \text{ cm}$$

Now, use the C.S.A. formula to find \(h\):

$$C.S.A. = 2 \pi r h$$ $$1540 = 2 \times \frac{22}{7} \times 7 \times h$$ $$1540 = 2 \times 22 \times h$$ $$1540 = 44h$$ $$h = \frac{1540}{44}$$ $$h = 35 \text{ cm}$$

The radius of the iron cylinder is 7 cm and the height is 35 cm.

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.

Given:

• External Radius \((R)\) = \(16 \text{ cm}\)
• Length (Height) \((h)\) = \(13 \text{ cm}\)
• Thickness = \(4 \text{ cm}\)

To find: Total Surface Area (T.S.A.) of the hollow wooden log.

First, calculate the Internal Radius \((r)\):

$$r = R - \text{thickness}$$ $$r = 16 - 4$$ $$r = 12 \text{ cm}$$

Formula for T.S.A. of a hollow cylinder:

$$T.S.A. = 2 \pi (R+r)(R-r) + 2 \pi (R+r)h$$

This can also be written as:

$$T.S.A. = 2 \pi (R+r) (R-r+h)$$

Substitute the values:

$$T.S.A. = 2 \times \frac{22}{7} \times (16+12) \times (16-12+13)$$ $$T.S.A. = 2 \times \frac{22}{7} \times (28) \times (4+13)$$ $$T.S.A. = 2 \times \frac{22}{7} \times 28 \times 17$$ $$T.S.A. = 2 \times 22 \times 4 \times 17$$ $$T.S.A. = 44 \times 68$$ $$T.S.A. = 2992 \text{ sq. cm.}$$

The Total Surface Area of the hollow wooden log is 2992 sq. cm.

4. A right angled triangle \(PQR\) where \(\angle Q=90^\circ\) is rotated about \(QR\) and \(PQ\). If \(QR=16 \text{ cm}\) and \(PR=20 \text{ cm}\), compare the curved surface areas of the right circular cones so formed by the triangle.

Given:

• Right-angled triangle \(PQR\), \(\angle Q=90^\circ\).
• \(QR = 16 \text{ cm}\) (base)
• \(PR = 20 \text{ cm}\) (hypotenuse)

To find: Compare C.S.A. when rotated about \(QR\) and \(PQ\).

First, find \(PQ\) using Pythagoras theorem:

$$PQ^2 + QR^2 = PR^2$$ $$PQ^2 + 16^2 = 20^2$$ $$PQ^2 + 256 = 400$$ $$PQ^2 = 400 - 256$$ $$PQ^2 = 144$$ $$PQ = 12 \text{ cm}$$

Case 1: Rotated about QR

When rotated about \(QR\):

• Height \(h_1 = QR = 16 \text{ cm}\)
• Radius \(r_1 = PQ = 12 \text{ cm}\)
• Slant height \(l_1 = PR = 20 \text{ cm}\)

Curved Surface Area \(C.S.A._1 = \pi r_1 l_1\)

$$C.S.A._1 = \pi \times 12 \times 20$$ $$C.S.A._1 = 240 \pi \text{ sq. cm.}$$

Case 2: Rotated about PQ

When rotated about \(PQ\):

• Height \(h_2 = PQ = 12 \text{ cm}\)
• Radius \(r_2 = QR = 16 \text{ cm}\)
• Slant height \(l_2 = PR = 20 \text{ cm}\)

Curved Surface Area \(C.S.A._2 = \pi r_2 l_2\)

$$C.S.A._2 = \pi \times 16 \times 20$$ $$C.S.A._2 = 320 \pi \text{ sq. cm.}$$

Comparison:

Ratio of C.S.A.s:

$$\frac{C.S.A._1}{C.S.A._2} = \frac{240 \pi}{320 \pi}$$ $$\frac{C.S.A._1}{C.S.A._2} = \frac{24}{32} = \frac{3}{4}$$

Thus, \(C.S.A._1 : C.S.A._2 = 3 : 4\).

The curved surface areas are in the ratio 3:4. The C.S.A. when rotated about QR is \(240 \pi \text{ sq. cm.}\) and when rotated about PQ is \(320 \pi \text{ sq. cm.}\).

5. 4 persons live in a conical tent whose slant height is 19 m. If each person require 22 m\(^2\) of the floor area, then find the height of the tent.

Given:

• Number of persons = 4
• Slant height \((l)\) of conical tent = \(19 \text{ m}\)
• Floor area required per person = \(22 \text{ m}^2\)

To find: Height \((h)\) of the tent.

Total floor area required = Number of persons \(\times\) Area per person

$$Total Area = 4 \times 22 = 88 \text{ m}^2$$

The floor area of a conical tent is the area of its circular base:

$$Area = \pi r^2$$ $$88 = \frac{22}{7} \times r^2$$ $$r^2 = \frac{88 \times 7}{22}$$ $$r^2 = 4 \times 7$$ $$r^2 = 28$$

Now, use the relationship between \(l, r, h\) for a cone: \(l^2 = r^2 + h^2\)

$$19^2 = 28 + h^2$$ $$361 = 28 + h^2$$ $$h^2 = 361 - 28$$ $$h^2 = 333$$ $$h = \sqrt{333}$$

To simplify \(\sqrt{333}\): \(333 = 9 \times 37\)

$$h = \sqrt{9 \times 37}$$ $$h = 3 \sqrt{37} \text{ m}$$

The height of the tent is \(3\sqrt{37} \text{ m}\).

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 cm\(^2\), how many caps can be made with radius 5 cm and height 12 cm.

Given:

• Total area of paper = \(5720 \text{ cm}^2\)
• For each cap:
  • Radius \((r)\) = \(5 \text{ cm}\)
  • Height \((h)\) = \(12 \text{ cm}\)

To find: Number of caps that can be made.

First, find the slant height \((l)\) for one cap using \(l^2 = r^2 + h^2\):

$$l^2 = 5^2 + 12^2$$ $$l^2 = 25 + 144$$ $$l^2 = 169$$ $$l = \sqrt{169}$$ $$l = 13 \text{ cm}$$

The area of paper required for one cap is its Curved Surface Area (C.S.A.):

$$C.S.A. = \pi r l$$ $$C.S.A. = \frac{22}{7} \times 5 \times 13$$ $$C.S.A. = \frac{1430}{7} \text{ sq. cm.}$$

Number of caps = \(\frac{\text{Total area of paper}}{\text{Area required for one cap}}\)

$$\text{Number of caps} = \frac{5720}{\frac{1430}{7}}$$ $$\text{Number of caps} = \frac{5720 \times 7}{1430}$$ $$\text{Number of caps} = \frac{572 \times 7}{143}$$

Since \(143 \times 4 = 572\):

$$\text{Number of caps} = 4 \times 7$$ $$\text{Number of caps} = 28$$

28 birthday caps can be made.

7. The ratio of the radii of two right circular cones of same height is 1:3. Find the ratio of their curved surface area when the height of each cone is 3 times the radius of the smaller cone.

Given:

• Ratio of radii \(r_1 : r_2 = 1 : 3\). Let \(r_1 = k\) and \(r_2 = 3k\).
• Both cones have the same height \(h\).
• Height of each cone \(h = 3 \times r_1 = 3k\).

To find: Ratio of their C.S.A.s, \(C.S.A._1 : C.S.A._2\).

For Cone 1:

• Radius \(r_1 = k\)
• Height \(h = 3k\)

Slant height \(l_1 = \sqrt{r_1^2 + h^2} = \sqrt{k^2 + (3k)^2}\)

$$l_1 = \sqrt{k^2 + 9k^2} = \sqrt{10k^2} = k\sqrt{10}$$

\(C.S.A._1 = \pi r_1 l_1 = \pi \times k \times k\sqrt{10} = \pi k^2 \sqrt{10}\)

For Cone 2:

• Radius \(r_2 = 3k\)
• Height \(h = 3k\)

Slant height \(l_2 = \sqrt{r_2^2 + h^2} = \sqrt{(3k)^2 + (3k)^2}\)

$$l_2 = \sqrt{9k^2 + 9k^2} = \sqrt{18k^2} = 3k\sqrt{2}$$

\(C.S.A._2 = \pi r_2 l_2 = \pi \times (3k) \times (3k\sqrt{2}) = 9 \pi k^2 \sqrt{2}\)

Ratio of C.S.A.s:

$$\frac{C.S.A._1}{C.S.A._2} = \frac{\pi k^2 \sqrt{10}}{9 \pi k^2 \sqrt{2}}$$ $$\frac{C.S.A._1}{C.S.A._2} = \frac{\sqrt{10}}{9\sqrt{2}} = \frac{\sqrt{5 \times 2}}{9\sqrt{2}} = \frac{\sqrt{5} \times \sqrt{2}}{9\sqrt{2}}$$ $$\frac{C.S.A._1}{C.S.A._2} = \frac{\sqrt{5}}{9}$$

The ratio of their curved surface areas is \(\sqrt{5} : 9\).

8. The radius of a sphere increases by 25%. Find the percentage increase in its surface area.

Given: Radius of a sphere increases by 25%.

To find: Percentage increase in its surface area.

Let the original radius be \(r\).

Original Surface Area \(S_1 = 4 \pi r^2\).

New radius \(r'\) after a 25% increase:

$$r' = r + 0.25r = 1.25r$$

New Surface Area \(S_2 = 4 \pi (r')^2\)

$$S_2 = 4 \pi (1.25r)^2$$ $$S_2 = 4 \pi (1.5625r^2)$$ $$S_2 = 1.5625 \times (4 \pi r^2)$$ $$S_2 = 1.5625 S_1$$

Increase in Surface Area = \(S_2 - S_1\)

$$\text{Increase} = 1.5625 S_1 - S_1 = 0.5625 S_1$$

Percentage increase = \(\frac{\text{Increase}}{S_1} \times 100\%\)

$$\text{Percentage increase} = \frac{0.5625 S_1}{S_1} \times 100\%$$ $$\text{Percentage increase} = 0.5625 \times 100\%$$ $$\text{Percentage increase} = 56.25\%$$

The percentage increase in its surface area is 56.25%.

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 ₹ 0.14 per cm\(^2\).

Given:

• Internal diameter = \(20 \text{ cm} \implies\) Internal radius \(r = 10 \text{ cm}\)
• External diameter = \(28 \text{ cm} \implies\) External radius \(R = 14 \text{ cm}\)
• Cost to paint = ₹ \(0.14 \text{ per cm}^2\)

To find: Total cost to paint the vessel all over.

The area to be painted is the Total Surface Area (T.S.A.) of the hollow hemisphere.

Formula for T.S.A. of a hollow hemisphere:

$$T.S.A. = \text{Outer C.S.A.} + \text{Inner C.S.A.} + \text{Area of the ring}$$ $$T.S.A. = 2 \pi R^2 + 2 \pi r^2 + \pi (R^2 - r^2)$$ $$T.S.A. = \pi (2R^2 + 2r^2 + R^2 - r^2)$$ $$T.S.A. = \pi (3R^2 + r^2)$$

Substitute the values:

$$T.S.A. = \frac{22}{7} (3 \times 14^2 + 10^2)$$ $$T.S.A. = \frac{22}{7} (3 \times 196 + 100)$$ $$T.S.A. = \frac{22}{7} (588 + 100)$$ $$T.S.A. = \frac{22}{7} (688)$$ $$T.S.A. = \frac{15136}{7} \text{ sq. cm.}$$

Total cost to paint = T.S.A. \(\times\) Cost per sq. cm.

$$\text{Cost} = \frac{15136}{7} \times 0.14$$ $$\text{Cost} = 15136 \times \frac{0.14}{7}$$ $$\text{Cost} = 15136 \times 0.02$$ $$\text{Cost} = 302.72$$

The total cost to paint the vessel all over is ₹ 302.72.

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 ₹ 2.

Given dimensions from the image for the frustum:

• Height \((h)\) = \(8 \text{ cm}\)
• Top diameter = \(6 \text{ cm} \implies\) Top radius \(r = 3 \text{ cm}\)
• Bottom diameter = \(12 \text{ cm} \implies\) Bottom radius \(R = 6 \text{ cm}\)
• Cost of painting = ₹ \(2 \text{ per sq. cm.}\)

To find: Total cost of painting the lamp (including the top part).

First, calculate the slant height \((l)\) of the frustum:

$$l = \sqrt{h^2 + (R-r)^2}$$ $$l = \sqrt{8^2 + (6-3)^2}$$ $$l = \sqrt{64 + 3^2}$$ $$l = \sqrt{64 + 9}$$ $$l = \sqrt{73} \text{ cm}$$

The area to be painted is the Curved Surface Area (C.S.A.) of the frustum plus the area of the top circular part.

Formula for C.S.A. of frustum: \(C.S.A. = \pi (R+r) l\)

$$C.S.A. = \pi (6+3) \sqrt{73}$$ $$C.S.A. = 9\pi \sqrt{73} \text{ sq. cm.}$$

Area of the top part = \(\pi r^2\)

$$\text{Area of top} = \pi (3^2) = 9\pi \text{ sq. cm.}$$

Total Area to be painted = C.S.A. of frustum + Area of top part

$$\text{Total Area} = 9\pi \sqrt{73} + 9\pi$$ $$\text{Total Area} = 9\pi (\sqrt{73} + 1) \text{ sq. cm.}$$

Using \(\pi \approx 3.14159\) and \(\sqrt{73} \approx 8.544\):

$$\text{Total Area} \approx 9 \times 3.14159 \times (8.544 + 1)$$ $$\text{Total Area} \approx 9 \times 3.14159 \times 9.544$$ $$\text{Total Area} \approx 269.44 \text{ sq. cm. (approx.)}$$

Total cost of painting = Total Area \(\times\) Cost per sq. cm.

$$\text{Cost} = 9\pi (\sqrt{73} + 1) \times 2$$ $$\text{Cost} = 18\pi (\sqrt{73} + 1)$$ $$\text{Cost} \approx 18 \times 3.14159 \times 9.544$$ $$\text{Cost} \approx 538.88$$

The total cost of painting the lamp is \(18\pi(\sqrt{73} + 1)\) Rupees (approximately ₹ 538.88).