AP State Syllabus AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.3 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 10th Lesson Surface Areas and Volumes Exercise 10.3

Question 1.

The base area of a cone is 38.5 cm Its volume is 77 cm^{3}. Find its height.

Solution:

Base area of the cone, πr^{2} = 38.5 cm^{2}

Volume of the cone, V = \(\frac{1}{3}\) πr^{2} h = 77

πr^{2} = 38.5

\(\frac{22}{7}\) r^{2} = 38.5

r^{2} = 38.5 x \(\frac{7}{22}\)

r^{2} = 12.25

r = \(\sqrt{12.25}\) = 3.5

V= \(\frac{1}{22}\) x \(\frac{22}{7}\) x 3.5 x 3.5 x h = 77

∴ h = \(\frac{77 \times 3 \times 7}{22 \times 12.25}\) = 6

∴ Height of the cone = 6 cm

Question 2.

The volume of a cone is 462 m^{3}. Its base radius is 7 m. Find its height.

Solution:

The volume of a cone”V’= \(\frac{1}{3}\) πr^{2} h = 462

Radius ‘r’ = 7 m

Height = h (say)

\(\frac{1}{3}\) x \(\frac{22}{7}\) x 7 x h = 462

h = \(\frac{462 \times 3}{22 \times 7}\) = 9

∴ Height = 9m

Question 3.

Curved surface area of a cone is 308 cm^{2} and its slant height is 14 cm. Find (1) radius of the base (ii) total surface area of the cone.

Solution:

C.S.A. of the cone, πrl = 308

Slant height, l = 14 cm

i) πrl = 308; l = 14 cm

\(\frac{22}{7}\) x r x 14 = 308

r = \(\frac{308}{44}\) = 7cm

ii) T.S.A. = πrl + πr^{2}

= πr (r + l) = \(\frac{22}{7}\) x 7 x (7 + 14)

= 22 x 21 = 462 cm^{3}

Question 4.

The cost of painting the total surface area of a cone at 25 paise per cm2 is ₹176. Find the volume of the cone, if its slant height is 25 cm.

Solution:

Slant height of the cone, l = 25 cm

Total cost at the rate of 25 p/cm^{2}

= ₹176

∴ Total surface area of the cone

= \(\frac{176}{25}\) x 100 = 176 x 4 = 704cm^{2}

But T.S.A. of the cone = πr (r + l) = 704

Thus \(\frac{22}{7}\)r(r + 25) = 704

r(r + 25) = \(\frac{704 \times 7}{22}\) = 224

r^{2} + 25r = 224

⇒ r^{2} + 32r – 7r – 224 = 0

⇒ r (r + 32) – 7 (r + 32) = 0

⇒ (r + 32) (r – 7) = 0

⇒ r = 7 (∵ ’r’ can’t be negative)

∴ Volume of the cone = \(\frac{1}{3}\) πr^{2}h

= \(\frac{1}{3}\) x \(\frac{22}{7}\) x 7 x 7 24

= 22 x 7 x 8 = 1232cm^{3}

Question 5.

From a circle of radius 15 cm, a sector with angle 216° is cut out and its bounding radii are bent so as to form a cone. Find its volume.

Solution:

Radius of the sector, ‘r’ = 15 cm

Angle of the sector, ‘x’ = 216°

∴ Length of the arc, l = \(\frac{x}{360}\) x 2πr

\(=\frac{216}{360} 2 \pi r=\frac{3}{5}(2 \pi r)\)

Perimeter of the base of the cone = Length of the arc

2πr of cone = \(=\frac{6}{5}\) πr of the circle

Radius of the cone ‘r’ = \(\frac{3}{5}\) x 15 = 9

Radius ‘r’ of the circle = slant height l of the cone = 9 cm

= 1018.3 cm^{3}

Question 6.

The height of a tent is 9 m. Its base diameter is 24 m. What is its slant height ? Find the cost of canvas cloth required if it costs ₹14 per sq.m.

Solution:

Height of a conical tent ‘h’ = 9 m

Base diameter = 24 m

Thus base radius ’r’= \(\frac{d}{2}=\frac{24}{2}\) = 12m

Cost of canvas = ₹ 14 per sq.m.

C.S.A. of the cone = πrl

Question 7.

The curved surface area of a cone is 1159\(\frac { 5 }{ 7 }\) cm^{2}. Area of its base is 254 \(\frac { 4 }{ 7 }\) cm^{2}. Find its volume.

Solution:

C.S.A. of the cone = πrl

Question 8.

A tent is cylindrical to a height of 4.8 m and conical above it. The ra¬dius of the base is 4.5 m and total height of the tent is 10.8 m. Find the canvas required for the tent in square

meters.

Solution:

Radius of cylinder, l = 4.5 m

Height of the cylinder = h = 4.8 m

∴ C.S.A. of the cylinder = 2πrh

= 2 x \(\frac{22}{7}\) x 4.5 x 4.8

= 135.771 m^{2}

Radius of the cone ‘r’ =

Radius of the cylinder = 4.5 m

Height of the cone ’h’ = 10.8 – 4.8 = 6 m

∴ Slant height of the cone

∴ C.S.A. of the cone = πrl

= \(\frac{22}{7}\) x 4.5 x 7.5

= \(\frac{742.5}{7}\) = 106.071m^{2}

∴ Total canvas required

= C.S.A of cylinder + C.S.A. of cone

= 135.771 + 106.071

= 241.842 m^{2}

Question 9.

What length of tarpaulin 3 m wide will be required to make a conical tent of height 8 m and base radius 6m? Assume that extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (use π = 3.14) [Note : Take 20 cm as 0.6 m^{2}]

Solution:

Radius of the cone, r = 6 m

Height of the cone, h = 8 m

∴ C.S.A. = πrl = 3.14 x 6 x 10 = 188.4 m^{2}

Let the length of the tarpaulin = l

∴ Area of the tarpaulin, lb = 188.4 + 0.6

= 189 m^{2}

⇒ 3l = 189

⇒ l = \(\frac{189}{3}\) = 63m

Question 10.

A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 27 cm. Find the area of the sheet required to make 10 such caps.

Solution:

Radius of the cone, r = 7 cm

Height of the cone, h = 27 cm

Slant height of the cone (l)

∴ Total area of the sheet required for

10 caps = 10 x 22 \(\sqrt{778}\)

= 6136.383 cm^{2}

Question 11.

Water is pouring into a conical vessel (as shown in the given figure), at the rate of 1.8 m^{3} per minute. How long will it take to fill the vessel?

Solution:

From the figure, diameter of the cone 5.2 m

Thus its radius ’r’ = \(\frac{5.2}{2}\) = 2.6 m

∴ Height of the cone = h = 6.8 m

Volume of the cone = \(\frac{1}{3}\) πr^{2} h

= \(\frac{1}{3} \times \frac{22}{7}\) x 2.6 x 2.6 x 6.8

= \(\frac{1011.296}{21}\)

= 48.156 m^{3}

Quantity of water that flows per minute

= 1.8 m^{3}

∴ Total time required = \(\frac{\text { Total volume }}{1.8}\)

= \(\frac{48.156}{1.8}\) = 26.753

27 minutes.

Question 12.

Two similar cones have volumes 12π CU. units and 96π CU. units. If the curved surface area of smaller cone is 15π sq.units, what is the curved surface area of the larger one?

Solution:

πrl = 15 π

\(r \sqrt{\left(r^{2}+h^{2}\right)}=15\)

Squaring on both sides

r^{2} (r^{2} + h^{2}) = 15 x 15

= 3 x 5 x 3 x 5

= 3 x 3 x 25

r^{2}(r^{2} + h^{2}) = 3^{2}(3^{2} + 4^{2})

∴ r = 3 cm, h = 4 cm

C.S.A. = π x 3 x \(\left(\sqrt{3^{2}+4^{2}}\right)\) = 15π

\(\frac{1}{3}\)πr^{2}H = 96π

\(\frac{3 \times 3 \times \mathrm{H}}{3}\) = 96

∴ H = \(\frac{96}{3}\) = 32 units