GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

   

Gujarat Board GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

Question 1.
The diameter of the base of the cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Solution:
∴ Diameter of the base = 10.5 cm
∴ Radius of the base (r) = \(\frac {10.5}{7}\) cm = 5.25 cm
Slant height (l) = 10 cm
∴ Curved surface area of the cone = πrl
= \(\frac {22}{7}\) x 5.25 x 10 = 165 cm2

GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

Question 2.
Find the total surface area of a cone, if its slant height is 21 m and the diameter of its base is 24 m.
Solution:
Slant height (l) = 21 m
Diameter of base = 24 m
∴ Radius of base (r) = \(\frac {24}{2}\) m = 12 m
∴ Total curved surface area of the cone
= πr (l + r) = \(\frac {22}{7}\) x 12 (21+ 12)
= \(\frac {22}{7}\) x 12 x 33
= \(\frac {8712}{7}\) = 1244 \(\frac {4}{7}\)m2

Question 3.
The curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find
1. the radius of the base and
2. the total surface area of the cone.
Solution:
1. Slant height (l) = 14 cm
Curved surface area = 308 cm2
πrl = 308
\(\frac {22}{7}\) x r x 14 = 308
r = \(\frac{308 \times 7}{22 \times 14}\) = r = 7cm
Hence, the total radius of the base is 7 cm.

GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

2. Total surface area of the cone
= πr (l + r) = \(\frac {22}{7}\) x 7 x (14 + 7)
= \(\frac {22}{7}\) x 7 x 21 = 462 cm2
Hence, the total surface area of the cone is 462 cm2.

Question 4.
A conical tent is 10 m high and the radius of its base is 24 m. Find:
1. the slant height of the tent
2. cost of the canvas required to make the tent, if the cost of 1 m2 canvas is ₹ 70.
Solution:
1. h = 1o m and r = 24
l = \(\sqrt{r^{2}+h^{2}}\) = \(\sqrt{(24)^{2}+(10)^{2}}\)
= \(\sqrt{576+100}\) = \(\sqrt{676}\) = 26m
Hence, the slant height of the tent is 26 m.

2. Curved surface area of the tent = πrl
= \(\frac {22}{7}\) x 24 x 26 m2

∴ Cost of the canvas required to make the tent, if the cost of 1 m2 canvas is ₹ 70
= ₹ \(\frac {22}{7}\) x 24 x 26 x 70
= ₹ 1,37,280
Hence, the cost of the canvas is 1,37,280.

GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

Question 5.
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use it = 3.14)
Solution:
For conical tent
h = 8m
r = 6 m
l = \(\sqrt{r^{2}+h^{2}}\) = \(\sqrt{(6)^{2}+(8)^{2}}\)
= \(\sqrt{36+64}\) = \(\sqrt{100}\) = 10m
∴ Curved surface area = itrl
= 3.14 x 6 x 10 = 188.4 m2
Width of tarpaulin = 3 m
∴ Length of tarpaulm = \(\frac{188.4}{3}\) = 62.8 m
Extra length of the material required
= 20 cm = 0.2 m
∴ Actual length of tarpaulin required
= 62.8 m + 0.2 m = 63 m

GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

Question 6.
The slant height and a base diameter of a conical tomb is 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹ 210 per 100 m2.
Solution:
Slant height (l) = 25 m
Base diameter (d) = 14 m
∴ Base radius (r) = \(\frac{14}{2}\)m = 7m
∴ Curved surface area of the tomb
= πrl = \(\frac {22}{7}\) x 7 x 25 = 550 m2
∴ Cost of white-washing the curved surface area of the tomb at the rate of 210 per 100 m2
= ₹ \(\frac {210}{100}\) x 550 = ₹ 1155

Question 7.
A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Solution:
Base radius (r) = 7 cm
Height (h) = 24 cm
∴ Slant height (l) = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{(7)^{2}+(24)^{2}}\) = \(\sqrt{49+576}\)
= \(\sqrt{625}\) = 25 cm
∴ Curved surface area of a cap = πrl
= \(\frac {22}{7}\) x 7 x 25 = 550 cm2
∴ Curved surface area of 10 caps
= 550 x 10 = 5500 cm2
Hence, the area of the sheet required to make 10 such caps is 5500 cm2

GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

Question 8.
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and a height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is 12 per m2, what will be the cost of painting all these cones?
(Use n = 3.14 and take \(\sqrt{1.04}\) = 1.02)
Solution:
Base diameter = 40 cm
∴ Base radius (r) = \(\frac {40}{2}\) cm = 20 cm
= \(\frac {20}{100}\) m = 0.2 m
Height(h) = 1 m
∴ l = \(\sqrt{r^{2}+h^{2}}\) = \(\sqrt{(0.2)^{2}+(1)^{2}}\)
= \(\sqrt{0.04+1}\) = \(\sqrt{1.04}\) = 1.02 m
∴ Curved surface area = πrl
= 3.14 x 0.2 x 1.02 = 0.64056 m2
∴ Curved surface area of 50 cones
= 0.64056 x 50 m2 = 32.028 m2
∴ Cost of painting all these cones at ₹ 12 per m2
= ₹ 32.028 x 12
= ₹ 384.336 = ₹ 384.34 (approximately)

GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

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