The document RD Sharma Solutions -Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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**Q 14: The diameters of two cones are equal. If their slant height are in the ratio 5:4,find the ratio of their curved surfaces.**

**Solution:**

It is given that

Diameters of two cones are equal

Therefore their radius are also equal i.e

r_{1} = r_{2}

Let the ratio of slant height be x

Therefore l_{1 }= 5x

l_{2 }= 4x

Therefore Ratio of curved surface area =

=

∴ Ratio of curved surface area is 5:4.

**Q 15: Curved surface area of a cone is 308 cm ^{2} and its slant height is 14cm. Find the radius of the base and total surface area of the cone.**

**Solution:**

(1) It is given that

Slant height of cone = 14cm

Let radius of circular end of cone = r

Curved surface area of cone = πrl

⇒ 308 cm^{2} = ∗r∗14

⇒ r = = 7cm

Thus radius of circular end of cone = 7 cm.

(ii) It is given that C.S.A = 308 cm^{2}

We know that total surface area of a cone

= curved surface area of a cone + Area of base

= πr_{l} + πr_{2}

= [308 + (∗7^{2})]

= 308 + 154

= 462cm^{2}

Thus total surface area of the cone is 462cm^{2}

**Q 16: The slant height and base diameter of a conical tomb are 25m and 14m respectively. Find the cost of white washing its curved surface area at the rate of Rs 210 per 100m ^{2}.**

**Solution:**

It is given that

Slant height of conical tomb(l) = 25m

Base radius (r) of tomb = = 7m

Curved surface area of conical length tomb = πrl

= ∗7∗25

= 550 m^{2}

Cost of white washing 100 m^{2 }area = Rs 210

Cost of white washing 550m^{2} area = = Rs 1155

Therefore the cost of white washing the whole tomb is Rs 1155.

**Q 17: A conical tent is 10 m high and the radius of it base is 24 m. Find the slant height of the tent. If the cost of 1m ^{2} canvas is Rs 70, find the cost of canvas required for the tent.**

**Solution:**

It is given that

Height of the conical tent(h) = 10m

Radius of conical tent(r) = 24m

Let slant height of conical tent be l

l^{2} = h^{2} + r^{2}

= 10^{2 }+ 24^{2} = 100 + 576

= 676 m^{2}

⇒ l = 26m

Thus,the slant height of the conical tent is 26m.

(ii) It is given that:

Radius(r) = 24 m

Slant height (l) = 26 m

C.S.A of tent = πrl

= ∗24∗26

Cost of 1m^{2} canvas = Rs 70

Cost of m^{2} canvas = Rs *70 = Rs 1,37,280

Thus the cost of canvas required to make the tent is Rs 1,37,280.

**Q 18: A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24 m. The height of the cylindrical portion is 11m while the vertex of the cone is 16 m above the ground. Find the area of the canvas required for the tent.**

**Solution:**

It is given that

Diameter of cylinder = 24 m

therefore radius =

= 12 cm

Also radius of cone = 12 m

Height of cylinder = 11 m

Height of cone = 16−11 = 5m

Slant height of cone

= 13m

Therefore area of canvas required for the tent = πrl + 2πrh

= [(12∗13) + (2∗12∗11)] = 490.286 + 829.714

= 1320m^{2}

**Q19. A circus tent is cylindrical to a height of 3 m and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent.**

**Solution:**

Given diameter = 105 m,

Radius = m = 52.5 m

Therefore curved surface area of circus tent = πrl + 2πrh

= (∗52.5∗53) + (2∗∗52.5∗3)

= 8745 + 990 = 9735 m^{2}

Therefore length of the canvas required for tent

= 1947 m

**Q20 The circumference of the base of a 10m height conical tent is 44m, calculate the length of canvas used in making the tent if width of canvas is 2m.**

**Solution:**

We know that

C.S.A of cone = πrl

Given circumference = 2πr

⇒ 2∗∗r = 44

⇒ r = 7m

Therefore l =

Therefore C.S.A of tent = πrl

Therefore the length of canvas used in making the tent

= 134.2 m

**Q 21: What length of tarpaulin 4 m wide will be required to make a conical tent of height 8 m and base radius 6 m?Assume that the extra length of material will be required for stitching margins and wastage in cutting is approximately 20 cm.**

**Solution:**

Given that,

Height of conical tent(h) = 8m

Radius of base of tent(r) = 6m

Slant height(l) =

C.S.A of conical tent = πrl

= (3.14*6*10) m^{2} = 188.4 m^{2}

Let the length of tarpaulin sheet required be l

As 20 cm will wasted ,so effective

Length will be (l – 0.2 m)

Breadth of tarpaulin = 3m

Area of sheet = C.S.A of sheet

[l*0.2*3]m = 188.4 m^{2} = l – 0.2 m = 62.8 m

Accounting extra for wastage:

⇒ l = 63 m

Thus the length of the tarpaulin sheet will be = 63 m

**Q 22: A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled card board. Each cone has a base diameter of 40 cm and height 1m. If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per m ^{2},what will be the cost of painting all these cones?**

**Solution:**

The area to be painted is the curved surface area of each cone.

The formula of the curved surface area of a cone with base radius and slant height 7 is given as

Curved Surface Area = πrl

For each cone, we’re given that the base diameter is 0.40 m.

Hence the base radius r = 0.20 m.

The vertical height l 1 = 1 m.

To find the slant height ‘l’ to be used in the formula for Curved Surface Area we use the following relation

Slant height ,

l = 1.02 m

Now substituting the values of r = 0.2 m and slant height 1 = 1.02 m and using pi = 3.14 in the formula of C.S.A.

We get Curved Surface Area = (3.14)(0.2)(1.02) = 0.64056 m^{2}

This is the curved surface area of a single cone.

Since we need to paint 50 such cones the total area to be painted is,

Total area to be painted = (0.64056) (50) = 32.028 m^{2}

The cost of painting is given as Rs. 12 per m^{2}

Hence the total cost of painting = (12) (32.028) = 384.336

Hence, the total cost that would be incurred in the painting is Rs. 384.336

**Q23. A cylinder and a cone have equal radii of their base and equal heights. If their curved surface area are in the ratio 8:5, show that the radius of each is to the height of each as 3:4.**

**Solution:**

It is given that the base radius and the height of the cone and the cylinder are the same.

So let the base radius of each is ‘ r ‘ and the vertical height of each is ‘h’.

Let the slant height of the cone be ‘ l ’.

The curved surface area of the cone = πrl

The curved surface area of the cylinder = Extra close brace or missing open brace

It is said that the ratio of the curved surface areas of the cylinder to that of the cone is 8:5

So,

But we know that l =

Squaring on both sides we get:

Hence it is shown that the ratio of the radius to the height of the cone as well as the cylinder is: 3 : 4

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