# Class 9 RD Sharma Solutions- Chapter 21 Surface Area and Volume of a Sphere – Exercise 21.1

**Question 1: **Find the surface area of a sphere of radius:

**(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm**

**Solution:**

(i)Radius (r) = 10.5 cmSurface area = 4á´¨r

^{2}

^{ }= 4 * (22/7) * (10.5)^{2}cm^{2}

^{ }= 4 *(22/7) * (21/2) * (21/2) cm^{2}= 1386 cm

^{2}

(ii)Radius (r) = 5.6 cmSurface area = 4á´¨r

^{2}= 4 * (22/7) * (5.6)

^{2}cm^{2}= 4 *(22/7) * (56/10) * (56/10) cm

^{2}= 39424/100 = 394.24 cm

^{2}

(iii)Radius (r) = 14 cmSurface area = 4á´¨r2

= 4 * (22/7) * (14)

^{2}cm^{2}= 4 *(22/7) * (14) * (14) cm

^{2}= 2464 cm

^{2}

**Question 2:** Find the surface area of a sphere of diameter:

### (i) 14 cm (ii) 21 cm (iii) 3.5 cm

**Solution:**

(i)Diameter of a sphere = 14 cmRadius (r) = 14/2 cm = 7 cm

Surface area = 4á´¨r

^{2}= 4 * (22/7) * (7)

^{2}cm^{2}= 4 *(22/7) * (7) * (7) cm

^{2}= 616 cm

^{2}

(ii)Diameter of a sphere = 21 cmRadius (r) = 21/2 cm

Surface area = 4á´¨r

^{2}= 4 *(22/7) * (21/2) * (21/2) cm

^{2}= 1386 cm^{2}

(iii)Diameter of a sphere = 3.5 cmRadius (r) = 3.5/2 = 7/4 cm

Surface area = 4á´¨r

^{2}= 4 *(22/7) * (7/4) * (7/4) cm

^{2}

^{ }= 77/2 = 38.5 cm^{2}

**Question 3:** Find the total surface area of a hemisphere and a solid hemisphere each of radius 10 cm. (Ï€=3.14)

**Solution:**

1.Radius of hemisphere = 10 cm

Total surface area of hemisphere = 2á´¨r

^{2 }

^{ }= 2 * 3.14 * 10 * 10 cm^{2}= 628 cm^{2}2. Total surface area of solid hemisphere = 3á´¨r

^{2}= 3 * 3.14 * 10 * 10 cm

^{2}= 942 cm^{2}

**Question 4: **The surface area of a sphere is 5544 cm2, find its diameter.

**Solution:**

Let r be the radius of a sphere, then surface area = 4á´¨r

^{2}So, 5544 = 4 * (22/7) * r

^{2}r

^{2}= (5544 * 7)/(4 * 22) = 63 * 7 cm^{2}= 441 = (21)

^{2}So, r = 21 cm

Now diameter = 2 * r = 2 * 21 = 42 cm

**Question 5:** A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin plating it on the inside at the rate of Rs.4 per 100 cm^{2}.

**Solution:**

Inner diameter of a hemispherical bowl = 10.5 cm

Radius (r) = 10.5/2 = 5.25 cm = 525/100 = 21/4 cm

Surface area of inner part of bowl = 2á´¨r

^{2 }= 2 *(22/7) * (21/4) * (21/4) cm^{2 }= 693/4 cm^{2}Rate of tin plating = Rs 4 per 100 cm

^{2}Total cost = (693 * 4) / (4 * 100) = Rs 693/100 = Rs 6.93

**Question 6:** The dome of a building is in the form of a hemisphere. Its radius is 63 dm. Find the cost of painting it at the rate of Rs. 2 per sq m.

**Solution:**

Radius of dome (hemispherical) = 63 dm

Area of curved surface = 2á´¨r

^{2}= 2 * (22/7) * 63 * 63 dm^{2}= 24948 dm^{2}Rate of painting = Rs 2 per sq. meter

Total cost = (24948 * 2) / 100 = Rs 249.48 * 2 = Rs 498.96

### Question 7: Assuming the earth to be a sphere of radius 6370 km, how many square kilometres is area of the land, if three-fourth of the earthâ€™s surface is covered by water?

**Solution:**

Radius of earth (sphere) = 6370 km

Water on earth = 3/4 % of total area

Required area = (1/4) * (4á´¨r

^{2}) = á´¨r^{2}= (22/7) * (6370)

^{2}km^{2}= 22 * 910 * 6310 km^{2 }= 127527400 km^{2}

**Question 8:** A cylinder of same height and radius is placed on the top of a hemisphere. Find the curved surface area of the shape if the length of the shape be 7 cm.

**Solution:**

Total height of the so formed shape = 7cm

Radius = height of cylinder = 7/2 cm

Curved surface area = 2á´¨rh + 2á´¨r

^{2}= 2á´¨r(h+r) cm^{2}= 2 * (22/7) * (7/2) * (7/2 + 7/2) cm

^{2}= 22 * 7 = 154 cm^{2}

**Question 9:** The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

**Solution:**

Diameter of moon = 1/4 of diameter of earth

Let radius of earth be r km

Then the radius of moon = (r / 4) km

Now, surface area of earth = 4á´¨r

^{2}Surface area of moon = 4á´¨(r/4)

^{2}= 4á´¨ * (1/16) r

^{2}= (1/4) * á´¨r^{2}Ratio between surface area of moon and earth = (1/4) * á´¨r

^{2}: 4á´¨r^{2}= (1/4) : 4 = 1/16

**Question 10: **A hemi-spherical dome of a building needs to be painted. If the circumference of the base of the dome is 17.6 m, find the cost of painting it, given the cost of painting is â‚¹5 per 100 cm^{2}. [NCERT]

**Solution:**

Circumference (c) of the base of dome (r) = 17.6 cm

Radius = c/2á´¨ = (17.6 * 7) / (2 * 22) = 2.8 m

Surface area = 2á´¨r

^{2}= 2 * (22/7) * (2.8)^{2}m^{2 }= 49.28 m^{2}Rate of painting the surface = Rs 5 per 100 cm

^{2}Total cost = (49.28 * 5 * 10000) / 100 = Rs 24640

**Question 11:** A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 16 cm and its height is 15 cm. Find the cost of painting the toy at â‚¹7 per 100 cm^{2}.

**Solution:**

Diameter of toy = 16 cm

Radius (r) = 16/2 = 8 cm

Height of conical part (h) =15 cm

Slant height (l) = sqrt. (r

^{2}+ h^{2})= sqrt. (8

^{2}+ 15^{2}) = sqrt (64 + 225) = sqrt (289) = 17 cmTotal surface area of the toy = á´¨rl + 2á´¨r

^{2}= (22/7) *8 * 17 + 2 * (22/7) * 8 * 8 cm

^{2}= (176/7) * 33 cm^{2}= (5808/7) cm^{2}Rate of painting the surface of the toy = Rs 7 per cm

^{2}Total cost = (5808/7) * (7/100) = Rs (5808/100) = Rs 58.08

**Question 12:** A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of â‚¹10 per m^{2}.

**Solution:**

Diameter of tank = 1.4 m

Radius (r) = 1.4/2 = 0.7 m

Height of cylindrical portion = 8m

Outer surface area of tank = 2á´¨rh + 2á´¨r

^{2}= 2á´¨r (h + r)= 2 * (22/7) * 0.7 * (8 + 0.7) m

^{2}= (44/10) * 8.7 m^{2}Rate of painting = Rs 10 per m

^{2}Total cost = (44 * 8.7 *10) /10 = Rs 382.80

**Question 13:** The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in the figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm^{2} and black paint costs 5 paise per cm^{2}. [NCERT]

**Solution:**

Diameter of each sphere = 21 cm

Radius (R) = 21/2 cm

Radius of each cylinder(r) = 1.5 cm and height (h) = 7 cm

Now surface area of one sphere = 4á´¨R

^{2}= 4 * (22/7) * (21/2) * (21/2) cm

^{2}= 1386 cm^{2}Surface area of one cylinder = 2á´¨rh

= 2 * (22/7) * 1.5 * 7 cm

^{2}= 66 cm^{2}Surface area of 8 spheres = 8 * 1386 cm

^{2}= 11088 cm^{2}Surface area of 8 cylinders tops = 8á´¨r

^{2}= 8 * (22/7) * 1.5 * 1.5 cm^{2}= 56.57 cm^{2}Surface area of 8 cylinders = 8 * 66 cm

^{2}Surface area of spheres excluding base area = 11088 – 56.57 = 11031.43 cm

^{2}Rate of silver painting the spheres = Rs 25 per cm

^{2}Total cost = Rs (11031.43 *25)/100 = Rs 2757.86

Rate of black painting the spheres = Rs 5 per cm

^{2}Total cost = Rs (528 * 5)/100 = Rs 26.40

Total cost of painting = Rs 2757.86 + Rs 26.40 = Rs 2784.26

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