Surface Area

Exercise 13.4 Part 2

Question 1: Find the surface area of a sphere of radius:
(i) 10.5 cm

Answer: Surface area of sphere `= 4π r^2`
`= 4 xx (22/7) xx 10.5 xx 10.5`
`= 132 xx 10.5 = 1386  sq  cm`

(ii) 5.6 cm

Answer: Surface area of sphere `= 4 π r^2`
`= 4 xx (22/7) xx 5.6 xx 5.6 = 394.24  sq  cm`

(iii) 14 cm

Answer: Surface area of sphere `= 4 π r^2`
`= 4 xx (22/7) xx 14 xx 14 = 2464  sq  cm`


Question 2: Find the surface area of a sphere of diameter:
(i) 14 cm

Answer: Surface area of sphere `= 4 π r^2`
`= 4 xx (22/7) xx 7 xx 7 = 616  sq  cm`

(ii) 21 cm

Answer: Surface area of sphere `= 4 π r^2`
`= 4 xx (22/7) xx 10.5 xx 10.5 = 1386  sq  cm`

(iii) 3.5 m

Answer: Surface area of sphere `= 4 π r^2`
`= 4 xx (22/7) xx 1.75 xx 1.75 = 38.5  sq  cm`


Question 3: Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

Answer: Total surface area of hemisphere `= 3 π r^2`
`= 3 xx 3.14 xx 10 xx 10 = 942  sq  cm`

Question 4: The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Answer: `r_1 = 7  cm, r_2 = 14  cm`
Surface area of smaller balloon `= 4 π r^2`
`= 4 π xx 7^2 = 4 π xx 49`

Surface area of bigger balloon `= 4 π r^2`
`= 4 π xx 14^2 = 4 π xx 196`
Ratio of areas `= (4 π xx 49)/(4 π xx 196) = 1 : 4`

Area of two similar shapes is in duplicate ratio of the ratio of their dimensions. This means when radius becomes double then surface area becomes four times, i.e. 22

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 16 per 100 cm2.

Answer: r = 5.25 cm
Curved surface area of hemisphere `= 2 π r^2`
`= 2 xx (22/7) xx 5.25 xx 5.25`
`= 33 xx 5.25 = 173.25  sq  cm`
`Co\st = ar\ea\ xx\ ra\te`
`= 173.25 xx (16/100) = Rs. 27.72`



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