Surface Area

Exercise 13.4

Part 1

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

Answer: Surface area of sphere `=4πr^2`

`=4xx(22)/7xx10.5xx10.5`

`=132xx10.5` = 1386 sq cm

(ii) 5.6 cm

Answer: Surface area of sphere `=4πr^2`

`=4xx(22)/7xx5.6xx5.6=394.24` sq cm

(iii) 14 cm

Answer: Surface area of sphere `=4πr^2`

`=4xx(22)/7xx14xx14=2464` sq cm


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

Answer: Surface area of sphere `=4πr^2`

`=4xx(22)/7xx7xx7=616` sq cm

(ii) 21 cm

Answer: Surface area of sphere `=4πr^2`

`=4xx(22)/7xx10.5xx10.5=1386` sq cm

(iii) 3.5 m

Answer: Surface area of sphere `=4πr^2`

`=4xx(22)/7xx1.75xx1.75=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 `=3πr^2`

`=3xx3.14xx10xx10=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 `=4πr^2`

`=4π\xx7^2=4π\xx49`

Surface area of bigger balloon `=4πr^2`

`=4π\xx14^2=4π\xx196`

Ratio of areas `=(4π\xx49)/(4π\xx196)` = 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:Given: r = 5.25 cm

Curved surface area of hemisphere `=2πr^2`

`=2xx(22)/7xx5.25xx5.25`

`=33xx5.25=173.25` sq cm

Cost = Area × Rate

`=173.25xx(16)/(100)` = Rs. 27.72



Copyright © excellup 2014