EXERCISE 13.4
Q 1, Ex 13.4, Page No 225, Surface Areas & Volumes - NCERT Class 9th Maths
Page No 225:
Question 1:
Find the surface area of a sphere of radius:(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm
$\left[\text { Assume } \pi=\frac{22}{7}\right]$
Answer:
(i) Radius (r) of sphere = 10.5 cmSurface area of sphere = 4πr2
$=\left[4 \times \frac{22}{7} \times(10.5)^{2}\right] \mathrm{cm}^{2}$
$=\left(4 \times \frac{22}{7} \times 10.5 \times 10.5\right) \mathrm{cm}^{2}$
$=(88 \times 1.5 \times 10.5) \mathrm{cm}^{2}$
$=1386 \mathrm{cm}^{2}$
Therefore, the surface area of a sphere having radius 10.5cm is 1386 cm2.
(ii) Radius(r) of sphere = 5.6 cm
Surface area of sphere = 4πr2
$=\left[4 \times \frac{22}{7} \times(5.6)^{2}\right] \mathrm{cm}^{2}$
$=(88 \times 0.8 \times 5.6) \mathrm{cm}^{2}$
$=394.24 \mathrm{cm}^{2}$
Therefore, the surface area of a sphere having radius 5.6 cm is 394.24 cm2.
(iii) Radius (r) of sphere = 14 cm
Surface area of sphere = 4πr2
$=\left[4 \times \frac{22}{7} \times(14)^{2}\right] \mathrm{cm}^{2}$
$=(4 \times 44 \times 14) \mathrm{cm}^{2}$
$=2464 \mathrm{cm}^{2}$
Therefore, the surface area of a sphere having radius 14 cm is 2464 cm2.
Q 2, Ex 13.4, Page No 225, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 2:
Find the surface area of a sphere of diameter:(i) 14 cm (ii) 21 cm (iii) 3.5 m
$\left[\text { Assume } \pi=\frac{22}{7}\right]$
Answer:
(i) Radius (r) of sphere =\frac{\text { Diameter }}{2}=\left(\frac{14}{2}\right) \mathrm{cm}=7cm
Surface area of sphere = 4πr2
$=\left(4 \times \frac{22}{7} \times(7)^{2}\right) \mathrm{cm}^{2}$
$=(88 \times 7) \mathrm{cm}^{2}$
$=616 \mathrm{cm}^{2}$
Therefore, the surface area of a sphere having diameter 14 cm is 616 cm2.
(ii) Radius (r) of sphere $=\frac{21}{2}=10.5 \mathrm{cm}$
Surface area of sphere = 4πr2
$=\left[4 \times \frac{22}{7} \times(10.5)^{2}\right] \mathrm{cm}^{2}$
$=1386 \mathrm{cm}^{2}$
Therefore, the surface area of a sphere having diameter 21 cm is 1386 cm2.
(iii) Radius (r) of sphere $=\frac{3.5}{2}$
=1.75m
Surface area of sphere = 4πr2
$=\left[4 \times \frac{22}{7} \times(1.75)^{2}\right] \mathrm{m}^{2}$
$=38.5 \mathrm{m}^{2}$
Therefore, the surface area of the sphere having diameter 3.5 m is 38.5 m2.
Q 3, Ex. 13.4, Page No 225, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 3:
Find the total surface area of a hemisphere of radius 10 cm. [Use π = 3.14]Answer:
Radius (r) of hemisphere = 10 cm
Total surface area of hemisphere = CSA of hemisphere + Area of circular end of hemisphere
$=2 \pi r^{2}+\pi r^{2}$
$=3 \pi r^{2}$
$=\left[3 \times 3.14 \times(10)^{2}\right] \mathrm{cm}^{2}$
$=942 \mathrm{cm}^{2}$
Therefore, the total surface area of such a hemisphere is 942 cm2.
Q 4, Ex. 13.4, Page No 225, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
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:
Radius (r1) of spherical balloon = 7 cmRadius (r2) of spherical balloon, when air is pumped into it = 14 cm
Required ratio $=\frac{\text { Initial surface area }}{\text { Surface area after pumping air into balloon }}$
$=\frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}=\left(\frac{r_{1}}{r_{2}}\right)^{2}$
$=\left(\frac{7}{14}\right)^{2}=\frac{1}{4}$
Therefore, the ratio between the surface areas in these two cases is 1:4.
Q 5, Ex. 13.4, Page No 225, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
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. $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Inner radius (r) of hemispherical bowl =\left(\frac{10.5}{2}\right) \mathrm{cm}=5.25cm
Surface area of hemispherical bowl = 2πr2
$=\left[2 \times \frac{22}{7} \times(5.25)^{2}\right] \mathrm{cm}^{2}$
$=173.25 \mathrm{cm}^{2}$
Cost of tin-plating 100 cm2 area = Rs 16
Cost of tin-plating 173.25 cm2 area $=\operatorname{Rs}\left(\frac{16 \times 173.25}{100}\right)$ = Rs 27.72
Therefore, the cost of tin-plating the inner side of the hemispherical bowl is Rs 27.72.
Q 6, Ex. 13.4, Page No 225, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 6:
Find the radius of a sphere whose surface area is 154 cm2. $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Let the radius of the sphere be r.Surface area of sphere = 154
∴ 4πr2 = 154 cm2
$r^{2}=\left(\frac{154 \times 7}{4 \times 22}\right) \mathrm{cm}^{2}=\left(\frac{7 \times 7}{2 \times 2}\right) \mathrm{cm}^{2}$
$r=\left(\frac{7}{2}\right) \mathrm{cm}$
=3.5cm
Therefore, the radius of the sphere whose surface area is 154 cm2 is 3.5 cm.
Q 7, Ex. 13.4, Page No 225, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 7:
The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface area.Answer:
Let the diameter of earth be d. Therefore, the diameter of moon will be$\frac{d}{4}$.Radius of earth $=\frac{d}{2}$
Radius of moon $=\frac{1}{2} \times \frac{d}{4}=\frac{d}{8}$
Surface area of moon $=4 \pi\left(\frac{d}{8}\right)^{2}$
Surface area of earth $=4 \pi\left(\frac{d}{2}\right)^{2}$
Required ratio $=\frac{4 \pi\left(\frac{d}{8}\right)^{2}}{4 \pi\left(\frac{d}{2}\right)^{2}}$
$=\frac{4}{64}=\frac{1}{16}$
Therefore, the ratio between their surface areas will be 1:16
Q 8, Ex. 13.4, Page No 225, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 8:
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.$\left[\text { Assume } \pi=\frac{22}{7}\right]$
Answer:
Inner radius of hemispherical bowl = 5 cmThickness of the bowl = 0.25 cm
∴ Outer radius (r) of hemispherical bowl = (5 + 0.25) cm
= 5.25 cm
Outer CSA of hemispherical bowl = 2πr2
$=2 \times \frac{22}{7} \times(5.25 \mathrm{cm})^{2}$
=173.25cm2
Therefore, the outer curved surface area of the bowl is 173.25 cm2.
Q 9, Ex. 13.4, Page No 225, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 9:
A right circular cylinder just encloses a sphere of radius r (see figure). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).
Answer:
(i) Surface area of sphere = 4πr2
(ii) Height of cylinder = r + r = 2r
Radius of cylinder = r
CSA of cylinder = 2πrh
= 2πr (2r)
= 4πr2
(iii) Required ratio $=\frac{\text { Surface area of sphere }}{\text { CSA of cylinder }}$
$=\frac{4 \pi r^{2}}{4 \pi r^{2}}$
$=\frac{1}{1}$
Therefore, the ratio between these two surface areas is 1:1
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