Surface Areas and Volumes CBSE Class 9 Mathematics Worksheets Set 06

Practice Set

Question 1. A room is half as long again as it is broad. Carpeting at Rs. 3.25 per m\( ^2 \) costs Rs. 175.50 and papering the walls at Rs. 1.40 per m\( ^2 \) costs Rs. 240.80. If one door and two windows take up 8 m\( ^2 \), find the dimensions of the room.
Answer:
Let the breadth be \( b \) and length \( 1.5b \). Floor area \( =\frac{175.50}{3.25}=54 \) so \( 1.5b^2=54 \), \( b=6 \) m, \( l=9 \) m.
Area papered \( =\frac{240.80}{1.40}=172 \) m\( ^2 \). Walls \( =172+8=180 \).
\( 2(9+6)h=180 \) gives \( h=6 \) m. The room is \( 9\times6\times6 \) m.

Question 2. The external dimensions of a closed wooden box are 18 cm, 10 cm and 6 cm and the wood is \( \frac12 \) cm thick. Empty, the box weighs 15 kg; filled with sand, 100 kg. Find the weight of one cubic cm of wood and of sand.
Answer:
Inner dimensions \( 17,9,5 \): inner volume \( 765 \) cm\( ^3 \); outer volume \( 1080 \); wood \( 315 \) cm\( ^3 \).
Wood: \( \frac{15}{315}=\frac1{21} \) kg per cm\( ^3 \).
Sand weighs \( 100-15=85 \) kg, so per cm\( ^3 \): \( \frac{85}{765}=\frac19 \) kg.

Question 3. If \( h \), \( C \) and \( V \) are the height, curved surface area and volume of a cone, prove that \( 3\pi Vh^3-C^2h^2+9V^2=0 \).
Answer:
\( V=\frac13\pi r^2h \), \( C=\pi rl \) with \( l^2=r^2+h^2 \).
\( 3\pi Vh^3=\pi^2r^2h^4 \), \( 9V^2=\pi^2r^4h^2 \), \( C^2h^2=\pi^2r^2(r^2+h^2)h^2=\pi^2r^4h^2+\pi^2r^2h^4 \).
So \( 3\pi Vh^3+9V^2-C^2h^2=\pi^2r^2h^4+\pi^2r^4h^2-\pi^2r^4h^2-\pi^2r^2h^4=0 \).

Question 4. A conical tent is 9 m high and the radius of its base is 12 m. Find (i) the cost of canvas at Rs. 10 per square metre, (ii) the number of persons it can hold if each needs 2 m\( ^2 \) of ground and 15 m\( ^3 \) of air (take \( \pi=3.14 \)).
Answer:
\( l=\sqrt{144+81}=15 \) m. Curved area \( =3.14\times12\times15=565.2 \) m\( ^2 \), cost \( =5652 \) rupees.
Volume \( =\frac13\times3.14\times144\times9=1356.48 \) m\( ^3 \); persons by air \( =\frac{1356.48}{15}=90.4 \), so 90.
Ground area \( =452.16 \) m\( ^2 \) allows \( 226 \) persons, so the limit is 90 persons.

Question 5. A solid is a cylinder with hemispherical ends. The total height is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and total surface area (\( \pi=\frac{22}7 \)).
Answer:
Cylinder height \( =19-7=12 \) cm and \( r=3.5 \).
Volume \( =\pi r^2(12)+\frac43\pi r^3=147\pi+\frac{343}{6}\pi=641.67 \) cm\( ^3 \).
Surface \( =2\pi r(12)+4\pi r^2=84\pi+49\pi=133\pi=418 \) cm\( ^2 \).

Question 6. A solid cylinder has total surface area 462 cm\( ^2 \) and its curved surface area is one third of the total. Find the volume (\( \pi=\frac{22}7 \)).
Answer:
Curved area \( =154 \) and the two bases \( =308 \), so \( 2\pi r^2=308 \) and \( r=7 \).
\( 2\pi rh=154 \) gives \( h=3.5 \).
\( V=\frac{22}7\times49\times3.5=539 \) cm\( ^3 \).

Question 7. A hollow hemispherical vessel has internal and external diameters 24 cm and 25 cm. Paint costs 7 paise per cm\( ^2 \). Find the cost of painting the vessel all over (ignore the rim, \( \pi=\frac{22}7 \)).
Answer:
Outer curved area \( =2\pi(12.5)^2=982.14 \) cm\( ^2 \); inner \( =2\pi(12)^2=905.14 \) cm\( ^2 \).
Total \( =1887.28 \) cm\( ^2 \); cost \( =1887.28\times0.07=132.11 \) rupees.

Question 8. A rectangular tank 150 m by 100 m is filled through a pipe of cross-section 2 dm by 1.5 dm at 15 km/h. In what time will the water be 3 m deep?
Answer:
Volume needed \( =150\times100\times3=45000 \) m\( ^3 \).
Cross-section \( =0.2\times0.15=0.03 \) m\( ^2 \); speed \( =15000 \) m/h; inflow \( =450 \) m\( ^3 \) per hour.
Time \( =\frac{45000}{450}=100 \) hours.