Class 10 Mensuration: Important Formulas & 20 Practice Problems
Important Mensuration Formulas
1. Cuboid: (Length $l$, Breadth $b$, Height $h$)
- Volume $= l \times b \times h$
- Total Surface Area (TSA) $= 2(lb + bh + hl)$
- Lateral Surface Area (LSA) $= 2h(l + b)$
2. Cube: (Side $a$)
- Volume $= a^3$
- Total Surface Area (TSA) $= 6a^2$
- Lateral Surface Area (LSA) $= 4a^2$
3. Right Circular Cylinder: (Radius $r$, Height $h$)
- Volume $= \pi r^2 h$
- Curved Surface Area (CSA) $= 2\pi r h$
- Total Surface Area (TSA) $= 2\pi r(r + h)$
4. Right Circular Cone: (Radius $r$, Height $h$, Slant Height $l$)
- Slant Height $l = \sqrt{r^2 + h^2}$
- Volume $= \frac{1}{3}\pi r^2 h$
- Curved Surface Area (CSA) $= \pi r l$
- Total Surface Area (TSA) $= \pi r(r + l)$
5. Sphere: (Radius $r$)
- Volume $= \frac{4}{3}\pi r^3$
- Surface Area $= 4\pi r^2$
6. Hemisphere: (Radius $r$)
- Volume $= \frac{2}{3}\pi r^3$
- Curved Surface Area (CSA) $= 2\pi r^2$
- Total Surface Area (TSA) $= 3\pi r^2$
Problem 1
Find the volume and total surface area of a cuboid whose length, breadth, and height are $12 \text{ cm}$, $8 \text{ cm}$, and $5 \text{ cm}$ respectively.
Solution:
Given: $l = 12 \text{ cm}$, $b = 8 \text{ cm}$, $h = 5 \text{ cm}$.
Volume $= l \times b \times h = 12 \times 8 \times 5 = 480 \text{ cm}^3$.
Total Surface Area $= 2(lb + bh + hl) = 2(12 \times 8 + 8 \times 5 + 5 \times 12)$
$= 2(96 + 40 + 60) = 2(196) = 392 \text{ cm}^2$.
Problem 2
A right circular cylinder has a radius of $7 \text{ cm}$ and a height of $14 \text{ cm}$. Find its curved surface area. (Take $\pi = \frac{22}{7}$)
Solution:
Given: $r = 7 \text{ cm}$, $h = 14 \text{ cm}$.
Curved Surface Area (CSA) $= 2\pi r h$
$= 2 \times \frac{22}{7} \times 7 \times 14 = 2 \times 22 \times 14 = 616 \text{ cm}^2$.
Problem 3
The volume of a right circular cone is $9856 \text{ cm}^3$. If the diameter of the base is $28 \text{ cm}$, find the height of the cone.
Solution:
Given: Volume $= 9856 \text{ cm}^3$, Diameter $= 28 \text{ cm} \implies r = 14 \text{ cm}$.
Volume of cone $= \frac{1}{3}\pi r^2 h$
$9856 = \frac{1}{3} \times \frac{22}{7} \times (14)^2 \times h$
$9856 = \frac{1}{3} \times \frac{22}{7} \times 196 \times h$
$h = \frac{9856 \times 3 \times 7}{22 \times 196} = 48 \text{ cm}$.The height of the cone is $48 \text{ cm}$.
Problem 4
Find the surface area of a sphere of radius $10.5 \text{ cm}$.
Solution:
Given: $r = 10.5 \text{ cm}$.
Surface Area $= 4\pi r^2$
$= 4 \times \frac{22}{7} \times (10.5)^2$
$= 4 \times \frac{22}{7} \times 110.25 = 1386 \text{ cm}^2$.
Problem 5
A solid metallic sphere of radius $8 \text{ cm}$ is melted and recast into a right circular cylinder of radius $4 \text{ cm}$. Find the height of the cylinder.
Solution:
Volume of Sphere = Volume of Cylinder
$\frac{4}{3}\pi R^3 = \pi r^2 h$
$\frac{4}{3} \times 8^3 = 4^2 \times h$
$\frac{4}{3} \times 512 = 16 \times h$
$h = \frac{4 \times 512}{3 \times 16} = \frac{128}{3} = 42.67 \text{ cm}$.
Problem 6
Three metallic solid spheres of radii $3 \text{ cm}$, $4 \text{ cm}$, and $5 \text{ cm}$ are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:
Let $R$ be the radius of the new sphere.
Volume of new sphere = Sum of volumes of three spheres
$\frac{4}{3}\pi R^3 = \frac{4}{3}\pi (3^3 + 4^3 + 5^3)$
$R^3 = 27 + 64 + 125$
$R^3 = 216$
$R = 6 \text{ cm}$.
Problem 7
A toy is in the form of a cone of radius $3.5 \text{ cm}$ mounted on a hemisphere of same radius. The total height of the toy is $15.5 \text{ cm}$. Find the total surface area of the toy.
Solution:
Radius ($r$) $= 3.5 \text{ cm}$. Total height $= 15.5 \text{ cm}$.
Height of cone ($h$) $= 15.5 - 3.5 = 12 \text{ cm}$.Slant height of cone ($l$) $= \sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + (12)^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5 \text{ cm}$.
Total Surface Area = CSA of Cone + CSA of Hemisphere
$= \pi r l + 2\pi r^2 = \pi r(l + 2r)$
$= \frac{22}{7} \times 3.5 \times (12.5 + 2(3.5))$
$= 11 \times (12.5 + 7) = 11 \times 19.5 = 214.5 \text{ cm}^2$.
Problem 8
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is $14 \text{ mm}$ and the diameter of the capsule is $5 \text{ mm}$. Find its surface area.
Solution:
Diameter $= 5 \text{ mm} \implies r = 2.5 \text{ mm}$.
Length of cylindrical part $= \text{Total Length} - 2(\text{radius of hemisphere})$
$h = 14 - 2(2.5) = 14 - 5 = 9 \text{ mm}$.Total Surface Area = CSA of Cylinder + $2 \times$ (CSA of Hemisphere)
$= 2\pi r h + 2(2\pi r^2) = 2\pi r(h + 2r)$
$= 2 \times \frac{22}{7} \times 2.5 \times (9 + 5)$
$= 2 \times \frac{22}{7} \times 2.5 \times 14 = 220 \text{ mm}^2$.
Problem 9
A 20 m deep well with diameter $7 \text{ m}$ is dug and the earth from digging is evenly spread out to form a platform $22 \text{ m}$ by $14 \text{ m}$. Find the height of the platform.
Solution:
Volume of earth dug out = Volume of well (cylinder)
$V = \pi r^2 h = \frac{22}{7} \times (3.5)^2 \times 20 = \frac{22}{7} \times 12.25 \times 20 = 770 \text{ m}^3$.The earth is spread to form a cuboidal platform.
Volume of platform $= \text{Area of base} \times \text{height} = 22 \times 14 \times H$
$22 \times 14 \times H = 770$
$308 \times H = 770$
$H = \frac{770}{308} = 2.5 \text{ m}$.
Problem 10
How many silver coins, $1.75 \text{ cm}$ in diameter and of thickness $2 \text{ mm}$, must be melted to form a cuboid of dimensions $5.5 \text{ cm} \times 10 \text{ cm} \times 3.5 \text{ cm}$?
Solution:
Coin is a cylinder. Radius $r = \frac{1.75}{2} = 0.875 \text{ cm}$.
Thickness (height) $h = 2 \text{ mm} = 0.2 \text{ cm}$.Volume of one coin $= \pi r^2 h = \frac{22}{7} \times 0.875 \times 0.875 \times 0.2 = 0.48125 \text{ cm}^3$.
Volume of cuboid $= l \times b \times h = 5.5 \times 10 \times 3.5 = 192.5 \text{ cm}^3$.
Number of coins $= \frac{\text{Volume of cuboid}}{\text{Volume of one coin}} = \frac{192.5}{0.48125} = 400$.
So, 400 coins must be melted.
Problem 11
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $2.1 \text{ m}$ and $4 \text{ m}$ respectively, and the slant height of the top is $2.8 \text{ m}$, find the area of the canvas used for making the tent.
Solution:
Radius of cylinder and cone $r = 2 \text{ m}$. Height of cylinder $h = 2.1 \text{ m}$. Slant height $l = 2.8 \text{ m}$.
Area of canvas = CSA of cylinder + CSA of cone
$= 2\pi r h + \pi r l = \pi r (2h + l)$
$= \frac{22}{7} \times 2 \times (2(2.1) + 2.8)$
$= \frac{44}{7} \times (4.2 + 2.8) = \frac{44}{7} \times 7 = 44 \text{ m}^2$.
Problem 12
Water is flowing at the rate of $3 \text{ km/hr}$ through a circular pipe of $20 \text{ cm}$ internal diameter into a circular cistern of diameter $10 \text{ m}$ and depth $2 \text{ m}$. In how much time will the cistern be filled?
Solution:
Pipe: Radius $r = 10 \text{ cm} = 0.1 \text{ m}$. Flow rate $h = 3 \text{ km/hr} = 3000 \text{ m/hr}$.
Volume of water flowing per hour $= \pi r^2 h = \pi \times (0.1)^2 \times 3000 = 30\pi \text{ m}^3\text{/hr}$.Cistern: Radius $R = 5 \text{ m}$. Depth $H = 2 \text{ m}$.
Volume of cistern $= \pi R^2 H = \pi \times 5^2 \times 2 = 50\pi \text{ m}^3$.Time required $= \frac{\text{Total volume}}{\text{Volume per hour}} = \frac{50\pi}{30\pi} = \frac{5}{3} \text{ hours} = 1 \text{ hour } 40 \text{ minutes}$.
Problem 13
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $1 \text{ cm}$ and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$.
Solution:
Given: $r = 1 \text{ cm}$, $h(\text{cone}) = 1 \text{ cm}$.
Volume of Solid = Volume of Cone + Volume of Hemisphere
$= \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$
$= \frac{1}{3}\pi (1)^2(1) + \frac{2}{3}\pi (1)^3$
$= \frac{1}{3}\pi + \frac{2}{3}\pi = \pi \text{ cm}^3$.
Problem 14
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are $15 \text{ cm} \times 10 \text{ cm} \times 3.5 \text{ cm}$. The radius of each of the depressions is $0.5 \text{ cm}$ and the depth is $1.4 \text{ cm}$. Find the volume of wood in the entire stand.
Solution:
Volume of cuboid $= 15 \times 10 \times 3.5 = 525 \text{ cm}^3$.
Volume of 1 conical depression $= \frac{1}{3}\pi r^2 h$
$= \frac{1}{3} \times \frac{22}{7} \times (0.5)^2 \times 1.4$
$= \frac{1}{3} \times \frac{22}{7} \times 0.25 \times 1.4 = \frac{1}{3} \times 22 \times 0.25 \times 0.2 = \frac{1.1}{3} = 0.366 \text{ cm}^3$.Volume of 4 depressions $= 4 \times 0.366 = 1.464 \text{ cm}^3$.
Volume of wood $= 525 - 1.464 = 523.536 \text{ cm}^3$.
Problem 15
A metallic sphere of radius $4.2 \text{ cm}$ is melted and recast into the shape of a cylinder of radius $6 \text{ cm}$. Find the height of the cylinder.
Solution:
Volume of Cylinder = Volume of Sphere
$\pi R^2 H = \frac{4}{3}\pi r^3$
$6^2 \times H = \frac{4}{3} \times (4.2)^3$
$36 \times H = \frac{4}{3} \times 74.088$
$36 \times H = 98.784$
$H = \frac{98.784}{36} = 2.744 \text{ cm}$.
Problem 16
A well of diameter $3 \text{ m}$ is dug $14 \text{ m}$ deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width $4 \text{ m}$ to form an embankment. Find the height of the embankment.
Solution:
Volume of earth dug out $= \pi r^2 h = \pi \times (1.5)^2 \times 14 = 31.5\pi \text{ m}^3$.
Embankment forms a hollow cylinder. Inner radius $r = 1.5 \text{ m}$. Outer radius $R = 1.5 + 4 = 5.5 \text{ m}$.
Area of embankment base $= \pi(R^2 - r^2) = \pi(5.5^2 - 1.5^2) = \pi(30.25 - 2.25) = 28\pi \text{ m}^2$.
Height of embankment $H = \frac{\text{Volume}}{\text{Area}} = \frac{31.5\pi}{28\pi} = 1.125 \text{ m}$.
Problem 17
Water in a canal, $6 \text{ m}$ wide and $1.5 \text{ m}$ deep, is flowing with a speed of $10 \text{ km/h}$. How much area will it irrigate in $30 \text{ minutes}$, if $8 \text{ cm}$ of standing water is needed?
Solution:
Speed of water $= 10 \text{ km/h} = 10000 \text{ m/h}$.
In 30 minutes ($\frac{1}{2}$ hour), length of water column $= \frac{10000}{2} = 5000 \text{ m}$.Volume of water flowing in 30 minutes $= l \times b \times h$
$= 5000 \times 6 \times 1.5 = 45000 \text{ m}^3$.Volume of water on field $= \text{Area} \times \text{Height of standing water}$
$45000 = \text{Area} \times \frac{8}{100}$
$\text{Area} = \frac{45000 \times 100}{8} = 562500 \text{ m}^2 = 56.25 \text{ hectares}$.
Problem 18
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is $10 \text{ cm}$, and its base is of radius $3.5 \text{ cm}$, find the total surface area of the article.
Solution:
Given: Cylinder height $h = 10 \text{ cm}$, radius $r = 3.5 \text{ cm}$.
Total Surface Area = CSA of cylinder + $2 \times$ (CSA of hemisphere)
$= 2\pi r h + 2(2\pi r^2) = 2\pi r(h + 2r)$
$= 2 \times \frac{22}{7} \times 3.5 \times (10 + 2 \times 3.5)$
$= 22 \times (10 + 7) = 22 \times 17 = 374 \text{ cm}^2$.
Problem 19
A cone of height $24 \text{ cm}$ and radius of base $6 \text{ cm}$ is made up of modeling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.
Solution:
Volume of Sphere = Volume of Cone
$\frac{4}{3}\pi R^3 = \frac{1}{3}\pi r^2 h$
$\frac{4}{3} R^3 = \frac{1}{3} \times 6^2 \times 24$
$4 R^3 = 36 \times 24$
$R^3 = \frac{36 \times 24}{4} = 36 \times 6 = 216$
$R = \sqrt[3]{216} = 6 \text{ cm}$.
Problem 20
A copper rod of diameter $1 \text{ cm}$ and length $8 \text{ cm}$ is drawn into a wire of length $18 \text{ m}$ of uniform thickness. Find the thickness (diameter) of the wire.
Solution:
Rod (Cylinder): $r_1 = 0.5 \text{ cm}$, $h_1 = 8 \text{ cm}$.
Wire (Cylinder): length $h_2 = 18 \text{ m} = 1800 \text{ cm}$. Let radius be $r_2$.Volume of Wire = Volume of Rod
$\pi r_2^2 h_2 = \pi r_1^2 h_1$
$r_2^2 \times 1800 = (0.5)^2 \times 8$
$r_2^2 \times 1800 = 0.25 \times 8 = 2$
$r_2^2 = \frac{2}{1800} = \frac{1}{900}$
$r_2 = \frac{1}{30} \text{ cm}$.Thickness (Diameter) $= 2 \times r_2 = 2 \times \frac{1}{30} = \frac{1}{15} \text{ cm}$ (or approx $0.67 \text{ mm}$).