9.3 Volume of Similar Solids

9.3 Volume of Similar Solids

Two solids are said to be similar if they have same shape but possibly different sizes. Two solids are similar if lengths of the corresponding sides are proportional i.e., the ratio of the corresponding lengths are equal. e.g.,

The two cylinders are similar if $\dfrac{r_1}{r_2} = \dfrac{h_1}{h_2}$

If $r_1 = 4 \text{ cm}, r_2 = 5 \text{ cm}, h_1 = 8 \text{ cm}$ and $h_2 = 10 \text{ cm}$, then we note that:

Volume of smaller cylinder, $V_1 = \pi r_1^2 h_1$
$= \pi \times 4^2 \times 8$
$= 128\pi \text{ cm}^3$

Volume of larger cylinder, $V_2 = \pi r_2^2 h_2$
$= \pi \times 5^2 \times 10$
$= 250 \pi \text{ cm}^3$

Ratio of volumes: $\dfrac{V_1}{V_2} = \dfrac{128\pi}{250\pi} = \dfrac{64}{125} = \left(\dfrac{4}{5}\right)^3$

So, $\dfrac{V_1}{V_2} = \left(\dfrac{r_1}{r_2}\right)^3$ or $\dfrac{V_1}{V_2} = \left(\dfrac{h_1}{h_2}\right)^3$

Hence the ratio of the volume of any two similar solids is equal to the cube of the ratio of any two corresponding lengths of the solids.

Since each length is $k$ times of the other, we take $\dfrac{\ell_1}{\ell_2} = k$, then $\dfrac{V_1}{V_2} = k^3$. i.e., Volume $V_1$ is $k^3$ times the volume $V_2$ and $k$ is called scale factor.

Since mass of a substance is proportional to its volume, the ratio of the mass of two similar solids is equal to the ratio of their volumes. If the masses of two similar solids are $w_1$ and $w_2$ and volumes are $V_1$ and $V_2$, then

Therefore, $\dfrac{w_1}{w_2} = \left(\dfrac{\ell_1}{\ell_2}\right)^3$

Example 9

Find the unknown volume in the following similar solids:
(i) [Diagram of two similar cones with volumes $V_1=200$ cm³, $V_2=?$, and radii $2.5$ cm, $4$ cm]
(ii) [Diagram of two similar cubes with volumes $V_1=?, V_2=686$ cm³, and sides $5$ cm, $7$ cm]

Solution: (i) $\ell_1 = 2.5 \text{ cm}, \ell_2 = 4 \text{ cm}$

(ii) Using formula $\dfrac{V_1}{V_2} = \left(\dfrac{\ell_1}{\ell_2}\right)^3$

Example 10

A solid cone $C$ is cut into two pieces $A$ and $B$ with sloping edges $6$ cm and $4$ cm. Find the ratio of:
(i) the diameters of the bases of the cones $A$ and $C$.
(ii) the area of the bases of the cones $A$ and $C$.
(iii) the volumes of the cones $A$ and $C$.
(iv) If volume of cone $A$ is $72 \text{ cm}^3$, find the volume of solid $B$.

Solution: Let diameter of cone $A = d_1$
Diameter of cone $C = d_2$

(i) The ratios of the corresponding lengths are equal because of similarity of the cones.

(ii) $\dfrac{\text{Area of cone } A}{\text{Area of cone } C} = \left(\dfrac{\ell_1}{\ell_2}\right)^2 = \left(\dfrac{3}{5}\right)^2 = \dfrac{9}{25}$

(iii) $\dfrac{\text{Volume of cone } A}{\text{Volume of cone } C} = \left(\dfrac{\ell_1}{\ell_2}\right)^3 = \left(\dfrac{3}{5}\right)^3 = \dfrac{27}{125}$

(iv) $V_1 = \text{Volume of cone } A = 72 \text{ cm}^3$
$V_2 = \text{Volume of cone } C = ?$

Volume of solid $B$ = Volume of cone $C$ - Volume of cone $A$

Example 11

The mass of sack of rice is $50$ kg and height $60$ cm. Find the mass of the similar sack of rice with height of $90$ cm.

Solution: Mass of the smaller sack of rice $w_1 = 50$ kg
Height of smaller sack of rice $h_1 = 60$ cm
Mass of larger sack of rice $w_2 = ?$
Height of larger sack of rice $h_2 = 90$ cm

Using formula $\dfrac{w_1}{w_2} = \left(\dfrac{h_1}{h_2}\right)^3$

Example 12

The ratio of the corresponding lengths of two similar cylindrical cans is $3 : 2$.
(i) The larger cylindrical can has surface area of $67.5$ square metres. Find the surface area of the smaller cylindrical can.
(ii) The smaller cylindrical can has a volume of $132$ cubic metres. Find the volume of larger tin can.

Solution: (i) Surface area of larger can $= A_1 = 67.5 \text{ m}^2$
Surface area of smaller can $= A_2 = ?$

Ratio of corresponding lengths is $\dfrac{\ell_1}{\ell_2} = \dfrac{3}{2}$

Using formula for areas of the similar figures:

(ii) Volume of smaller can $= V_2 = 132 \text{ m}^3$
Volume of larger can $= V_1 = ?$

Using formula for volume of similar figures: $\dfrac{V_1}{V_2} = \left(\dfrac{\ell_1}{\ell_2}\right)^3$