Question

12 find:
a the height of a rectangular prism with base 5 cm by 3 cm and volume 40 cm³
b the side length of a cube of butter with volume 34.01 cm³
c the height of a glass cone with base diameter 24.6 cm and volume 706 cm³
d the radius of a spherical weather balloon with volume 73.62 m³
e the radius of a steel cylinder with height 4.6 cm and volume 43.75 cm³
f the base radius of a conical bin with height 6.2 m and volume 203.9 m³.

Explanation:

Part (a)

Step1: Recall Volume of Rectangular Prism

The volume \( V \) of a rectangular prism is given by \( V = l\times w\times h \), where \( l \) is length, \( w \) is width, and \( h \) is height. Here, \( l = 5\space\text{cm} \), \( w = 3\space\text{cm} \), \( V = 40\space\text{cm}^3 \). We need to solve for \( h \). Rearranging the formula: \( h=\frac{V}{l\times w} \).

Step2: Substitute Values

Substitute \( V = 40 \), \( l = 5 \), \( w = 3 \) into the formula: \( h=\frac{40}{5\times3}=\frac{40}{15}=\frac{8}{3}\approx2.67\space\text{cm} \).

Step1: Recall Volume of Cube

The volume \( V \) of a cube with side length \( s \) is \( V = s^3 \). We need to find \( s \) when \( V = 34.01\space\text{cm}^3 \). So, \( s=\sqrt[3]{V} \).

Step2: Calculate Cube Root

\( s=\sqrt[3]{34.01}\approx3.24\space\text{cm} \) (since \( 3.24^3 = 3.24\times3.24\times3.24\approx34.01 \)).

Step1: Recall Volume of Cone

The volume \( V \) of a cone is \( V=\frac{1}{3}\pi r^2h \), where \( r \) is radius and \( h \) is height. The diameter \( d = 24.6\space\text{cm} \), so radius \( r=\frac{d}{2}=\frac{24.6}{2}=12.3\space\text{cm} \), \( V = 706\space\text{cm}^3 \). We solve for \( h \): \( h=\frac{3V}{\pi r^2} \).

Step2: Substitute Values

Substitute \( V = 706 \), \( r = 12.3 \), \( \pi\approx3.14 \): \( h=\frac{3\times706}{3.14\times(12.3)^2}=\frac{2118}{3.14\times151.29}=\frac{2118}{475.0506}\approx4.46\space\text{cm} \).

Answer:

\(\frac{8}{3}\space\text{cm}\) (or approximately \( 2.67\space\text{cm} \))

Part (b)