Question

two similar prisms have widths as shown below. (note: the figures are not drawn to scale.) (a) if the volume of the prism on the left is 343 cm³, find the volume of the prism on the right. (b) if the surface - area of the prism on the left is 441 cm², find the surface - area of the prism on the right. (c) if the height of the prism on the left is 35 cm, find the height of the prism on the right.

Explanation:

Step1: Find the ratio of side - lengths

For two similar prisms, if the ratio of corresponding side - lengths is $k$, the ratio of volumes is $k^{3}$ and the ratio of surface - areas is $k^{2}$. Let the side - length of the left prism be $a_1$ and that of the right prism be $a_2$. Given $a_1 = 7$ cm and $a_2=5$ cm, the ratio of side - lengths $k=\frac{a_1}{a_2}=\frac{7}{5}$.

Step2: Solve part (a)

The ratio of volumes of two similar prisms is $k^{3}$. Let $V_1$ be the volume of the left prism and $V_2$ be the volume of the right prism. We know $V_1 = 343$ cm³ and $k=\frac{7}{5}$. Since $\frac{V_1}{V_2}=k^{3}=(\frac{7}{5})^{3}=\frac{343}{125}$, then $V_2=\frac{343\times125}{343}=125$ cm³.

Step3: Solve part (b)

The ratio of surface - areas of two similar prisms is $k^{2}$. Let $S_1$ be the surface - area of the left prism and $S_2$ be the surface - area of the right prism. We know $S_1 = 441$ cm² and $k=\frac{7}{5}$. Since $\frac{S_1}{S_2}=k^{2}=(\frac{7}{5})^{2}=\frac{49}{25}$, then $S_2=\frac{441\times25}{49}=225$ cm².

Step4: Solve part (c)

The ratio of heights of two similar prisms is equal to the ratio of side - lengths. Let $h_1$ be the height of the left prism and $h_2$ be the height of the right prism. We know $h_1 = 35$ cm and $k=\frac{7}{5}$. Since $\frac{h_1}{h_2}=k=\frac{7}{5}$, then $h_2=\frac{35\times5}{7}=25$ cm.

Answer:

(a) $125$ cm³
(b) $225$ cm²
(c) $25$ cm