these figures are similar. the area of one is given. find the area of the other. area = 32 in² 9 in 12 in ?in²

Explanation:

Step1: Find the ratio of the side - lengths

The ratio of the side - lengths of the two similar figures is $\frac{9}{12}=\frac{3}{4}$.

Step2: Recall the relationship between the ratio of side - lengths and areas of similar figures

For two similar figures, if the ratio of their side - lengths is $a:b$, the ratio of their areas is $a^{2}:b^{2}$. Here, the ratio of the areas is $(\frac{3}{4})^{2}=\frac{9}{16}$. Let the area of the unknown - area figure be $A_1$ and the given area be $A_2 = 32$ in². We have $\frac{A_1}{A_2}=\frac{9}{16}$.

Step3: Solve for the unknown area

$A_1=\frac{9}{16}\times A_2$. Substitute $A_2 = 32$ into the equation: $A_1=\frac{9}{16}\times32$. $A_1 = 18$ in².

Answer:

18 in²