Question

triangle cde is similar to triangle fgh. find the measure of side fg. round your answer to the nearest tenth if necessary. triangle mno is similar to triangle pqr. find the measure of side pq. round your answer to the nearest tenth if necessary.

Explanation:

Step1: Set up proportion for first pair of similar triangles

Since $\triangle CDE\sim\triangle FGH$, the ratios of corresponding sides are equal. Let $FG = x$. We have $\frac{CD}{FG}=\frac{DE}{GH}$. Substituting the given values: $\frac{17}{x}=\frac{13.2}{59}$.

Step2: Cross - multiply

Cross - multiplying gives us $13.2x=17\times59$.

Step3: Calculate the right - hand side

$17\times59 = 1003$, so $13.2x = 1003$.

Step4: Solve for $x$

$x=\frac{1003}{13.2}\approx76.0$.

For the second pair of similar triangles $\triangle MNO\sim\triangle PQR$. Let $PQ = y$. Assume the side lengths of $\triangle MNO$ are $MO = 3$ and $NO = 6$, and for $\triangle PQR$, $QR = 44$. We set up the proportion $\frac{MO}{PQ}=\frac{NO}{QR}$, so $\frac{3}{y}=\frac{6}{44}$. Cross - multiplying gives $6y=3\times44$, $6y = 132$, and $y=\frac{132}{6}=22$.

Answer:

For the first pair, the length of $FG$ is approximately $76.0$.
For the second pair, the length of $PQ$ is $22$.