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2 - 108. solve for the missing lengths. show all work. homework help a.…

Question

2 - 108. solve for the missing lengths. show all work. homework help
a. △ghi ∼ △pqr
b. △abc ∼ △xyz

Explanation:

Step1: Recall similarity - ratio property

For similar triangles $\triangle ABC\sim\triangle XYZ$, the ratios of corresponding sides are equal. That is, $\frac{AB}{XY}=\frac{AC}{XZ}=\frac{BC}{YZ}$.

Step2: Find the ratio for $\triangle ABC\sim\triangle XYZ$

We know that $\frac{AB}{XY}=\frac{AC}{XZ}$. Given $AB = 7$, $AC=23$, and $XZ = 49$. Let $XY=m$. Then $\frac{7}{m}=\frac{23}{49}$.

Step3: Cross - multiply

Cross - multiplying gives us $23m=7\times49$.

Step4: Solve for $m$

$m=\frac{7\times49}{23}=\frac{343}{23}\approx14.91$.

For $\triangle GHI\sim\triangle PQR$, assume the corresponding sides are in proportion. Let the side corresponding to $GI = 2$ in $\triangle PQR$ be $x$ and the side corresponding to $GH = 7$ be $112$. Then $\frac{GI}{x}=\frac{GH}{112}$. Substituting the values, $\frac{2}{x}=\frac{7}{112}$. Cross - multiplying gives $7x=2\times112$, so $x=\frac{2\times112}{7}=32$.

Answer:

For $\triangle ABC\sim\triangle XYZ$, $m=\frac{343}{23}$; for $\triangle GHI\sim\triangle PQR$, the side corresponding to $GI$ is $32$.