The 2 triangles share common vertex upper R.

Common vertex upper R lies on line segment upper P upper T.

Question

$△ Q P R$ is similar to $△ S T R$ . The lengths represented by $\bar{S T}$ , $\bar{Q P}$ , $\bar{P R}$ , and $\bar{Q R}$ in the figure are $14$ , $15$ , $20$ , and $25$ , respectively. What is the length of $\bar{S R}$ ?

Accepted Answer

Choice A is correct. The figure shows that angle $P$ in $△ Q P R$ and angle $T$ in $△ S T R$ are right angles. It follows that angle $P$ is congruent to angle $T$ . The figure also shows that the measures of angle $Q R P$ and angle $S R T$ are both $x °$ . Therefore, angle $Q R P$ is congruent to angle $S R T$ . It’s given that $△ Q P R$ is similar to $△ S T R$ . Since angle $P$ is congruent to angle $T$ , and angle $Q R P$ is congruent to angle $S R T$ , it follows that $\bar{Q R}$ corresponds to $\bar{S R}$ , and $\bar{Q P}$ corresponds to $\bar{S T}$ . Since corresponding sides of similar triangles are proportional, it follows that $\frac{S R}{Q R} = \frac{S T}{Q P}$ . It’s also given that the lengths of $\bar{S T}$ , $\bar{Q P}$ , and $\bar{Q R}$ are $14$ , $15$ , and $25$ , respectively. Substituting $14$ for $S T$ , $15$ for $Q P$ , and $25$ for $Q R$ in the equation $\frac{S R}{Q R} = \frac{S T}{Q P}$ yields $\frac{S R}{25} = \frac{14}{15}$ . Multiplying each side of this equation by $25$ yields $S R = (\frac{14}{15}) (25)$ , or $S R = \frac{350}{15}$ . Thus, the length of $\bar{S R}$ is $StartFraction 350 Over 15 EndFraction$ .

Choice B is incorrect. This is the result of solving the equation $\frac{S R}{25} = \frac{14}{20}$ , not $\frac{S R}{25} = \frac{14}{15}$ .

Choice C is incorrect. This is the result of solving the equation $\frac{S R}{14} = \frac{15}{20}$ , not $\frac{S R}{25} = \frac{14}{15}$ .

Choice D is incorrect. This is the result of solving the equation $\frac{S R}{14} = \frac{15}{25}$ , not $\frac{S R}{25} = \frac{14}{15}$ .

Question the-2-

The 2 triangles share common vertex upper R. Common vertex upper R lies on line segment

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