in right triangle prt, $m\\angle p=90^{\\circ}$, altitude pq drawn to hypotenuse rt, rt=17, and…

in right triangle prt, $m\\angle p=90^{\\circ}$, altitude pq drawn to hypotenuse rt, rt=17, and pr=15\ndetermine and state, to the nearest tenth, the length of rq.

in right triangle prt, $m\\angle p=90^{\\circ}$, altitude pq drawn to hypotenuse rt, rt=17, and pr=15\ndetermine and state, to the nearest tenth, the length of rq.

Answer

Explanation:

Step1: Identify similar triangles

In right triangle $PRT$ with altitude $PQ$ to hypotenuse $RT$, $\triangle PRT \sim \triangle QRP$. This means corresponding sides are proportional: $\frac{PR}{RT} = \frac{RQ}{PR}$

Step2: Rearrange to solve for $RQ$

Rearrange the proportion to isolate $RQ$: $RQ = \frac{PR^2}{RT}$

Step3: Substitute given values

Substitute $PR=15$ and $RT=17$: $RQ = \frac{15^2}{17} = \frac{225}{17}$

Step4: Calculate and round

Compute the value and round to the nearest tenth: $\frac{225}{17} \approx 13.2$

Answer:

13.2