Question

question 25 of 28
during a game of pool, kyle hit a ball and it followed the path from b to c and then to d. the angles formed by the path are congruent. to the nearest tenth of an inch, which of the following represents the total distance the ball traveled before it stopped at point d?
a. 68.1 in
b. 57.0 in
c. 43.0 in
d. 21.1 in

Explanation:

Step1: Prove similar triangles

Since $\angle BCA=\angle DCE$ and $\angle BAC = \angle DEC=90^{\circ}$, $\triangle BAC\sim\triangle DEC$ by AA (angle - angle) similarity criterion.

Step2: Set up proportion

We know that $\frac{DE}{BA}=\frac{EC}{AC}$. Let $AC = x$ and $EC=y$. Given $BA = 21.5$ in, $DE = 9$ in, and $BC = 48$ in. Using the Pythagorean theorem in $\triangle BAC$, $AC=\sqrt{BC^{2}-BA^{2}}=\sqrt{48^{2}-21.5^{2}}=\sqrt{(48 + 21.5)(48 - 21.5)}=\sqrt{69.5\times26.5}=\sqrt{1841.75}\approx42.9$.
From $\frac{DE}{BA}=\frac{EC}{AC}$, we have $\frac{9}{21.5}=\frac{y}{42.9}$, so $y=\frac{9\times42.9}{21.5}=\frac{386.1}{21.5}\approx17.96$.
Using the Pythagorean theorem in $\triangle DEC$, $DC=\sqrt{DE^{2}+EC^{2}}=\sqrt{9^{2}+17.96^{2}}=\sqrt{81 + 322.56}=\sqrt{403.56}\approx20.1$.

Step3: Calculate total distance

The total distance the ball traveled is $BC + DC=48+20.1 = 68.1$ in.

Answer:

A. 68.1 in