Question

sarafina and her teammates signed up for an outdoor obstacle course. the course consists of a 18 m vertical climb, a zip line, a horizontal crawl under a net, and a 25 m rope descent down a cliff side, as shown in the figure. the vertical climb and cliff side are parallel.

figure: climb (18 m from start), start to crawl begins is 42 m, crawl begins to cliff is? m, cliff to end is 25 m vertical

sarafina can monitor the progress of her teammates through a line of sight from the top of the zip line, through the location where the crawl begins, to the bottom of the cliff.

what is the distance, in meters, of the crawl to the cliff? round to the nearest unit.

answer box m

image of person on zip line

Explanation:

Step1: Identify Similar Triangles

The vertical climb (18 m) and cliff side (25 m) are parallel, so the two triangles formed are similar by the AA (Angle-Angle) similarity criterion (right angles and common angle from the line of sight). Let the unknown distance (crawl to cliff) be \( x \). The ratios of corresponding sides are equal: \( \frac{18}{25} = \frac{42}{42 + x} \)? Wait, no—correct ratio: the horizontal segment from start to crawl start is 42 m, and from crawl start to cliff is \( x \). The vertical sides are 18 m (climb) and 25 m (descent). So similar triangles: \( \frac{18}{25} = \frac{42}{42 + x} \)? Wait, no, actually, the triangles are similar with vertical sides 18 and 25, and horizontal sides 42 and \( x \)? Wait, no, let's re-examine. The top triangle has vertical leg 18, horizontal leg 42. The bottom triangle has vertical leg 25, horizontal leg \( x \). Since they are similar, \( \frac{18}{25} = \frac{42}{x} \)? Wait, no, that's not right. Wait, the line of sight creates two similar right triangles. The first triangle (top) has height 18 m and base 42 m. The second triangle (bottom) has height 25 m and base \( x \) (the distance from crawl start to cliff). Wait, no, actually, the two triangles are similar because the vertical climb and cliff are parallel, so the angles are equal. So the ratio of vertical to horizontal should be equal. So \( \frac{18}{42} = \frac{25}{x} \)? Wait, no, that's inverted. Wait, let's define the triangles: Triangle 1: vertical side 18, horizontal side 42. Triangle 2: vertical side 25, horizontal side \( x \). Since they are similar, corresponding sides are proportional. So \( \frac{18}{25} = \frac{42}{x} \)? No, wait, the vertical sides are 18 and 25, and the horizontal sides are 42 and \( x \). Wait, no, the angle at the crawl start is common, and the right angles are equal, so the triangles are similar. So the ratio of vertical to horizontal for the first triangle (climb side) is \( \frac{18}{42} \), and for the second triangle (cliff side) is \( \frac{25}{x} \). Wait, no, that's not. Wait, the top triangle: from start to climb top (18 m up) to crawl start (42 m right). The bottom triangle: from crawl start (right) to cliff (x m right) to end (25 m down). So the two triangles are similar, so \( \frac{18}{25} = \frac{42}{x} \)? Wait, no, cross-multiplying: \( 18x = 25 \times 42 \). Wait, let's calculate that. \( 25 \times 42 = 1050 \). Then \( x = \frac{1050}{18} \approx 58.33 \)? Wait, that can't be. Wait, maybe I mixed up the sides. Let's draw it: Start is at (0,0), Climb top at (0,18), Crawl start at (42,0), Cliff at (42 + x, 0), End at (42 + x, -25). The line of sight is from (0,18) to (42,0) to (42 + x, -25). So the slope from (0,18) to (42,0) is \( \frac{0 - 18}{42 - 0} = -\frac{18}{42} = -\frac{3}{7} \). The slope from (42,0) to (42 + x, -25) should be the same, so \( \frac{-25 - 0}{(42 + x) - 42} = \frac{-25}{x} = -\frac{3}{7} \). So \( \frac{25}{x} = \frac{3}{7} \)? Wait, no, that's not matching. Wait, no, the slope is \( -\frac{18}{42} = -\frac{3}{7} \), so the slope from (42,0) to (42 + x, -25) is \( \frac{-25}{x} = -\frac{3}{7} \), so \( \frac{25}{x} = \frac{3}{7} \)? No, that would mean \( 3x = 25 \times 7 = 175 \), \( x = \frac{175}{3} \approx 58.33 \). But that seems off. Wait, maybe the triangles are similar with vertical sides 18 and 25, and horizontal sides 42 and \( x \), but actually, the correct proportion is \( \frac{18}{42} = \frac{25}{x} \)? Wait, no, that would be \( 18x = 42 \times 25 \), \( x = \frac{42 \times 25}{18} \). Let's compute that: \( 42 \times 25 = 1050 \), \( 1050 \div 18 = 58.333... \). So \( x \approx 58 \) when rounded? Wait, but let's check again. The two triangles: one with height 18, base 42; another with height 25, base \( x \). Since they are similar, \( \frac{height1}{height2} = \frac{base1}{base2} \), so \( \frac{18}{25} = \frac{42}{x} \)? No, that's \( height1/base1 = height2/base2 \), so \( 18/42 = 25/x \), which is the same as \( 18x = 42 \times 25 \), so \( x = (42 \times 25)/18 \). Calculating: 42 divided by 18 is 7/3, so 7/3 * 25 = 175/3 ≈ 58.33, so rounded to nearest unit is 58. Wait, but let's confirm the similar triangles. The vertical segments are 18 and 25, horizontal segments are 42 and \( x \). The angle between the line of sight and the horizontal is the same for both triangles, so tangent of the angle is 18/42 for the first triangle and 25/x for the second. Since the angle is the same, 18/42 = 25/x, so x = (25 * 42)/18 = (25 * 7)/3 = 175/3 ≈ 58.33, so 58.

Step2: Solve for \( x \)

Using the proportion from similar triangles: \( \frac{18}{42} = \frac{25}{x} \) (tangent of the angle is equal for both triangles). Cross-multiplying: \( 18x = 42 \times 25 \). Calculate \( 42 \times 25 = 1050 \). Then \( x = \frac{1050}{18} = \frac{175}{3} \approx 58.33 \). Round to the nearest unit: 58.

Answer:

58