Question

4.3 similar triangles and slope
question 10 of 15 (1 point) | question attempt: 1 of 5

graph with points a, b, c, d, e, f forming right triangles abc, def on a coordinate grid

complete the parts below.
(a) find the rise, run, and slope given by triangle abc.
rise:
run:
slope:
(b) find the rise, run, and slope given by triangle def.

Explanation:

Step1: Analyze Triangle ABC

First, identify coordinates of points A, B, C. From graph, A is at (3, 0) (assuming x=3, y=0), B at (11, 0), C at (11, 10). Rise is vertical change: \( 10 - 0 = 10 \). Run is horizontal change: \( 11 - 3 = 8 \)? Wait, no, AB is horizontal from A(3,0) to B(11,0)? Wait, no, looking at the graph, A is at (3,0), B at (11,0)? Wait, no, the x-axis: A is at x=3, B at x=11? Wait, no, the right angle at B, so AB is horizontal, BC is vertical. So A: (3, 0), B: (11, 0), C: (11, 10). So rise (BC) is \( 10 - 0 = 10 \), run (AB) is \( 11 - 3 = 8 \)? Wait, no, maybe A is at (3,0), B at (11,0)? Wait, no, the grid: let's check the x-coordinates. A is at x=3 (since between 2 and 4), B at x=11 (since the right angle at B, x=11, y=0), C at (11, 10). So rise (vertical change from A to C via B) is \( 10 - 0 = 10 \), run (horizontal change from A to B) is \( 11 - 3 = 8 \)? Wait, no, slope is rise over run. Wait, maybe I misread. Wait, A is at (3,0), B at (11,0)? No, the distance from A to B: let's count the grid. From x=3 to x=11: that's 8 units? Wait, no, 11 - 3 = 8? Wait, 3 to 11 is 8? 11 - 3 = 8. Then BC is from y=0 to y=10, so rise is 10, run is 8? Wait, but slope is rise/run = 10/8 = 5/4? Wait, maybe I made a mistake. Wait, let's check again. A: (3, 0), B: (11, 0), C: (11, 10). So rise (vertical change) is \( 10 - 0 = 10 \), run (horizontal change) is \( 11 - 3 = 8 \). So slope is \( \frac{10}{8} = \frac{5}{4} \). Wait, but maybe A is at (3,0), B at (11,0), so run is 8, rise is 10.

Step2: Wait, maybe I messed up the coordinates. Let's re-express. The right angle at B, so AB is horizontal, BC is vertical. So A: (3, 0), B: (11, 0) (since x=11, y=0), C: (11, 10) (x=11, y=10). So rise (BC) is 10 - 0 = 10, run (AB) is 11 - 3 = 8. Slope is rise/run = 10/8 = 5/4.

For part (a) of triangle ABC:
Rise: 10 (vertical change from A to C: 10 - 0 = 10)
Run: 8 (horizontal change from A to B: 11 - 3 = 8)
Slope: \( \frac{10}{8} = \frac{5}{4} \)

Now part (b) triangle DEF: D, E, F. D: (15, 15)? Wait, no, E is at (19, 15), D at (15, 15), F at (19, 20). So right angle at E. So DE is horizontal from D(15,15) to E(19,15), EF is vertical from E(19,15) to F(19,20). So rise (EF) is \( 20 - 15 = 5 \), run (DE) is \( 19 - 15 = 4 \). Slope is \( \frac{5}{4} \), same as ABC, since triangles are similar.

But let's focus on part (a) first.

Step1 (Correcting): Wait, maybe A is at (3,0), B at (11,0), C at (11,10). So rise (vertical change) is 10 (from y=0 to y=10), run (horizontal change) is 8 (from x=3 to x=11). So slope is 10/8 = 5/4.

Step2: For triangle ABC:

Rise: 10 (vertical difference between A and C: 10 - 0 = 10)
Run: 8 (horizontal difference between A and B: 11 - 3 = 8)
Slope: \( \frac{10}{8} = \frac{5}{4} \)

Answer:

(a) rise: 10, run: 8, slope: \( \frac{5}{4} \)
(For part (b), similar logic: D, E, F. D: (15,15), E: (19,15), F: (19,20). Rise: 20 - 15 = 5, Run: 19 - 15 = 4, Slope: \( \frac{5}{4} \))