Question

use the triangle shown to answer the question.
what is the length of \\( \overline{pq} \\)?
\\( \bigcirc \\) a. 8
\\( \bigcirc \\) b. 9
\\( \bigcirc \\) c. 10
\\( \bigcirc \\) d. 14

Explanation:

Step1: Find coordinates of points

From the graph, \( P(1,1) \), \( R(10,1) \), \( Q(10,7) \). So \( PR \) length: \( 10 - 1 = 9 \), \( RQ \) length: \( 7 - 1 = 6 \).

Step2: Apply Pythagorean theorem

For right triangle \( PRQ \), \( PQ^2 = PR^2 + RQ^2 \). Substitute: \( PQ^2 = 9^2 + 6^2 = 81 + 36 = 117 \)? Wait, no, wait coordinates: Wait \( P(1,1) \), \( Q(10,7) \)? Wait no, \( R \) is at \( (10,1) \), \( Q \) at \( (10,7) \), so \( PR \) is horizontal from \( x=1 \) to \( x=10 \), so length \( 9 \), \( RQ \) vertical from \( y=1 \) to \( y=7 \), length \( 6 \). Then \( PQ \) is hypotenuse. Wait \( 9^2 + 6^2 = 81 + 36 = 117 \)? No, wait maybe I misread coordinates. Wait \( P \) is at \( (1,1) \), \( R \) at \( (10,1) \), \( Q \) at \( (10,7) \). So \( PR = 10 - 1 = 9 \), \( RQ = 7 - 1 = 6 \). Then \( PQ = \sqrt{9^2 + 6^2} = \sqrt{81 + 36} = \sqrt{117} \)? No, that's not matching options. Wait maybe \( P \) is at \( (1,1) \), \( Q \) at \( (10,7) \)? Wait no, maybe I messed up. Wait the graph: \( x \)-axis from 0 to 11, \( y \)-axis 0 to 11. \( P \) at (1,1), \( R \) at (10,1), \( Q \) at (10,7). So \( PR \) is 9 units (10-1), \( RQ \) is 6 units (7-1). Then Pythagorean theorem: \( PQ = \sqrt{9^2 + 6^2} = \sqrt{81 + 36} = \sqrt{117} \approx 10.8 \), but options are 8,9,10,14. Wait maybe \( P \) is at (1,1), \( Q \) at (9,7)? Wait the graph: \( Q \) is at (9,7)? Wait the original graph: let's check again. The triangle: \( P \) at (1,1), \( R \) at (9,1), \( Q \) at (9,7). Oh! I misread \( R \)'s x-coordinate. So \( R \) is at (9,1), not 10. So \( PR \) length: \( 9 - 1 = 8 \)? No, wait \( P(1,1) \), \( R(9,1) \), so \( PR = 8 \), \( RQ = 7 - 1 = 6 \). Then \( PQ = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \). Ah, that's option C. So correct coordinates: \( P(1,1) \), \( R(9,1) \), \( Q(9,7) \). So \( PR = 9 - 1 = 8 \)? No, 9-1=8? Wait 1 to 9 is 8 units? No, 9-1=8? Wait 1,2,3,4,5,6,7,8,9: 8 intervals, so length 8? No, distance between (1,1) and (9,1) is \( |9 - 1| = 8 \). Distance between (9,1) and (9,7) is \( |7 - 1| = 6 \). Then \( PQ = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \). Yes, that's option C.

Answer:

C. 10