9514 1404 393
Answer:
a) 191 yards
b) 40 yards
c) 1.5 minutes
Explanation:
Given:
ΔPQR is a right triangle with the right angle at Q
PQ = 40 yards; QR = 70 yards
M is the midpoint of PR
jogging rate is 150 yd/min
Find:
a) the perimeter of ΔPQR to the nearest yard
b) QM to the nearest yard
c) the time to jog the path PQMRQP to the nearest tenth minute
Solution:
a) The length of PR can be found from the Pythagorean theorem:
PR² = PQ² +QR²
PR² = 40² +70² = 6500
PR = √6500 ≈ 80.62 . . . . yards
Then the perimeter is ...
perimeter = PQ +QR +PR = 40 +70 +80.62 ≈ 191 yards
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b) Segment QM divides triangle PQR into two isosceles triangles: PMQ and RMQ. The lengths MP, MQ, and MR are all equal to half the length of PR.
QM = PR/2 = (80.62 yd)/2 = 40.31 yd
QM ≈ 40 yd
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c) The segments of path PQMRQP can be grouped to simplify the computation of the path length
PQMRQP = PQ +QM +MR +RQ +QP = PQ +(PQ +QR +(QM +MR))
Since QM +MR = PR, the length in the outer parentheses is the perimeter of the triangle. Then the jogger's total path length is ...
PQ + perimeter(ΔPQR) = 40 yd +191 yd = 231 yd
The time it takes to jog that path is ...
time = distance/speed
time = 231 yd/(150 yd/min) = 1.54 min ≈ 1.5 min
It takes about 1.5 minutes.