Question

big ideas math
29:00 total time
a path goes around a triangular park, as shown.
39
a. find the distance around the park to the nearest yard.
the distance is about (square) yards.
b. a new path and a bridge are constructed from point q to the mid - point m of pr. find qm to the nearest yard.
qm=(square) yd
c. a man jogs from p to q to m to r to q and back to p at an average speed of 150 yards per minute. to the nearest tenth of a minute, about how long does it take him to travel the entire distance?
it takes about (square) minutes.

Explanation:

Step1: Find coordinates of points

Assume $Q=(0,0)$, $P=(0,50)$, $R=(80,0)$.

Step2: Calculate distance between $P$ and $Q$

Using distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, for $P(0,50)$ and $Q(0,0)$:
$PQ=\sqrt{(0 - 0)^2+(50 - 0)^2}=50$

Step3: Calculate distance between $Q$ and $R$

For $Q(0,0)$ and $R(80,0)$:
$QR=\sqrt{(80 - 0)^2+(0 - 0)^2}=80$

Step4: Calculate distance between $P$ and $R$

For $P(0,50)$ and $R(80,0)$:
$PR=\sqrt{(80 - 0)^2+(0 - 50)^2}=\sqrt{6400 + 2500}=\sqrt{8900}\approx94.3$

Step5: Find perimeter for part a

Perimeter $=PQ + QR+PR=50 + 80+94.3 = 224.3\approx224$ yards

Step6: Find mid - point $M$ of $PR$

Mid - point formula $M=(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$, for $P(0,50)$ and $R(80,0)$:
$M=(\frac{0 + 80}{2},\frac{50+0}{2})=(40,25)$

Step7: Calculate distance $QM$ for part b

For $Q(0,0)$ and $M(40,25)$:
$QM=\sqrt{(40 - 0)^2+(25 - 0)^2}=\sqrt{1600 + 625}=\sqrt{2225}\approx47.2\approx47$ yards

Step8: Calculate total distance for part c

Total distance $=PQ+QM+MR+RQ+QP=50 + 47+47+80+50 = 274$ yards

Step9: Calculate time for part c

Time $t=\frac{d}{v}$, where $d = 274$ yards and $v = 150$ yards per minute
$t=\frac{274}{150}\approx1.8$ minutes

Answer:

a. 224
b. 47
c. 1.8