Question

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 yards
b. a new path and a bridge are constructed from point q to the mid - point m of (overline{pr}). find qm to the nearest yard.
qm = 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 minutes

Explanation:

Step1: Identify coordinates

Let \(Q=(0,0)\), \(P=(0,50)\), \(R=(80,0)\)

Step2: Calculate length of \(PQ\)

Using distance formula for two - points \((x_1,y_1)\) and \((x_2,y_2)\) \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For \(P(0,50)\) and \(Q(0,0)\), \(d_{PQ}=\sqrt{(0 - 0)^2+(50 - 0)^2}=50\)

Step3: Calculate length of \(QR\)

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

Step4: Calculate length of \(PR\)

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

Step5: Find perimeter of \(\triangle PQR\) (answer for part a)

\(P=d_{PQ}+d_{QR}+d_{PR}=50 + 80+94.3 = 224.3\approx224\) 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 length of \(QM\) (answer for part b)

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

Step8: Calculate total distance of jogging path

The path is \(P - Q - M - R - Q - P\). The total distance \(D = 50+47+47+80+50 = 274\) yards

Step9: Calculate time taken (answer for part c)

Using \(t=\frac{d}{v}\), 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