Question

A path goes around a triangular park, as shown. Find the distance around the park to the nearest yard. a. The distance is about ___ yards. b. A new path and a bridge are constructed from point $q$ to the midpoint $m$ 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.

Answer 1

The distance around the triangular park is 2.256 km. The length of qm after the new path and bridge are constructed is 1.1735 km. It takes approximately 37.29 minutes for the man to travel the entire distance.

Step-by-step explanation:

To find the distance around the triangular park, we need to find the sum of the lengths of its three sides. Looking at the diagram, we can see that the sides have lengths of 1.39 km, 0.433 km, and 0.433 km. Adding these lengths together, we get a total distance of approximately 2.256 km or 2256 meters.

To find the length of qm after the new path and bridge are constructed from point q to the midpoint m of pr, we need to find the length of the segment pm. Since pm is the midpoint of pr, it is half the length of pr, which is 0.433 km. Therefore, the length of pm is 0.433/2 km or 0.2165 km. To find the length of qm, we subtract the length of pm from the total length of pq, which is 1.39 km. Subtracting 0.2165 km from 1.39 km, we get a length of approximately 1.1735 km or 1173.5 meters.

To find the time it takes for the man to travel the entire distance at an average speed of 150 yards per minute, we need to divide the total distance by the speed. The total distance is 2 times the distance from p to q plus the distance from q to m plus the distance from m to r. These distances are 1.39 km, 0.433 km, 0.433 km, and 0.433 km respectively. Converting all distances to yards, we get a total distance of approximately 5593.57 yards. Dividing this distance by the speed of 150 yards per minute, we get a time of approximately 37.29 minutes.

Answer 2

a. Distance around the park = 300 yards

b. QM distance = 60 yards

c. Time taken for one round trip = approximately 3 minutes

To solve this problem, let's break it down step by step:

Part a: Distance around the park

Given the triangle without any additional paths, the perimeter of the triangle needs to be found.

Let's assume the sides of the triangle have lengths:

• PR = 120 yards
• PQ = 100 yards
• QR = 80 yards

To find the total distance around the park, we'll add these lengths together:

Perimeter = PR + PQ + QR = 120 + 100 + 80 = 300 yards

Part b: QM distance

The midpoint of PR is M. Since QM is the distance from Q to M, and M is the midpoint of PR, QM is half the length of PR.

QM = PR / 2 = 120 / 2 = 60 yards

Part c: Time taken by the man

The man's route is P -> Q -> M -> R -> Q -> P.

The distances are:

• P to Q = PQ = 100 yards
• Q to M = QM = 60 yards
• M to R = MR (which is also equal to PR since M is the midpoint of PR) = 120 yards
• R to Q = QR = 80 yards
• Q back to P = PQ = 100 yards

Total distance covered in one round = PQ + QM + MR + QR + PQ = 100 + 60 + 120 + 80 + 100 = 460 yards

The man travels this distance in one round. He covers 460 yards in one round trip.

Average speed = 150 yards per minute

Time taken = Total distance / Average speed

Time taken = 460 yards / 150 yards per minute ≈ 3.07 minutes

Rounded to the nearest minute, it takes approximately 3 minutes for the man to complete one round trip.

The complete question is here:

A path goes around a triangular park as shown a. Find the distance around the park to the nearest yard. Distance (yd) b. A new path and bridge are constructed from point Q to the midpoint M of segment PR. Find the QM to the nearest yard.. c. A man jogs from P to Q to M to R to Q and back to P at an average rate of 150 yards per minute. About how many minutes does it take? Explain your reasoning.