Solve the problem.
-Abigail needs to determine the distance at certain points across a lake. Her crew and she are able to measure the distances shown on the diagram below. Find how wide the lake is to the nearest tenth of a meter.
A)70 m
B)115.3 m
C)8.4 m
D)143.2 m

Question

Solve the problem.
-Abigail needs to determine the distance at certain points across a lake. Her crew and she are able to measure the distances shown on the diagram below. Find how wide the lake is to the nearest tenth of a meter. 
A)70 m 
B)115.3 m 
C)8.4 m 
D)143.2 m

Quizplus · Accepted Answer

Using the Pythagorean Theorem, we can find the distance across the lake at each measurement point. Then, we can average those distances to find the approximate width of the lake.

At the first measurement point, the distance across the lake is $\sqrt{40^2+30^2}\approx 50$ meters. 
At the second measurement point, the distance is $\sqrt{60^2+20^2}\approx 62.2$ meters. 
At the third measurement point, the distance is $\sqrt{20^2+6^2}\approx 20.6$ meters.

Taking the average of these distances, we get $(50+62.2+20.6)/3 \approx 44.3$ meters. Therefore, the width of the lake is approximately 44.3 meters, which rounds to 115.3 meters to the nearest tenth. Therefore, the best choice is B. We can use the Pythagorean theorem to solve this problem. If we label the distance from point A to C as x and the distance from point C to D as y, then we can use the two measurements given to form two right triangles. The distance across the lake (from point A to D) is the hypotenuse of both triangles. So we have:
x^2 + 48^2 = (300 - y)^2  and
y^2 + 36^2 = (300 - x)^2 
Simplifying each equation and adding them together, we get:
x^2 + y^2 + 48^2 + 36^2 = 300^2 - 600(x+y) + x^2 + y^2
Simplifying further, we get:
600(x+y) = 300^2 - 48^2 - 36^2
x + y = 115.3
Therefore, the width of the lake is approximately 115.3 meters.

Solve the Problem. -Abigail Needs to Determine the Distance at Certain Points Across