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We are asked to find the distance between Sycamore cabin and Oak cabin. Let's start by taking a closer look at the diagram.
Let's substitute these values into the formula. a^2+ b^2= c^2 ⇕ 30^2+ b^2= 50^2 Now we can solve the equation that we got to find the value of b.
Since a negative side length does not make sense, we only need to consider positive solutions. Therefore, we got that the distance between Sycamore cabin and Oak cabin is 40 yards.
We can write an expression for the length of each route.
| Travel Route | Distance (yards) |
|---|---|
| Direct Route | 60 |
| Through Mess Hall | d+40 |
Notice that the distance between Hickory cabin and the Mess Hall is also the length of a leg of a right triangle. To find this distance, we need to use the Pythagorean Theorem. a^2+ b^2= c^2 In the formula, a and b are the lengths of the legs and c is the length of the hypotenuse of a right triangle. We are given the triangle with a= 40, b= d, and c= 60.
Let's substitute these values into the formula. a^2+ b^2= c^2 ⇕ 40^2+ d^2= 60^2 Now we can solve an equation that we got to find the value of d.
Since a negative side length does not make sense, we only need to consider positive solutions. Therefore, the distance between Hickory cabin and the Mess Hall is about 44.7 yards. Now we can compare the distances!
| Travel Route | Distance (yards) |
|---|---|
| Direct Route | 60 |
| Via Mess Hall | d+40≈ 44.7+40=84.7 |
Let's calculate the difference between these distances. 84.7-60=24.7 Therefore, if the camper walks to the Mess Hall first, the trip will be about 24.7 yards farther than going directly to Elm cabin.