David runs out of gas so he gets out of his car and walks 32 yards north and 85 blocks yards east. How far is he from his car if he walked back in a straight line? Round your answer to the nearest tenth. Click Here for Help Video.
(1 point)
David needs to walk about
yards
David walks 32 yards north and 85 yards east, forming a right triangle with his car as the right angle.
Using the Pythagorean theorem, we can find the distance between David's starting point and his ending point (his car).
The formula is as follows:
Distance = sqrt((32^2) + (85^2))
Distance = sqrt(1024 + 7225)
Distance = sqrt(8249)
Distance ≈ 90.8 yards
Therefore, David needs to walk about 90.8 yards to get back to his car.
To find the distance David is from his car, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In this case, the distance David walked north and east form the legs of a right triangle, and the distance from his car to his final location is the hypotenuse.
Using the Pythagorean theorem, we can calculate the distance as follows:
Distance = √((32)^2 + (85)^2)
Simplifying the equation:
Distance = √(1024 + 7225)
Distance = √(8249)
Distance ≈ 90.9 yards
Therefore, David needs to walk about 90.9 yards to reach his car.
To find the distance between David's car and his current location, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
In this case, the distance north that David walked is the opposite side, and the distance east that he walked is the adjacent side of a right triangle. Let's call the distance he is from his car "d".
Using the Pythagorean theorem, we have:
d^2 = (32^2) + (85^2)
Simplifying the equation:
d^2 = 1024 + 7225
d^2 = 8249
To solve for d, we take the square root of both sides:
d = √8249
Calculating the square root:
d ≈ 90.9
Therefore, David needs to walk approximately 90.9 yards to get back to his car.