Math Problem requiring Pythagorean theorem?

Question

This is the question that I'm stumped by:

A 50-foot wire runs from the roof of a building to the top of a 10-foot pole 14 feet across the street. How much taller would the pole have to be if the street were 16 feet wider and the wire remained the same length? I know the answer to this question because of the answer key, but I don't know the process of figuring out this problem.

Answer 1

There's a right triangle using the building, the wire, and a line straight across from the top of the pole where the wire ends. The wire is the hypotenuse.

You're told that the street is 14 feet wide, and the wire is 50 feet long. Using "b" for the height of the building above the 10-foot pole:

a^2 + b^2 = c^2
14^2 + b^2 = 50^2
b^2 = 2,500 - 196
b^2 = 2,304
b = sqrt(2,304)
b = 48

You have calculated that the top of the building is 48 feet higher than the top of the pole. So, now you know the height of the building. It's (48 + 10 = 58) 58 feet tall.

Now, if the street were 16 feet wider (14+16 = 30), and the same wire (50 feet) were used, you'd have another right triangle where the wire is again the hypotenuse.

You know the hypotenuse (50 feet) and one leg (30 feet)... and you have to figure out the other leg (the distance below the roof of the building the wire will reach across the street). Using "b" for the distance down the side of the building:

a^2 + b^2 = c^2
30^2 + b^2 = 50^2
900 + b^2 = 2500
b^2 = 1600
b = sqrt(1600)
b = 40

The wire will reach 40 feet below the top of the building, across the street.

Since you know that the building is 58 feet high (from previous calculation), the wire will reach to within (58 - 40 = 18) 18 feet of the ground.

So... if the street were 16 feet wider, and the wire remained the same length, the pole would have to be 18 feet high, which is (18 - 10 = 8) "8 feet taller" than it was.

The answer is: 8 feet taller.

Answer 2

I find it helps to draw these problems. Draw a building and a pole. Draw a right triangle from the top corner of the building to the top of the pole and back to the side of the building, 10 ft up (the height of the pole). The distance to the pole is 14'. The hypotunse or wire is 50'. Apply Pythagorus' theorem. and you can solve for part of the height of the building. Add the 10 feet for the height of the pole and you get 58' for the building height. Now draw another pole, taller and 16 feet further out. Draw another triangle. The hypotnuse of this one is still 50 (the wire length hasn't changed) and the distance to the pole is now 30 feet. Again using Pythagorus' theorem, the portion of the triangle against the building has shrunk to 40 feet. Since we know the height of the building is 58', 58-40=18 feet. This is the height of the further pole. 18'-10'=8' The pole must be 8' taller if it is 16' further away.

Good luck on your homework and with your test

Answer 3

case 1

hypotenus = 50 ft.
base = 14 ft.
let h = height of building
other leg = height of building - height of pole = h - 10

so that

14^2 + (h-10)^2 = 50^2

which gives h = 58

case 2

hypotenus = 50
base = 14+ 16 = 30
last leg of triangle = L

L^2 + 30^2 = 50^2
L = 40

since building is 58 feet high, the new height of the pole must be 58 - 40 = 18 feet

since the pole was originally 10 feet high, the pole must be 18 - 10 or 8 feet taller...

Answer 4

a^2 + b^2 = c^2

a=unknown
b=30 (14+16)
c=500

x^2 + 900 = 2500
x^2 = 2500-900
x^2 = 1600
x = 40

40 - 10 = 30

Answer 5

50 Foot Pole

Answer 6

doing this really quick with basic calculator my guess would be 18 feet higher total of 28 feet high. if that is right then i will gladly explain how. if i am wrong i won't confuse you more.

Answer 7

hypotenuse squared = a squared + b squared
so if the hypotenuse is 50, a is 10, and b is 14
if the hypotenuse stays the same but b changes then you do 50 squared - 16 squared = a squared
so u square root the answer.

Answer 8

can u plz post this at http://www.tutorbuddy.org i will answer when u post