Question 1120601: A straight pole PR is leaning against a vertical wall MQ.
I) Given that PQ=4.2m and RQ=1.1m, find the length of the pole.
II) the upper end of the pole, P, slides down 0.9m to a point on the wall, X. Calculate YR, the distance the lower end of the pole has slid away from its original position R.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! pr is leaning against the wall mq.
pq = 4.2 meters.
rq = 1.1 meters.
pr is the hypotenuse of the right triangle formed by pqr.
as such, 4.2^2 + 1.1^2 = pr^2.
solve for pr to get pr = 4.341658669 meters.
if the pole slides down .9 meters to point x, then xq = 4.2 - .9 = 3.3.
the pole is still 4.341658669 meters.
the ground side of the pole goes from qr to qy, where qy is larger than qr.
the size of qr is equal to sqrt(4.34165866^2 - 3.3^2) = 2.821347196 meters.
the distance from point y to point r would be 2.821347196 minus 1.1 = 1.721347196 meters.
the top of the pole slid down .9 meters.
the bottom of the pole slid out 1.721347196 meters.
here's my diagram.
what you are using is the pythagorean formula of c^2 = a^2 + b^2, where c is the hypotenuse of a right triangle and a and b are the legs.
in your problem, you find pr by letting a = pq = 4.2 and b = qr = 1.1 and solving for c = pr to get c = pr = 4.341658669
you then drop pq to xq and extend qr to qy.
you use the pythagorean formula again and let a = xq = 4.2 - .9 = 3.3 and b = qy and c = xy = 4.341658669 and solve for b to get b = qy = 2.821347196.
you then subtract qy from qr to get ry which is the amount that the bottom of the pole has slid away from its original position.
|
|
|
