A seesaw K feet long

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A seesaw K feet long

by figs » Mon May 25, 2009 4:20 am
A seesaw K feet long is supported at its center by a frame N feet high, where 2N is less than k. What is the greatest number of feet either end of the seesaw can rise above the ground?
A. k/N
B. N
C. k/2 +N
D. 2N
E. KN

I think that i don't understand the quetion...
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by scoobydooby » Mon May 25, 2009 6:14 am
am getting 2N

the length of the seesaw on either side of the frame: K/2
let the left side of the seesaw be at the bottom and the right side go up, we get a triangle
the base on which the frame is fixed is of equal length on either side of the frame.
applying mid-point theorem (a line joining the mid-points of two sides of a triangle, parallel to the third side is 1/2 the length of the 3rd side)
the max height is the 3rd side here, N joins the midpoints, so the max height must be 2N

hence, D

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Re: A seesaw K feet long

by dtweah » Mon May 25, 2009 7:20 am
figs wrote:A seesaw K feet long is supported at its center by a frame N feet high, where 2N is less than k. What is the greatest number of feet either end of the seesaw can rise above the ground?
A. k/N
B. N
C. k/2 +N
D. 2N
E. KN

I think that i don't understand the quetion...
You need some instruments to convincingly derive this result. You can use:
1) Vertical Angle Equality and alternative interior angle equality or corresponding angle equality when two lines are ||
2) Side angle side SAS theorem of similarity

If you let one end of the ladder hit the bottom and the other you can draw derive 2 triangles from the figure. Connect two parralell lines one passing through the fulcrum and another passing through the point where ladder hits the ground. one bottom right triangle with hypotenuse k/2, height N and base X. A top right triangles with these same dimensions. Since the ladder and the line thrugh the fulcrum make vertical angles, we know they are equal. Call this angle Y. Under altertate interior angles, this angle equals the angle formed by bottom hypotenuse and bottom base. So in these two triangles we know SAS will apply. And since under SAS they are similar we know their respective heights must be proportional. So
X/X= N/H, where H is height of upper triangle. This means

H=N

Hence height above ground is H+N=2N

Choose D.