# Thread: [SOLVED] SAT page 718 #7

1. ## [SOLVED] SAT page 718 #7

Note: Figure not drawn to scale

In the figure below, line segment AE and line segment CD are each perpendicular to line segment CE. If x=y, the length of line segment AB is 4, and the length of line segment BD is 8, what is the length of line segment CE?

2. Originally Posted by fabxx
Note: Figure not drawn to scale

In the figure below, line segment AE and line segment CD are each perpendicular to line segment CE. If x=y, the length of line segment AB is 4, and the length of line segment BD is 8, what is the length of line segment CE?

note that x and y are 45 degrees. note further that we have similar triangles here. the sides of triangle BCD are all twice the corresponding sides of ABE. angle CDB = angle BAE and angle CBD = angle ABE

now, two questions: how did i know all these facts? and can you finish the problem now that you know them?

Originally Posted by Jhevon
note that x and y are 45 degrees. note further that we have similar triangles here. the sides of triangle BCD are all twice the corresponding sides of ABE. angle CDB = angle BAE and angle CBD = angle ABE

now, two questions: how did i know all these facts? and can you finish the problem now that you know them?
HI,
so knowing that an isoceles rt triangle with angles 45-45-90
have sides 1-1-sqrt2,
would you then take the hypotenuse of 4 and work it back to get sqrt2
and then take this multiplying factor and multiply the other sides by it?
you could have 2 sqrt2 = 4 or 4/sqrt2 = 4 (same thing)
the other 2 sides are 2*sqrt2

Then 8 is 2 times 4, triangles are the same type, so multiply 2 times
the smaller triangels sides to get 4sqrt2

4. Originally Posted by eri
HI,
so knowing that an isoceles rt triangle with angles 45-45-90
have sides 1-1-sqrt2,
would you then take the hypotenuse of 4 and work it back to get sqrt2
and then take this multiplying factor and multiply the other sides by it?
you could have 2 sqrt2 = 4 or 4/sqrt2 = 4 (same thing)
the other 2 sides are 2*sqrt2

Then 8 is 2 times 4, triangles are the same type, so multiply 2 times
the smaller triangels sides to get 4sqrt2
it is even easier. knowing these are similar triangles (since all the angles are proportional) we know that CE = 3BE. and we can find BE using trigonometry. $\displaystyle \sin x = \frac {BE}{AB} \implies BE = AB \sin x = 4 \sin 45 = 4 \cdot \frac {\sqrt{2}}2 = 2 \sqrt{2}$
so finally, $\displaystyle CE = 3 BE = 6 \sqrt{2}$