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In the diagram, square ABCD has sides of length 4, and triangle ABE is equilateral. Line segments BE and AC intersect at P. Point Q is on BC so that PQ is perpendicular to BC and PQ=x.  Compute AP.

 

 Apr 20, 2021
 #1
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Wow! This problem really is hard! I don't really know how to figure it out, but it might not be too hard once you've used a bunch of theorems about right triangles/cosines/sines and whatnot.

 

This question has already been answered though, so I'll leave you a link to the answer:

https://web2.0calc.com/questions/in-the-diagram-square-abcd-has-sides-of-length-4

 Apr 20, 2021
 #3
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That is asking for the area of APE, and he/she is asking for AP.

SparklingWater2  Apr 20, 2021
 #2
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triangle PQC is a 45-45-90, so QC=x

triangle PQB is a 30-60-90 triangle, so BQ = x$\sqrt3$

QC+BQ=4=x+x$\sqrt3$

x($\sqrt3$+1)=4

 

x=$\dfrac4{\sqrt3+1}$

 

we can multiply top and bottom of the fraction by $\sqrt3-1$:

 

x=$\dfrac{4\sqrt3-4}{3-1}=2\sqrt3-2$

 

PC=x$\sqrt2=2\sqrt6-2\sqrt2$

AP=AC-PC=$4\sqrt2-2\sqrt6+2\sqrt2=\boxed{6\sqrt2-2\sqrt6}$

 Apr 20, 2021
edited by SparklingWater2  Apr 20, 2021

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