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by truplayer256 » Sat Sep 26, 2009 8:06 am
asqrt(3)-a+asqrt(2)+2a or asqrt(3)+asqrt(2)+a.

If that's the correct answer, please let me know and I'll explain.
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by ssmiles08 » Sat Sep 26, 2009 9:35 am
I got asqrt(3) + asqrt(2) + a

DA = a
AD = a

traingle ADE would be a 90-45-45 triangle. so AE = asqrt(2)

Line CD bisects AB so since ABC is an equilateral triangle, AC = 2a.

Line CD = asqrt(3) since triangle ADC is a 30-60-90 triangle.

CD = asqrt(3) ED = a so CE = asqrt(3) - a

triangle ACE = aqrt(3) - a + asqrt(2) + 2a

asqrt(3) +asqrt(2) + a
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by fruti_yum » Wed Sep 30, 2009 9:34 am
ssmiles08 wrote:I got asqrt(3) + asqrt(2) + a

DA = a
AD = a

traingle ADE would be a 90-45-45 triangle. so AE = asqrt(2)

Line CD bisects AB so since ABC is an equilateral triangle, AC = 2a.

Line CD = asqrt(3) since triangle ADC is a 30-60-90 triangle.

CD = asqrt(3) ED = a so CE = asqrt(3) - a

triangle ACE = aqrt(3) - a + asqrt(2) + 2a

asqrt(3) +asqrt(2) + a
How can you just assume Line CD bisects AB?? It only said it forms a 90 degree angle!.. doesn't say ad= ab? anywhere?
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by xcusemeplz2009 » Wed Sep 30, 2009 9:58 am
fruti_yum wrote:
ssmiles08 wrote:I got asqrt(3) + asqrt(2) + a

DA = a
AD = a

traingle ADE would be a 90-45-45 triangle. so AE = asqrt(2)

Line CD bisects AB so since ABC is an equilateral triangle, AC = 2a.

Line CD = asqrt(3) since triangle ADC is a 30-60-90 triangle.

CD = asqrt(3) ED = a so CE = asqrt(3) - a

triangle ACE = aqrt(3) - a + asqrt(2) + 2a

asqrt(3) +asqrt(2) + a
How can you just assume Line CD bisects AB?? It only said it forms a 90 degree angle!.. doesn't say ad= ab? anywhere?
in an equilateral triangle a perpendicular drawn from a vertex to a side will bisect the side and h=sqrt3/2*(side).[theorem]
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