geometry problem

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.

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

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?

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]