geometry problem
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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.
If that's the correct answer, please let me know and I'll explain.
- ssmiles08
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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
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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How can you just assume Line CD bisects AB?? It only said it forms a 90 degree angle!.. doesn't say ad= ab? anywhere?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
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in an equilateral triangle a perpendicular drawn from a vertex to a side will bisect the side and h=sqrt3/2*(side).[theorem]fruti_yum wrote:How can you just assume Line CD bisects AB?? It only said it forms a 90 degree angle!.. doesn't say ad= ab? anywhere?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
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