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christoph
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in right triangle ACD, AD = 12, CD = 5, and BD is 21 Feb 2005, 01:25
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in right triangle ACD, AD = 12, CD = 5, and BD is perpendicular to AC. the lenght, BD, equals

a 12/5
b 144/25
c 4
d 60/13
e 6



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in right triangle ACD, AD = 12, CD = 5, and BD is 21 Feb 2005, 05:18
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ARGH I GET 0 for it
something's really wrong
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banerjeea_98
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in right triangle ACD, AD = 12, CD = 5, and BD is 21 Feb 2005, 10:02
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BG
AC=13 then (BD*AC)/2=30 so BD=60/13
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"D"

good approach.....got the same ans, a bit longer.

if X is AC....and Z is BD then

SQRT(c2+a2) = SQRT(a2-z2) + SQRT(c2-z2)

solve u get.....z = ac/SQRT(a2+c2)......60/13
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in right triangle ACD, AD = 12, CD = 5, and BD is 21 Feb 2005, 20:57
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christoph
in right triangle ACD, AD = 12, CD = 5, and BD is perpendicular to AC. the lenght, BD, equals

a 12/5
b 144/25
c 4
d 60/13
e 6
Show more


This should be simple.

Remember: Area of triangle = (1/2) * (base) * (height)

Now Area of triangle ADC = (1/2) * AD * CD = (1/2) * AC * BD

AC = 13 from Pythogores Theorm. Hence BD can be found out.

12 * 5 = 13 * BD

=> BD = 60/13

Answer is (D)
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MA
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in right triangle ACD, AD = 12, CD = 5, and BD is 21 Feb 2005, 22:45
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nice approach. actually i was doing similarly, but the compuer crashed so i could not post the same.

BG
AC=13 then (BD*AC)/2=30 so BD=60/13
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
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Thank you for understanding, and happy exploring!
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