Asked by nazz
ABCD is a parallelogram. AB = 3√2, AD = 8, ∠BAD = 45◦ M is a point on CD, CM : MD = 1 : 2. N is a point on AD, AN : ND = 3 : 1. What's the length of of AM and BN?
Answers
Answered by
mathhelper
Did you make a reasonable sketch?
Since we have a parallogram, AB = CD and we have to split CD so
that CM : MD = 1 : 2 = x : 2x
so 3x = 3√2
x = √2, and 2x = 2√2
in the same way we have to split AD int AN : ND = 3:1 = 3y : y
4y = 8
y = 2, so AN = 6 and ND = 2
Furthermore , angle D = 135° , properties of ||gram
by the cosine law for triangle AMD
AM^2 = AD^2 + MD^2 - 2(AD)(MD)cos 135°
= 8^2 + (2√2)^2 - 2(8)(2√2)cos135
= 64 + 8 - 32
= 40
AM = √40 = 2√10 or appr 6.325
Find BN in the same way using my results from above.
Since we have a parallogram, AB = CD and we have to split CD so
that CM : MD = 1 : 2 = x : 2x
so 3x = 3√2
x = √2, and 2x = 2√2
in the same way we have to split AD int AN : ND = 3:1 = 3y : y
4y = 8
y = 2, so AN = 6 and ND = 2
Furthermore , angle D = 135° , properties of ||gram
by the cosine law for triangle AMD
AM^2 = AD^2 + MD^2 - 2(AD)(MD)cos 135°
= 8^2 + (2√2)^2 - 2(8)(2√2)cos135
= 64 + 8 - 32
= 40
AM = √40 = 2√10 or appr 6.325
Find BN in the same way using my results from above.
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