Draw an altitude from D to AB. It has length h.
Let AB = x, AK=KD=y, BD=z
since m∠A = 60º, h=y√3
Now you have 4 equations in x,y,z,h:
2x+4y=24
x^2 = y^2+z^2
hx=2yz (area=base*height)
h=√3 y
Solve those and you will see that the parallelogram is a rhombus of side 6, with both altitudes 3√3
Given: ABCD is a parallelogram
m∠A = 60º ; BK ⊥AD AK = KD; Perimeter of ABCD = 24
Find: BD.
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