The two diagonals of a
cyclic quadrilateral EFGH are EG and FH while
the respective lengths of its adjacent sides EF and FG are 2 and 5.
It is known that Angle EFG = 60
o and the area of the quadrilateral is 4√3.
Determine the sum of the lengths GH + HE.
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Submitted by K Sengupta
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Rating: 3.5000 (2 votes)
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Solution:
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Angle EHG = 180o - 60o = 120o, and::
Area(EFGH) = Area (EFG) + Area(EHG)
Now, Area(EHG) = (½)*2s*sin 60o = 10V3/4, and:
Area(EHG) = pq*V3/4, where (HG, EH) = (p, q) (say)
Thus, 4 = (10+pq)/4, giving:
pq = 16-10 = 6
Now, in triangle EFG, we have:
EG^2 = EF^2 + FG^2 – 2*EF*FG*cos 60o
= 4+ 2s-20*(1/2)
= 19
Similarly, for triangle EHG, we have:
EG^2 = p^2 + q^2 – 2pq*(cos 120o)
= p^2 + q^2 + pq
= p^2+q^2 + 6
Thus, p^2+q^2+6 = 19, giving:
p^2+q^2 = 13
Solving for (p^2+q^2, pq) = (13, 6), we obtain:
(p, q) = (2, 3) or (3, 2), so that:
p+q = 5
Thus, the required sum of the lengths GH + HE is 5.
**** Also, refer to the methodology submitted by Bractals in this location.
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