if PQ=4.12, QR=11 and PR=12.65 , find the QS=?

Question

PQ=4.12 , PR=12.65 QR=11 QS=?

Answer 1

PQRS appears to be a parallelogram, so PR and QS are diagonals.

Cosine Rule: PR²=PQ²+QR²-2PQ.QRcos(PQ̂R),

12.65²=4.12²+11²-2(4.12)(11)cos(PQ̂R),

160.0225=16.9744+121-90.64cos(PQ̂R),

22.0481=-90.64cos(PQ̂R),

cos(PQ̂R)=-22.0481/90.64=-0.24325, PQ̂R=104.08° approx. 

QP̂S=QR̂S=75.92° (supplementary angles).

Cosine Rule: QS²=PQ²+PS²-2(PQ)(PS)cos(QP̂S).

PS=QR=11, so QS²=16.9744+121-90.64cos(75.92)=115.9263, QS=10.77 approx.

Note that cos(x)=-cos(180-x) so the only difference between the lengths of the diagonals is the sign of the cosine term.

Using a²=b²+c²-2bccosA, the absolute difference between the squares of the diagonals (lengths a₁ and a₂) is given by:

|a₁²-a₂²|=4bccosA where A is the acute angle of the parallelogram and b and c are the lengths of the sides.