Question

Let ABCD be a parallelogram whose equations for the diagonals AC and BD are x+2y = 3 and 2x+y = 3, respectively. If length of diagonal AC = 4 units and area of parallelogram ABCD = 8 sq.units. then

The length of side AB is equal to

(a) \(\frac{2\sqrt{58}}3\)

(b) \(\frac{4\sqrt{58}}9\)

(c) \(\frac{3\sqrt{58}}9\)

(d) \(\frac{5\sqrt{58}}9\)

Answer 1

cos(π–θ ) \(= \frac{AP^2+PB^2-AB^2}{2AP\times PB}\)

–cosθ \(=\frac{-4}5=\frac{4+\frac{100}{9}-AB^2}{2\times 2\times \frac{10}3}\)

AB = \(\frac{2\sqrt{58}}{3}\)