9.In the following figure AC = 10 &BD = 14, then area of quadrilateral

1.	9.In the following figure AC = 10 &BD = 14, then area of quadrilateral ABCD -A35 V250 20250(A) 70/3(B) 14013bKO(C)35 13(D) 70
Answer» just we have to use formula, AREA of quadrilateral = $\frac{1}{2}P_1P_2sin\theta$ where $P_1$ and $P_2$ are length of DIAGONALS of quadrilateral , $\theta$ is angle between diagonals. Let diagonals intersect at O. from ∆BOC, $\angle OBC+\angle BCO+\angle COB=180^{\circ}$ or, $25^{\circ}+35^{\circ}+\theta=180^{\circ}$ or, $\theta=120^{\circ}$ also $P_1=10$ and $P_2=<klux>14</klux>$ now area of quadrilateral = 1/2 × 10 × 14 × sin120° = 70 × sin120° = 70 × √3/2 = 35√3 sq unit. hence, option (C) is correct CHOICE .