Let A, B, C and D be the four points A = ( 3/ 2 , 2 ) , B = ( − 3/ 2 , 2 ) , C = ( − 3/ 2 , − 2 ) , D = ( 3 cos θ , 2 sin θ ) , 3 π /2

Question

Let A, B, C and D be the four points A = ( 3/ 2 ​ , 2 ​ ) , B = ( − 3/ 2 ​ , 2 ​ ) , C = ( − 3/ 2 ​ , − 2 ​ ) , D = ( 3 cos θ , 2 sin θ ) , 3 π /2 < θ < 2 π . ​ The value of θ that maximizes the area of quadrilateral ABCD is α. Choose the correct option(s).

TestHub · Accepted Answer

Split ABCD along diagonal AC. Triangle ABC has area 6. In the given fourth-quadrant interval, the distance of D from AC is largest at 7pi/4. Hence B and D are correct.

$A C : 3 y - 2 x = 0, ∣ A C ∣ = 26 . p = \frac{6}{13} ∣ sin θ - cos θ ∣. S_{A C D} = 32 ∣ sin θ - cos θ ∣. ∣ sin θ - cos θ ∣ \leq 2 . α = 7 π /4, p_{m a x} = 6 2/13 . S_{m a x} = 6 + 6 = 12.$

Mathematics - Area Under Curves Question with Solution | TestHub

Options:(select one or more)

Doubts & Discussion