Let A(4, -2), B(1, 1) and C(9, -3) be the vertices of a triangle ABC. Then the maximum area of the parallelogram AFDE, formed with vertices D, E and F on the sides BC, CA and AB of the triangle ABC respectively, is ______________.
Triangle Area Calculation
The vertices of triangle ABC are given: A(4, -3), B(1, 1), C(9, -3).
Use the area formula for a triangle with coordinates $(x_1, y_1), (x_2, y_2), (x_3, y_3)$: $ \text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| $
Substituting the values:
Area(ABC) $= \frac{1}{2} |4(1 - (-3)) + 1(-3 - (-3)) + 9(-3 - 1)|$
Area(ABC) $= \frac{1}{2} |4(4) + 1(0) + 9(-4)|$
Area(ABC) $= \frac{1}{2} |16 - 36| = \frac{1}{2} |-20| = 10$ square units.
Maximum Parallelogram Area Formula
A key geometric principle states that the maximum area of a parallelogram inscribed within a triangle, sharing one vertex with the triangle (like vertex A here), is exactly half the triangle's area.
Maximum Area of Parallelogram $= \frac{1}{2} \times \text{Area of Triangle ABC}$
Result Determination
Calculate the maximum possible area for parallelogram AFDE:
Maximum Area $= \frac{1}{2} \times 10 = 5$ square units.
The problem specifies that the correct value lies between 3 and 3.
Similar Questions
A line passing through the point $P(\sqrt{5}, \sqrt{5})$ intersects the ellipse $\frac{x^2}{36} + \frac{y^2}{25} = 1$ at A and B such that $(PA).(PB)$ is maximum. Then $5(PA^2 + PB^2)$ is equal to :
- If $\theta \in [ - 2\pi, 2\pi]$, then the number of solutions of $2\sqrt{2}\cos^2\theta+(2-\sqrt{6}) \cos\theta-\sqrt{3}=0$, is equal to:
- The sum of the infinite series $\cot^{-1} \left(\frac{7}{4}\right) + \cot^{-1} \left(\frac{19}{4}\right) + \cot^{-1} \left(\frac{39}{4}\right) + \cot^{-1} \left(\frac{67}{4}\right) + \dots$ is:
- The number of solutions of the equation $(4-\sqrt{3}) \sin x - 2\sqrt{3} \cos^2 x = -\frac{4}{1+\sqrt{3}}, x \in [-2\pi, \frac{5\pi}{2}]$ is
Consider the following two statements :-
Statement p :
The value of $sin120^\circ$ can be derived by taking $\theta = 240^\circ$ in the equation $2sin \frac{\theta}{2} = \sqrt{1+sin \theta} - \sqrt{1-sin \theta}$
Statement q :
The angles A, B, C and D of any quadrilateral ABCD satisfy the equation $cos \left( \frac{1}{2}(A+C) \right) + cos \left( \frac{1}{2}(B+D) \right) = 0$
Then the truth values of p and q are respectively :-
- A tower $T_1$ of height 60 m is located exactly opposite to a tower $T_2$ of height 80 m on a straight road. From the top of $T_1$, if the angle of depression of the foot of $T_2$ is twice the angle of elevaion of the top of $T_2$, then the width (in m) of the road between the feet of the towers $T_1$ and $T_2$ is :-
The number of solutions of the equation $\cos 2\theta \cos \frac{\theta}{2} + \cos \frac{5\theta}{2} = 2\cos^3 \frac{5\theta}{2}$ in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$is :
The number of solutions of $sin3x = cos2x$, in the interval $\left[ \frac{\pi}{2}, \pi \right]$ is :-
If $y=\cos\left(\frac{\pi}{3}+\cos^{-1}\frac{x}{2}\right)$, then $(x-y)^2+3y^2$ is equal to ____________.
- The angle of elevation of the top $P$ of a vertical tower for a person standing at a point $A$ on the horizontal ground due north of the tower is $45^\circ$. Another person $B$ is standing $50$ m west of $A$ on the ground. If the angle of elevation of $P$ at $B$ is $30^\circ$ then the height (in meters) of the tower is
Important Questions from Trigonometry
A line passing through the point $P(\sqrt{5}, \sqrt{5})$ intersects the ellipse $\frac{x^2}{36} + \frac{y^2}{25} = 1$ at A and B such that $(PA).(PB)$ is maximum. Then $5(PA^2 + PB^2)$ is equal to :
- If $\theta \in [ - 2\pi, 2\pi]$, then the number of solutions of $2\sqrt{2}\cos^2\theta+(2-\sqrt{6}) \cos\theta-\sqrt{3}=0$, is equal to:
- The sum of the infinite series $\cot^{-1} \left(\frac{7}{4}\right) + \cot^{-1} \left(\frac{19}{4}\right) + \cot^{-1} \left(\frac{39}{4}\right) + \cot^{-1} \left(\frac{67}{4}\right) + \dots$ is:
- The number of solutions of the equation $(4-\sqrt{3}) \sin x - 2\sqrt{3} \cos^2 x = -\frac{4}{1+\sqrt{3}}, x \in [-2\pi, \frac{5\pi}{2}]$ is
Consider the following two statements :-
Statement p :
The value of $sin120^\circ$ can be derived by taking $\theta = 240^\circ$ in the equation $2sin \frac{\theta}{2} = \sqrt{1+sin \theta} - \sqrt{1-sin \theta}$
Statement q :
The angles A, B, C and D of any quadrilateral ABCD satisfy the equation $cos \left( \frac{1}{2}(A+C) \right) + cos \left( \frac{1}{2}(B+D) \right) = 0$
Then the truth values of p and q are respectively :-
