Let a point A lie between the parallel lines | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the side length of the equilateral triangle $ABC$ be $a$. The distance between the parallel lines $L_1$ and $L_2$ is $6 + 3 = 9$ units.Let $

Vedclass · Accepted Answer

$21 \sqrt{3}$

Vedclass · Answer

(C) Let the side length of the equilateral triangle $ABC$ be $a$. The distance between the parallel lines $L_1$ and $L_2$ is $6 + 3 = 9$ units.Let $	heta$ be the angle that the side $BC$ makes with the line $L_2$. Since $C$ lies on $L_2$,the vertical distance from $A$ to $L_2$ is $9$. In the right-angled triangle formed by the projection of $A$ onto $L_2$,we have $\sin 	heta = \frac{3}{a}$ and $\sin(60^{\circ} + 	heta) = \frac{9}{a}$.Expanding $\sin(60^{\circ} + 	heta) = \sin 60^{\circ} \cos 	heta + \cos 60^{\circ} \sin 	heta = \frac{\sqrt{3}}{2} \cos 	heta + \frac{1}{2} \sin 	heta = \frac{9}{a}$.Substituting $\sin 	heta = \frac{3}{a}$ and $\cos 	heta = \sqrt{1 - \frac{9}{a^2}}$,we get $\frac{\sqrt{3}}{2} \sqrt{1 - \frac{9}{a^2}} + \frac{1}{2} \cdot \frac{3}{a} = \frac{9}{a}$.$\frac{\sqrt{3}}{2} \sqrt{\frac{a^2 - 9}{a^2}} = \frac{9}{a} - \frac{3}{2a} = \frac{15}{2a}$.$\sqrt{3} \sqrt{a^2 - 9} = 15 \implies 3(a^2 - 9) = 225 \implies a^2 - 9 = 75 \implies a^2 = 84$.The area of the equilateral triangle is $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} 	imes 84 = 21 \sqrt{3}$ sq. units.

Let a point $A$ lie between the parallel lines $L_1$ and $L_2$ such that its distances from $L_1$ and $L_2$ are $6$ and $3$ units,respectively. Then the area (in sq. units) of the equilateral triangle $ABC$,where the points $B$ and $C$ lie on the lines $L_1$ and $L_2$ respectively,is:

Similar Questions

If the three lines $x - 3y = p$,$ax + 2y = q$,and $ax + y = r$ form a right-angled triangle,then:

Let the equations of two adjacent sides of a parallelogram $ABCD$ be $2x - 3y = -23$ and $5x + 4y = 23$. If the equation of its one diagonal $AC$ is $3x + 7y = 23$ and the distance of $A$ from the other diagonal is $d$,then $50d^2$ is equal to $........$.

Let $A \equiv (3, 2)$ and $B \equiv (5, 1)$. An equilateral triangle $ABP$ is constructed on the side of $AB$ remote from the origin. The orthocentre of triangle $ABP$ is:

If the incentre and the circumcentre of the triangle formed by the lines $x=2$,$4x+3y+7=0$ and $y=3$ are $I$ and $S$ respectively,then $IS=$

The length of the altitude through $A$ of the triangle $ABC$,where $A \equiv (-3, 0)$,$B \equiv (4, -1)$,and $C \equiv (5, 2)$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module