ABCD is a tetrahedron with $Δ B C D$ being equilateral, having each side equal to 12 units. H is foot of perpendicular from A to $Δ B C D$ and it lies inside the triangle. If area of $Δ C D H$ is three times the area of $Δ B C H$ and area of $Δ D B H$ is two times the area of $Δ B C H$ , AH = 3 and M is midpoint of side BD. then which of the following statement(s) is(are) TRUE?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Perpendicular distance of point H from side BD is $2 \sqrt{3}$

Perpendicular distance of point H from side BC is $\sqrt{3}$

Perpendicular distance of point A from line CM is $\sqrt{15}$

Perpendicular distance of point A from line CM is $\sqrt{13}$

answer is A, B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

consider $H = t \bar{c} + s \bar{d}, (t, s \in R)$
Area of $Δ l e$ CDH = 3 Area of $Δ l e$ BCH
$\Rightarrow 1 - t - s = 3 s$ ____________(1)
Area of $Δ l e$ DBH = 2 Area of $Δ l e$ BCH
$\Rightarrow t = 2 s$ ____________(2)
solving 1 and 2, $t = \frac{1}{3}, s = \frac{1}{6}$
$H (\frac{1}{3} \bar{c} + \frac{1}{6} \bar{d}) \bar{a} - \frac{1}{3} \bar{c} - \frac{1}{6} \bar{d} = λ (\bar{c} \times \bar{d})$