ABCD is a tetrahedron with ΔBCD being equilateral, having each side equal to 12 units. H is foot of perpendicular from A to ΔBCD and it lies inside the triangle if area of ΔCDH is three times the area of ΔBCH and area of ΔDBH is two times the area of ΔBCH,AH=3 and M is midpoint of side BD. Choose the correct option(s)

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

ABCD is a tetrahedron with $Δ BCD$ being equilateral, having each side equal to 12 units. H is foot of perpendicular from A to $Δ BCD$ and it lies inside the triangle if area of $Δ CDH$ is three times the area of $Δ BCH$ and area of $Δ DBH$ is two times the area of $Δ BCH, AH = 3$ and M is midpoint of side BD. Choose the correct option(s)

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

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

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

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

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

answer is A, B, C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Let ⊥ distance from H to BC,BD and CD be $p, 2 p, 3 p$

$\frac{1}{2} \times 12 (p + 2 p + 3 p) = \frac{1}{2} 12^{2} sin 60^{\circ} p = \sqrt{3} \frac{1}{2} \times 6 \times 2 \sqrt{3} + \frac{1}{2} \times 12 \sqrt{3} + \frac{1}{2} \times h_{0} 6 \sqrt{3}$