AI Mentor

Check Your IQ

Free Expert Demo

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

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 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)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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

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

$= \frac{1}{2} (\frac{1}{2} \times 12^{2} sin 60^{\circ}) ∴ h_{0} = 2$ So $⊥$ distance of A from $C M = \sqrt{2^{2} + 3^{2}} = \sqrt{13}$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

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

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!