ABCD is a trapezoid with ∠DAB=∠ABC=π/2, AD=1 and PA=AB=BC=2; PA is perpendicular to plane ABCD. M is the midpoint of PC. Find the volume of P–ABD, then the distance from A to plane PBD.

01 / Standard#Your turn

Official paper · jm02-2021 · 1(a)(ii) · PDF 3

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Triangle ABD is right at A.

Hint 2

The same tetrahedron can use PBD as its base and A as its apex.

Worked solution

Compute the base area and the original perpendicular height.
$A_{ABD}=\tfrac12(2)(1)=1,\quad V=\tfrac13(1)(2)=\tfrac23$
Changing the base does not change the volume.
$\tfrac13 A_{PBD}d=\tfrac23\implies \tfrac{\sqrt6}{3}d=\tfrac23$
Use a positive perpendicular distance.
$d=2/\sqrt6=\sqrt{2/3}=\sqrt6/3$

Volume 2/3; distance √(2/3).

Checks and common pitfalls: The official √(2/3) has the entire fraction under the radical.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Compute the base area and the original perpendicular height.
$A_{ABD}=\tfrac12(2)(1)=1,\quad V=\tfrac13(1)(2)=\tfrac23$
Changing the base does not change the volume.
$\tfrac13 A_{PBD}d=\tfrac23\implies \tfrac{\sqrt6}{3}d=\tfrac23$
Use a positive perpendicular distance.
$d=2/\sqrt6=\sqrt{2/3}=\sqrt6/3$

Think first. Reveal a hint when the class is ready.

2021 JM02