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 area of triangle PBD.

01 / Standard#Your turn

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

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Compute PD, BD and PB from three right triangles.

Hint 2

Use PB as the base of isosceles triangle PBD.

Worked solution

The given perpendicularities allow Pythagoras.
$PD^2=PA^2+AD^2=5,\quad BD^2=AB^2+AD^2=5,\quad PB^2=PA^2+AB^2=8$
The altitude to PB bisects this base.
$h=\sqrt{5-(\sqrt8/2)^2}=\sqrt3$
Multiply half the base by its corresponding height.
$A_{PBD}=\tfrac12\sqrt8\sqrt3=\sqrt6$

Area √6.

Checks and common pitfalls: PA is the pyramid height, not the altitude to PB inside triangle PBD.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
The given perpendicularities allow Pythagoras.
$PD^2=PA^2+AD^2=5,\quad BD^2=AB^2+AD^2=5,\quad PB^2=PA^2+AB^2=8$
The altitude to PB bisects this base.
$h=\sqrt{5-(\sqrt8/2)^2}=\sqrt3$
Multiply half the base by its corresponding height.
$A_{PBD}=\tfrac12\sqrt8\sqrt3=\sqrt6$

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

2021 JM02