E–ABCD has rhombus base AB=2√2 and ∠DAB=π/3. EB=EC, ∠BEC=π/2, DE=√5. F is the midpoint of BC; G is the perpendicular foot from E to DF. Find AE.

01 / Standard#Your turn

Official paper · jm02-2023 · 1(c) · PDF 3

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Find DG=DF−FG.

Hint 2

AD⊥DF and EG⊥the base yield two Pythagorean steps.

Worked solution

The angle ADG equals π/3+π/6=π/2.
$DG=\sqrt6-\sqrt6/4=3\sqrt6/4,\quad AG^2=AD^2+DG^2=8+54/16=91/8$
Use right triangle AGE.
$AE^2=AG^2+EG^2=91/8+13/8=13\implies AE=\sqrt{13}$

AE=√13.

Checks and common pitfalls: Use the perpendicular height EG; EF is a sloping segment.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
The angle ADG equals π/3+π/6=π/2.
$DG=\sqrt6-\sqrt6/4=3\sqrt6/4,\quad AG^2=AD^2+DG^2=8+54/16=91/8$
Use right triangle AGE.
$AE^2=AG^2+EG^2=91/8+13/8=13\implies AE=\sqrt{13}$

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

2023 JM02