A square pyramid V–ABCD has base side a>0. Face VAD is equilateral and perpendicular to the base; M is the midpoint of AD. Find VM.

01 / Standard#Your turn

Official paper · jm02-2025 · 1(a) · PDF 3

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

The median of an equilateral triangle is an altitude.

Hint 2

Apply Pythagoras with hypotenuse a and one leg a/2.

Worked solution

The altitude bisects AD.
$AM=\frac a2,\quad VA=a,\quad VM\perp AD$
Take the positive square root.
$VM=\sqrt{a^2-\left(\frac a2\right)^2}=\frac{\sqrt3}2a$

VM=√3a/2.

Checks and common pitfalls: Use the altitude, not the sloping side VA, for VM.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
The altitude bisects AD.
$AM=\frac a2,\quad VA=a,\quad VM\perp AD$
Take the positive square root.
$VM=\sqrt{a^2-\left(\frac a2\right)^2}=\frac{\sqrt3}2a$

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

2025 JM02