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 the total surface area.

01 / Standard#Your turn

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

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Account for the base and all four triangular faces.

Hint 2

The two right faces are congruent; VBC is isosceles.

Worked solution

Compute the base, equilateral face, and two right faces.
$A_{ABCD}=a^2,\quad A_{VAD}=\frac{\sqrt3}4a^2,\quad A_{VAB}=A_{VDC}=\frac12a^2$
For VBC, its equal sides are √2a and its base is a.
$h_{VBC}=\sqrt{2a^2-\frac{a^2}4}=\frac{\sqrt7}2a,\quad A_{VBC}=\frac{\sqrt7}4a^2$
Sum all five faces.
$A_{\rm total}=\left(2+\frac{\sqrt3+\sqrt7}4\right)a^2$

Total area [2+(√3+√7)/4]a².

Checks and common pitfalls: Total surface area includes the square base.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Compute the base, equilateral face, and two right faces.
$A_{ABCD}=a^2,\quad A_{VAD}=\frac{\sqrt3}4a^2,\quad A_{VAB}=A_{VDC}=\frac12a^2$
For VBC, its equal sides are √2a and its base is a.
$h_{VBC}=\sqrt{2a^2-\frac{a^2}4}=\frac{\sqrt7}2a,\quad A_{VBC}=\frac{\sqrt7}4a^2$
Sum all five faces.
$A_{\rm total}=\left(2+\frac{\sqrt3+\sqrt7}4\right)a^2$

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

2025 JM02