A regular tetrahedron has edge length a. A plane cuts through the midpoints of three edges that meet at a common vertex. What is the ratio of the area of this triangular cross-section to the total surface area of the tetrahedron?

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EEJ MAIN Aptitude QUESTION #10990

Question 1

A regular tetrahedron has edge length $a$. A plane cuts through the midpoints of three edges that meet at a common vertex. What is the ratio of the area of this triangular cross-section to the total surface area of the tetrahedron?

$\frac{1}{8}$
$\frac{1}{12}$
$\frac{1}{16}$✔️
$\frac{1}{24}$

Correct Answer Explanation

The cross-section is an equilateral triangle with side $\frac{a}{2}$. Its area is $\frac{\sqrt{3}}{4}(\frac{a}{2})^2 = \frac{\sqrt{3}a^2}{16}$. Total surface area of tetrahedron: $4 \times \frac{\sqrt{3}a^2}{4} = \sqrt{3}a^2$. Ratio: $\frac{\sqrt{3}a^2/16}{\sqrt{3}a^2} = \frac{1}{16}$.

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