Problem 07 Find the area of the triangle whose vertex is at the midpoint of an upper edge of a cube of edge a and whose base coincides with the diagonally opposite edge of the cube.

Solution 07

$d^2 = 2a^2$

$d = a\sqrt{2}$

Area of triangle ABC $A_{ABC} = \frac{1}{2}ad$

$A_{ABC} = \frac{1}{2}a(a\sqrt{2})$

$A_{ABC} = \frac{1}{2}\sqrt{2} \, a^2$ answer

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