Question

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

Answer 1

Step 1

The area of a triangle is given as

$\frac{1}{2}\times base\times height$

From the question,

The base = a

The height= √(2a²)

Step 2

Find the area

$\begin{gathered} \text{Area= }\frac{1}{2}\times a\times\sqrt[]{2a^2} \\ \end{gathered}$$\begin{gathered} \text{Area}=\frac{1}{2}\times a\times\sqrt[]{2}\times\sqrt[]{a^2} \\ \text{Area}=\frac{1}{2}\times a\times\sqrt[]{2}\times a \end{gathered}$$\begin{gathered} \text{Area}=\frac{1}{2}\times a^2\times\sqrt[]{2} \\ \text{Area}=\frac{a^2\sqrt[]{2}}{2}unit^2 \end{gathered}$