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

`(1)/(2)* base* height`

From the question,

The base = a

The height= √(2a²)

Step 2

Find the area

`\begin{gathered} \text{Area= }(1)/(2)* a*\sqrt[]{2a^2} \\ \end{gathered}`
`\begin{gathered} \text{Area}=(1)/(2)* a*\sqrt[]{2}*\sqrt[]{a^2} \\ \text{Area}=(1)/(2)* a*\sqrt[]{2}* a \end{gathered}`
`\begin{gathered} \text{Area}=(1)/(2)* a^2*\sqrt[]{2} \\ \text{Area}=\frac{a^2\sqrt[]{2}}{2}unit^2 \end{gathered}`