Chapter 8.3, Problem 10WE

To determine

To prove: that $\Delta BCD$ must be obtuse.

Explanation of Solution

Given information:

The given figure is as follows:

  

  $\overline{AC}\perp$ plane $P$ .

Proof:

Consider the right angle triangle, $\Delta ABC$

Applying Pythagorean theorem,

  $\begin{array}{l}A{B}^2=A{C}^2+B{C}^2\\ {10}^2=8^2+B{C}^2\\ B{C}^2=100-64\\ B{C}^2=36\\ BC=6\end{array}$

Consider the triangle, $\Delta BCD$

  $\begin{array}{l}C{D}^2+C{B}^2=7^2+6^2\\ =49+36\\ =85\end{array}$

From the figure,

  $B{D}^2={11}^2=121$

So, $B{D}^2>C{D}^2+C{B}^2$

That is, the sum of the squares of the two shorter sides of a triangle is smaller than the square of the longest side, which is the property of an obtuse triangle.

Therefore, $\Delta BCD$ must be obtuse.

Hence, Proved.