Chapter 15, Problem 44CR

To determine

To find: the perimeter of the quadrilateral HFEB .

Answer to Problem 44CR

  $HFBB=16\sqrt{2}-8$ .

Explanation of Solution

Given:

ABCD is a square in which $HF\perp AC$ as shown in the given figure.

  

The perimeter of the square is 32 units and $FC=BC$ .

Calculation:

According to the question, ABCD is a square in which $HF\perp AC$ . Now,

  $\begin{array}{l}BC=\frac{1}{4}\left(32\right)=8\\ \therefore AB=8\end{array}$

Now, by Pythagoras theorem,

  $\begin{array}{l}A{C}^2=A{B}^2+B{C}^2\\ =8^2+8^2\\ =128\\ \therefore AC=\sqrt{128}=8\sqrt{2}\end{array}$ .

Now, $AF=AC-FC=8\sqrt{2}-8$

  $\begin{array}{l}BE=CE=4\sqrt{2}\\ FE=FC-CE=8-4\sqrt{2}\\ HB=AB-AH=8-\left(16-8\sqrt{2}\right)=8\sqrt{2}-8\end{array}$

Now, the perimeter is

  $\begin{array}{l}HFBB=HF+FE+BE+HF\\ =8\sqrt{2}-8+8-4\sqrt{2}+4\sqrt{2}+8\sqrt{2}-8\\ =16\sqrt{2}-8\end{array}$

Hence, the perimeter of $HFBB=16\sqrt{2}-8$ .