Chapter 3.6, Problem 48STP

(a)

To determine

To Draw: The figure.

Explanation of Solution

Given:Segment $ABandCD$ are perpendicular to each other and bisect each other at point $X$ .

  

(b)

To determine

To Find: The $BDifAB=12andCD=16$ .

Answer to Problem 48STP

  $BD=10$

Explanation of Solution

Given:Segment $ABandCD$ are perpendicular to each other and bisect each other at point $X$ .

Using Pythagoras theorem

  $\begin{array}{l}B{D}^2=B{X}^2+D{X}^2\\ BD=\sqrt{B{X}^2+D{X}^2}\\ now\\ AB=AX+XB\\ AB=AX+AX(AX=XB)\\ 12=2AX\\ AX=6\\ similarly\\ CD=CX+DX\\ CD=CX+CX\\ 16=2CX\\ CX=8\\ Then\\ BD=\sqrt{B{X}^2+D{X}^2}\\ BD=\sqrt{6^2+8^2}\\ BD=\sqrt{36+64}\\ BD=10\end{array}$

(c)

To determine

To Find: The $BDifAB=24andCD=18.$

Answer to Problem 48STP

  $BD=15$

Explanation of Solution

Given:Segment $ABandCD$ are perpendicular to each other and bisect each other at point $X$ .

  

  $\begin{array}{l}B{D}^2=B{X}^2+D{X}^2\\ BD=\sqrt{B{X}^2+D{X}^2}\\ now\\ AB=AX+XB\\ AB=AX+AX(AX=XB)\\ 24=2AX\\ AX=12\\ similarly\\ CD=CX+DX\\ CD=DX+DX\\ 18=2DX\\ DX=9\\ Then\\ BD=\sqrt{B{X}^2+D{X}^2}\\ BD=\sqrt{{12}^2+9^2}\\ BD=\sqrt{144+81}\\ BD=15\end{array}$