Chapter 8.3, Problem 6PSA

To determine

To calculate: The coordinates of B.

Answer to Problem 6PSA

Coordinates of B is $(0,10)$ and length of BD is $26$ units.

Explanation of Solution

Given information : Here, we have given that $\Delta OAC\sim \Delta OBD$ .

Formula used : Distance formula: d = $\sqrt{{\left(x_2-x_1\right)}^2-{(y_2-y_1)}^2}$

Calculation: According to the distance formula, d = $\sqrt{{\left(x_2-x_1\right)}^2-{(y_2-y_1)}^2}$ ,

  $\begin{array}{l}\overline{OA}=\sqrt{{\left(0-0\right)}^2-{(5-0)}^2}\\ \Rightarrow \overline{OA}=\sqrt{{(5)}^2}\\ \Rightarrow \overline{OA}=5\end{array}$

Now, as point B is on y axis, hence, coordinates of B is $(0,y)$ ,

  $\begin{array}{l}\overline{OB}=\sqrt{{\left(0-0\right)}^2-{(y-0)}^2}\\ \Rightarrow \overline{OB}=\sqrt{{(y)}^2}\\ \Rightarrow \overline{OB}=y\end{array}$

Similarly,

  $\begin{array}{l}\overline{OC}=\sqrt{{\left(12-0\right)}^2-{(0-0)}^2}\\ \Rightarrow \overline{OC}=\sqrt{{(12)}^2}\\ \Rightarrow \overline{OC}=12\end{array}$

And,

  $\begin{array}{l}\overline{OD}=\sqrt{{\left(24-0\right)}^2-{(0-0)}^2}\\ \Rightarrow \overline{OD}=\sqrt{{(24)}^2}\\ \Rightarrow \overline{OD}=24\end{array}$

Now, as $\Delta OAC\sim \Delta OBD$ ,

  $\Rightarrow \frac{OA}{OB}=\frac{OC}{OD}$ (Corresponding sides of similar triangles are equal in ratio)

  $\begin{array}{l}\Rightarrow \frac{5}{y}=\frac{12}{24}\\ \Rightarrow \frac{5}{y}=\frac{1}{2}\\ \Rightarrow y=10\end{array}$

  $\therefore$ Coordinate of B is $(0,10)$ .

Now, in $\Delta OAC$ and $\Delta OBD$ ,

  $\frac{OA}{OB}=\frac{OC}{OD}$ (Given)

  $\angle BOD=\angle BOD$ (Common)

  $\Rightarrow \Delta OAC\sim \Delta OBD$ (By SAS Similarity rule)

According to the distance formula,

  $\begin{array}{l}\overline{BD}=\sqrt{{(24-0)}^2+{(0-10)}^2}\\ \Rightarrow \overline{BD}=\sqrt{{(24)}^2+{(10)}^2}\\ \Rightarrow \overline{BD}=\sqrt{576+100}\\ \Rightarrow \overline{BD}=\sqrt{676}\\ \Rightarrow \overline{BD}=26\end{array}$