Chapter 9.7, Problem 26PSC

To determine

To calculate: Distance between Origin and Point D

Answer to Problem 26PSC

Distance between point D and origin is $12.08$ units.

Explanation of Solution

Given information:Here, we have given that area of rectangle ABCE is eight times that of $\Delta BCD$ .

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

Calculation: Co-ordinates of A are $(-4,0)$ and C $(16,5)$ .

Now, Length of rectangle ( $\overline{AB}$ ) = $\sqrt{{[16-(-4)]}^2+{(0-0)}^2}$

  $\begin{array}{l}\Rightarrow \sqrt{{(16+4)}^2}\\ \Rightarrow \sqrt{{(20)}^2}\\ \Rightarrow 20\end{array}$

Similarly, Breadth of rectangle ( $\overline{AE}$ ) = $\sqrt{{[-4-(-4)]}^2+{(5-0)}^2}$

  $\begin{array}{l}\Rightarrow \sqrt{{(-4+4)}^2+{(5)}^2}\\ \Rightarrow \sqrt{{(-4+4)}^2+{(5)}^2}\\ \Rightarrow \sqrt{{(5)}^2}\\ \Rightarrow 5\end{array}$

Now, Area of rectangle = $l\times b$

  $\begin{array}{l}\Rightarrow 20\times 5\\ \Rightarrow 100\end{array}$

Now, $\overline{AE}=\overline{BC}$ (Opposite sides of rectangle are equal)

  $\Rightarrow \overline{BC}$ (Height of triangle) = $5$

Now, as Area of rectangle is eight time of area of triangle,

  $\begin{array}{l}\Rightarrow 100=8\left[\frac{1}{2}\times 5\times CD\right]\\ \Rightarrow \frac{100}{8}=\frac{1}{2}\times 5\times CD\\ \Rightarrow \frac{200}{8}=5\times CD\\ \Rightarrow 25=5\times CD\\ \Rightarrow CD=\frac{25}{5}\\ \Rightarrow CD=5\end{array}$

Now, distance of D from origin $(0,0)$ = $\sqrt{{(0-11)}^2+{(0-5)}^2}$

  $\begin{array}{l}\Rightarrow \sqrt{{(-11)}^2+{(-5)}^2}\\ \Rightarrow \sqrt{121+25}\\ \Rightarrow \sqrt{146}\\ \Rightarrow 12.08\end{array}$