Chapter 9.3, Problem 29PPS

(a)

To determine

To Find: The equation that represents the boundary of the sprinkler area.

Answer to Problem 29PPS

  $x^2+y^2=400$

Explanation of Solution

Given: End points of the diameter are $(-12,16),(12,-16)$

Find the center using the Mid-Point formula

  $(h,k)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Here

  $\begin{array}{l}{x}_1=-12\\ y_1=16\\ x_2=12\\ y_2=-16\end{array}$

Now

  $\begin{array}{l}(h,k)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ (h,k)=\left(\frac{-12+12}{2},\frac{16-16}{2}\right)\\ (h,k)=\left(0,0\right)\end{array}$

Now

Find the radius using the distance formula

  $r=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Here

  $\begin{array}{l}{x}_1=-12\\ y_1=16\\ x_2=0\\ y_2=0\end{array}$

  $\begin{array}{l}r=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}\\ r=\sqrt{{\left(0-12\right)}^2+{\left(0+16\right)}^2}\\ r=\sqrt{144+256}\\ r=\sqrt{400}\\ r=20\end{array}$

Thus

Equation of the circle is

  $\begin{array}{l}{\left(x-0\right)}^2+{\left(y-0\right)}^2={\left(20\right)}^2\\ x^2+y^2=400\end{array}$

(b)

To determine

To Find: The area of the lawn that the sprinkler waters.

Answer to Problem 29PPS

  $1256sq.unit$

Explanation of Solution

Given: A sprinkle waters a circular lawn.

Since the lawn is circular

Area of the lawn

  $\begin{array}{l}A=\pi r^2\\ where\\ r=20\\ i.e.\\ A=\pi r^2\\ $ A=3.14\times {(20)}^2$A=3.14\times 400\\ A=1256sq.unit\end{array}$