Chapter 8.6, Problem 43PPS

To determine

To calculate:

The perimeter of the new garden.

Answer to Problem 43PPS

The perimeter of new garden is $96.2$ ft.

Explanation of Solution

Given information:

  

Concept used:

Pythagorean theorem:

  $a^2+b^2=c^2$

Where,

  $a,b=$ side of right triangle

  $c=$ hypotenuse

Law of cosines:

  $c^2=a^2+b^2-2ab\cos C$

Where,

  $a,b=$ sides

  $\cos C=$ angle

Law of sines:

  $a=b\frac{\sin \alpha }{\sin \beta }$

Calculation:

  

First use the Pythagorean theorem to find the length of the hypotenuse of the right triangle.

Pythagorean theorem:

  $\begin{array}{l}{a}^2+b^2=c^2\\ {18}^2+{20}^2=x^2\\ 324+400=x^2\\ 724=x^2\\ x\approx 26.9\end{array}$

Now get the triangle as follows:

  

  $m\angle C={180}^{\circ }-{52}^{\circ }-{74}^{\circ }={54}^{\circ }$

Since, having the one side and two angles, to get the values of $BC$ use the law of sines.

Law of sines:

  $a=b\frac{\sin \alpha }{\sin \beta }$

First substitute,

  $\frac{\sin 52}{BC}=\frac{\sin 54}{26.9}$

Now use the cross products property

  $BC\cdot \left(\sin 54\right)=26.9\cdot \left(\sin 52\right)$

Now divide each side by $\sin 54$

  $BC=\frac{26.9\cdot \sin 52}{\sin 54}$

Now get the value of $BC$ ,

  $BC\approx 26.2$

Now, use the law of cosines to get the value of $AC$ , as having the two sides and an included angle.

Law of cosines:

  $c^2=a^2+b^2-2ab\cos C$

First substitute,

  ${\left(AC\right)}^2={\left(26.2\right)}^2+{\left(26.9\right)}^2-2\left(26.2\right)\left(26.9\right)\cos \left(7.4\right)$

Now simplify,

  ${\left(AC\right)}^2=1022$

Now, take the square root of each side

  $AC=\sqrt{1022}$

Now, get the value of $AC$

  $AC=32.0$

Therefore, the perimeter of the garden will be about $18+20+26.2+32.0=96.2$ ft.

Hence, the perimeter of new garden is $96.2$ ft.