Chapter 8.6, Problem 48PPS

To determine

To calculate:

The perimeter of the given diagram.

Answer to Problem 48PPS

The perimeter of the quadrilateral is $14.7$ units.

Explanation of Solution

Given information:

  

Concept used:

Law of sines:

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

Calculation:

  

First name the quadrilateral $ABCD$ , the sum of the angles of a triangle is $180$ so,

  $m\angle ADB=180-\left(98+27\right)=55$ and $m\angle C=180-\left(68+40\right)=72$

Use the law of sines to get the value of the missing sides.

Law of sines:

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

First substitute,

  $\frac{\sin {27}^{\circ }}{2.3}=\frac{\sin {55}^{\circ }}{AB}$

Use the cross-product property,

  $AB\cdot \sin 27=\sin 55\cdot \sin 2.3$

Now, divide each side by $\sin 27$

  $AB=\frac{\sin 55\cdot 2.3}{\sin 27}$

Now get the value of $AB$ ,

  $AB=4.1$

Similarly, solve for $BD$ , using the law of sines:

First substitute,

  $\frac{\sin {27}^{\circ }}{2.3}=\frac{\sin {98}^{\circ }}{BD}$

Use the cross products property,

  $BD\cdot \sin 27=2.3\cdot \sin 98$

Divide each side by $\sin 27$

  $BD=\frac{2.3\cdot \sin 98}{\sin 27}$

Now get the value of $BD$ ,

  $BD=5.0$

Similarly, solve for $BC$ , using the law of sines:

First substitute,

  $\frac{\sin {72}^{\circ }}{5}=\frac{\sin {40}^{\circ }}{BC}$

Use the cross products property,

  $BC\cdot \sin 72=5\cdot \sin 40$

Divide each side by $\sin 72$

  $BC=\frac{5\cdot \sin 40}{\sin 72}$

Now get the value of $BC$ ,

  $BC=3.4$

Similarly, solve for $DC$ , using the law of sines:

First substitute,

  $\frac{\sin {72}^{\circ }}{5}=\frac{\sin {68}^{\circ }}{DC}$

Use the cross products property,

  $DC\cdot \sin 72=5\cdot \sin 68$

Divide each side by $\sin 72$

  $DC=\frac{5\cdot \sin 68}{\sin 72}$

Now get the value of $DC$ ,

  $DC=4.9$

Therefore, the perimeter of the quadrilateral is about $2.3+4.1+3.4+4.9=14.7$ units.