Chapter 9.5, Problem 46PS

Solution

a.

To find: the area of the $△ABD$ .

  $Area\approx 29.23{\text{ in}}^2$

Given information:

The image of quadrilateral $ABCD$:

  

Definition Used:

Area of a triangle

The area of any triangle is given by one half the product of the lengths of two sides times the sine of their included angle. For $△ABC$

shown, there are three ways to calculate the area:

  

  $\begin{array}{l}Area=\frac{1}{2}bc\sin A\\ Area=\frac{1}{2}ac\sin B\\ Area=\frac{1}{2}ab\sin C\end{array}$

Explanation:

Let $A=54°,b=8.5\text{ in},d=8.5\text{ in}$

Since, two sides of a triangle and the angle between them is given, so using the above-mentioned formula, the area of the given triangle is:

  $\begin{array}{c}Area=\frac{1}{2}bd\sin A\\ =\frac{1}{2}\cdot 8.5\cdot 8.5\cdot \sin 54°\\ \approx 29.23{\text{ in}}^2\end{array}$

b.

To find: the area of the $△BCD$ .

  $Area\approx 51.93{\text{ in}}^2$

Given information:

The image of quadrilateral $ABCD$:

  

Definition Used:

Area of a triangle

The area of any triangle is given by one half the product of the lengths of two sides times the sine of their included angle. For $△ABC$

shown, there are three ways to calculate the area:

  

  $\begin{array}{l}Area=\frac{1}{2}bc\sin A\\ Area=\frac{1}{2}ac\sin B\\ Area=\frac{1}{2}ab\sin C\end{array}$

Explanation:

Let $C=32°,d=14\text{ in},e=14\text{ in}$

Since, two sides of a triangle and the angle between them is given, so using the above-mentioned formula, the area of the given triangle is:

  $\begin{array}{c}Area=\frac{1}{2}de\sin C\\ =\frac{1}{2}\cdot 14\cdot 14\cdot \sin 32°\\ \approx 51.93{\text{ in}}^2\end{array}$

c.

To find: the area of the kite ABCD.

  $Area\approx 81.16{\text{ in}}^2$

Given information:

The image of quadrilateral $ABCD$:

  

Definition Used:

Area of a triangle

The area of any triangle is given by one half the product of the lengths of two sides times the sine of their included angle. For $△ABC$

shown, there are three ways to calculate the area:

  

  $\begin{array}{l}Area=\frac{1}{2}bc\sin A\\ Area=\frac{1}{2}ac\sin B\\ Area=\frac{1}{2}ab\sin C\end{array}$

Explanation:

From the given figure it is clear that,

  $Area\left(kite\right)=Area\left(△ABD\right)+Area\left(△BCD\right)$

Using values from part (a) and (b),

  $\begin{array}{c}Area\left(kite\right)=Area\left(△ABD\right)+Area\left(△BCD\right)\\ \approx 29.23+51.93\\ =81.16{\text{ in }}^2\end{array}$