Directions: Solve each problem. Your answer may be your work includes the formula you are using and the 1) Find the area. 5 m d 60° 11 m

Elementary Geometry For College Students, 7e
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ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
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**Directions**: Solve each problem. Your answer may be rounded to the nearest hundredth if necessary. Make sure your work includes the formula you are using and the measurements for each variable.

### 1) Find the area.

An irregular quadrilateral is depicted with the following measurements:
- One side of the shape is 11 meters long.
- The height perpendicular to the 11-meter side from the opposite vertex is 5 meters.
- One of the angles formed by the height and the side is 60 degrees.

### 2) Find the area.

A parallelogram is depicted with the following characteristics:
- Two of the sides are marked with a length of 12 units.
- The angle between these two sides is 60 degrees.
- The height perpendicular to one of the sides forms a right angle (90 degrees) and is indicated for calculation purposes.

For both shapes, to find the area you will use different formulas based on the geometry.

**For the first shape (trapezoid/irregular quadrilateral)**:
- Recognize that the given dimensions and angle can be used to find the base and height of a simpler geometric shape, such as a parallelogram.
- The area \( A \) of a trapezoid can be found using:
  \[
  A = \text{Base} \times \text{Height}
  \]
  Substitute the given dimensions:
  \[
  A = 11 \, \text{m} \times 5 \, \text{m} = 55 \, \text{m}^2
  \]

**For the second shape (parallelogram)**:
- The area \( A \) of a parallelogram can be given by the formula:
  \[
  A = \text{Base} \times \text{Height}
  \]
- The base is 12 units and the height is derived using trigonometric relationships with the height and the angle. If \( b \) is the base and \( h \) is the height:
  \[
  h = 12 \sin(60^\circ)
  \]
  Thus, you will use:
  \[
  A = 12 \times (12 \sin(60^\circ))
  \]
  \[
  \sin(60^\circ) = \frac{\sqrt{3}}{2}
  \]
  \[
  A = 12
Transcribed Image Text:**Directions**: Solve each problem. Your answer may be rounded to the nearest hundredth if necessary. Make sure your work includes the formula you are using and the measurements for each variable. ### 1) Find the area. An irregular quadrilateral is depicted with the following measurements: - One side of the shape is 11 meters long. - The height perpendicular to the 11-meter side from the opposite vertex is 5 meters. - One of the angles formed by the height and the side is 60 degrees. ### 2) Find the area. A parallelogram is depicted with the following characteristics: - Two of the sides are marked with a length of 12 units. - The angle between these two sides is 60 degrees. - The height perpendicular to one of the sides forms a right angle (90 degrees) and is indicated for calculation purposes. For both shapes, to find the area you will use different formulas based on the geometry. **For the first shape (trapezoid/irregular quadrilateral)**: - Recognize that the given dimensions and angle can be used to find the base and height of a simpler geometric shape, such as a parallelogram. - The area \( A \) of a trapezoid can be found using: \[ A = \text{Base} \times \text{Height} \] Substitute the given dimensions: \[ A = 11 \, \text{m} \times 5 \, \text{m} = 55 \, \text{m}^2 \] **For the second shape (parallelogram)**: - The area \( A \) of a parallelogram can be given by the formula: \[ A = \text{Base} \times \text{Height} \] - The base is 12 units and the height is derived using trigonometric relationships with the height and the angle. If \( b \) is the base and \( h \) is the height: \[ h = 12 \sin(60^\circ) \] Thus, you will use: \[ A = 12 \times (12 \sin(60^\circ)) \] \[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \] \[ A = 12
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