Square, Rectangle, Parallelogram Problems, 5

Category: Analytic Geometry, Plane Geometry

"Published in Newark, California, USA"

Show that the points (1, 1), (4, 5), (0, 8), and (-3, 4) are the vertices of a square, and find its area.

Solution:

To illustrate the problem, it is better to draw the figure as follows:

Photo by Math Principles in Everyday Life

The first thing that we have to do is to get the slope of each sides of the parallelogram using the two point formula as follows

For the slope of AB:

For the slope of BC:

For the slope of CD:

For the slope of AD:

Since the slopes are negative reciprocals to each other

then a parallelogram is either a rectangle or a square. All sides of a rectangle or a square are perpendicular to each other.

Next, we need to get the length of the sides of a rectangle or a square using the distance formula of two points as follows

For the length of AB:

For the length of BC:

For the length of CD:

For the length of AD:

Since the sides of a parallelogram are perpendicular to each other and congruent to each other as well,

then it is a square. A square is a closed figure or quadrilateral whose sides are perpendicular and equal to each other.

Since we know the length of the sides of a square, we can calculate the area as follows

Therefore, the area of a square is

Irregular Polygon Problems

Category: Analytic Geometry, Plane Geometry

"Published in Newark, California, USA"

Find the area of the pentagon connecting the points (3, 0), (2, 3), (-1, 2), (-2, -1), and (0, -3).

Solution:

To illustrate the problem, it is better to draw the figure as follows

Photo by Math Principles in Everyday Life

The area of any closed and convex polygon is given by the formula

Substitute the values of the coordinates of the vertices of a pentagon to the above equation, we have

Solve for the value of matrix, we have

Therefore, the area of a pentagon is

More Triangle Problems, 3

Category: Trigonometry, Plane Geometry

"Published in Vacaville, California, USA"

For the triangle shown, find ∠BCD and ∠DCA.

Photo by Math Principles in Everyday Life

Solution:

Consider the given figure above

Photo by Math Principles in Everyday Life

The unknown angles of two adjacent triangles can be solved by using Sine Law.

Consider ΔABC:

Use Sine Law in order to solve for ∠ABC, we have

 

 

 

 

 

 

Consider ΔBCD:

Since BC ≅ CD = 20, then it follows that ΔBCD is an isosceles triangle. If the two sides of an isosceles triangle are equal, then the two base angles are equal also. In this case,
∠DBC ≅ ∠CDB = 44.427°. Therefore,

 

 

Consider ΔACD:

Since ∠BDC and ∠ADC are supplementary angles, then we can solve for ∠ADC as follows

 

 

Therefore,