Area of the Side of a Goal

I noticed that the side of the soccer goal was almost a trapezoid and I decided to use it to find area.

To find the area of the side I extended the lines to make a trapezoid than subtracted the area of the small triangle.

 

Using GSP measurements the area of the shape made by the side of the goal is 45.2395cm^2.
(NW5)

Measuring Lateral Area and Volume of a Cone

After finding a three dimensional cone in the art room, I decided to put it to use to find its circumference and area, as well as use Theorem 12-7 and 12-8 to find its lateral area and volume.

First, I began by measuring the cone's height (h) and slant height (l).

Next, I measured the radius, and used that to calculate both the circumference of the base and the area.

After collecting all the needed information, I put it to use to calculate lateral area using Theorem 12-7, and volume using Theorem 12-8.

Theorem 12-7: The lateral area of a cone equals half the circumference of the base times the slant height.

Theorem 12-8: The volume of a cone equals one third the area of the base times the height of the cone.

(~RO5)

Circumference, Area, and Pie

After taking a picture of this fruit pie (because everyone takes a picture of their food before they eat it these days) I noticed that this pie was a perfect example of a circle, and using some simple formulas, I could find its circumference and area.

After measuring a radii of the circle, I was able to calculate the area and circumference using the formulas shown above. 

But, why stop there? It also became apparent that I could inscribe a polygon inside the circle, and calculate the area of that as well. In this case, I used a trapezoid.

As shown above, I was able to calculate the area of this trapezoid using Theorem 11-5.

It's seems that everything, even a fruit pie, has some relation to math, and more specifically: geometry. 

-RO5

Parallelogram in Railing

I noticed that the railing next to some stairs formed a parallelogram.

This parallelogram can be proved as a parallelogram since opposite angles are congruent.

If you draw a line from angle A to angle C, you can show theorem 5-11 which states that the segment that joins the midpoints of two sides of a triangle is 1. parallel to the third side and 2. half as long as the third side.

In the diagram this theorem is proved because 1. EF is parallel to BA which is shown by corresponding angles being congruent and 2. EF=15.60cm which is half as long as the 31.20cm length of BA.

Finally,  the area of a parallelogram is bh.  To find the height I drew an altitude from angle C down and used  sin of angle B

(∼NW5)

Arch

This is an example of two area's being added up to create one area. For example one of the problems  we had in the book that was about a free throw line. 

ABCD= 12

semicircle P1= 8

ABCD+P1= 20

(~CB5)

Home Decorating

My mom wanted to buy fabric to cover our semicircle window, and she wanted to know how many feet of the fabric to buy, and how much adhesive she needed in order to put it up. In order to find the area, I used ½(3.14)(30^2), which ended up to be 1413.71 in, or about 118 ft. Consequently, I told my mom to buy around 150 ft of fabric, in order to have enough. Next, I found half of the circle's circumference, plus the diameter, which ended up in resulting in 154.25 in, or about 12.85 feet.
(~KR2)

Bent Shower Rod

This past week I went to Monterrey with my mom for a little vacation. We stayed in a small hotel and while there I noticed the bent shape of the shower rod. I immediately thought of geometry.

After drawing in a straight bar and calculating the area of the triangle formed versus the increased area with a bent shower rod:

I concluded that the bent rod greatly increases the space within the shower.
37.31-21.63= 15.68 cm-squared more space.

This is probably one of the simplest yet one of the most useful inventions for everyday life. (~LU2)

Geometry in Action

 

My dad recently asked me what would be larger: a circle in which you increase the circumference by 2, or a circle in which you increase the radius by two. Using a few formulas (C=2πr; A=πr^2), it became apparent that the circle in which you increase the radius by two is considerably larger. I was able to illustrate this in GSP so you could get a dramatic visual of the question. (TF-2)