This is a free lesson from our course in Geometry In this lesson you�ll explore on applying the learnt concepts, formulas, properties and relationship for Real World Applications-Area of Polygons and Circles, an important part of the geometric applications. The presentation covering such content will be done by the instructor in own handwriting, using video and with the help of several examples and solution. This will help you understand important geometric relationships to solve home work problems and also use them for real life applications. (More text below video...)
Other useful lessons:
 Area of a Rectangle Area of a Triangle - Areas of Polygons and Circles Area of a Square Area of a Rhombus Area of a Parallelogram Area of a Trapezoid Area of a Circle Effect of dimension changes on Area
(Continued from above) Area is always a positive number. It represents the number of square units needed to cover a shape, such as a polygon or a circle. Generally formulas are used to find measures such as area in the real-world application contexts. Further on you�ll learn in relation to Area of Polygons and Circles applying. Pythagorean Theorem, angle relationships to find the related arc length given the measure of a central angle in degrees, radius of the circle. This all will be used to solve practical real-world problems involving properties of polygons and circles.
For example, you are required to determine the number of bricks for an application here in: The method to work out would be to calculate the square foot area to be covered and to do it, multiply the length by the width. If the walkway (40 ft x 3 ft) is straight, the number of bricks needed enough to cover (40 x 3) = 120 ft2. In a different situation when walkway is not straight, additional step would be to break the walkway up into parts, calculate the square footage of each part and sum up them all together. Now the total area of the walkway is known. Then you can go to the home store, decide the pattern and color of the brick. Calculate how many number of bricks you would need for the above summed up area.
Here after you�ll explore with the help of examples, how to apply and work on area of polygons and Circles to solve practical real-world problems:
Example 1: Given- Erin is building a pen for dogs near the barn. The planned pen is hexagon in shape with sides of 3.5 m. Find the area of the pen.
Step 1: You may recall the formula for area A of regular hexagon i.e. A = 3/2( 3* s2), where s is one of the sides of hexagon.
Step 2: s is 3.5 m (given), plug in value of s in the formula.
Thus, = 3/2[ 3* (3.5)2] = 31.8 m2
The area of the pen is 31.8 m2 (rounded), as the final answer.

Example 2: Given- The well known building Pentagon in Arlington has the dimensions as: each side of the building is 921 ft in length, and the apothem is 633 ft (approx.). Calculate the approx footage enclosed at ground level of the building.
Step 1: Perimeter (P) of the building = 5 * 921 = 4605 ft.
Step 2: Area A = 1/2 * (P) * Apothem
= 1/2 * 4606 * 633 = 1457799
The footage enclosed at ground level of the building is approx 1457799 ft2, as the final answer.

Example 3: Given- Lisa wants new carpeting for her bedroom. The bedroom is in rectangular shape wit dimensions, 10 m by 7.5 m rectangle. If carpeting costs \$3/m2 , how much will it cost to lay the carpet for her bedroom? You may understand the problem that involves two parts: first to find out how much carpet is needed and second the cost of the carpet.
Step 1: Calculate the area A i.e. it equals to: (10 * 7.5) = 75 m2
Step2: The carpet required is 75 m2.
The cost of carpeting is \$3/m2
Step 3: Total cost = \$ (75 * 3) = \$225.
The cost of carpeting is \$225, as the final answer.

Example 4: Given- Rob had a 10 ft by 15 ft wall painted. For a wall 2.5 times as wide, the painter charged him two and a half times as much. Is this reasonable?
Step 1: The area of the wall equals (10 x 15) ft
Step 2: The other wall is dimensions are 2.5 times. When the dimensions of a figure are changed, the area also changes proportionally. Thus area of the other wall is two and a half times that of the first. The painter has charged 2.5 times as much. It is reasonable.
Yes, the charges are reasonable, as the final answer.

Example 5: Given- a rectangular table is 48 in wide and 66 in long. A tablecloth hangs over the table 8 in at each edge. Calculate the area of the required table cloth.
Step 1: Area of the table able cloth hangs over 8 in on each edge.
Thus dimnsions of the table cloth are: w = (48 +16) = 64 in,and l= (66 +16) = 82 in.
Step 3: Area of the table cloth equals to (82 * 64) = 5248 in2
The area of the required table cloth is 5248 in2, is the final answer.

Example 6: Given- Kelly wants to add a triangular deck in the yard behind her house. Each side is to be 25 feet long. Find the length of the railing that will fit around the deck, and also the area of the deck.
Step 1: The railing that will fit around the deck will equal to the perimeter (P) of the triangle.
Each side is 25 ft i.e. it is equilateral triangle. P = 3 x 25= 75 ft
Step 2: The formula for area of the triangle A= (1/4) * 3 * s2, where s is the length of the side of the triangle.
Thus A = (1/4) * 3 * 252 = 270.6 ft2 (rounded), as the final answer.

Example 7: Given- Alex uses a lawn sprinkler to water the circular region of the lawn having 30 ft diameter. Calculate the area watered on the lawn.
Step 1: The shape of watered region is circular. Formula for area A = r2 where r is the radius of the circle.
Step 2: The diameter of the circular region is 30 ft i.e. r = 30/2 = 15ft
Area A = r2 = * 152
= (22/7) * 225
= 706 ft2

Example 8: Given- A city has two perfectly circular and concentric ring roads, the outer ring road (OR) being twice as long as the inner ring road (IR). There are also four (straight line) chord roads from E1, the east end point of OR to N2, the north end point of IR; from N1, the north end point of OR to W2, the west end point of IR; from W1, the west end point of OR, to S2, the south end point of IR; and from S1 the south end point of OR to E2, the east end point of IR. Find the ratio of the sum of lengths of all chord roads to the length of outer ring road? Step 1: Say the radius of inner road (IR) 'r' and radius of outer road (OR) be 'R'. Therefore, the length of outer road is 2 R and that of inner road is 2 r.
Step 2: Outer ring road (OR) is twice as long as the inner ring road (IR): Given.
=> 2 R = 2(2 r)
=> R = 2r ..................(1)
Step 3: Using the Pythagorean Theorem for right triangle E1ON2: E1N2 = (r2 + R2)
Step 4: Sum of the length of all the chord = E1N2 + N1W2 + W1S2 + S1E2 = 4(E1N2 ) = 4 (r2 + R2)
Therefore, the ratio of the sum of lengths of all chord roads to the length of outer ring road is (4 (r2 + R2)) / 2 R,
Step 5: You know in (1) above, R = 2r,
Therefore, (4 (r2 + R2)) / 2 R
= (4 (r2 + (2r)2)) / 2 (2r)
= 5 / or 5 : Thus the ratio of the sum of lengths of all chord roads to the length of outer ring road is: 5: ., as the final answer.

Example 9: Given- refer the figure below- three horses are grazing within a semi-circular field. In the diagram given below, AB is the diameter of the semi-circular field with centre at O. Horses are tied up at P, R and S such that PO and RO are the radii of semi-circles with centers at P and R respectively, and S is the centre of the circle touching the two semi-circles with diameters AO and OB. The horses tied at P and R can graze within the respective semi -circles and the horse tied at S can graze within the circle centered at S. The percentage of the area of the semi-circle with diameter AB that cannot be grazed by the horses is nearest to?  Draw the diagram (as above on right) to understand the problem and arrive at the required answer.
Step 1: say the radius of the field is r, then the total area of the field = ( r2)/2.
The radius of the semi-circles with centre�s P and R = r/2.
Thus their total area = ( r2)/4
Step 2: Let the radius if the circle with centre S be x.
Thus, OS = (r � x), OR = r/2 and RS = (r/2 + x)
Step 3: Apply Pythagoras theorem, it gives: (r � x)2 + (r/2)2 = (r/2 + x)2
On simplifying it, you get x = r/3.
Step 4: Thus the area of the circle with centre S = ( r2)/9.
Step 5: The total area that can be grazed = r2(1/4 + 1/9) = r2(13/36).
Thus the fraction of the field that can be grazed = 26/36 (area that can be grazed / area of the field).
Therefore, fraction that cannot be grazed = 10/36 = 28% (approx.)

The video above will explain more in detail about Area of polygons and Circles to solve practical real-world problems, and how to apply the concepts in solving real-world problems. This is explained with the help of several examples and done watching video. This also helps you doing the Math Geometry home work.
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