In this lesson you’ll explore on applying the learnt concepts, formulas, properties and relationship for
Real World ApplicationsArea 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...)
(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
realworld 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 realworld 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 ft^{2}. 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 realworld 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* s^{2}), 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 m^{2}
The area of the pen is 31.8 m^{2} (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 ft^{2}, 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/m^{2} , 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 m^{2}
Step2: The carpet required is 75 m^{2}.
The cost of carpeting is $3/m^{2}
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,andl= (66 +16) = 82 in.
Step 3: Area of the table cloth equals to (82 * 64) = 5248 in^{2}
The area of the required table cloth is 5248 in^{2}, 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 * s^{2}, where s is the length of the side of the triangle.
Thus A = (1/4) * 3 * 252 = 270.6 ft^{2} (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 = r^{2} 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 = r^{2} = * 15^{2}
= (22/7) * 225
= 706 ft^{2} (rounded), as the final answer.
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 E_{1}, the east end point of OR to N_{2}, the north end point of IR; from N_{1}, the north end point of OR to W_{2}, the west end point of IR; from W_{1}, the west end point of OR, to S_{2}, the south end point of IR; and from S_{1} the south end point of OR to E_{2}, 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 2R and that of inner road is 2r.
Step 2: Outer ring road (OR) is twice as long as the inner ring road (IR): Given.
=> 2R = 2(2r) => R = 2r ..................(1)
Step 3: Using the Pythagorean Theorem for right triangle E_{1}ON_{2}:
E_{1}N_{2} = (r^{2} + R^{2})
Step 4: Sum of the length of all the chord = E_{1}N_{2} + N_{1}W_{2} + W_{1}S_{2} + S_{1}E_{2} = 4(E_{1}N_{2} ) = 4(r^{2} + R^{2})
Therefore, the ratio of the sum of lengths of all chord roads to the length of outer ring road is
(4(r^{2} + R^{2})) / 2R,
Step 5: You know in (1) above, R = 2r,
Therefore,
(4(r^{2} + R^{2})) / 2R
= (4(r^{2} + (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 semicircular field. In the diagram given below,
AB is the diameter of the semicircular field with centre at O. Horses are tied up at
P, R and S such that PO and RO are the radii of semicircles with centers at
P and R respectively, and S is the centre of the circle touching the two semicircles 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 semicircle 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 = (r^{2})/2.
The radius of the semicircles with centre’s P and R = r/2.
Thus their total area = (r^{2})/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 = (r^{2})/9.
Step 5: The total area that can be grazed = r^{2}(1/4 + 1/9) = r^{2}(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 realworld problems, and how to apply the concepts in solving realworld problems. This is explained with the help of several examples and done watching
video. This also helps you doing the Math Geometry home work.
Winpossible's online math courses and tutorials have gained rapidly popularity since
their launch in 2008. Over 100,000 students have benefited from Winpossible's courses...
these courses in conjunction with free unlimited homework help serve as a very effective
mathtutor for our students.

All of the Winpossible math tutorials have been designed by topnotch instructors
and offer a comprehensive and rigorous math review of that topic.

We guarantee that any student who studies with Winpossible, will get a firm grasp
of the associated problemsolving techniques. Each course has our instructors providing
stepbystep solutions to a wide variety of problems, completely demystifying the
problemsolving process!

Winpossible courses have been used by students for help with homework and by homeschoolers.

Several teachers use Winpossible courses at schools as a supplement for inclass
instruction. They also use our course structure to develop course worksheets.