In this lesson you’ll explore the Real world problems application, and apply the
formulas and relationship between perimeter and area for problem solving
in real life experiences. The presentation covering such content will be
done by the instructor in own handwriting, and with the help of several
examples with solution and using video. This will help you understand and
apply the relationship for perimeter related problem solving.
Most geometry word problems are interesting but a bit more
involved i.e. simple onestep exercises like the one learnt
above. More often you will be having at least two pieces of
information, like a statement about a circle' are not
perimeter and then a question about its area. To find the
solution, you will need to identify the formula or equations
that can be connected to the given information. Now use
the formula or solve one of the equations for the relevant
information, and plug this result into another of the
equation(s). Then you'll almost end up with a valid answer.
(More text below video...)
(Continued from above)
E.g. a simple illustration, if the spray from a lawn sprinkler makes
a circle of 30 feet radius, what are the diameter and circumference
of the circle of lawn watered? You can use the formula: the diameter is two times the radius
= 60 ft; and circumference
= 2r = 2 * 3.14 * 30 = 188.4 ft, as the final answer.
Further a few uses covered here may relate to building a swimming pool economically,
fencing off an area to plot a crop, building a barn with box stalls for horse,
and choosing buying the wood economically in available options etc.
The objective would be to find the solution to minimize the building
material used and/or build the largest area, given the planned amount
for building material or similar ones.
Real world Examples with solutions: Geometry
In this section you’ll find examples with solutions
of the common shapes in use in the real world and you may see them in everyday life.
Example 1: Given Bob had a 10 ft by 15 ft wall painted. For a wall twice as wide,
the painter charged him twice as much. Is this reasonable?
If you read the problem
carefully and analyze, will notice that here logic and formula learnt about
‘Effects of changing Dimensions proportionally’ can be applied.
Since second
wall is twice as wide, its area is twice the area of the first wall.
Therefore, the material used and efforts by the painter is double.
Thus the amount charged by the painter is reasonable.
Example 2: Jose has a rectangular rose garden that measures 6 m by 15 m.
One bag of fertilizer can cover 10 m^{2}. How many bags will he need to cover
the entire garden?
Step 1: Find the area first and then calculate how many bags are needed.
Area = 6 m * 15 m = 90 m^{2}
Step 2: 10 m^{2} area is covered using one bag. Therefore; to cover 90 m^{2} area
will need 90/10 bags.
Number of bags = 9, as the final answer.
Example 3: Given Rectangular tiles each of size 70 cm by 30 cm must be laid horizontally
on a rectangular floor of size 110 cm by 130 cm, such that the tiles do not overlap.
A tile can be placed in any orientation so long as its edges are parallel to the
edges of the floor. No tile should overshoot any edge of the floor. What is the
maximum number of tiles that can be accommodated on the floor?
Step 1: The size of tile is 70cm by 30cm respectively.
Step 2: Area of a tile = (70 * 30) = 2100 cm2.
Step 3: Given the rectangular floor size (110 cm by 130 cm).
Step 4: Area of rectangular floor = 130 * 110 = 14300 cm2.
To find the maximum number of tiles that can be accommodated on the floor,
without overshooting, we have to divide the area of rectangular floor
by area of a tile i.e. total area of rectangular floor / area of a tile
=> 14300 / 2100 => 6.81
Thus 6 tiles can be placed on the floor, without overshooting any edge of the floor.
Therefore, the maximum numbers of tiles that can be accommodated on the floor are 6, as the final answer.
Example 4: Given Casey decided to fence off the area to plot a crop.
If one side of the existing square plot is doubled in length and the adjacent
side is decreased by two meters, the area of the resulting rectangular plot is
96 square meters larger than that of the original square. Find the dimensions
of the rectangular plot.
Step 1: Notice that from the square plot with sides of some unknown length,
sides of the rectangle are related in terms of this unknown length.
So you pick up a variable for the unknown sidelength, create expressions
for the sides of the rectangle, and then work for the solution.
Step 2: Say side length of the square is x.
Step 3: Given that one side is doubled i.e. it is 2x, adjacent side becomes (x 
2).
Step 4: Area of existing square is x^{2},
and area of the rectangle shall be (2x)(x  2) i.e. 2x^{2}
 4x.
Step 5: As given the new area is 96 more than old area i.e.
2x^{2}  4x = x^{2} + 96 i.e.
2x^{2}  4x = x^{2}
+ 96
x^{2}  4x  96 = 0
(x  12) (x + 8) = 0
x = 12 or x = 8
Since; 8 can’t be the logical answer; side of the square is 12 meters.
Thus dimensions of the rectangle shall be, 24 by 14 meters, as the final answer.
Example 5: Given Area of a walkway path in the garden; having outside length 25m and
width 20m and the width of walkway path being 2m, is 164 m^{2}.
Determine area of the path.
Step 1: Find the area of rectangle with outside dimensions = 25 * 20 = 500 m2
Step 2: The area of small rectangle = (25  2  2)
* (20  2  2) = 336 m^{2}
Step 3: Area of the path = Area of large rectangle  Area of small rectangle = 164 m^{2}
Area of the path is 164 m^{2}, as the final answer.
Example 6: Given A farmer plans to level a triangular field. The length of sides of the farm is  32 m, 30 m and 24 m. If indicated cost of leveling is
$ 1.50/m^{2}, find out the estimated total cost involved.
See the figure below (Not to Scale), where a = 32, b = 30 and c = 24.
Step 1: Length of sides, a = 32 m, b = 30 m, c = 24 m. You know the relationship between area and three sides of a triangle.
Step 2: p =(a + b + c) / 2 = (32 + 30 + 24) / 2 = 43
Step 3: Calculate area
Step 4: Estimated cost of leveling = 341.8 * 1.50 = $512.70, as the final answer.
Example 7: Given Rob has a triangle shape land plot on which he wants to grow Beans and Peas. The plot has a separator along the line PD, such that D is the midpoint of QR. He grows Beans in one part and Peas in other part. What is the area in hectares used for cultivation of Beans?
See the figure below (Not to Scale),
Step 1: Find the area of PQR.
p = (410+350+220) / 2 = 980 / 2 = 490 m
Step 2: Area = 38494m^{2} = 3.849 hectares (10000 m^{2} = 1 hectare) Step 3: Since PD is the median of PQR, hence it divides the triangle in two equal parts i.e. the area used for cultivation is divided into two equal parts.
Area in hectares used for cultivating Beans equals to (3.849/2) = 1.92 hectares, as the final answer.
Example 8: Given Amy has decided to order another table top, where circumference of the wooden circular table top is increased by 50% over the existing one on order. The carpenter has offered to make it at the same rates per unit area, so long thickness is same. How much she will be required to pay more for the second table top?
Step 1: Circumference of circular table top = 2r, where r is radius of circle in
question and area of circle =
* r^{2}
Notice that circumference is proportional to
radius.
Step 2: When circumference is increased by 50% ,the radius will also go up by 50%
i.e. the new radius will be 1.5 r
Step 3: Then new area =
* (1.5 r)^{2} =* 2.25 r^{2}
Step 4: Increase in area =(1.5 r)^{2
}
* r^{2} = (Step3  Step1) =
* 1.25 r^{2}
Step 5: % increase in area = (increase in area for second table top/area of first table top) * 100
=
(
* 1.25 r^{2} /
* r^{2} ) * 100 = 125%
Since area of second table top is 125% more, she needs to pay for it 1.25 times than the first one, as the final answer.
The video above will explains in more detail on solving Real world application problems, including watch video lessons and practice questions with solutions.
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