This lesson explains the basics and concepts of Effect of dimension changes on Area,
an important part of the geometric applications. Here you’ll explore how a change in an
object’s dimensions affects its area (for example, if one or more dimension is changed for
polygon how it affects the area or how a change in radius or diameter will affect the area of a circle). The important geometric relationship and formulas will be reviewed and the concept elaborates how these can be applied to solve real world application problems. 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 relationships to solve home work problems and also use them for real life applications.
(More text below video...)
(Continued from above)Area is the inside space occupied by an object or shapes like rectangle, square, circle etc. The value of area is denoted using square units. Dimensions of area: The quantities which are required to calculate the area are called the dimensions of area. E.g. length and width are the dimensions of area for rectangle as they are needed to calculate the area and also the value of area is directly proportional to them.
COMMON SHAPES
DIMENSIONS OF AREA
Rectangle
Length and width
Square
Length of side
Triangle
Base and height
Circle
Radius or Diameter
Further on you’ll explore with the help of examples, what effect does changing one or more dimensions have on the area of polygons and circle twodimensional object ?
Example 1: Given if the length and width of rectangle is reduced to one third of its value then what is the change in its area?
Step 1: Original area A of the rectangle = l * w
Step 2: Reduced length = l/3 and Reduced width = w/3
Thus new area = (l/3)*(w/3)
= l * w/9
= A/9
Hence the area is reduced to 1/9 times of its original value, as the final answer.
Example 2: Given if width of the rectangle (8 in x 5in) is tripled, what will be its effect on the area?
Step 1: Area A = l * w = 8 * 5 = 40 sq in
Step 2: Increased area = l * w = 8 * 15 = 120 sq in i.e. three times.
If width of the rectangle is tripled; the area is also tripled, as the final answer.
Example 3: Given A photographer needs to frame a photo of 6 in by 8 in. The frame is twice as long and twice as wide as the photo. Find how many photos of same size can be fitted in the given frame.
Interpret it as the problem having change in the dimensions of frame/photo and its effect in the resulting area.
Step 1: Area of Photo = l * w, where
l  length and w  width
= 8 * 6 = 48 sq in
Area of frame = length of frame * width of frame
Step 2: Remember the length and width of the frame is double of the photo
Area of frame = (2 * 6) * (2 * 8) = 12 * 16 = 192 sq in
Step 3: Now number of photo can be fitted = Area of frame / Area of photo
= 192/48 = 4
Thus four photos can be fitted into the given frame, as the final answer.
Example 4: Given A square has an area of 64 in^{2}. If the area is quadrupled, find out the increased side length?
Step 1: Area of a square (A) = s^{2} = 64.
Step 2: Solving for s, it gives s = 8
Step 3: The quadrupled area is 64 * 4 =256 in^{2}
Step 4: (side)^{2} = 256.
Taking sq root on both sides gives, s = 16 i.e. increased side length is 16 in.
Thus the increased side length is 16 in, as the final answer.
Example 5: Given the length of one side of a square is increased by 3 units and an adjacent side is decreased by 3 units. How much the new rectangle area differs than the areas of the original square in square units?
Step 1: Say the length of a side of the original square is x.
Thus Area of the square (A) = x^{2}.
Step 2: When the length of one side of a square is increased by 2 units and an adjacent side is decreased by 2 units, the length of rectangle is (x + 3) and width or height is (x  3).
Step 3: The area is l * w =(x + 3) * (x  3)= x^{2}  9 i.e. the new rectangle area differs by 9 sq units, as the final answer.
Example 6: Given Mark is helping his grandfather with a garden. Last year the garden plot was in the shape of a square. This year they decided to change the shape of the garden by adding 2 m on the east side and taking away 2 m from the south side. Mark's grandfather wonders if the garden will be the same size, larger, or smaller than last year's garden. If it is not the same, how much larger or smaller will it be?
Refer to the figure (NTS) on right. ABCD is the original square garden. By adding 2 m on the east side and taking away 2 m from the south side, it will shape like
ABCPC’D’.
Step 1: Say square was with side
x. So last year garden area is x^{2}
Step 2: The changed shape is (as you see in the diagram).
Step 3: The changed area shall be AQC’D’ =(ABCD + PC’D’D  BCPQ) = [x^{2} + 2(x  2) 2x] = (x^{2} + 2x  4  2x) = x^{2}  4
Thus garden will be the smaller than last year's garden (x^{2}) by 4 m^{2}, as the final answer.
Example 7: Given the base and height of the triangle with vertices
P (2, 5), Q (2, 1), and R (7, 1) are tripled. What will be the effect on its area?
Recall the formula for area of triangle i.e. A =1/2 (bh).
Step 1: The base and height are tripled (given)
Step 2: It is known from the relationships and triangle properties that when the dimensions of a figure are changed proportionally, the figure will be similar to the original figure.
Thus the area will be proportionally increased. Since both base and height are tripled, the area will be 3^{2} times i.e. original area is multiplied by 9, as the final answer.
Example 8: Given area of a triangular plot is 40 m^{2}. With available of additional land the height of plot has become 6 times. What is the change in area of the plot?
It is known from the relationships and triangle properties that when the dimensions of a figure are changed proportionally, the figure will be similar to the original figure. Thus the area will go up proportionally with the increase.
Since in this case height of the triangular plot is multiplied by 6 i.e. increased 6 times, the area A of the plot will also become six times.
The area of new plot will be 6 * 40 = 240 m^{2}, as the final answer.
Example 9: Given refer the figure (NTS) below where ABCDEF is be a regular hexagon.
G, H, I, J, K,
and L are respectively the midpoints of sides AB, BC, CD,
DE, EF, and AF. Joining a vertex and
midpoint created segments are AH, BI, CJ, DK, EL and
FG which result in a smaller regular hexagon.
If the ratio of the area of the smaller hexagon to the area of ABCDEF is expressed as a fraction
a/b, where a and b are prime positive integers, find the value of (a +
b).
Step 1: AH and BI intersect at M .As
ABH and
BCI
are rotational images of one another, you may recall from the properties learnt earlier, that
MBH =
HAB and hence
ABH ~ BMH. By the cosine law, AH 7/4.
Step 2: Use the property of proportion of similarity, which gives BH/AH =>
Thus the ratio of areas ABH & BMH is 1/7.
Step 3: Notice that each of the 6 triangles congruent to ABH shall have an area of 1/12 of the hexagon
ABCDE.
Thus each of the six triangles congruent to BMH has an area of (1/7)* (1/12).
Step 4: Total area of six triangles congruent to ABH gives the area in the larger hexagon (ABCDEF) around the smaller. As each triangle congruent to
BMH has been counted twice, therefore, the area around the small hexagon but in the larger is equal to:
[6(1/12)  6(1/12) * (1/7)] = 3/7.
Thus the area of the smaller hexagon is (1 – 3/7) = 4/7 of the hexagon
ABCDEF (larger hexagon), that gives (a + b) = 4 + 7= 11, as the final answer.
Example 10: Given a circle has a circumference of 32 in. If the area is multiplied by 4, what will be change in the radius?
Step 1: The formula for circumference of a circle = 2r
Step 2: Circumference = 32 (given)
Thus 32 = 2r.
Therefore r =16 units.
Step 3: Area A = r^{2} = 16^{2} = * 256
Step 4: If the area is multiplied by 4, it equals * 1024 units (new area).
Step 5: To find new radius,
r^{2} =
* 1024
Divide both sides by
, you get r^{2} = 1024.
Taking sq root on both sides, r = 32 units.Thus the radius is doubled, as the final answer.
Example 11: Given What will be the change in the area of a circle, when its radius is doubled?
Step 1: Area of circle A =
r^{2}
Step 2: If radius is doubled, changed area =
(2r)^{2}
= 4r^{2} = 4A i.e. 4 times the original area.
Thus area increases to 4 times, as the final answer.
The video above will explains in more detail about Effect of dimension changes on Area, including watch
video lessons and practice questions with solutions.
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