Solving poblems related to area and perimeter of 2d shapes.

In this section we will be learning how to solve some problems related to area and perimeter of complex 2 dimensional shapes.

Before starting we should have a thorough idea about the area and perimeter equations of the basic shapes that are combined to these complex shapes. So following is a summarized table of equations of some 2d shapes.

Now let us try to find some answer to some common problems.

Finding area/perimeter of a given shape

When solving these kinds of problems, the important point to know is how to break/rearrange the given complex shape into one or more basic shapes.

For example,
Shown below are some complex shapes and the ways they can be broken and/or rearranged in to one or simpler shapes.

Figure I

Figure II

After breaking and/or rearranging we can find out the area/perimeter quite easily.

So when considering the shape in figure 1 and taking the radius of the large circle is r,the area of the above shape would be ½ * πr² (Half a circle’s area).

And taken the height and width of the rectangle as 2x and y respectively, the area of the shape would be
= rectangles area + half of a big circle’s area – 1 small circle’s area.
= (2x*y ) + ½ π (2x)² - πx²


When an area/perimeter is given finding one or more unknown value/s (height, radius).

Well, these types of questions eventually lead to complex equations, so you should be familiar with equation solving (which is very easy once you get the hang of it.). Oh and remember, if there’s n number of unknown variables, you should have n number of equations as well.


For example
Question :
Check the image below

Figure III

Let say, the above rectangle shape has a height and length of x and y respectively and a boarder of 3 units and if the area of the inner rectangle is 40 and the area with the border is 154 what’s the value of x and y

Answer :
To solve this problem we first need to create 2 equations (because there is 2 unknown variables x and y)
So here we go,
As the inner rectangle's area is given as 40 we can write the following equation:

x*y = 40

And as the total area is given as 154 we can write the following equation:

(x+3+3) * (y+3+3) = 154
(x +6)*(y+6) = 154
xy + 6x +6y +36 = 154

Now the geometry part is over, from here on, it is up to your equation solving capabilities to find the value for x and y.

But to make thing more exciting I’ll solve the equation.

x*y = 40
xy + 6x +6y +36 = 154

substituting x*y we get the equation

40 +6x +6y+36 = 154
6x+6y = 154 – 76
6(x+y) = 78
x+y = 13
y = 13 – x

Substituting y’s value to the fist equation

x(13-x) =40
-x² +13x = 40
x² -13x = -40 (multiply by -1)
x²-13x+40 =0
(x-5)(x-8) = 0

So x should be 5 or 8.
If x was 5 then y would be 8 (as x+y=13) and if x was 8 then y would be 5,
but as x should be the lesser value we take x as 5 and y as 8.
So x = 5 and y = 8
Problem solved.



A complex example,

Figure IV

Question:
If the shape (colored in black) in the above image has an area of 350 units and has a border (colored in green) of 2 units thick, how much area does the border cover?

Anwser:
First we will need to find the value r.
As we can see the black color area consists of 2 half circles and a square (because height and width is equal)
So we create an equation and solve it to find r

2* (½ π r²) + (2r)² = 350
πr² + 4r2 = 350
r² (π + 4) = 350
r² (22/7 + 28/7 ) = 350
r² (50/7) = 350
r² = 350 *7 /50
r² = 7*7
r = 7 (as a length cannot be negative, we neglect the negative value)

Now we know the value of “r”,
Now we need to find the total area covered (border + shape) and then reduce the shapes area, so that we get the answer for the question.
So we first calculate the total area

=2* (½ π (r+2)²) + (2r)*(2r+4)

(2r)*(2r+4) comes as the earlier square is now a rectangle, as only the height has been affect from the border.

=π (r+2)² + (2r)*(2r+4)

So solve the equation to get the total area and subtract 350 from it to get the area covered from the boarder.