Chapter 21 - Circle Measures - Perimeter

In the previous section we completed the discussion on polynomials. In this chapter we will see Circle measures.

In this section, we will try to find a method for calculating the perimeter of any given circle. We know that perimeter of any polygon can be calculated if it’s sides are known. All we have to do is add the sides. For example, all the sides of a 5 sided polygon is given in fig.21.1(a). 

Fig.21.1

It’s perimeter is equal to 3 + 5 + 4 + 5 + 6 = 23 cm. In fig.b, another polygon with 6 sides are given. It is a ‘regular hexagon’. So all it’s 6 sides will be equal. Thus the calculation of perimeter becomes easier. We will get perimeter as 6 × 3 cm = 18 cm

Now we will consider perimeter of circles. Suppose there is a circular ground. The owner of the ground wants to make a fence around it. He will want to know how much barb wire will be needed to make the fence. In such a case there should be a method to calculate the perimeter of circles. One method is to use a simple procedure as follows:

Fig.21.2

1. Fix a peg on the periphery of the circle (fig.21.2)

2. Tie one end of a rope to the peg.

3. Place the rope along the periphery to make one complete circle upto the peg.

4. Put a mark on the rope where it meets the peg

5. Measure the length of the rope from the mark upto the peg. This will give the perimeter.

But in maths we want actual calculations. Mathematicians from very early days have tried to derive a method for calculating perimeter of circles. We will now try to understand how they derived the method that we now use commonly in Science and Engineering.

Consider fig.21.3. Some polygons are drawn inside circles. All the circles in the fig.21.3 have the same diameter. 

Fig.21.3

1. The first fig.a shows the polygon with the smallest possible number of sides. The smallest possible number of sides to form a polygon is of course three. Because, with two2 sides, we cannot form a closed figure. So, in fig.a, we have a triangle inside a circle. Note that, it is a regular polygon. So all sides have to be equal. Thus it is an equilateral triangle.

2. Next, in fig.b, we have a regular polygon with 4 sides. That is., we have a square. Note that, a rectangle, though have 4 sides, is not a regular polygon.

3. Continuing like this, in fig c, d, e and f, we have regular pentagon, hexagon, septagon and octagon. The number of sides 'n' increase by 1 in each successive fig.

4. We can see that, as the number of sides n increases, the polygon inside gets closer and closer to the circle.

5. If n is very large, the polygon will become so close to the circle that, It will be difficult to distinguish between the two. Such a polygon is shown in the fig.21.4(a) below. It has n = 14.

Fig.21.4(a)

A small portion of this fig.a is enlarged and is shown in fig.21.4(b) below:

Fig.21.4(b)

In this enlarged fig.b, we can distinguish between the two. But if we increase n further, even an enlarged fig. will not show much difference.

■ So we can say this:

1. We want to calculate the perimeter of a circle.

2. For that, we draw a regular polygon inside the circle.

3. This regular polygon has a very large ‘n’ that, it is hard to distinguish between the circle and the polygon

4. In such a situation, if we calculate the perimeter of that regular polygon, that perimeter will be approximately equal to the perimeter of the circle.

This seems to be a very good method to calculate the perimeter of circles. But before proceeding further, we must be sure of one thing: We must make sure that, we are able to calculate the perimeter of any regular polygon drawn inside a circle of known diameter. We will start with the polygon with the lowest possible n. That is., the triangle. We will do it as a solved example.

Solved example 21.1

Fig.21.5(a) below shows an equilateral triangle drawn inside a circle of diameter 1m.

Fig.21.5

Calculate the perimeter of the equilateral triangle.

Solution:

We have only two information:

• The triangle is equilateral

• The diameter of the circle is 1 m

With these, we have to calculate the perimeter of the triangle.

1. Let us draw two medians as in fig.b

• Median AE (E is the midpoint of BC)

• Median CD (D is the midpoint of AB)

Now look at the two medians carefully.

• The median CD is drawn from the vertex C to the midpoint D of the opposite side AB. This median will be perpendicular to the side AB. This is because ABC is an equilateral triangle. Also, CD passes through the midpoint D. So CD is the perpendicular bisector of AB

• Similarly, AE is the perpendicular bisector of BC.

So we can write:

The 'medians' in any equilateral triangle will serve another purpose also:

They will bisect the side perpendicularly. In other words, they are perpendicular bisectors also. 

• We know that the point of intersection of any two perpendicular bisectors of a triangle is it’s circumcentre. (details here)

2. So O is the circumcentre of the given equilateral triangle. Let us mark the radius. OB is the radius marked with a dashed green arrow. OB = half of diameter = ¹⁄₂ of 1 m = ¹⁄₂ m = 0.5 m

3. Now we use one important property of medians: The point of intersection splits the medians in the ratio 2:1, measured from the vertex. (Theorem 18.6)

4. So we can write: OC:OD = 2:1

5. That means: If we divide CD into 3 equal parts, OC will constitute 2 such parts, and OD will constitute 1 such part

6. When we divide CD into 3 equal parts, each part will be ^CD⁄₃

7. Two such parts = ^2CD⁄₃

8. So OC = ^2CD⁄₃

9. But OC = radius = ¹⁄₂ m

10. Substituting this in (8) we get: ¹⁄₂ m = ^2CD⁄₃. So we get CD = ³⁄₄ m

11. Now, AD = ^AB⁄₂. But AB = AC (∵ ABC is an equilateral triangle)

12. So AD = ^AC⁄₂

13. Now consider the right triangle ADC

14. Applying Pythagoras theorem, we get: AC² = AD² + CD².

⇒ AC² = (^AC⁄₂)² + (³⁄₄)². ⇒ AC² = (^AC2⁄₄) + (⁹⁄₁₆).
⇒ AC² - (^AC2⁄₄) = (⁹⁄₁₆) ⇒ (^3AC2⁄₄) = (⁹⁄₁₆)
⇒ AC² = ³⁄₄ ⇒ AC = ^√3⁄₂.
15. Thus we got one side of the equilateral triangle.
So perimeter = 3 × (^√3⁄₂) = ^(3√3)⁄₂

Solved example 21.2

Fig.21.6(a) below shows a square drawn inside a circle of diameter 1 m. 

Fig.21.6

Calculate the perimeter of the square.

Solution:

We have only two information:

• The polygon is a square

• The diameter of the circle is 1 m

With these, we have to calculate the perimeter of the square.

1. Let us draw the two diagonals AC and BD of the square.

2. Length of each of these diadonals are same as the diameter of the circle = 1 m

3. The diagonals bisect each other at right angles. So OA = OB = OC = OD = 0.5 m = ¹⁄₂ m

4. As the diagonals are perpendicular to each other, angle at O = 90^o

5. Consider any of the right triangles. Say ⊿OCD. Applying Pythagorus theorem we get:
CD² = OC² + OD²..
⇒ CD² = (¹⁄₂)² + (¹⁄₂)² = ¹⁄₄ + ¹⁄₄ = ¹⁄₂
⇒ CD = ¹⁄_√2
6. So perimeter of the square = 4 × ¹⁄_√2  = ⁴⁄_√2  = ^{(2×√2×√2)}⁄_(√2) = 2√2

Solved example 21.3

Fig.21.7(a) below shows a regular hexagon drawn inside a circle of diameter 1m. 

Fig.21.7

Calculate the perimeter of the regular hexagon.

Solution:

We have only two information:

• The polygon is a regular hexagon

• The diameter of the circle is 1 m

With these, we have to calculate the perimeter of the regular hexagon.
1. Draw the three diagonals of the hexagon
2. The diagonals will interset at the centre O of the circle
3. The diagonals will give 6 equilateral triangles
4. Consider any one of those 6 triangles. Say OAB
5. OB is half of diameter = ¹⁄₂ of 1 m = ¹⁄₂ m = 0.5 m
6. So OA and OB will also be equal to ¹⁄₂ m
7. So perimeter of hexagon = 6 × ¹⁄₂ = 3 m

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We have calculated the perimeter of a triangle, a square and a hexagon in a circle of 1 m diameter. In this way we can calculate perimeters of polygons with any number of sides.  In the next section we will see how we can put it to use.

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Chapter 14.3 - Polygon to Triangle - Solved examples

In the previous section we saw how a quadrilateral, or even a polygon, can be transformed into a triangle having the same area. In this section, we will see some solved examples.

Solved example 14.8
Draw the two quadrilaterals shown in fig.14.15(a) and 14.16(a) below. For each of them, draw a triangle of same area, and calculate the area. Lengths required for the area calculations may be measured

Fig.14.15

Solution:
■ Fig.14.15(a) is a rough sketch. But it contains all the details required to construct the actual quadrilateral. The construction of ABCD is done in fig.b
• Once the construction is done, we can proceed to do the transformation:
• Draw diagonal BD 
• Draw MN parallel to BD through C
• On MN, mark E, in line with AB
• Join B and D to E. This gives the transformed shape: ΔAED
• Now we have to find the area of ΔAED
• From the actual construction, the required lengths can be measured. Base AE is already drawn. To obtain height, draw a perpendicular DF to AB
• By measurement, base AE = 10.6 cm and Height DF = 2.9 cm
• So Area = ¹⁄₂ × b × h = ¹⁄₂ × 10.6 × 2.9 = 15.66 cm²

Fig.14.16

■ Fig. 14.16(a) is a rough sketch. But it contains all the details required to construct the actual quadrilateral. The construction of ABCD is done in fig.b
• Once the construction is done, we can proceed to do the transformation:
• Draw diagonal BD 
• Draw MN parallel to BD through C
• On MN, mark E, in line with AB
• Join B and D to E. This gives the transformed shape: ΔAED
• Now we have to find the area of AED
• From the actual construction, the required lengths can be measured. Base AE is already drawn. To obtain height, draw a perpendicular DF to AB
• By measurement, base AE = 12.6 cm and Height DF = 5.2 cm
• So Area = ¹⁄₂ × b × h = ¹⁄₂ × 12.6 × 5.2 = 32.76 cm² 

Solved example 14.9
Draw a rhombus of sides 6 cm, and one angle 60^o. Then draw a right triangle of the same area
Solution:

Fig.14.17

• We have learned to construct a rhombus in lower classes. In fig.14.17, ABCD is the required rhombus. It's construction details are not shown here.
• Once the construction is done, we can proceed to do the transformation:
• Draw diagonal BD 
• Draw MN parallel to BD through C
• On MN, mark E, in line with AB
• Join B and D to E. This gives the transformed shape: ΔAED
• But we want a right triangle of the same area as ABCD
• So we need to transform ΔAED into a right triangle
• For that, extend CD to the left. Draw a perpendicular AF to this extended line
• Join A and E to F. ΔAEF is the required right triangle

Solved example 14.10
Draw a regular pentagon, and then a triangle of the same area. Calculate the area of the triangle.
Solution:

Fig.14.18

• We have learned to construct a regular pentagon in lower classes. In this problem, we can draw a regular pentagon of any side. In fig.14.18, ABCDE is the required regular pentagon with sides 4 cm. It's construction details are not shown here.
• Once the construction is done, we can proceed to do the transformation:
• Draw diagonal BD 
• Draw MN parallel to BD through C
• On MN, mark F, in line with AB
Join B and D to F. The pentagon ABCDE has now become a quadrilateral AFDE. Now work on the left side:
• Draw diagonal AD 
• Draw M'N' parallel to AD through E
• On M'N', mark G, in line with AB
Join A and D to G. GFD is the required triangle. Now we want it's area
• From the actual construction, the required lengths can be measured. Base GF is already drawn. To obtain height, draw a perpendicular DH to AB
• By measurement, base AE = 8 cm and Height DF = 6.5 cm
• So Area = ¹⁄₂ × b × h = ¹⁄₂ × 8 × 6.5 = 26 cm²

Solved example 14.11
Fig.14.19(a) shows a rectangle ABCD. It is split into an yellow part and a green part. The lines EF and FG does this separation. Instead of the two lines, draw a single line to make the separation in such a way that the areas of the two parts remain the same. Calculate the area of the two parts

Fig.14.19

Solution:
• We are given the fig.14.19(a). It has all the required details for us to draw it in our note book:
• The rectangle has length 5 cm, and width 3 cm. So the rectangle can be easily drawn
• Mark a point G on CD, 1 cm from D. Mark a point E on AB 2 cm from A. So we get G and E. The only remaining point is F. It's details are also given
• Point F satisfies two conditions: It is 3 cm from side AD and 2 cm from side AB
• So draw a line UV (see fig.b), parallel to AD, at a distance of 3 cm from AD
• Draw line XY, parallel to AB, at a distance of 2 cm from AB
• The point of intersection of UV and XY is the point F
• Now we can proceed to solve the actual problem
• AEFGD is a pentagon. Two of it's sides, EF and FG is separating the rectangle into two parts.
• We want the two lines to become one. Then the pentagon will become a quadrilateral.
• The area of the new quadrilateral should be same as the original pentagon
• For that, draw the diagonal EG. Draw MN parallel to EG through F
• MN will intersect AB at H. Join H to G. AHGD is the required quadrilateral.
• The final two areas are shown in fig.(d). We can now proceed to find the areas:
• AHGD is a quadrilateral. Not just a quadrilateral, it is a non isosceles trapezium. We have seen how to find it's area
• We have, Area = ¹⁄₂ × (a + b)d.  In this problem, a = 1, b = 4 (by measurement), and d = 3 cm
• Area = ¹⁄₂ × (1 + 4) × 3 = 7.5 cm²
• Consider the green portion HBCG. It is an inverted shape of the yellow portion (∵ it has the parallel sides equal to 1 cm and 4 cm, and the height 3 cm)
• So the area of the green portion is also 7.5

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In the next section we will see unchanging areas of parallelograms.

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Chapter 14.2 - Triangle with the same area as a Quadrilateral

In the previous section we completed the discussion on 'unchanging areas of triangles between parallel lines'. In this section, we will see the relation between the areas of quadrilaterals and triangles.

• Fig.14.11(a) below, shows a quadrilateral ABCD

Fig.14.11

• In the fig.b, a diagonal BD is drawn. This diagonal splits the quadrilateral into two triangles: ΔABD and ΔCBD. Through C, draw a line MN, parallel to the diagonal BD. This is shown in fig.b. 
• Extend the side AB towards the right. It will meet MN at E. This is shown in the fig.c
• Join B and D to E. We will get a new ΔEBD as shown in fig.d
• What is the peculiarity of this new ΔEBD ?
ΔEBD and ΔCBD has the same base BD. Their third vertices E and C lies on the same line MN parallel to the base BD, and passing through C. So ΔEBD has the same area as ΔCBD
• Now ignore the original ΔCBD. Because, we have it's area in another shape, which is ΔEBD.
We split the original quadrilateral into two triangles, and now, one of them has transformed into another shape with the same area. So the total area of the quadrilateral remains unchanged
• Besides 'keeping the area the same', we achieve one more thing in this transformation:
Look at the final shape in fig.e: AED is a triangle. This was achieved by marking 'E'  in line with the side AB. When one vertex comes in line with another two vertices, one side will disappear. This is shown in the animation in fig.14.12 below:

Fig.14.12

In fig.b, the vertex 'C' comes in line with side AB. So two sides: AB and BC becomes one side AE
• So we transformed a '4 sided quadrilateral' into a '3 sided triangle' with the same area.

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Figs. (a) to (e) in 14.11 were drawn to show each step of the transformation. In an actual construction, We can do all the steps in a single fig. This is shown in the fig. 14.13 below:

Fig.14.13

The fig.14.13 was drawn quickly by following the steps below:
• Draw diagonal BD 
• Draw MN parallel to BD through C
• On MN, mark E, in line with AB
• Join B and D to E. This gives the transformed shape: ΔAED

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So now we know how to transform a quadrilateral into a triangle of same area. The same method can be used to transform any polygon into a triangle of same area. But we may have to do the above process many number of times. The 'number of times' will depend on the number of sides of the polygon. For example, if we are given a pentagon, we will apply the method once to get a quadrilateral. So the five sides of the pentagon will become four sides of a quadrilateral. Then we apply the method a second time to change the quadrilateral into a triangle. Like wise, if we are given a hexagon, we will have to apply the method 3 times. Fig.14.14 below shows the transformation of a pentagon.

• Fig.14.14(a) shows the original pentagon
• In fig.(b), the pentagon is split into a quadrilateral and a triangle, by drawing the diagonal BD
• In fig(c), one side of the pentagon is reduced, by bringing the vertex C to F. Thus we get a quadrilateral AFDE. This quadrilateral has the same area as the original pentagon ABCDE
• In fig(d), the method is applied again. This time, on the left side of the quadrilateral. (This is because, if we continue working on the right side, the resulting triangle will be too elongated towards the right. The reader may try that too). In fig.(d), the quadrilateral is split into two triangles.   
• In fig(e), the vertex E is brought down to G. We get the final ΔGFD

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■ While doing the actual construction, the above steps can be done in a single fig. This is shown in fig(f). The steps involved are:
• Draw the diagonal BD. Draw MN through C, parallel to BD
• Extend AB towards the right, to meet MN at F. Join B and D to F
• Now work on the left side of the pentagon. Draw diagonal AD. Draw M'N' through E, parallel to AD
• Extend AB towards the left, to meet M'N' at G. Join A and D to G. ΔGFD is the required triangle

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In the next section we will see some solved examples.

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