InterMath | Geometry | Triangles
Title
Biggie Size It!
Problem Statement
If you double the lengths of each of the sides of a triangle, what happens to the perimeter and the area? Explain why.
How would the results change if you triple the lengths of each of the sides? What if you make the sides ten times their original size? Explain your reasoning.
Problem setup
In any given triangle, what happens to the perimeter and area of the triangle when you double the lengths of each of the sides of the triangle? What happens if you triple the length of each of the sides instead of doubling them? What if you increase the length of the sides by 10 times?
Plans to Solve/Investigate the Problem
First I am going to construct a triangle in GSP. Then I will create an excel chart with heron’s formula for area. I will use excel to try different triangle dimensions including doubling and tripling the lengths of the triangle.
Investigation/Exploration of the Problem
Construct a triangle:
1.
Draw three points in GSP. Connect them with line segments.
A
B
C
2.
Use the measurement function to measure the size of the lengths of the triangle.
3.
Add the lengths up to determine the perimeter of the triangle.

AC=x
AE=2x
AB=x
AD=2x x
A
2x C x
B 2x
E D
The above figure shows a smaller triangle ( ABC ) inscribed within a triangle ( ADE ) whose lengths are twice the lengths of ABC .
In Excel:
1.
Investigate what happens when you double the lengths of a triangle.
2.
Plug in lengths for a triangle
3.
Use the equation A

( x
 
)(
 
)(
 
)(
  z ) s

AB

BC

AC
2
4.
Use Heron’s formula to find the area of the triangle. (
A

(

)(

)(

AC ) )
5.
Use the perimeter formula (P=AB+BC+AC) to determine the perimeter of the triangle
6.
Double the lengths of the sides of the triangle and use the same formulas in steps 3, 4, and 5.
AB
7
14
6
12
BC
9
18
7
14
AC
5
10
5
10 s Area
10.5 17.41228
21 69.64912
9 14.69694
18 58.78775
Perimeter
4
21
42
18
36
When you divide the area of the second triangle by the area of the first triangle you get 4.
Therefore, the area increased by 4 times when the lengths of the sides were doubled.
This occurs because:
AB
 x
BC
 y
AC
 z

s
 x y z
2
A= x y z x
  z
 x
2 2
A x y z
 y
2 x y z
 z
2 x
  z x
  
2 x x
  
2 y x
  
2 z
2 2 2 2
A x y z y
  x x
  y x
  z
2 2 2 2
A

A

( x
 
)(
 
)(
 
)(
  z )
16
( x
 
)(
 
)(
 
)(
  z )
4
When you double the lengths you get: AB=2x, BC=2y, and AC=2z s

2 x

2 y
2 s x y z

2 z
A

( x
 
)(
  
2 )(
  
2 )(
  
2 )
A

( x
 
)(
 
)(
 
)(
  z )
Now you need to compare the two area formulas you determined:
1. A

( x
 
)(
 
)(
 
)(
  z )
4
2. A

( x
 
)(
 
)(
 
)(
  z )
As you can see the only difference in the two equations is that the first equation is divided by 4.
Therefore, the area of the triangle whose lengths are doubled is four times the area of the other triangle.
The perimeter of a triangle whose sides had been doubled will be twice the perimeter of the original triangle. This occurs because perimeter is equal to the measurement of the sides of the triangles added together. Otherwise known as:
P
   z
When the sides are doubled: x = 2x, y = 2y, and z = 2z.
Therefore, P

2 x

2 y

2 z
This can be rewritten as: P

2( x y z ) if you factor out a 2. Therefore, the perimeter of the triangle with sides twice as long has a perimeter that is twice as big.

Triple the sides: x

3 ,

3 ,

3 z
P

3 x

3 y

3 z
P

3( x y z )
Therefore, the perimeter of a triangle with triple the length of the sides is 3 times the perimeter of the first triangle ( P x y z ).
As found earlier the area of the original triangle is:
( x
 
)(
 
)(
 
)(
  z )
A

4
The area of the triangle with triple the length of the sides is: s s

 x y z
3( s
2 x y z
2

3 x

) s

3
2
(
3
2 x y
 y
3 z z )
A

A

3( x
2 y z y
  x ) 3( x y z )

3( x y z )
2
16
2
81( x
 
)(
 
)(
 
)(
  z )
2
A

3 x

3
2 y

3 z 3 x

3
2 y

3 z

6
2 x 3 x

3
2 y

3 z

6
2 y 3 x

3
2 y

3 z

6
2 z
A

3 x

3 y

3 z 3 y

3 z

3 x 3 x

3 y

3 z 3 x

3 y

3 z
2 2 2 2
A
A


3( x y z
2
) 3(
 y
  x ) 3( x
  z ) 3( x y z )
2 2 2
81( x
 
)(
 
)(
 
)(
  z )
16
A

9
4
( ( x
 
)(
 
)(
 
)(
  z )
Now we compare the area formula for the original triangle and the area formula for the triangle with side lengths tripled.
1. A

( x
 
)(
 
)(
 
)(
  z )
4
2.
A

9
4
( ( x
 
)(
 
)(
 
)(
  z )

As you can see the only difference is the 9 times the second formula. Therefore, when the side lengths of a triangle are tripled, the area is multiplied by 9 times.
If you multiply the lengths of the sides by 10 times…
X = 10x, y = 10y, z = 10z
Perimeter = 10X + 10Y + 10Z
Perimeter = 10(x + y + z)
The perimeter will be 10 times the perimeter of the original triangle.
Area: s

10 x

10 y

2 s

5 x

5 y

5 z
10 z
A

(5 x

5 y

5 )(5 x

5 y

5 z

10 )(5 x

5 y

5 z

10 )(5 x

5 y

5 z

10 )
A

A

(5 x

5 y

5 )(5 y

5 z

5 )(5 x

5 z

5 )(5 x

5 y

5 )

5( x y z y z x
  x
  y
  x y z )

A

25 ( x
 
)(
 
)(
 
)(
  z )
Let’s review what the original area formula was to compare.
1. A

( x
 
)(
 
)(
 
)(
  z )
4
2.
A

25 ( x
 
)(
 
)(
 
)(
  z )
As you can see the area of the triangle whose side lengths were multiplied by 10 has an area multiplied by 100. You can see from above that you have to multiple formula 1 by 4 to get rid of the denominator. Since you did it to one you have to do it to the other and 25 multiplied by 4 equals 100.
After looking at these three examples, it seems that whatever you do to the length of the sides (if you triple them for example) the perimeter will be multiplied by this number. So the perimeter would be 3 times the perimeter of the original triangle in the example. The area though seems to increase by the square of how it changes. So in the example of the tripling, the area would be increased by 9 times.
For example, look at the case of a triangle ABC whose lengths have been tripled in size.

2x j' x
C k
A j
B j = 2.08 cm l = 2.25 cm k = 2.12 cm
Area of triangle ABC:
A

(

)(

)(

)
A
 s
A


17.93072628
a b c
2 s

2 s

3.225
The triangle whose lengths have been triple: k' k' = 6.36 cm j' = 6.23 cm l' = 6.75 cm s

2 l' s

2 s

9.67
Area of triangle ABC:
A

(

)(

)(

)
A

A

1.994578116
Area of the triangle with tripled lengths:
A

(

)(

)(

)
A

A

17.93072628
The area of the second triangle divided by the area of triangle ABC = 9. Therefore, the difference in the area between a triangle and one that had triple the lengths is correct as can be seen in this example.
Extensions of the Problem
If you double the area of a triangle, explain what happens to the side lengths and perimeter.
By how many times would you need to increase the area in order for the new side length and new perimeter to be 5 times the original measures?

As I deduced earlier the area is the square of the change in the sides. Therefore, if you are going the other way you would take the square root of the area to get the length of the sides. So in this example, if the area is doubled then the lengths of the sides are multiplied
5 x

5 y

5 z s

2
A

5 x

5 y

5 z 5 x

5 y

5 z

5 x
2 2
5 x

5 y

5 z

2
5 y
5 x

5 y

5 z

5 z
2 by
A

A

5 x

5
2 y

5 z 5 x

5
2 y

5 z

10
2 x 5 x

5
2 y

5 z

10
2 y 5 x

5 y

5 z

10
2 2 z
5 x

5 y

5 z 5 z

5 y

5 x 5 x

5 y

5 z 5 x

5 y

5 z
2 2 2 2
.
A

5( x y z )

2 
5 y z
2 x



5  x y
2 z



5  x y
2 z


A

25 ( x
 
)(
 
)(
 
)(
  z )
4
The following is proof of why it happens.
A

( x
 
)(
 
)(
 
)(
  z )
4
Therefore, if you double the area you get…
2 A

2
( x
 
)(
 
)(
 
)(
 
4
( x
 
)(
 
)(
 
)(
  z ) z )
2 A

2

s

A

2 x

2 x

2 y

2
2
2 y

2 z
2 z


2 x

2 y

2
2 z

2 2
2 x


2 x

2 y

2
2 z

2 2
2 y


2 x

2 y

2
2 z

2 2
2 z


A

2 x

2 y

2 z


2 z

2 y

2 x


2 x

2 y

2 z


2 x

2 y

2 z


2 2 2 2

2( x y z )

2( z
  x )

2( x y z )

2( x y z )

A

A

2 ( x
 
)(
 
4
4
)(
 
)(
  z )
A

( x
 
)(
 
)(
 
)(
  z )
2
This equals the formula I found to be equal to what doubling the area would equal.
Therefore, when the area is doubled the lengths of the sides are multiplied by 2 .
In order to get the lengths of the sides of a triangle and the perimeter of the triangle to be 5 times the original measurements you would have to increase the area by 25 times. This is because you take the square root of the increase of the area to find the increase of the sides and perimeter. 5 is the square root of 25; therefore, the lengths of the sides and the perimeter would need to be increased by 5 times in order for the area to be increased by 25 times.
If you need to see it worked out:
X=5x, y = 5y, z = 5z s

5 x

5 y

5 z
2
A

5 x

5 y

5 z 5 x

5 y

5 z

5 x
2 2
A

5 x

5 y

5 z

5 y
2
5 x

5
2 y

5 z

5 z
5 x

5
2 y

5 z 5 x

5
2 y

5 z

10
2 x 5 x

5
2 y

5 z

10
2 y 5 x

5
2 y

5 z

10
2 z
A

5 x

5 y

5 z 5 z

5 y

5 x 5 x

5 y

5 z 5 x

5 y

5 z
2 2 2 2
A

5( x y z )

2 
5 y z
2 x



5  x y
2 z



5  x y
2 z


A

25 ( x
 
)(
 
)(
 
)(
  z )
4

Since the original formula for the area of a triangle is:
A

( x
 
)(
 
)(
 
)(
  z )
4
The area is increased by 25 times when the lengths of the sides and the perimeter are increased by 5 times. This supports my hypothesis I made.
Author & Contact
Tabitha Davis tabbydavis240@yahoo.com
Link(s) to resources, references, lesson plans, and/or other materials
Link 1
Link 2