a perimeter of 15 cm two or more equal sides and each side a whole number of centimeters. Prove that there are only four of these triangles. You do no need to construct the triangles.
If two or more sides are equal, then just run through the possibilities:
1 1 13
2 2 11
3 3 9
4 4 7
5 5 5
6 6 3
7 7 1
Now, are any of these impossible? Yes, if you recall that any side must be less than the sum of the other two sides. That kicks out
1 1 13
2 2 11
3 3 9
leaving only four remaining
To prove that there are only four triangles with a perimeter of 15 cm, two or more equal sides, and each side a whole number of centimeters, let's consider all the possible combinations.
Let's assume the three sides of the triangle are a, b, and c, with a ≤ b ≤ c.
Since two or more sides are equal, we have three cases to consider:
Case 1: Two equal sides and one distinct side
If two sides are equal, let's assume a = b and c be the distinct side length.
Now, the perimeter is given as 15 cm: a + a + c = 15.
Simplifying this equation, we have 2a + c = 15.
Since a and c are whole numbers, the values of a and c can be selected from the set of positive integers.
To find the possible values of a and c such that 2a + c = 15, we can iterate through all possible values for a and determine the corresponding value of c.
Possible values for a and c are as follows:
a = 1, c = 13
a = 2, c = 11
a = 3, c = 9
a = 4, c = 7
a = 5, c = 5
From these values, it can be observed that there are only two triangles that satisfy the conditions: (1, 1, 13) and (5, 5, 5).
Case 2: All three sides are equal
If all three sides of the triangle are equal, let's assume a = b = c.
Now, the perimeter is given as 15 cm: a + a + a = 15.
Simplifying this equation, we have 3a = 15.
Solving for a, we find that a = 5.
Therefore, the only triangle that satisfies this condition is (5, 5, 5).
Case 3: Three distinct side lengths
Since the perimeter is given as 15 cm, we have a + b + c = 15.
To find the possible values of a, b, and c, we can iterate through all possible values for a and b and determine the corresponding value of c.
Possible values for a, b, and c that satisfy a + b + c = 15 are as follows:
a = 1, b = 2, c = 12
a = 1, b = 3, c = 11
a = 1, b = 4, c = 10
a = 1, b = 5, c = 9
a = 1, b = 6, c = 8
a = 1, b = 7, c = 7
a = 2, b = 3, c = 10
a = 2, b = 4, c = 9
a = 2, b = 5, c = 8
a = 2, b = 6, c = 7
a = 3, b = 4, c = 8
a = 3, b = 5, c = 7
a = 4, b = 5, c = 6
From these values, it can be observed that there are ten triangles that satisfy the condition, but they do not have two or more equal sides.
Therefore, combining all the cases, we have a total of four triangles with a perimeter of 15 cm, two or more equal sides, and each side a whole number of centimeters.
These four triangles are: (1, 1, 13), (5, 5, 5), (1, 7, 7), and (3, 4, 8).
Hence, the proof is complete.
To prove that there are only four triangles that satisfy the given conditions, you can use a systematic approach to evaluate all possible combinations mathematically.
Let's call the length of each side of the triangle "a." Since there are two or more equal sides, we have three possible scenarios:
Scenario 1: Two sides are equal, and the third side is different.
In this case, the possible lengths for the equal sides can range from 1 cm to 7 cm (since the perimeter is 15 cm and the sides are whole numbers). The remaining side will be the remaining length to reach the total perimeter of 15 cm minus twice the length of the equal sides.
The combinations for this scenario are:
- Equal sides: 1 cm, 1 cm, remaining side: (15 - 2*1) = 13 cm
- Equal sides: 2 cm, 2 cm, remaining side: (15 - 2*2) = 11 cm
- Equal sides: 3 cm, 3 cm, remaining side: (15 - 2*3) = 9 cm
- Equal sides: 4 cm, 4 cm, remaining side: (15 - 2*4) = 7 cm
- Equal sides: 5 cm, 5 cm, remaining side: (15 - 2*5) = 5 cm
- Equal sides: 6 cm, 6 cm, remaining side: (15 - 2*6) = 3 cm
- Equal sides: 7 cm, 7 cm, remaining side: (15 - 2*7) = 1 cm
So, in this scenario, we have seven possible triangles.
Scenario 2: Three sides are equal.
In this case, all three sides of the triangle have the same length. The length of each side, "a," can range from 1 cm to 5 cm (since the perimeter is 15 cm and the sides are whole numbers).
The combinations for this scenario are:
- All sides equal: 1 cm, 1 cm, 1 cm
- All sides equal: 2 cm, 2 cm, 2 cm
- All sides equal: 3 cm, 3 cm, 3 cm
- All sides equal: 4 cm, 4 cm, 4 cm
- All sides equal: 5 cm, 5 cm, 5 cm
So, in this scenario, we have five possible triangles.
Scenario 3: Only two sides are equal, and corresponding angles are unequal.
In this case, the possible lengths for the equal sides can range from 1 cm to 7 cm, just like in Scenario 1. However, the remaining side must have a length smaller than the difference between the lengths of the equal sides.
The combinations for this scenario are:
- Equal sides: 1 cm, 1 cm, remaining side: less than (1 - 1) = 0 cm (not possible)
- Equal sides: 2 cm, 2 cm, remaining side: less than (2 - 2) = 0 cm (not possible)
- Equal sides: 3 cm, 3 cm, remaining side: less than (3 - 3) = 0 cm (not possible)
- Equal sides: 4 cm, 4 cm, remaining side: less than (4 - 4) = 0 cm (not possible)
- Equal sides: 5 cm, 5 cm, remaining side: less than (5 - 5) = 0 cm (not possible)
- Equal sides: 6 cm, 6 cm, remaining side: less than (6 - 6) = 0 cm (not possible)
- Equal sides: 7 cm, 7 cm, remaining side: less than (7 - 7) = 0 cm (not possible)
In this scenario, no valid triangles exist.
Therefore, by analyzing all three scenarios, we find that there are only four triangles that satisfy the given conditions.