Troublesome Triangle: Solution

We are given a triangle whose perimeter is 40 cm and whose area is 60 sq. cm. The lengths of the sides are positive integers. Let us determine the side lengths.

This solution may raise a few eyebrows because the solution is not entirely mathematical. But I can't help it. I know no other way of finding the solution. My dilemma arose from the fact that I don't know what type of triangle I am dealing with. For instance, had I been told that the triangle was right-angled, the approach would have been different. I will come back to this at the end.

Step 2: Substituting Known Values

Substituting s = 20:

A² = 20(20 − a)(20 − b)(20 − c)

The area of the triangle is 60 sq. cm, so:

60² = 20(20 − a)(20 − b)(20 − c)

Simplifying:

3600 = 20(20 − a)(20 − b)(20 − c)

Dividing both sides by 20:

180 = (20 − a)(20 − b)(20 − c)

Step 3: Making a Substitution

Let us define new variables:

• x = 20 − a
• y = 20 − b
• z = 20 − c

The equation becomes:

xyz = 180

Now consider the sum:

x + y + z = (20 − a) + (20 − b) + (20 − c)

Since a + b + c = 40:

x + y + z = 60 − 40 = 20

Step 4: Solving by Inspection

We now have two equations:

• Equation I: xyz = 180
• Equation II: x + y + z = 20

Factorizing 180 gives:

180 = 1 × 2 × 2 × 3 × 3 × 5

A close inspection shows that:

• 12 × 3 × 5 = 180
• 12 + 3 + 5 = 20

Thus, x = 12, y = 3, and z = 5.

Step 5: Finding the Sides

Recall that:

• a = 20 − x
• b = 20 − y
• c = 20 − z

Substituting the values:

• a = 20 − 12 = 8
• b = 20 − 3 = 17
• c = 20 − 5 = 15

The sides of the triangle are 8 cm, 17 cm, and 15 cm.

Step 6: Verifying the Triangle

To confirm that these lengths form a triangle, we apply the triangle inequality theorem:

• 8 + 17 > 15
• 8 + 15 > 17
• 17 + 15 > 8

All conditions are satisfied.

Moreover, the triangle is right-angled, since:

17² = 8² + 15²(Pythagorean check)

Problem solved.

Alternative Solution

If Ashish had mentioned the triangle was right-angled, I would have solved the problem in a different way.

The problem, restated with the additional information, will be: A right-angled triangle has an area of 60 sq. cm and the sum of the lengths of its sides is 40 cm. Find the lengths of its sides.

Let the sides be a and b, and let the hypotenuse be h.

So, a² + b² = h²

Given, a + b + h = 40

We also have ab/2 = 60 (the area of the triangle).

ab = 120

The (a, b> pairs of values which satisfy this condition are (1, 120), (2, 60), (3, 40), (4, 30), (5, 24), (6, 20), (8, 15), (10, 12).

Now, a + b + h = 40, which implies a + b is smaller than 40. So, we can rule out (1, 120), (2, 60), and (3, 40).

Consider the next pair (4, 30); 4 + 30 + h = 40, so h = 6. The hypotenuse cannot be smaller than any of the other sides. So, we can rule out this pair as well. The pairs (5, 24) and (6, 20) also give smaller hypotenuse, which is not possible.

That leaves us with the pairs (8, 15) and (10, 12). In the first case, the hypotenuse works out to be 17 and in the second case 18. So, the likely triplets are (8, 15, 17) and (10, 12, 18). A right-angled triangle has to obey the Pythagorus Rule. For (8, 15, 17), 17² = 289, and 8² + 15² = 64 + 225 = 289. Pythagorus Rule is satisfied.

For the triplet (10, 12, 18), (18² = 324, and 10² + 12² = 100 + 144 = 244. Pythagorus Rule is not satisfied.

So the only solution is (8, 15, 17).

Of course, this again is not a mathematically-sound solution. But it is Ashish's fault. He should also have provided the length of one of the sides, say the hypotenuse. In that case I am sure I would have worked for hours, but come out with a fool-proof and mathematically-sound solution.

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