To find the third side and the area of the triangle, let's proceed step by step.

Given:

• Two sides: $a = 20 \, \text{cm}, b = 29 \, \text{cm}$
• Perimeter: $P = 60 \, \text{cm}$

Step 1: Find the third side

The perimeter of a triangle is the sum of all its sides. Let $c$ be the third side. We know:

$a + b + c = P$

Substituting the given values:

$20 + 29 + c = 60$

$49 + c = 60$

Solving for $c$:

$c = 60 - 49 = 11 \, \text{cm}$

Thus, the third side is $11 \, \text{cm}$.

Step 2: Find the area using Heron's formula

To find the area, we will use Heron’s formula, which is:

$A = \sqrt{s(s - a)(s - b)(s - c)}$

Where $s$ is the semi-perimeter of the triangle, given by:

$s = \frac{P}{2} = \frac{60}{2} = 30 \, \text{cm}$

Now, applying Heron’s formula:

$A = \sqrt{30(30 - 20)(30 - 29)(30 - 11)}$

$A = \sqrt{30 \times 10 \times 1 \times 19}$

$A = \sqrt{30 \times 190} = \sqrt{5700}$

Approximating the square root:

$A \approx 75.50 \, \text{cm}^2$

Thus, the area of the triangle is approximately 75.50 cm².

Summary:

• Third side $c = 11 \, \text{cm}$
• Area $A \approx 75.50 \, \text{cm}^2$

Would you like more details or further clarifications?

Related questions:

1. How do you apply Heron's formula for an isosceles triangle?
2. Can you find the height of this triangle using trigonometry?
3. What happens if the triangle is obtuse? Does Heron's formula still work?
4. How can we verify that a triangle with these side lengths exists (triangle inequality)?
5. How does Heron’s formula change for a right triangle?

Tip:

Always ensure the sum of the lengths of any two sides is greater than the third side when dealing with triangles (triangle inequality).