find the other sides and area of the triangle . if two sides of the triangle are 20 cm and 29 cm , then perimeter is 60 cm

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).

Learn how to find the third side and calculate the area of a triangle using Heron’s formula. Given two sides of 20 cm and 29 cm, and a perimeter of 60 cm, this step-by-step solution is ideal for students in Grades 9-10.

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