The perimeter of an isosceles triangle is 80 cm and the altitude to its base is 20 cm. Find area of the triangle.

To find the area of the isosceles triangle, we'll use the given information: the perimeter is 80 cm, and the altitude to the base is 20 cm.

### Step 1: Understand the triangle structure
Let's denote:
- \( a \) as the length of the two equal sides of the isosceles triangle.
- \( b \) as the length of the base.

The perimeter of the triangle is given by:
\[
2a + b = 80 \, \text{cm}
\]
So,
\[
b = 80 - 2a
\]

### Step 2: Use the altitude to find the base
The altitude \( h \) divides the base \( b \) into two equal halves, each of length \( \frac{b}{2} \). The altitude forms a right triangle with half of the base and one of the equal sides. Using the Pythagorean theorem:
\[
a^2 = \left(\frac{b}{2}\right)^2 + h^2
\]
Given that \( h = 20 \, \text{cm} \), we substitute \( h \) and \( b = 80 - 2a \):
\[
a^2 = \left(\frac{80 - 2a}{2}\right)^2 + 20^2
\]
\[
a^2 = \left(\frac{80 - 2a}{2}\right)^2 + 400
\]

### Step 3: Solve for \( a \)
Simplify the equation:
\[
a^2 = \left(\frac{80 - 2a}{2}\right)^2 + 400
\]
\[
a^2 = \left(40 - a\right)^2 + 400
\]
Expanding the square:
\[
a^2 = (40 - a)^2 + 400
\]
\[
a^2 = 1600 - 80a + a^2 + 400
\]
Simplifying, we get:
\[
0 = 2000 - 80a
\]
\[
80a = 2000
\]
\[
a = 25 \, \text{cm}
\]

### Step 4: Find the base \( b \)
Now, substitute \( a = 25 \) into the equation for the perimeter:
\[
b = 80 - 2(25)
\]
\[
b = 80 - 50 = 30 \, \text{cm}
\]

### Step 5: Calculate the area
The area \( A \) of the triangle is given by:
\[
A = \frac{1}{2} \times \text{base} \times \text{height}
\]
\[
A = \frac{1}{2} \times 30 \, \text{cm} \times 20 \, \text{cm}
\]
\[
A = \frac{1}{2} \times 600 \, \text{cm}^2 = 300 \, \text{cm}^2
\]

So, the area of the isosceles triangle is \( \boxed{300 \, \text{cm}^2} \).

Would you like more details or have any other questions?

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Here are some related questions you might find interesting:

1. What would be the area if the perimeter was 100 cm with the same altitude?
2. How do you find the area of an equilateral triangle given only its perimeter?
3. How does the area change if the altitude is doubled while keeping the perimeter constant?
4. What is the significance of the altitude in determining the properties of an isosceles triangle?
5. How can the Pythagorean theorem be used to solve other properties of an isosceles triangle?

**Tip:** Understanding the relationship between the sides, altitude, and area in triangles is key to solving more complex geometric problems.

Learn how to calculate the area of an isosceles triangle with a perimeter of 80 cm and an altitude of 20 cm. This problem involves using the perimeter formula and the Pythagorean theorem to find the dimensions of the triangle. Suitable for students in Grades 9-10.

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