Math Problem Statement

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

Solution

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:

  • aa as the length of the two equal sides of the isosceles triangle.
  • bb as the length of the base.

The perimeter of the triangle is given by: 2a+b=80cm2a + b = 80 \, \text{cm} So, b=802ab = 80 - 2a

Step 2: Use the altitude to find the base

The altitude hh divides the base bb into two equal halves, each of length b2\frac{b}{2}. The altitude forms a right triangle with half of the base and one of the equal sides. Using the Pythagorean theorem: a2=(b2)2+h2a^2 = \left(\frac{b}{2}\right)^2 + h^2 Given that h=20cmh = 20 \, \text{cm}, we substitute hh and b=802ab = 80 - 2a: a2=(802a2)2+202a^2 = \left(\frac{80 - 2a}{2}\right)^2 + 20^2 a2=(802a2)2+400a^2 = \left(\frac{80 - 2a}{2}\right)^2 + 400

Step 3: Solve for aa

Simplify the equation: a2=(802a2)2+400a^2 = \left(\frac{80 - 2a}{2}\right)^2 + 400 a2=(40a)2+400a^2 = \left(40 - a\right)^2 + 400 Expanding the square: a2=(40a)2+400a^2 = (40 - a)^2 + 400 a2=160080a+a2+400a^2 = 1600 - 80a + a^2 + 400 Simplifying, we get: 0=200080a0 = 2000 - 80a 80a=200080a = 2000 a=25cma = 25 \, \text{cm}

Step 4: Find the base bb

Now, substitute a=25a = 25 into the equation for the perimeter: b=802(25)b = 80 - 2(25) b=8050=30cmb = 80 - 50 = 30 \, \text{cm}

Step 5: Calculate the area

The area AA of the triangle is given by: A=12×base×heightA = \frac{1}{2} \times \text{base} \times \text{height} A=12×30cm×20cmA = \frac{1}{2} \times 30 \, \text{cm} \times 20 \, \text{cm} A=12×600cm2=300cm2A = \frac{1}{2} \times 600 \, \text{cm}^2 = 300 \, \text{cm}^2

So, the area of the isosceles triangle is 300cm2\boxed{300 \, \text{cm}^2}.

Would you like more details or have any other questions?


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.

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Math Problem Analysis

Mathematical Concepts

Geometry
Triangle Properties
Pythagorean Theorem

Formulas

Area of a Triangle = 0.5 * base * height

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 9-10