In isoceles triangle height is 8cm and perimeter vof triangle is 64cm ...
Problem Statement
In an isosceles triangle, the height is 8cm and the perimeter of the triangle is 64cm. Find the area of the triangle.
Solution
Let's label the triangle as ABC, where AC = BC and AD is the height of the triangle.
We know that the perimeter of the triangle is the sum of all its sides, so:
Perimeter = AB + AC + BC = 2AC + AB = 64cm
Since we know that AC = BC, we can substitute AC for BC in the equation:
Perimeter = 2AC + AB = 64cm
2AC + AB = 64cm
Now, we need to use the fact that the height of the triangle is 8cm to find the length of AB. The height of the triangle divides the base AB into two equal parts. Let's label the midpoint of AB as D.
Using the Pythagorean theorem on triangle ABD, we get:
AD^2 + BD^2 = AB^2
8^2 + (AB/2)^2 = AB^2
Simplifying the equation, we get:
64 + AB^2/4 = AB^2
3AB^2/4 = 64
AB^2 = 256/3
AB = sqrt(256/3)
AB ≈ 9.237cm
Now that we know the length of AB, we can find the length of AC:
Perimeter = 2AC + AB = 64cm
2AC + 9.237 ≈ 64cm
2AC ≈ 54.763
AC ≈ 27.3815cm
Now that we know the lengths of AB and AC, we can calculate the area of the triangle using the formula:
Area = (1/2) x base x height
Area = (1/2) x AB x AD
Area = (1/2) x 9.237 x 8
Area ≈ 36.948cm^2
Answer
The area of the isosceles triangle is approximately 36.948cm^2.