Find the area of largest isoscless triangle having perimeter 18 inches

that will be an equilateral triangle of side 6.

maa chuda fir

To find the area of the largest isosceles triangle with a perimeter of 18 inches, we first need to understand some properties of isosceles triangles.

An isosceles triangle has two sides of equal length. Let's assume that the two equal sides have a length of x, and the remaining side has a length of y.

According to the perimeter formula for a triangle, the perimeter (P) of an isosceles triangle is given by:

P = x + x + y
P = 2x + y

In this case, we know that the perimeter is 18 inches. So we can write the equation:

18 = 2x + y

To maximize the area of the triangle, we need to maximize the base (y) while keeping the perimeter constant.

To maximize y, we need to minimize x. Therefore, let's assume that x = 0. Then we can rewrite the equation as:

18 = 2(0) + y
18 = y

So, the two equal sides have a length of 0 inches, and the remaining side (the base) has a length of 18 inches.

Now, we can find the area (A) of the triangle using the formula:

A = (1/2) * base * height

In this case, the base is given as 18 inches, and since the triangle is isosceles, the height is the distance from the top vertex to the base, which we can find using the Pythagorean theorem.

Let's assume the height is h. Using the Pythagorean theorem, we can write:

h^2 = x^2 - (1/2 * y)^2
h^2 = 0^2 - (1/2 * 18)^2
h^2 = -9^2
h^2 = 81

Since h cannot be negative in this situation, we can conclude that the height is 9 inches.

Finally, we can calculate the area:

A = (1/2) * base * height
A = (1/2) * 18 * 9
A = 9 * 9
A = 81 square inches

Therefore, the area of the largest isosceles triangle with a perimeter of 18 inches is 81 square inches.