Question

Find the area of the largest isosceles triangle having perimeter 18 metres-

Answer 1

Answer:

Note that all equilateral triangles are isosceles triangles

Find the length that will give the largest possible area:

Length = Perimeter ÷ 3

Length = 18 ÷ 3

Length = 6 inches

Find the area:

Perimeter = 18 inches

Semiperimeter = 18 ÷ 2 = 9 inches

Area = √P(P - A)(P - B)(P- C)

Area = √9(9 - 6)³

Area = √243

Area = 15.6 in²

Answer: The largest area is 15.6 in²

Answer 2

Answer:

$9\sqrt{3}$

Step-by-step explanation:

As per the given information that we have an isosceles triangle with a perimeter of $18m$.

Let us assume one of the isosceles sides is $x$.

The perimeter of a triangle = Sum of the sides.

Thus, the third side of the isosceles triangle will be $18-2x$.

With the use of Heron's method

Area of isosceles triangle=$\sqrt{\frac{s}{2}(s-a)(s-b)(s-c) }$, where $s$ demotes to the semi perimeter of the triangle and $a,b,c$ are the sides.

$=\sqrt{s(s-a)(s-b)(s-c) }\\=\sqrt{9(9-x)(9-x)[9-(18-2x)]}\\=\sqrt{9 (9-x)^{2}(9-(18-2x)) }\\=\sqrt{9 (9-x)^{2}(2x-9)) }\\=3(9-x)\sqrt{2x-9} (Equation 1)$

Now, we can

$x$∈$(9,\frac{9}{2} )$

∴$\frac{dA}{dx} =\frac{9(6-x)}{\sqrt{2x-9}}$

$\frac{dA}{dx} =0\\\frac{9(6-x)}{\sqrt{2x-9}}=0\\6-x=0\\x=6$

We get the isosceles side as $6$. Put it in Equation 1

$3(9-x)\sqrt{2x-9} =3(9-6)\sqrt{2*6-9} =3*3*\sqrt{3} =9\sqrt{3}$