Question

The perimeter of an isosceles triangle is 24 cm. What is the maximum area possible?

Answer 1

Answer:

9.94cm

Step-by-step explanation:

y²=x²+x²

y²=2x²

y=√2x----------(equation 1)

x+x+y=24

2x+y=24--------(equation 2)

substitute values of y from eq 1 and 2

2x+x√2=24

x(2+√2)=24

x=24÷2+√2

x=24/3.4142

x=7.03 cm

y=7.03* √2

y=9.94cm.

hope it helps ....

Answer 2

Given:

Perimeter of an isosceles triangles$=24cm$

To find: the maximum area possible.

Solution:

Know that, in an isosceles triangle two sides are equal to each other, so assume that the two similar sides are x cm, x cm respectively and the third side be y cm.

Therefore,

$\[\begin{array}{l}{y^2} = {x^2} + {x^2}\\ \Rightarrow {y^2} = 2{x^2}\\ \Rightarrow y = \sqrt 2 x\end{array}\]$    ------(1)

Understand that, from the question the perimeter of the triangle is $24 cm$.

$\[\begin{array}{l}x + x + y = 24\\ \Rightarrow 2x + y = 24\end{array}\]$ ------(2)

Solve equation (1) and (2) to find the value of x.

$\[\begin{array}{l}2x + \sqrt 2 x = 24\\ \Rightarrow x\left( {2 + \sqrt 2 } \right) = 24\\ \Rightarrow x = \frac{{24}}{{\left( {2 + \sqrt 2 } \right)}}\end{array}\]$

$\[\begin{array}{l} \Rightarrow x = \frac{{24}}{{3.414}}\\ \Rightarrow x = 7.03cm\end{array}\]$

Find the value of y.

$\[\begin{array}{l}y = \sqrt 2 x\\ \Rightarrow y = \sqrt 2 \times 7.03\\ \Rightarrow y = 9.94cm\end{array}\]$

Understand that for maximum area all the sides of the triangle should be equal to each other.

Therefore,

Assume that all the sides of the triangle is $9.94 cm$.

Find the maximum area of the triangle.

$\[\begin{array}{l}A = \frac{{\sqrt 3 }}{4} \times 9.94 \times 9.94\\ \Rightarrow A = 42.78c{m^2}\end{array}\]$

Hence, the maximum area of the triangle is $\[42.78c{m^2}\]$.