Question

find the area of the largest isosceles triangle having the perimeter 18cm.

class 12th​

Answer 1

Answer:

What is the maximum area of an isosceles triangle whose perimeter is 18 inches?

For all rectangle with the same perimeter, the one that has maximum area is a square. Similarly, for all triangles with the same perimeter, the one that has maximum area is an equilateral triangle. In fact, when comparing all regular polygon (all sides equal, all angles equal) with same perimeter, the more sides the polygon has, the larger its area. Taking this to its natural conclusion, of all shapes with the same perimeter, the one with maximum area is a circle.

Let △ABC be isosceles with vertex C. AB=2x,AC=AB=9−x

Drop perpendicular from C to side AB at point D. By Pythagorean theorem:

CD2=(9−x)2−x2⟹CD=81–18x−−−−−−√,0≤x≤4.5

Now we can express area of triangle in terms of variable x:

A=12×AB×CD=12(2x)81–18x−−−−−−√=81x2–18x3−−−−−−−−−√

A′=12(81x2–18x3)−12⋅(162x−54x^281x−27x281x2–18x3−−−−−−−−−√=0

81x−27x2=0

27x(3−x)=0

x=0,x=3

We check values of A where A′=0 and at endpoints

A(0)=81(0)2–18(0)3−−−−−−−−−−−√=0

A(3)=81(3)2–18(3)3−−−−−−−−−−−√=243−−−√=93–√

A(4.5)=81(4.5)2–18(4.5)3−−−−−−−−−−−−−−√=0

Maximum Area =93−−√ in2 when x=3

AB=2x=6;AC=BC=9−x=6⟹△ABC

Answer 2

$\large\underline{\sf{Solution-}}$

Let assume that the three sides of triangle be x cm, x cm and y cm respectively.

As it is given that, Perimeter of isosceles triangle is 18 cm

$\rm \: x + x + y = 18$

$\rm \: 2x + y = 18$

$\rm\implies \:y = 18 - 2x$

So, three sides of triangle be x, x, 18 - 2x

and

Semi Perimeter of triangle, s = 9 cm

Now, area of triangle using Heron's Formula is given by

$\rm \: A = \sqrt{9(9 - x)(9 - x)(9 - 18 + 2x)}$

$\rm \: A = \sqrt{9(9 - x)(9 - x)( 2x - 9)}$

$\rm \: A = 3(9 - x) \sqrt{2x - 9}$

On differentiating both sides w. r. t. x, we get

$\rm \:\dfrac{d}{dx} A = 3\dfrac{d}{dx}(9 - x) \sqrt{2x - 9}$

$\rm \: \dfrac{dA}{dx} = 3\bigg( \sqrt{2x - 9}\dfrac{d}{dx}(9 - x) + (9 - x)\dfrac{d}{dx} \sqrt{2x - 9} \bigg)$

$\rm \: \dfrac{dA}{dx} = 3\bigg( - \sqrt{2x - 9} + (9 - x)\dfrac{1}{2 \sqrt{2x - 9} }(2) \bigg)$

$\rm \: \dfrac{dA}{dx} = 3\bigg( - \sqrt{2x - 9} + \dfrac{9 - x}{ \sqrt{2x - 9} } \bigg)$

$\rm \: \dfrac{dA}{dx} = 3\bigg(\dfrac{ - 2x + 9 + 9 - x}{ \sqrt{2x - 9} } \bigg)$

$\rm \: \dfrac{dA}{dx} = 3\bigg(\dfrac{ - 3x + 18}{ \sqrt{2x - 9} } \bigg)$

$\rm \: \dfrac{dA}{dx} = 9\bigg(\dfrac{ - x + 6}{ \sqrt{2x - 9} } \bigg)$

For maxima and minima, we get

$\rm \: \dfrac{dA}{dx} = 0$

$\rm \: 9\bigg(\dfrac{ - x + 6}{ \sqrt{2x - 9} } \bigg) = 0$

$\rm \: - x + 6= 0$

$\rm\implies \:x = 6$

Now, to check whether A is maxima or minima at x = 6, we use first derivative test

$\begin{gathered}\boxed{\begin{array}{c|c} \bf When & \bf Sign \: of \: \dfrac{dA}{dx} \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x < 6 & \sf + \\ \\ \sf x > 6 & \sf - \end{array}} \\ \end{gathered}$

As sign of changes from (+) to (-), it means A is maximum.

So, Maximum area is given by

$\rm \: A = 3(9 - x) \sqrt{2x - 9}$

$\rm \: A = 3(9 - 6) \sqrt{12 - 9}$

$\rm \: A = 3 \times 3 \sqrt{3}$

$\rm\implies \:A \: = \: 9 \sqrt{3} \: {cm}^{2}$

$\rule{190pt}{2pt}$

Heron's Formula :-

The area of triangle having sides a, b and c respectively, is given by

$\boxed{ \rm{ \: \triangle \: = \: \sqrt{s(s - a)(s - b)(s - c)} \: \: }}$

where,

$\rm \: s \: = \: \dfrac{a + b + c}{2}$

Additional Information :-

Other method to find maxima or minima :-

Let y = f(x) be a given function.

To find the maximum and minimum value, the following steps are follow :

1. Differentiate the given function.

2. For maxima or minima, put f'(x) = 0 and find critical points.

3. Then find the second derivative, i.e. f''(x).

4. Apply the critical points ( evaluated in second step ) in the second derivative.

5. Condition :-

• The function f (x) is maximum when f''(x) < 0.

• The function f (x) is minimum when f''(x) > 0.