# Find the maximum area of a rectangular triangle with perimeter 4.

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A rectangular triangle or a right triangle is one where the sides a, b and c are related as a^2 + b^2 = c^2.

The perimeter of the triangle is 4.

The area of the triangle is (1/2)*a*b. This area is maximum when a and b are constrained if a and b are equal.

=> `a^2 + a^2 = (4 - 2a)^2`

=> `2a^2 = 16 + 4a^2 - 16a`

=> `2a^2 - 16a + 16 = 0`

=> `a^2 - 8a + 8 = 0`

The roots of the equation are `4-2^(3/2)` and `2^(3/2)+4` but a cannot be greater than 4.

**This gives the maximum area as `(1/2)*(4 - 2^(3/2))^2` = `12-2^(7/2)` **

The area of a "rectangular" or right-angled triangle is maximum when the two side are eaqual.

Let each side of the triagle be x

then length of the hypotenuse = sqrt(x^2+x^2) = x.sqrt(2)

The perimeter of the triangle = x + x + x.sqrt(2) = x*{2+sqrt(2)}

x*{2+sqrt(2)} = 4 .... given

x = 4/{2+sqrt(2)} = 4*{2-sqrt(2)}/[{2+sqrt(2)}*{2-sqrt(2)}]

x = 4*{2-sqrt(2)}/[4-2] = 4-2*sqrt(2)

Area of triangle = (x^2)/2 = {4-2*sqrt(2)}^2/2

={16+8-16*sqrt(2)}/2

= {24-16*sqrt(2)}/2

= 12-8*sqrt(2)

**The maximum area **of the right-angled triangle with perimeter 4** is equal to 12-8*sqrt(2)**