If three sides of a triangle are less than 1 iIn length, show that its area is less than \dfrac{\sqrt{3}}{4}

#### Question

#### Answer 1

If three sides of a triangle are given values, the area of the triangle can be calculated with Heron’s formula.

A = \sqrt{s(s-a)(s-b)(s-c)} ,

in which s =\dfrac{a+b+c}{2} ,

or directly in terms of a, b, c

A = \dfrac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}

Since

a <1, b<1, b< 1

Addition of the three inequalities gives

Therefore,

A < \dfrac{\sqrt{3} }{4}\cdotp \sqrt{(-a+b+c)(a-b+c)(a+b-c)}

Using AM-GM inequality

A < \dfrac{\sqrt{3} }{4}\cdotp\dfrac{(-a+b+c)+(a-b+c)+(a+b-c)}{3}

= \dfrac{\sqrt{3} }{4}\cdotp\dfrac{a+b+c}{3}

< \dfrac{\sqrt{3} }{4}\cdotp \dfrac{3}{3}

= \dfrac{\sqrt{3} }{4}

Now we have verified the area of the triangle is less than \dfrac{\sqrt{3} }{4} if the length of its three sides is less than 1

#### Answer 2

Since the sum of three internal angles of a triangle is equal to 180\degree , if C is the angle with the smallest measure, C is less than 60\degree .

Given two sides and the Included angle of the two sides, the area of a triangle can be expressed as,

A = \dfrac{1}{2}ab\sin C

\because a<1, b<1 and C<60\degree

A<\dfrac{1}{2} \cdotp 1\cdotp 1\cdotp \sin 60\degree =\dfrac{\sqrt{3} }{4}