Area of Triangle

Few classes back, we learned that \( \text {area of a triangle} = \frac {1} {2} base \times height,\) when the measurements of base and height are given.

Similarly, the area of the triangle can be found when the length of sides is given.
Let the three sides of triangle ABC are a, b and c respectively. As AD⊥BC and AD is the height (h) of the triangle ABC.

Suppose the length of DC = x
Then the length of BD= a - x

Now, In the right-angled triangle ADB,

AB² = BD² + AD²
AB² = BD² + AD²
or, AD²= AB²- BD²
or, h²= c² - (a - x)² ............. (1)

Again, In the right-angled triangle ADC,

AC² = AD²+ DC²
or, AD²= AC²- DC²
or h²= b²- x²..................... (2)

From equation (1) and (2) we get,

c²- (a-x)²= b²- x²
or, c²- (a - x)²= b² - x²
or, c² - ( a² -2ax + x² ) = b²- x²
or, c² - a² + 2ax -x²= b² - x²
or, 2ax = b²- x² - c²+ a²+ x²
or, 2ax = a² + b²- c²

\( or, x = \frac {a^2 + b^2 - c^2} {2a} \).................. (3)

Substituting the value of x from equation (3) in equation (2), we get

\( h^2 = b^2 - \left (\frac {a^2 + b^2 - c^2} {2a} \right) ^ 2 \)

\(or, h^2 = \left (b +\frac {a^2 + b^2 - c^2} {2a} \right) \left(b - \frac {a^2 + b^2 - c^2} {2a} \right) \)

\(or, h^2 = \left ( \frac {2ab + a^2 + b^2 - c^2} {2a}\right)\left ( \frac {2ab - a^2 - b^2 + c^2} {2a}\right)\)

\(or, h^2 = \frac {\{(a + b)^2 -c^2\}} {2a} \) \( \frac {\{c^2 - (a -b)^2\}} {2a}\)

\(or, h^2 = \frac {(a + b + c) (a + b - c)} {2a} \)\(\frac {(c - a + b ) (c + a - b)} {2a} \)

\(or, h^2 = \frac {(a + b + c) (a + b - c)(c - a + b ) (c + a - b)} {4a^2}\)

\(or, h^2 = \frac {(a + b + c) (a + b + c - 2c)(c - a + b ) (c + a + b - 2b)} {4a^2}\) ............... (4)

Now, Substituting , a+b+c = 2s, we get,

\( h^2 = \frac {2s ( 2s - 2c )( 2s - 2b )( 2s - 2a )} {4a^2}\)

\(or, h^2=\frac {2s.2( s - c ).2(s - b ).2(s - a )} {4a^2}\)

\(or, h^2=\frac {16s( s - a ) (s - b ) (s - c )} {4a^2}\)

\(or, h = \sqrt {\frac {4s( s - a ) (s - b ) (s - c )} {a^2} }\)

\(or, h = \frac{2}{a}\sqrt {s( s - a ) (s - b ) (s - c )}\)

Now, \(\text{Area of}\; \Delta \text {ABC} = \frac {1}{2} base \times height \)

\( = \frac {1} {2} a \times \frac {2} {a} \sqrt {s( s - a ) (s - b ) (s - c )} \)

\(\boxed {\therefore \text {Area of}\; \Delta \text {ABC} =\sqrt {s( s - a ) (s - b ) (s - c )}}\)

\(\text { Semi-perimeter of triangle } = \frac {a + b + c} {2}\)

\(\text { Area of equilateral triangle } = \frac {\sqrt {3} } {4} a^2 \)

Area of isosceles triangle = \(\frac{b} {4} \) \(\sqrt{4a^2-b^2}\)[ where a is the base and b is the lenght of equal sides.]