Question

find area of triangle formed by (1,1) (2,3) (4 ,5) using heron's formula ​

Answer 1

Answer:

The area of a triangle is:

• A = 1 unit²

Step-by-step explanation:

Given the points of a triangle

A(1, 1)

B(2, 3)

C(4, 5)

$\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

Distance between AB:

$AB=\sqrt{\left(2-1\right)^2+\left(3-1\right)^2}$

     $=\sqrt{1+4}$

     $=\sqrt{5}$

so

• $a=\sqrt{5}$

Distance between BC:

$BC=\sqrt{\left(4-2\right)^2+\left(5-3\right)^2}$

     $=\sqrt{2^2+2^2}$

     $=\sqrt{2^3}$

     $=2\sqrt{2}$

so

• $b=2\sqrt{2}$

Distance between AC:

$AC=\sqrt{\left(4-1\right)^2+\left(5-1\right)^2}$

    $=\sqrt{3^2+4^2}$

   $=\sqrt{5^2}$

    $=5$

so

• $c=5$

Semiperimeter = sum of sides ÷ 2            

                       $s=\frac{\sqrt{5}+2\sqrt{2}+5}{2}=5.03224$

Area of the triangle:

$A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

• $a=\sqrt{5}$

• $b=2\sqrt{2}$

• $c=5$

$A=\sqrt{s\left(s-\sqrt{5}\right)\left(s-2\sqrt{2}\right)\left(s-5\right)}$

$A=\sqrt{5.03224\dots \left(5.03224\dots -\sqrt{5}\right)\left(5.03224\dots -2\sqrt{2}\right)\left(5.03224\dots -5\right)}$

$A=1$

Therefore, the area of a triangle is:

• A = 1 unit²