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KATRIN_1 [288]
3 years ago
10

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

Mathematics
1 answer:
MAVERICK [17]3 years ago
7 0

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
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