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pogonyaev
3 years ago
5

On a coordinate plane, triangle A B C is shown. Point A is at (0, 0), point B is at (3, 4), and point C is at (3, 2). What is th

e area of triangle ABC?
Mathematics
2 answers:
Elena L [17]3 years ago
7 0

Answer:

The area of triangle for the given coordinates is  1.5\sqrt{4.6}

Step-by-step explanation:

Given coordinates of triangles as

A = (0,0)

B = (3,4)

C = (3,2)

So, The measure of length AB = a = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

Or, a = \sqrt{(3-0)^{2}+(4-0)^{2}}

Or, a =  \sqrt{9+16}

Or, a =   \sqrt{25}

∴ a = 5 unit

Similarly

The measure of length BC = b = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

Or, b = \sqrt{(3-3)^{2}+(2-4)^{2}}

Or, a =  \sqrt{0+4}

Or, b =   \sqrt{4}

∴ b = 2 unit

And

So, The measure of length CA = c = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

Or, c = \sqrt{(3-0)^{2}+(2-0)^{2}}

Or, c =  \sqrt{9+4}

Or, c =   \sqrt{13}

∴ c = \sqrt{13} unit

Now, area of Triangle written as , from Heron's formula

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

and s = \frac{a+b+c}{2}

I.e  s = \frac{5+2+\sqrt{13}}{2}

Or. s =  \frac{7+\sqrt{13}}{2}

So, A = \sqrt{(\frac{(7+\sqrt{13})}{2})\times ((\frac{(7+\sqrt{13})}{2})-5)\times (\frac{7+\sqrt{13}}{2}-2)\times (\frac{7+\sqrt{13}}{2}-\sqrt{13})}

Or, A = \sqrt{(\frac{(7+\sqrt{13})}{2})\times (\frac{(\sqrt{13}-3)}{2})\times (\frac{4+\sqrt{13}}{2})\times (\frac{7-\sqrt{13}}{2})}

Or, A = \frac{3}{2} × \sqrt{1+\sqrt{13} }

∴  Area of triangle = 1.5\sqrt{4.6}

Hence The area of triangle for the given coordinates is  1.5\sqrt{4.6}  Answer

Naily [24]3 years ago
3 0

Answer:

3

Step-by-step explanation:

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