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A triangle has vertices at (-2,-3),(4, -3), and (3,5). What is the area of the triangle?

Sagot :

Answer:

this triangle is ABC with A(-2 ; -3), B(4; - 3) and C(3;5)

=> [tex]AB=\sqrt{(4-(-2))^{2}+\sqrt{(-3 -(-3))^{2} } }=\sqrt{6^{2} }=6\\\\AC=\sqrt{(3-(-2))^{2}+(5-(-3))^{2} }=\sqrt{5^{2}+8^{2} }=\sqrt{89}\\\\BC =\sqrt{1^{2}+8^{2} }=\sqrt{65}[/tex]

using Heron theorem, we have:

[tex]S=\sqrt{p(p-AB)(p-AC)(p-BC)}\\\\S=\sqrt{(\frac{6+\sqrt{89}+\sqrt{65} }{2})(\frac{\sqrt{89}+\sqrt{65}-6 }{2} )(\frac{6+\sqrt{65}-\sqrt{89} }{2})(\frac{6+\sqrt{89}-\sqrt{65} }{2}) }\\\\S=24 \\\\[/tex]

with S is the area of the triangle

       [tex]p=\frac{AB+AC+BC}{2}=\frac{6+\sqrt{89}+\sqrt{65} }{2}[/tex]

Step-by-step explanation:

Area of the triangle with given vertices is equals to 24 square units.

What is Area?

"Area is the surface which represents the total space occupied by any two dimensional object."

What is triangle?

"Triangle is a  polygon which has three sides and three vertices. Sum of all the interior angles of a triangle is equals to 180°."

Formula used

In ΔABC , A(x₁ , y₁) , B(x₂, y₂) and C(x₃, y₃)

Area of a triangle =  [tex]\frac{1}{2} [ x_{1} (y_{2} -y_{3} + x_{2} (y_{3} -y_{1} +x_{3} (y_{1} -y_{2} ]\\[/tex]

According to the question,

Let ABC be the triangle with A(-2,-3) , B(4,-3) and C(3,5)

Area of the triangle = [tex]\frac{1}{2}[/tex] [ -2(-3-5) + 4(5+3) +3 (-3 +3)]

                                 = [tex]\frac{1}{2}[/tex] [ 16 +32 + 0]

                                 = 24 square units

Hence Area of the triangle with given vertices is equals to 24 square units.

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