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katovenus [111]
3 years ago
9

Triangle ABC has vertices at A(-2,1), B(-2,-3), and C(1,-2). What is the area of the triangle? a 18 square units b 9 square unit

s c 6 square units d 12 square units
Mathematics
1 answer:
natka813 [3]3 years ago
7 0

Answer:

Option C. 6 square units

Step-by-step explanation:

we know that

Heron's Formula is a method for calculating the area of a triangle when you know the lengths of all three sides.  

Let  

a,b,c be the lengths of the sides of a triangle.  

The area is given by:

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

where

p is half the perimeter

p=\frac{a+b+c}{2}

we have

Triangle ABC has vertices at A(-2,1), B(-2,-3), and C(1,-2)

the formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

step 1

Find the distance AB

d=\sqrt{(-3-1)^{2}+(-2+2)^{2}}

d=\sqrt{(-4)^{2}+(0)^{2}}

dAB=4\ units

step 2

Find the distance BC

d=\sqrt{(-2+3)^{2}+(1+2)^{2}}

d=\sqrt{(1)^{2}+(3)^{2}}

dBC=\sqrt{10}\ units

step 3

Find the distance AC

d=\sqrt{(-2-1)^{2}+(1+2)^{2}}

d=\sqrt{(-3)^{2}+(3)^{2}}

dBC=\sqrt{18}\ units

step 4

a=AB=4\ units

b=BC=\sqrt{10}\ units

c=AC=\sqrt{18}\ units

Find the half perimeter p

p=\frac{4+\sqrt{10}+\sqrt{18}}{2}=5.70\ units

Find the area

A=\sqrt{5.7(5.7-4)(5.7-3.16)(5.7-4.24)}

A=\sqrt{5.7(1.7)(2.54)(1.46)}

A=\sqrt{35.93}

A=6\ units^{2}

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