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scZoUnD [109]
4 years ago
9

Zoe draws ΔABC on the coordinate plane. What is the approximate perimeter of ΔABC

Mathematics
2 answers:
lord [1]4 years ago
6 0

Answer:

Option C is correct.

Step-by-step explanation:

Given: Δ ABC on coordinate plane.

From Coordinate plane, Coordinates are A( 1 , 1 ) , B( 3 , 5 ) and C( 5 , 1 )

We need to find Approximate perimeter of the triangle.

we use the distance formula of two point (x_1,y_1)\:and\:(x_2,y_2)

Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

So,

AB=\sqrt{(3-1)^2+(5-1)^2}=\sqrt{(2)^2+(4)^2}=\sqrt{4+16}=\sqrt{20}=4.47\:(approx.)

CB=\sqrt{(5-3)^2+(1-5)^2}=\sqrt{(2)^2+(-4)^2}=\sqrt{4+16}=\sqrt{20}=4.47\:(approx.)

AC=\sqrt{(5-1)^2+(1-1)^2}=\sqrt{(4)^2+(0)^2}=\sqrt{16}=4

Thus, Perimeter = AB + AC + CB = 4.47 + 4.47 + 4 = 12.94 units

Therefore, Option C is correct.

Mama L [17]4 years ago
5 0
B. 12 units because when you add the lengths together you get 12
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<h3>Equation</h3>

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<u>First, let's find the surface area of the triangular prism.</u>

  • 2(1/2 x 8 x 12) + 2(10 x 6) + (12 x 6) = Surface area of triangular prism
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<u>Now, let's find the surface area of the rectangular prism.</u>

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<u>Now, add both the surface areas to find the surface area of the figure.</u>

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