Question

You design a bean bag toss board using a coordinate plate. You plot vertices of the board at C (3,2) D (3,6) E (5,2) F (5,6). What is the perimeter of the bean bag toss board?

Answer 1

The perimeter of the bean bag toss board is 12 units

What is the perimeter of the bean bag toss board?

The points are given as:

C (3,2) D (3,6) E (5,2) F (5,6)

Calculate the distance between adjacent points using:

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

So, we have:

$CD = \sqrt{(3 - 3)^2 + (2 - 6)^2} = 4$

$DF = \sqrt{(3 - 5)^2 + (6 - 6)^2} = 2$

$FE = \sqrt{(5 - 5)^2 + (2 - 6)^2} = 4$

$CE = \sqrt{(3 - 5)^2 + (2 - 2)^2} = 2$

The perimeter is then calculated as:

P = CD + DF + FE + CE

This gives

P = 4 + 2 + 4 + 2

Evaluate

P =12

Hence, the perimeter of the bean bag toss board is 12 units

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