Question

Identify the polygon with vertices A(5,0), B(2,4), C(−2,1), and D(1,−3), and then find the perimeter and area of the polygon. HELP ASAP!

Answer 1

Part 1) The polygon is a square

Part 2) The perimeter is equal to
`20\ units`

Part 3) The area is equal to
`25\ units^(2)`

Explanation:

we have

`A(5,0), B(2,4), C(-2,1),D(1,-3)`

Plot the points

see the attached figure

we know that

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

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

Find the distance AB

`A(5,0),B(2,4)`

substitute in the formula

`d=\sqrt{(4-0)^(2)+(2-5)^(2)}`

`d=\sqrt{(4)^(2)+(-3)^(2)}`

`d=√(25)`

`AB=5\ units`

Find the distance BC

`B(2,4), C(-2,1)`

substitute in the formula

`d=\sqrt{(1-4)^(2)+(-2-2)^(2)}`

`d=\sqrt{(-3)^(2)+(-4)^(2)}`

`d=√(25)`

`BC=5\ units`

Find the distance CD

`C(-2,1),D(1,-3)`

substitute in the formula

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

`d=\sqrt{(-4)^(2)+(3)^(2)}`

`d=√(25)`

`CD=5\ units`

Find the distance AD

`A(5,0),D(1,-3)`

substitute in the formula

`d=\sqrt{(-3-0)^(2)+(1-5)^(2)}`

`d=\sqrt{(-3)^(2)+(-4)^(2)}`

`d=√(25)`

`AD=5\ units`

we have that

AB=BC=CD=AD

Find the distance BD (diagonal)

`B(2,4),D(1,-3)`

substitute in the formula

`d=\sqrt{(-3-4)^(2)+(1-2)^(2)}`

`d=\sqrt{(-7)^(2)+(-1)^(2)}`

`BD=√(50)\ units`

Verify if the polygon is a square

If the triangle BDA is a right triangle, then the polygon is a square

Applying the Pythagoras theorem

`BD^(2)=AD^(2)+AB^(2)`

substitute

`(√(50))^(2)=5^(2)+5^(2)`

`50=50` -----> is true

so

The triangle BDA is a right triangle

therefore

The polygon is a square

Find the Area of the polygon

The area of a square is equal to

`A=b^(2)`

we have

`b=5\ units`

`A=5^(2)=25\ units^(2)`

Find the perimeter of the polygon

The perimeter of a square is equal to

`P=4b`

we have

`b=5\ units`

`P=4(5)=20\ units`