Question

Identify the polygon that has vertices P(−6,2), A(−3,3), T(0,2), and H(−3,1), and then find the perimeter and area of the polygon.

rhombus; P=6 units; A=4√ 1 0 units2

parallelogram; P=6 units; A=4√ 10 units2

rhombus; P=4√ 10√ units; A=6 units2
rhombus; P = 40 units; A = 12 units2

Answer 1

`p=4√(10)units`

`A=6units`
`square`

Explanation:

Given,

`P\left ( -6,2 \right ),A\left ( -3,3 \right ),T\left ( 0,2 \right ),H\left ( -3,1 \right )`

Distance between two points

`h=\sqrt{\left ( x_(1)-x_(2) \right )^(2)+\left ( y_(1) -y_(2)\right )^(2)}`

`PA=\sqrt{\left ( -6+3 \right )^(2)+\left ( 2-3 \right )^(2)}`

`=√(9+1)=√(10)units`

`AT=\sqrt{\left ( -3+0 \right )^(2)+\left ( 3-2 \right )^(2)}`

`=√(9+1)=√(10)units`

`TH=\sqrt{\left ( 0+3 \right )^(2)+\left ( 2-1 \right )^(2)}`

`=√(9+1)=√(10)units`

`HP=\sqrt{\left ( -3+6 \right )^(2)+\left ( 1-2 \right )^(2)}`

`=√(9+1)=√(10)units`

`PA=AT=TH=HP` (i)

Length of diagonal

`d_(1) =PT=\sqrt{\left ( -6+0 \right )^(2)+\left ( 2-2 \right )^(2)}`

`=√(36+0)=6`

`d_(2)=AH=\sqrt{\left ( -3+3 \right )^(2)+\left ( 3-1 \right )^(2)}`

`=√(0+4)=2units`

Length of diagonals are not equal

`d_(1) \\eq d_(2)` (ii)

From above conditions this polygon is rhombus

Perimeter of rhombus =4×length of side

`p=4√(10)units`

Area of rhombus

`A=1/2* d_(1)* d_(2)`

`a=1/2* 6* 2`

`A=6units`
`square`