Respuesta :
Answer:
[tex]p=4\sqrt{10}units[/tex]
[tex]A=6units[/tex] [tex]square[/tex]
Step-by-step explanation:
Given,
[tex]P\left ( -6,2 \right ),A\left ( -3,3 \right ),T\left ( 0,2 \right ),H\left ( -3,1 \right )[/tex]
Distance between two points
[tex]h=\sqrt{\left ( x_{1}-x_{2} \right )^{2}+\left ( y_{1} -y_{2}\right )^{2}}[/tex]
[tex]PA=\sqrt{\left ( -6+3 \right )^{2}+\left ( 2-3 \right )^{2}}[/tex]
[tex]=\sqrt{9+1}=\sqrt{10}units[/tex]
[tex]AT=\sqrt{\left ( -3+0 \right )^{2}+\left ( 3-2 \right )^{2}}[/tex]
[tex]=\sqrt{9+1}=\sqrt{10}units[/tex]
[tex]TH=\sqrt{\left ( 0+3 \right )^{2}+\left ( 2-1 \right )^{2}}[/tex]
[tex]=\sqrt{9+1}=\sqrt{10}units[/tex]
[tex]HP=\sqrt{\left ( -3+6 \right )^{2}+\left ( 1-2 \right )^{2}}[/tex]
[tex]=\sqrt{9+1}=\sqrt{10}units[/tex]
[tex]PA=AT=TH=HP[/tex] (i)
Length of diagonal
[tex]d_{1} =PT=\sqrt{\left ( -6+0 \right )^{2}+\left ( 2-2 \right )^{2}}[/tex]
[tex]=\sqrt{36+0}=6[/tex]
[tex]d_{2}=AH=\sqrt{\left ( -3+3 \right )^{2}+\left ( 3-1 \right )^{2}}[/tex]
[tex]=\sqrt{0+4}=2units[/tex]
Length of diagonals are not equal
[tex]d_{1} \neq d_{2}[/tex] (ii)
From above conditions this polygon is rhombus
Perimeter of rhombus =4×length of side
[tex]p=4\sqrt{10}units[/tex]
Area of rhombus
[tex]A=1/2\times d_{1}\times d_{2}[/tex]
[tex]a=1/2\times 6\times 2[/tex]
[tex]A=6units[/tex] [tex]square[/tex]
