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hudsonlilliana

hudsonlilliana

04.03.2020 •

Mathematics

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

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Ответ:

georgehall9515

20.01.2022

$p=4\sqrt{10}units$

A=6units square

Step-by-step 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

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

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

$=\sqrt{9+1}=\sqrt{10}units$

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

$=\sqrt{9+1}=\sqrt{10}units$

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

$=\sqrt{9+1}=\sqrt{10}units$

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

$=\sqrt{9+1}=\sqrt{10}units$

PA=AT=TH=HP (i)

Length of diagonal

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

$=\sqrt{36+0}=6$

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

$=\sqrt{0+4}=2units$

Length of diagonals are not equal

$d_{1} \neq d_{2}$ (ii)

From above conditions this polygon is rhombus

Perimeter of rhombus =4×length of side

$p=4\sqrt{10}units$

Area of rhombus

$A=1/2\times d_{1}\times d_{2}$

$a=1/2\times 6\times 2$

A=6units square

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Ответ:

anniekwilbourne

12.01.2023

d

Step-by-step explanation:

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