Question

Can someone help me with this problem ​

Answer 1

Answer:

A = 20

P = 6√10

Step-by-step explanation:

The formula of a distance between two points:

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

A(-3, 0) , B(3, 2)

$AB=\sqrt{(3-(-3))^2+(2-0)^2}=\sqrt{6^2+2^2}=\sqrt{36+4}=\sqrt{40}=\sqrt{4\cdot10}=\sqrt4\cdot\sqrt{10}=2\sqrt{10}$

A(-3, 0), D(-2, -3)

$AD=\sqrt{(-2-(-3))^2+(-3-0)^2}=\sqrt{1^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

AB = CD and AD = BC

The area of a rectangle:

$A=(AB)(AD)$

Substitute:

$A=(2\sqrt{10})(\sqrt{10})=(2)(10)=20$

The perimeter of a rectangle:

$P=2(AB+AD)$

Substitute:

$P=2(2\sqrt{10}+\sqrt{10})=2(3\sqrt{10})=6\sqrt{10}$