Question

A rectangle in the xy-coordinate plane is formed by the points (1,−4), (1,6), (3,−4), and (3,6). What is the perimeter of this rectangle?

Answer 1

Answer:

Perimeter of the rectangle=$20+2\sqrt{104}$

Step-by-step explanation:

The perimeter of the rectangle with the given coordinates is the sum of its all four sides:

Let the points be A(1,-4), B(1,6), C(3,-4) and D(3,6)

Finding the sides of the rectangle using the Distance formula:

AB=

               $\sqrt{(1-1)^2+(6-(-4))^2} \\\\=\sqrt{0+10^2} \\\\=\sqrt{100}\\\\ =10$

BC

            $=\sqrt{(3-1)^2+(-4-6)^2} \\\\=\sqrt{2^2+(-10)^2}\\\\ =\sqrt{4+100} \\\\=\sqrt{104}$

CD=

             $\sqrt{(3-3)^2+(6-(-4))^2} \\\\=\sqrt{0+10^2} \\\\=\sqrt{100} \\\\=10$

AD=

            $\sqrt{(3-1)^2+(6-(-4))^2} \\\\=\sqrt{2^2+10^2} \\\\=\sqrt{4+100}\\\\ =\sqrt{104}$

 Perimeter of the rectangle=AB+BC+CD+AD

                                    =$10+\sqrt{104}+ 10+\sqrt{104}$

Perimeter of the rectangle=$20+2\sqrt{104}$