Question

The coordinates of the vertices of a rectangle are (−3, 4) , (7, 2) , (6, −3) , and (−4, −1) . What is the perimeter of the rectangle? Round each step to the nearest tenth. Enter your answer as a decimal in the box.

Answer 1

Answrer

Find out the what is the perimeter of the rectangle .

To prove

Now as shown in the figure.

Name the coordinates as.

A(−3, 4) ,B (7, 2) , C(6, −3) , and D(−4, −1) .

In rectangle opposite sides are equal.

Thus

AB = DC

AD = BC

Formula

$Disatnce\ formula = \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2}}$

Now the points  A(−3, 4) and  B(7, 2)

$AB = \sqrt{(7- (-3))^{2} +(2- 4)^{2}}$

$AB = \sqrt{(10)^{2} +(-2)^{2}}$

$AB = \sqrt{100+4}$

$AB = \sqrt{104}$

$AB = 2\sqrt{26}\units$

Thus

$CD= 2\sqrt{26}\units$

Now the points

A (−3, 4) , D (−4, −1)

$AD = \sqrt{(-4 - (-3))^{2} +(-1- 4)^{2}}$

$AD = \sqrt{(-1)^{2} +(-5)^{2}}$

$AD = \sqrt{1 + 25}$

$AD = \sqrt{26}\units$

Thus

$BC = \sqrt{26}\units$  

Formula

Perimeter of rectangle = 2 (Length + Breadth)

Here

$Length = 2\sqrt{26}\ units$

$Breadth = \sqrt{26}\ units$  

$Perimeter\ of\ rectangle = 2(2\sqrt{26} +\sqrt{26})$

$Perimeter\ of\ rectangle = 2(3\sqrt{26})$

$Perimeter\ of\ rectangle = 6\sqrt{26}$

$\sqrt{26} = 5.1 (Approx)$

$Perimeter\ of\ rectangle = 6\times 5.1$

Perimeter of a rectangle = 30.6 units.

Therefore the perimeter of a rectangle is 30.6 units.