Question

A rectangle has vertices located at (–4, 4), (2, 4), (2, –4) and (–4, –4). What is the length of a diagonal of the rectangle?

Answer 1

Solution:-

1. First draw a rectangle with vertices ABCD

• A = ( -4 4)
• B = (2 ,4)
• C = ( 2 , -4)
• D = (-4 , -4 )

2) Draw the two diagonals

If we see the diagonals It is nothing but AC and BD

If we find distance between AC or BD That is called diagonal

So, diagonal is nothing but Distance b/w A & C (OR) Distance b/w B & D

We find distance between A&C

A= ( -4 ,4)

C = (2, -4)

${x_1 = -4}$

${x_2 = 2}$

${y_1 = 4}$

${y_2 = -4}$

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

AC = $\sqrt{(-4-2)^2 +(4-(-4))^2}$

AC = $\sqrt{(-6)^2 +(8)^2}$

AC = $\sqrt{36+64}$

AC = $\sqrt{100}$

AC = 10

So, the length of diagonal is 10 units

In a rectangle diagonals are equal bisect each other

So, BD also 10 units

Hence length of Diagonal is 10units