MATH SOLVE

2 months ago

Q:
# Find the area of rectangle ABCD

Accepted Solution

A:

Find the length of the rectangle with pythagoras

You can use point (0,0) and (3,3) as the reference points.

l = √(Δx² + Δy²)

l = √((3-0)² + (3-0)²)

l = √(3² + 3²)

l = √18

Find the width of the rectangle with pythagoras

You can use point (0,0) and (2,-2) as the reference points.

w = √(Δx² + Δy²)

w = √((-2-0)² + (2-0)²)

w = √((-2)² + 2²)

w = √(4 + 4)

w = √8

Find the area of rectangle

A = l × w

A = √18 × √8

A = √9 × √2 × √4 × √2

A = 3 × √2 × 2 × √2

A = 12

The area of the rectangle is 12 unit length

You can use point (0,0) and (3,3) as the reference points.

l = √(Δx² + Δy²)

l = √((3-0)² + (3-0)²)

l = √(3² + 3²)

l = √18

Find the width of the rectangle with pythagoras

You can use point (0,0) and (2,-2) as the reference points.

w = √(Δx² + Δy²)

w = √((-2-0)² + (2-0)²)

w = √((-2)² + 2²)

w = √(4 + 4)

w = √8

Find the area of rectangle

A = l × w

A = √18 × √8

A = √9 × √2 × √4 × √2

A = 3 × √2 × 2 × √2

A = 12

The area of the rectangle is 12 unit length