HELP ASAP! The vertices of a quadrilateral ABCD are A(4, 8), B(10, 10), C(10, 4), and D(4, 4). The vertices of another quadrilateral EFCD are E(4, 0), F(10, -2), C(10, 4), and D(4, 4). Which conclusion is true about the quadrilaterals?

My conclusion about Quadrilaterals ABCD and EFCD is that both quadrilaterals are similar to each other.

THE REASON IS BECAUSE THE SIDES OF QUADRILATERAL ABCD ARE THE SAME AS THE SIDES OF QUADRILATERAL EFCD.

Step-by-step explanation:

When we are given vertices, (x1, y1) , (x2 ,y2), we use the formula:

√(x2 - x1)² + (y2 - y1)²

For quadrilateral ABCD are A(4, 8), B(10, 10), C(10, 4), and D(4, 4)

Side AB: A(4, 8), B(10, 10)

√(x2 - x1)² + (y2 - y1)²

√(10 - 4)² + (10 - 8)²

= √6² + 2²

= √40

Side BC: B(10, 10), C(10, 4)

√(x2 - x1)² + (y2 - y1)²

= √(10 - 10)² + ( 4 - 10)²

= √ 0² + (-6)²

= √36

= 6

Side CD: C(10, 4), D(4, 4)

√(x2 - x1)² + (y2 - y1)²

= √ (4 - 10)² + ( 4 - 4)²

= √-6² + 0²

= √36

= 6

Side AD: A(4, 8), D(4, 4)

=√(x2 - x1)² + (y2 - y1)²

= √(4 - 4)² + (4 - 8)²

= √0² + (-4²)

= √16

= 4

Therefore, for Quadrilateral ABCD

Side AB = √40

Side BC = 6

Side CD = 6

Side AD = 4

For quadrilateral EFCD are E(4, 0), F(10, -2), C(10, 4), and D(4, 4).

Side EF: E(4, 0), F(10, -2)

√(x2 - x1)² + (y2 - y1)²

= √(10 - 4)² + (-2 - 0)²

= √6² + 2²

= √40

Side FC: F(10, -2), C(10, 4)

√(x2 - x1)² + (y2 - y1)²

= √(10 - 10)² + (4 -(-2))²

= √ 0² + 6²

= √36

= 6

Side CD: C(10, 4), D(4, 4)

√(x2 - x1)² + (y2 - y1)²

= √(10 - 4)² + (4 - 4)²

= √6² + 0²

= √36

= 6

Side ED: E(4, 0), D(4, 4)

√(x2 - x1)² + (y2 - y1)²

= √(4 - 4)² + (4 - 0)²

= √0² + 4²

= √16

= 4

Therefore, for Quadrilateral EFCD

Side EF = √40

Side FC = 6

Side CD = 6

Side ED = 4

My conclusion about Quadrilaterals ABCD and EFCD is that both quadrilaterals are similar to each other.

THE REASON IS BECAUSE THE SIDES OF QUADRILATERAL ABCD ARE THE SAME AS THE SIDES OF QUADRILATERAL EFCD.

1/8 3/16 3/4

step-by-step explanation:

you put them as all the same denominator and multiply