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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?

User Atle
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Answer:

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.

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.

User Alessio Cantarella
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