Question

Lucinda is writing a coordinate proof to show that a diagonal of a parallelogram partitions the parallelogram into two equal areas.

A parallelogram graphed on a coordinate plane. The vertices of rectangle are labeled as K L M and N. The vertex labeled as K lies on begin ordered pair 0 comma 0 end ordered pair. The vertex labeled as L lies on begin ordered pair x comma 2 y end ordered pair. The coordinate of vertex M is left blank. The vertex labeled as N lies on begin ordered pair 3 x comma 0 end ordered pair. A diagonal is drawn between points K and M.



Enter your answers in the boxes to complete Lucinda's proof.

Since KLMN is a parallelogram and a parallelogram's opposite sides are parallel and congruent, the coordinates for M are (4x, 2y).


In △KMN, the length of the base is and the height is . So an expression for the area of △KMN is .


In △KLM, the length of the base is 3x and the height is 2y. So an expression for the area of △KLM is .


Comparing the area of the two triangles that are formed by a diagonal of the parallelogram shows that a diagonal of a parallelogram partitions the parallelogram into two equal areas.

Answer 1

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Answer:

• ΔKMN: base, 3x; height, 2y; area, 3xy
• ΔKLM: area, 3xy

Explanation:

ΔKMN

The base length is the length of the horizontal line segment KN. That length is the difference of the x-coordinates: 3x -0 = 3x.

The height is the difference of the y-coordinate of point M and the y-coordinate of horizontal segment KN. That difference is 2y -0 = 2y.

The area is half the product of base and height:

A = (1/2)bh

A = 1/2(3x)(2y) = 3xy

In ΔKMN, the length of the base is 3x and the height is 2y. So an expression for the area of ΔKMN is 3xy.

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ΔKLM

In ΔKLM, the length of the base is 3x and the height is 2y. So an expression for the area of ΔKLM is 3xy.