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PLEASE HELP WITH THESE TWO

4. Tyler is writing a coordinate proof to show that a diagonal of a rectangle divides the rectangle into two triangles that have equal areas. He starts by assigning coordinates to a rectangle. Then he uses these coordinates to write an expression for the area of each triangle formed by the diagonal.

What is the area of one of the triangles formed by the diagonal of the rectangle?

Enter an expression in the box for the area of the triangle.

5. Amira is writing a coordinate proof to show that the area of a triangle created by joining the midpoints of an isosceles triangles is one-fourth the area of the isosceles triangle. She starts by assigning coordinates as given.

Enter your answers, in simplest form, in the boxes to complete the coordinate proof.

Point Q is the midpoint of DE , so the coordinates of point Q are (a, b).

Point R is the midpoint of FE , so the coordinates of point R are ([ ], b).

In ΔDEF , the length of the base, DF , is [ ], and the height is 2b, so its area is [ ].

In ΔQRP , the length of the base, QR , is [ ], and the height is b, so its area is ab.

Comparing the expressions for the areas proves that the area of the triangle created by joining the midpoints of an isosceles triangle is one-fourth the area of the triangle.

Thanks in advance. Please explain.

PLEASE HELP WITH THESE TWO 4. Tyler is writing a coordinate proof to show that a diagonal of a rectangle divides the rectangle into two triangles that have equa class=
PLEASE HELP WITH THESE TWO 4. Tyler is writing a coordinate proof to show that a diagonal of a rectangle divides the rectangle into two triangles that have equa class=

Answer :

princessanon
Question 4.
 ab/2

Question 5.
(3a,b)
4a
area is 4ab
2a

Answer:

#4) 1/2(ab); #5) 3a, 4a, 4ab, 2a

Step-by-step explanation:

#4) Take the top triangle.  The height will be the difference in the y-coordinates of J and M:

b-0 = b

The base of the triangle would be the difference in the x-coordinates of J and K:

a-0 = a

The area is given by the formula A=1/2bh, where b is the base and h is the height; substituting our height and base, we have

A=1/2(b)(a) = 1/2(ab)

#5) To find the midpoint of a section, we add together the x-coordinates and divide by to, and add together the y-coordinates and divide by 2.

The endpoints of FE are (4a, 0) and (2a, 2b).  This makes R:

[tex](\frac{2a+4a}{2},\frac{2b+0}{2})\\\\=(\frac{6a}{2},\frac{2b}{2})\\\\=(3a, b)[/tex]

The length of DF can be found by subtracting the x-coordinates:

4a-0 = 4a

The area is then found by multiplying 1/2 by the base and the height:

A=1/2(4a)(2b) = 1/2(8ab) = 4ab

The length of QR can be found by subtracting the x-coordinates:

3a-a = 2a

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