Question

Ay 82a25) F C(2a + 2c, 26) The area of a parallelogram is given by the formula A = bh, where A is the area, b is the length of a base, and h is the height perpendicular to the base. ABCD is a parallelogram. E, F, G, and Hare the midpoints of the sides. 7. Show that the area of EFGH is half the area of ABCD. G A(0,0) H D(2c, 0)

Answer 1

First let's calculate the area of ABCD.

Using AD as the base, the height can be calculated with the difference of y-coordinates of the points B and A, and the base is the difference of x-coordinates of the points D and A:

`\begin{gathered} h=2b-0=2b \\ base=2c-0=2c \end{gathered}`

So the area of ABCD is:

`A=2b\cdot2c=4bc`

Now, finding the coordinates of the midpoints E, F, G and H, we have:

`\begin{gathered} E_x=(2a+0)/(2)=a \\ E_y=(2b+0)/(2)=b \\ \\ F_x=(2a+2c+2a)/(2)=2a+c \\ F_y=(2b+2b)/(2)=2b \\ \\ G_x=(2a+2c+2c)/(2)=a+2c \\ G_y=(2b+0)/(2)=b \\ \\ H_x=(0+2c)/(2)=c \\ H_y=(0+0)/(2)=0 \end{gathered}`

Using EG as a base of the triangles EGF and EGH, we can calculate the areas of these triangles.

The height of these triangles are the difference in y-coordinates of the point F and E:

`\begin{gathered} h=2b-b=b \\ \text{base}=a+2c-a=2c_{} \\ \text{Area}=(2c\cdot b)/(2)=bc \end{gathered}`

Adding the area of the two triangles, we have the area of EFGH:

`A=bc+bc=2bc`

So comparing the area of EFGH and ABCD, we have:

`(2bc)/(4bc)=(1)/(2)`

Therefore the area of EFGH is half the area of ABCD.