Question

can somebody please help me with this problem and can you not only give me thwe answer but also show me how you got it? the following figure, NDBA is a rectangle, AN = NG, BF = DB, GD = 11, and DF = 8√2. What is the area of parallelogram GDFA?

Answer 1

the area parallelagrams are equal to the average of the bases * height

Answer 2

with this picture the top and bottom measures are the same so the average length of the top and bottom is a pointless step

Answer 3

the information tells us that the little triangles on the end are 45 degrees; and also that we subtract the "height" from the "width" to get the bases for the parallelagram

Answer 4

the information tells us that the base = GD = 11

we just need to determine the height now

Answer 5

the pythag thrm states that the length of the hypotenuse squared is equal to the sum of the square of the legs .... we know that the legs are equal in length so lets assign a variable to them such as "s" for side.

s^2 + s^2 = (8sqrt(2))^2

Answer 6

2s^2 = 64(2)

s^2 = 64

s = 8; well, this tell us our height is 8, so lets use that :)

the area of our parallelagram is determined by:

Area = base*height
= 11*8
= 88 .... if I did it right

Answer 7

DB=FB
in tri DBF,
by using pythagoras theorem
DF^2=DB^2+FB^2
(8√2)^2 = DB^2+DB^2(DB=FB)
128 = 2DB^2
64 = DB^2
√64 = DB
DB = 8

Area of llgm = b*h
= GD*DB
= 11*8
= 88