SOLUTION: Please Help Me (I could not copy and paste the picture): The diagram is a picture of a square. Within the square are smaller triangles, squares, and a parallelogram. The larger

Algebra ->  Formulas -> SOLUTION: Please Help Me (I could not copy and paste the picture): The diagram is a picture of a square. Within the square are smaller triangles, squares, and a parallelogram. The larger       Log On


   



Question 467675: Please Help Me (I could not copy and paste the picture):
The diagram is a picture of a square. Within the square are smaller triangles, squares, and a parallelogram. The larger square has a diagonal line. The top left hand corner of the square is a triangle. Beneath that is another triangle, square, triangle, parallelogram. The bottom half of the larger square is divided into two triangles. One triangle is shaded, above that is the smaller triangle, which is also shaded.
Find the area of the shaded region in the figure. Round results to the nearest unit. Use = 3.14.
The hypotenuse of the smaller triangular portion of the shaded region has length The sides of the outer square have length

A. 123 ft2
B. 61 ft2
C. 49 ft2
D. Not enough data

The answer is B. I have tried to figuare this question out but keep coming up with a different answer everytime. Please Help!


Answer by ccs2011(207) About Me  (Show Source):
You can put this solution on YOUR website!
First you did not include any given lengths of the sides to determine area so I will give the solution in general terms.
I hope I depicted the figure accurately from your description.
Assuming the bottom half of square is cut into 2 Equal triangles.
Then area of 1 of those triangles is one-fourth area of outer square.
Also notice this triangle is a special 45-45-90 right triangle.
*Diagonal line across square cuts corners into two 45 degree angles*
Thus smaller triangle above is also a special 45-45-90 right triangle.
The sides of such a triangle are (x,x, and sqrt%282%29x)
The area of such a triangle is:
A = x%5E2%2F2
In terms of the hypotenuse, h where h = sqrt%282%29x then x = h%2Fsqrt%282%29
A = %28%28h%2Fsqrt%282%29%29%5E2%29%2F2 = h%5E2%2F4
Now given the side length of the outer square (s) and the hypotenuse of smaller triangle (c).
**Notice that s is the hypotenuse of the larger shaded triangle**
Use the above formula to obtain the area of the two shaded triangles
Area = %28s%5E2%2F4%29+%2B+%28c%5E2%2F4%29 = %28s%5E2+%2B+c%5E2%29%2F4