SOLUTION: Find the area of the shaded polygons: b. (pic: http://homework.russianschool.com/resource?key=17347ioj8c4pw ) c. (pic: http://homework.russianschool.com/resource?key=17348ioj8dmg

Algebra ->  Surface-area -> SOLUTION: Find the area of the shaded polygons: b. (pic: http://homework.russianschool.com/resource?key=17347ioj8c4pw ) c. (pic: http://homework.russianschool.com/resource?key=17348ioj8dmg      Log On


   

Question 1120555: Find the area of the shaded polygons:
b. (pic: http://homework.russianschool.com/resource?key=17347ioj8c4pw )
c. (pic: http://homework.russianschool.com/resource?key=17348ioj8dmgo )
e. (pic: http://homework.russianschool.com/resource?key=17350ioj8fsbq )
f. (pic: http://homework.russianschool.com/resource?key=17351ioj8gsb0 )
h. (pic: http://homework.russianschool.com/resource?key=17353ioj8iv1b )
i. (pic: http://homework.russianschool.com/resource?key=17354ioj8jxv6 )
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Enclose the figure in a rectangle; then the area of the polygon is the area of the rectangle minus the area of the triangular regions that are outside the polygon and inside the rectangle.

The area of each triangular region is one-half base times height.

For the first one, the rectangle will be 4 wide and 3 high, with an area of 4*3=12.

The three triangular regions have legs of 1 and 1 (area 1/2), 2 and 2 (area 2), and 4 and 1 (area 2).

The area of the polygon is then 12 - (1/2+2+2) = 12 - 4 1/2 = 7 1/2

The others can be solved the same way....

Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.
1)  Too many questions to answer in one post.


2)  I will answer very first question only (question b).


    
    The general principle of solving such problems is to conclude the given figure into the MINIMAL RECTANGLE on the given grid.


    Then you will see some number of triangles that COMPLEMENT your figure to this minimal rectangle.


    Usually, it is EASY TASK to determine the area of these complementary triangles.


    As soon as you complete this auxiliary task, you will be able to determine  the area of the given figure.


For the student of the Russian Math School this hint (this instruction) is more than enough.

Boldly go ahead and good luck in solving this problem.