6 ,
The area of parallelogram is b*h=3*4=12 , Rectangle inside parallelogram and on its 2 sides (parralel sides) therefore equal to 1/2 its are so 1/2*12=6
In the first place, the base of the rectangle is 6, not 3. Second, I dispute the assumption that the inner rectangle is half the area of the larger.
+36915 Guest is referrring to the parallellgram which contains the rectangle ....not the lareger 6 x 4 rectangle
so area of the parallelogramI IS base x height = 3 x 4 = 12
I do not know about the second part of their solution though it may be correct.....
Oh, I see what you mean now. Thanks for that. However, wouldn't the base of that parallelogram, as a hypotenuse, equal the square root of 32+42 ? I'm just not getting it, maybe I ought to just move on. Thanks again for the parallelogram explanation.
+36915 Yah....you could use that as the base (it would be 5) , but then you would have to calculate the height (which is NOT 3)...
___________________ ___
/ / |
/ 3 / h
/ / |
________5_________ ___|__
+128406 The triangle on the left forms a 3 - 4 - 5 right triangle
Note that the area of this triangle = (1/2) (3)(4) = 6
And drawing a segment from the lower left vertex of the large rectangle perpendicular to the side of the shaded rectangle will also form an altitude of this right triangle
So
(1/2) (base) (altitude) = 6
(1/2) (5) (altitude) = 6
(5/2) ( altitude) = 6
altitude = 12/5 = 2.4
And this perpendicular also foms a right triangle with a hypotenuse of 3 and one leg = 2.4
So....the other leg = sqrt (3^2 - 2.4^2 ) = 1.8
And the area of this triangle = (1/2) (product of the leg lengths) =
(1/2) (2.4)(1.8) = 2.16 units^2
And, by AAS, this triangle is congruent to the smaller right triangle at the lower right of the figure
And we have two of these smaller right triangles and two of the larger right triangles
So....their total area = 2 ( 6 + 2.16) = 2 ( 8.16) = 16. 32
And the area of the large rectangle = 6 * 4 = 24
So.....the shaded area = 24 - 16.32 = 7.68 units^2
