# Solution to Square Triangle Area

i have question again about the same link because i have national exam in one week i am trying to know way of solution of such recriational geometry problem see please below link of stated problem and it's solution.i have one question : is it correct?

Problem: find the area of the shaded region:

http://web.gnowledge.org/wiki/index.php/Square%2C_Triangle_and_Area#

Solution: See how the region has been restructured. The right triangle that has been removed and the new one that has been added are both conguent by the property ASA (prove it!). Thus the two triangles have the same areas.

It is now easy to find the area of the shaded region: 9 square units.

http://web.gnowledge.org/wiki/index.php/Solution_to_Square_Triangle_Area

is here everything correct? if not please can anybody tell me how to solve it?

Update: thanks guys everything is clear just i am curious what would be result if angle will not be right but anything else? could we find required area or it would be impossible?

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The solution is correct. –  joriki Jul 7 '11 at 17:30
@ joriki it is because triangle is right angle yes?it's sides are 10 and 8 and one of angle is right am i correct? –  dato datuashvili Jul 7 '11 at 17:33
The ping doesn't work if you put a space between the '@' and the name. The $10$ and $8$ don't enter into it; the small triangles that cancel each other are congruent because the angle of the big triangle is right. –  joriki Jul 7 '11 at 17:37
great thanks very much –  dato datuashvili Jul 7 '11 at 17:39
my only question is: what about if angle will be different from 90? –  dato datuashvili Jul 7 '11 at 17:43

After a while, you will have rotated the triangle so that its "legs" are horizontal and vertical. Then the yellow area has become a $3 \times 3$ square.
Comment: Mathematicians usually prefer counterclockwise rotation. I would like a mathematical watch whose hands rotate couunterclockwise. Of course noon and midnight would be at the right end of the watch, not on top. And there would not be a $12$ there, instead there would be a $0$.