Note: Figure not drawn to scale
In the figure, what is the area of the square with the darken borders?
page 721 #16
Hello, fabxx!
In the figure, what is the area of the inner square?Code:2 1 *-----------*-----* 1 | * * | | * * | * * | 2 |* * | | * *| 2 | * * | * * | | * * | 1 *-----*-----------* 1 2
It is a square within a square.
. . The four right triangles are congruent.
In a right triangle, the legs are 1 and 2.
. . Hence, the hypotenuse is $\displaystyle \sqrt{5}$
Therefore, the area of the inner square is: .$\displaystyle \left(\sqrt{5}\right)^2 \;=\;5 $ square units.
Area of outside square=9
Now find the area of the triangles
they all have an area of 1
so all triangles put together have an area of 4
9-4=5
the area of the inside square is 5
instead of using pythagoras, you could just understand that the area of a right triangle is (length x height)/2