+1161 An isosceles right triangle is removed from each corner of a square piece of paper, as shown, to create a rectangle. If \(AB=12\) units, what is the combined area of the four removed triangles, in square units?
An isosceles right triangle is removed from each corner of a square piece of paper, as shown, to create a rectangle.
"as shown" huh? There's nothing shown.
As long as this is a guessing game, my submission is either 36 or 144. Pick one.
+128407 Call the equal sides of the small isosceles triangle , a
Call the equal sides of the larger isosceles triangle, b
The short side of the rectangle in the middle = sqrt (a^2 + a^2) = sqrt (2a^2) = a sqrt (2)
The long side of the rectangle in the middle = sqrt (b^2 + b^2) =sqrt (2b^2) = b sqrt (2)
By the Pythagorem Theorem
[ asqrt (2)]^2 + [b sqrt (2) ]^2 = AB^2
[a sqrt (2) ] ^2 + [ b sqrt (2) ] ^2 = 12^2
2a^2 + 2b^2 = 144 divide by 2
a^2 + b^2 = 72
But a^2 = the area of the two smaller isosceles triangles
And b^2 = the area of the larger two isosceles triangles
So....their combined area = 72
