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?

nerdiest Jul 6, 2022

#5**+2 **

*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.

Guest Jul 6, 2022

#8**+2 **

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

CPhill Jul 6, 2022