A sheet of paper ABDE is a 12-by-18-inch rectangle, as shown

The diagrams are confusing, because they aren't to scale -- in the second diagram, AF is suddenly much longer than in the first, even though the length AF hasn't changed.

Regardless, after folding the paper, the angle at E remains a 90 degree angle, the length of DE is still 12, and the lengths AF and FE must add to 18. So the triangle at the bottom, AEF, is a right triangle. If we call its shortest side, EF, "c", then its hypotenuse is 18-c, because AF and FE add to 18. So by Pythagoras,

c^2 + 12^2 = (18 - c)^2

Seeing that the answers all work out to integers, you might, seeing the "12", guess that this is a 5-12-13 triangle. That turns out to be the case. Or you could solve the equation:

c^2 + 12^2 = 18^2 - 36c + c^2
36c = 18^2 - 12^2
36c = (18 + 12)(18 - 12)
36c = 30*6
c = 5

Since c is the length of AF, and since the lengths of AF and EF sum to 18, the length of EF is 13. So in the shaded triangle, if we take the horizontal line AF as the base, the base is 13, and the height is 12, so the area is (13*12)/2 = 78.

BTGmoderatorDC wrote:



A sheet of paper ABDE is a 12-by-18-inch rectangle, as shown in Figure 1. The sheet is then folded along the segment CF so that points A and D coincide after the paper is folded, as shown in Figure 2 (The shaded area represents a portion of the back side of the paper, not visible in Figure 1). What is the area, in square inches, of the shaded triangle shown?

A) 72
B) 78
C) 84
D) 96
E) 108

OA B

Source: Manhattan Prep

We can let BC = x and thus CD = 18 - x. We see that triangle DBC (in figure 2) is a right triangle and thus we have:

(DB)^2 + (BC)^2 = (CD)^2

12^2 + x^2 = (18 - x)^2

144 + x^2 = 324 - 36x + x^2

36x = 180

x = 5

Likewise, if we let FE = y and thus AF = 18 - y. We see that triangle AEF (in figure 2) is a right triangle. Since side AE of triangle AEF is now same as DE, AE = 12 and we have:

(AE)^2 + (FE)^2 = (AF)^2

12^2 + y^2 = (18 - y)^2

This is equivalent to the equation above, so y = 5.

Now we can argue that:

Area of triangle DBC + Area of triangle AEF + 2(Area of triangle CAF) = Area of rectangle ABDE

(½)(12)(5) + (½)(12)(5) + 2(Area of triangle CAF) = 12(18)

30 + 30 + 2(Area of triangle CAF) = 216

2(Area of triangle CAF) = 156

Area of triangle CAF = 78

But the area of triangle CAF is the area of the shaded region, so the area of the shaded region is 78.

Answer: B