A sheet of paper ABDE is a 12-by-18-inch rectangle, as shown
This topic has expert replies
-
- Moderator
- Posts: 7187
- Joined: Thu Sep 07, 2017 4:43 pm
- Followed by:23 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
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
GMAT/MBA Expert
- Ian Stewart
- GMAT Instructor
- Posts: 2621
- Joined: Mon Jun 02, 2008 3:17 am
- Location: Montreal
- Thanked: 1090 times
- Followed by:355 members
- GMAT Score:780
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.
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.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
ianstewartgmat.com
ianstewartgmat.com
GMAT/MBA Expert
- Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7261
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
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: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
(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
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews