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12 Oct 2020, 03:00
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
42% (02:11) correct
58%
(02:29)
wrong
based on 52
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Not Attempted Yet
Two identical rectangles overlap, as shown above. Each rectangle has sides of 1 and 2. What is the area of the region where the two rectangles overlap?
A. 1
B. 5/4
C. 4/3
D. 3/2
E. 7/4
Are You Up For the Challenge: 700 Level Questions
Attachment:
Untitled.png [ 19.52 KiB | Viewed 7357 times ]
16 Oct 2020, 08:46
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The four triangles are identical -- each shares a side of length 1 with one of the rectangles, and using parallel lines and the fact lines cross in an X, we can see the angles are the same in all four.
Since the four triangles are identical, their hypotenuses are the same length, and the blue shape in the middle is a rhombus (a parallelogram with four equal sides). Say each side is x. Then looking at any triangle, say the one at the bottom, its vertical side has length 2-x, since it's just the remaining part of the side of the rectangle of length 2. So we know its three sides are 1, 2-x and x, and using Pythagoras we have
1^2 + (2 - x)^2 = x^2
1 + 4 - 4x + x^2 = x^2
4x = 5
x = 5/4
so our rhombus has sides of length 5/4, and since the area of any parallelogram (rhombus or not) is base*height, and since the height of our rhombus is the short side of a rectangle, the area is (5/4)(1) = 5/4.
I suspect there's a much faster way to do the problem that I'm not seeing at a glance.
Since the four triangles are identical, their hypotenuses are the same length, and the blue shape in the middle is a rhombus (a parallelogram with four equal sides). Say each side is x. Then looking at any triangle, say the one at the bottom, its vertical side has length 2-x, since it's just the remaining part of the side of the rectangle of length 2. So we know its three sides are 1, 2-x and x, and using Pythagoras we have
1^2 + (2 - x)^2 = x^2
1 + 4 - 4x + x^2 = x^2
4x = 5
x = 5/4
so our rhombus has sides of length 5/4, and since the area of any parallelogram (rhombus or not) is base*height, and since the height of our rhombus is the short side of a rectangle, the area is (5/4)(1) = 5/4.
I suspect there's a much faster way to do the problem that I'm not seeing at a glance.
19 Oct 2020, 10:56
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Quote:
Step 1: Understanding the question
As the two identical rectangles overlap, AB = BF = DC = DE = 1 and AG = HF = HC = EG = x
(refer the attachment)
Step 2: Calculation
Applying Pythagoras th. in triangle GAB
\(x^2 + 1 = (2-x)^2\)
Solving, x = \(\frac{3}{4}\)
Area of triangle BFH = \(\frac{1}{2}\) * 1*\(\frac{3}{4}\) = \(\frac{3}{8}\)
Area of the overlap region = Area of the rectangle EBFD - Area of the two green triangles = 2 - 2*\(\frac{3}{8}\) = \(\frac{5}{4}\\
\)
B is correct
Attachments
Untitled (1).png [ 24.82 KiB | Viewed 6317 times ]
