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# What is the area of the region enclosed by the figure above?

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Joined: 02 Sep 2009
Posts: 49496
What is the area of the region enclosed by the figure above?  [#permalink]

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04 Oct 2017, 00:30
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Difficulty:

15% (low)

Question Stats:

95% (01:00) correct 5% (01:06) wrong based on 25 sessions

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What is the area of the region enclosed by the figure above?

(A) 66
(B) 68
(C) 72
(D) 76
(E) 80

Attachment:

2017-10-04_1121.png [ 3.44 KiB | Viewed 561 times ]

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What is the area of the region enclosed by the figure above?  [#permalink]

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04 Oct 2017, 10:28
Bunuel wrote:

What is the area of the region enclosed by the figure above?

(A) 66
(B) 68
(C) 72
(D) 76
(E) 80

Attachment:
The attachment 2017-10-04_1121.png is no longer available

Attachment:

2017-10-04_1121ed.png [ 9.54 KiB | Viewed 392 times ]

Abbreviated version without explanation

The enclosed region consists of a square and an isosceles right triangle.

The square's area is 8*8 = 64
The triangle's area is (4*4)/2 = 8

64 + 8 = 72

Version with a bit of explanation

The enclosed region's area is the sum of the areas of a square and an isosceles right triangle.

From the right angle marks on the diagram:

1) AB and EF are perpendicular to AF

2) Hence AB and EF are parallel
(two lines that are perpendicular to a third line are parallel to each other)

Draw a line, CG.

Because 4 is the bisection point for AF, BE || AF, and CG is also parallel to AB and EF

From the properties of parallel lines cut by transversals, where BE is a transversal:

1) Angle A = Angle BCG = Angle BEF (three light green shaded areas marked as 90 degrees)
--Angle A = Angle BCG because they are alternate interior angles
--Angle BCG = Angle BED because they are corresponding angles
--Angle A = Angle BED by transitive property a = b = c , a = c

2) ABEF is a square

3) ∆ CDE is a right isosceles triangle
--Angle CED = 90 because it lies on a straight line with another right angle
--angle D is given as 45 degrees
--the third angle must be 45 degrees

4) Perpendicular bisector CG means side CE of the triangle is 4. Legs opposite equal angles are equal. So DE also = 4

Area of square = $$s^2 = 8^2 = 64$$
Area of triangle = $$\frac{s^2}{2} = \frac{4^2}{2} = 8$$

Total area of enclosed region: $$64 + 8 = 72$$

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What is the area of the region enclosed by the figure above? &nbs [#permalink] 04 Oct 2017, 10:28
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