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# In the trapezoid above, BC is parallel to AD, angle A is 45 degrees

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Manager
Joined: 26 Dec 2015
Posts: 246
Location: United States (CA)
Concentration: Finance, Strategy
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In the trapezoid above, BC is parallel to AD, angle A is 45 degrees  [#permalink]

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20 Feb 2018, 20:45
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Difficulty:

65% (hard)

Question Stats:

54% (02:13) correct 46% (02:39) wrong based on 48 sessions

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In the trapezoid above, BC is parallel to AD, angle A is 45 degrees and the length of line segment CD is 20. If the height of the trapezoid is 10, then which of the following is the value of x+y?

A) 180

B) 255

C) 270

D) 285

E) 300

Attachments

E GEO.png [ 14.27 KiB | Viewed 625 times ]

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Joined: 07 Jan 2016
Posts: 1090
Location: India
GMAT 1: 710 Q49 V36
Re: In the trapezoid above, BC is parallel to AD, angle A is 45 degrees  [#permalink]

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20 Feb 2018, 21:58
1
LakerFan24 wrote:
In the trapezoid above, BC is parallel to AD, angle A is 45 degrees and the length of line segment CD is 20. If the height of the trapezoid is 10, then which of the following is the value of x+y?

A) 180

B) 255

C) 270

D) 285

E) 300

given cd = 20

hieght is 10 , we have a right angle triangle.. hypotunese 20 side= 10

now we know side1^2 + side2^2 = hypotunese^2 ( pythagoras therom)

10^2 + x^2 = 20^2 i.e x^2 = 400 -100 = 300

X= rt300 = 10 rt 3

now sides are 10, 10rt3 and 20 ratio 1:rt3:2 .. the angles are 90-60-30

the smaller angle has the smaller side i.e angle D = 30

now x+y+ angle a + angle b = 360 ( angle in a quadilateral)

x+y = 360 - a - b = 360 - 45 -30 =285

(D) imo
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Joined: 31 Jul 2017
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In the trapezoid above, BC is parallel to AD, angle A is 45 degrees  [#permalink]

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20 Feb 2018, 22:15
1
LakerFan24 wrote:
In the trapezoid above, BC is parallel to AD, angle A is 45 degrees and the length of line segment CD is 20. If the height of the trapezoid is 10, then which of the following is the value of x+y?

A) 180

B) 255

C) 270

D) 285

E) 300

We have -

$$X + A = 180$$. Given, $$A = 45 --> X = 135$$
$$Y + D = 180..... Y = 30$$ as Sin Y = $$\frac{10}{20}. Hence Y = 150$$
Hence, $$X + Y = 150 + 135 = 285$$
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In the trapezoid above, BC is parallel to AD, angle A is 45 degrees  [#permalink]

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22 Feb 2018, 21:02
LakerFan24 wrote:
In the trapezoid above, BC is parallel to AD, angle A is 45 degrees and the length of line segment CD is 20. If the height of the trapezoid is 10, then which of the following is the value of x+y?

A) 180

B) 255

C) 270

D) 285

E) 300

Attachment:

E GEO ed.png [ 16.23 KiB | Viewed 502 times ]

Find $$x$$:

A base angle and upper angle on the same side of a trapezoid sum to 180°
(Parallel lines cut by a transversal create supplementary adjacent interior angles.)
45 + x = 180
x = 135

Find y:
Drop an altitude from C to base AD at X.

∆ CDX is a right triangle
Trapezoid height = 10
For right ∆ CDX, Leg CX = 10
Hypotenuse CD = 20
Let Leg DX = $$L$$

$$L^2 + 10^2 = 20^2$$
$$L^2 = 300$$
$$\sqrt{L^2} =\sqrt{100*3}$$
$$L = DX= 10\sqrt{3}$$

If a right triangle has sides in ratio $$a : a\sqrt{3} : 2a$$,
angles opposite those sides, respectively, have measures 30°-60°-90°

$$10$$ corresponds with $$a$$
Leg CX must be opposite a 30° angle
$$10\sqrt{3}$$ corresponds with $$a\sqrt{3}$$.
Leg DX must be opposite a 60° angle

At vertex C, two angles form ∠$$y$$
∠XCD = 60
∠BCX = 90
$$y$$ = (90 + 60) = 150

$$x + y = (135 + 150) = 285$$

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In the trapezoid above, BC is parallel to AD, angle A is 45 degrees   [#permalink] 22 Feb 2018, 21:02
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