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In the figure above, side lengths AB, BD, and DC are all equ

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sandy
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In the figure above, side lengths AB, BD, and DC are all equ [#permalink] New post   26 Sep 2018, 15:31
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In the figure above, side lengths AB, BD, and DC are all equal.

Quantity A
Quantity B
x
y


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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B
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chetan2u
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Re: In the figure above, side lengths AB, BD, and DC are all equ [#permalink] New post   22 Oct 2018, 08:11
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tigran wrote:
Can i know how the right answer is B?



Look at the attached sketch..
In ∆BCD, BD=DC, so an isosceles triangle..y=180-(2*65)=180-130=50
In ∆ABD, AB=BD, so aga8n an isosceles triangle..
Angle ABD=180-65=115.. so x+x+115=180...x=32.5

So y>x or B>A

B
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Re: In the figure above, side lengths AB, BD, and DC are all equ [#permalink] New post   24 Oct 2018, 03:02
sandy wrote:
In the figure above, side lengths AB, BD, and DC are all equal.

Quantity A
Quantity B
x
y


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


In the \(\triangle BDC\)

\(BD = DC\), so it is an isosceles \(\triangle\)

i.e. \(\angle DBC = \angle DCB = 65°\)

and \(\angle y\)= \(180° - (2 * 65°) = 50°\)



Now in \(\triangle ADB\)

\(\angle DAB = \angle BDC = x°\) ( since AB = BD)

therefore \(\angle DBC = \angle DAB + \angle BDC\) (exterior angle is sum of two opposite interior angle)

or\(65° = 2x°\)

or \(x° = 32.5°\)

therefore QTY A < QTY B
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Re: In the figure above, side lengths AB, BD, and DC are all equ [#permalink] New post   11 Jan 2019, 21:25
x + angl ADB=65

y=50

Definitely B is greater
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B
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Re: In the figure above, side lengths AB, BD, and DC are all equ [#permalink] New post   18 Dec 2022, 18:54
pranab223 wrote:
sandy wrote:
In the figure above, side lengths AB, BD, and DC are all equal.

Quantity A
Quantity B
x
y


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


In the \(\triangle BDC\)

\(BD = DC\), so it is an isosceles \(\triangle\)

i.e. \(\angle DBC = \angle DCB = 65°\)

How do we know that ∠DBC should equal ∠DCB?

and \(\angle y\)= \(180° - (2 * 65°) = 50°\)



Now in \(\triangle ADB\)

\(\angle DAB = \angle BDC = x°\) ( since AB = BD)

therefore \(\angle DBC = \angle DAB + \angle BDC\) (exterior angle is sum of two opposite interior angle)

or\(65° = 2x°\)

or \(x° = 32.5°\)

therefore QTY A < QTY B
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gmatclubot
Re: In the figure above, side lengths AB, BD, and DC are all equ [#permalink]
18 Dec 2022, 18:54
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