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In the figure above, side lengths AB, BD, and DC are all equ
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26 Sep 2018, 15:31
26 Sep 2018, 15:31
Expert Reply
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Question Stats:
50% (01:49) correct
49% (01:31) wrong based on 67 sessions
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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.
Re: In the figure above, side lengths AB, BD, and DC are all equ
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22 Oct 2018, 08:11
22 Oct 2018, 08:11
3
Expert Reply
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
Attachments
PicsArt_10-22-09.37.09.jpg [ 9.21 KiB | Viewed 4320 times ]
Re: In the figure above, side lengths AB, BD, and DC are all equ
[#permalink]
24 Oct 2018, 03:02
24 Oct 2018, 03:02
sandy wrote:
In the figure above, side lengths AB, BD, and DC are all equal.
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.
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
Attachments
Capture.JPG [ 12.84 KiB | Viewed 4233 times ]
