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In the figure above, SQRE is a square, AB = AC, and AS = AQ.

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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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New post 21 Jul 2016, 06:55
Bunuel wrote:
guireif wrote:
Hi guys,

I know this is an old question, but I'm not sure about one thing.

Should I assume by looking at the figure that CE=RB? Because in the first statement it just say that RE is less than twice the length of BR, but how do I know that the triangle is "centered" with the square, by not looking the second statement?

thanks


Since AB = AC (triangle ABC is isosceles), and AS = AQ, then CE must be equal to RB.


Now I got it. Haven't noticed that AS=AQ. Perfect. Thank you so much
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In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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New post 22 Jul 2016, 07:35
AccipiterQ wrote:
Attachment:
The attachment 2967-1.gif is no longer available
In the figure above, SQRE is a square, AB = AC, and AS = AQ. What is the difference between the perimeter of triangle ABC and the perimeter of square SQRE?

(1) The length of RE is 4 times the length of BR.

(2) The area of triangle ABC is 75% of the area of square SQRE.

(1)4BR=RE
2(BR+CE)=RE (since AC=AB && AS=AQ)
CB=RE/2+RE=3/2RE
Take triangle ACD
AC^2(3/4RE)^2+RE^2 (CD=1/2CB &&CB=3/2RE)
AC^2=25/16RE^2
AC=5/4RE
PERIMETER OF ABC=5/4RE+5/4RE+3/2RE=4RE
PERIMETER OF SQ.=4RE
Perimeters are equal.
suff.
(2) 1/2 CB*AD=3/4 RE^2 (75%=3/4)
as AD=RE
CB=3/2RE
same info as (1)

suff.

Ans D
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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ. &nbs [#permalink] 24 Oct 2018, 07:10

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