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Difficult Geometry

Expert replies
Source: — Data Sufficiency |

by prateek_guy2004 » Thu Sep 01, 2011 11:31 am
Hey akhil

This is a Gmat prep ques....Are you sure that the answer is C
Don't look for the incorrect things that you have done rather look for remedies....

https://www.beatthegmat.com/motivation-t90253.html
Join the discussion

by akhilsuhag » Thu Sep 01, 2011 11:35 am
Yeah the answer is C
I just checked!!
Please press "thanks" if you think my post has helped you.. Cheers!!
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by jbivins » Thu Sep 01, 2011 11:46 am
If we knew the length of the angles on either side of x we could determine the value of x since all three angles would sum to 180. I'm having a hard time trying to figure out how to explain this....with out a picture. Say that both of the angles next to x can be represented by Z and Y

1. Z+X+Y=180 - straight line so the angles sum to 180

All the angels of the triangle sum to 180 as well and we know one of the angels is 90 since it's a right triangle. There for the sum of the largest triangles remaining angles is 90.
I will represent those angles as A and B.
2. A+B=90

With both pieces of information we now know that we have two isosceles triangles. on either side of x so know the properties of isosceles triangles we get the following equations
3. 2Z+A=180 z=(180-A)/2
4. 2Y+B=180 y=(180-B)/2

combining equations 3 4 and 1 we get

x+((180-A)/2)+((180-B)/2)=180

some basic algebra and we get

2x+360-(A+B)=360 from equation 2 we know A+B equals 90.

At this point we know we can solve for x so no point in going on
I believe the answer is c with 45 being the explicit answer
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by GMATGuruNY » Thu Sep 01, 2011 12:46 pm
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by prateek_guy2004 » Thu Sep 01, 2011 1:16 pm
Mitch Hunt excellent approach ...even i thought at first glance about 180 degree straight line but then after spending sometime on it i came to a conclusion that its E.

But yes great way to solve this...........
Don't look for the incorrect things that you have done rather look for remedies....

https://www.beatthegmat.com/motivation-t90253.html
Join the discussion