gmat prep geometry question

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ifthyder: Master | Next Rank: 500 Posts; Posts: 118; Joined: Fri Apr 18, 2008 4:58 am; Thanked: 1 times

gmat prep geometry question

by ifthyder » Tue Sep 23, 2008 1:52 am

need explaination

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Source: Beat The GMAT — Problem Solving | Tue Sep 23, 2008 1:52 am

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 am; Thanked: 2256 times; Followed by:1535 members; GMAT Score:800

by lunarpower » Tue Sep 23, 2008 2:03 am

RIGHT OFF THE BAT, MARK SIDES BD AND BC AS ISOSCELES.

TAKEAWAY #1: if they show a triangle containing equal angles, then it's the resultant equal sides that are important. if they show you a triangle containing equal sides, then it's the resultant equal angles that are important.

first, you should be able to eliminate choices (b) and (d) fairly quickly, because it should be fairly clear that statement 2 is insufficient. this is the case because, if only statement 2 is true, then you have NO lengths to work with at all; the sides of the figure could be numbers in the millions, tiny decimals, or anything in between. this situation is clearly insufficient for determining the length of one of the sides.

TAKEAWAY #2: you can't determine lengths/sizes unless you know the length/size of at least one thing in the picture.

now examine statement one.

here are two different approaches:

1. ALGEBRAIC APPROACH: (this is the preferred approach - this algebra is straightforward, albeit a bit tricky, but none of it requires any creative thinking at all)
just take advantage of the fact that there are 180 degrees in a triangle, and also 180 degrees in a straight line.

TAKEAWAY #3: almost every problem mixing variables and angles comes down to exactly the same two things: 180° in a triangle, and 180° in a straight line.

look at angle ADB. since the angle adjacent to it is 2x°, and the two angles together make a straight line, angle ADB must be (180 - 2x)°.

now look at angle ABD. this is a bit trickier, but you use the fact that a triangle has 180 degrees. therefore, add the other two angles together, and then subtract that total from 180°.
ABD = 180° - (x° + (180 - 2x)°)
= 180° - (180° - x°)
= 180° - 180° + x°
= x°

TAKEAWAY #4: don't forget to distribute subtraction when you have quantities in parentheses.

therefore, triangle ABD is isosceles.
therefore, AD = BD. but we already know BD = BC, so AD = BD = BC = 6.
sufficient.

approach #2 follows.[/list]

Ron has been teaching various standardized tests for 20 years.

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 am; Thanked: 2256 times; Followed by:1535 members; GMAT Score:800

by lunarpower » Tue Sep 23, 2008 2:08 am

lunarpower wrote:RIGHT OFF THE BAT, MARK SIDES BD AND BC AS ISOSCELES.

TAKEAWAY #1: if they show a triangle containing equal angles, then it's the resultant equal sides that are important. if they show you a triangle containing equal sides, then it's the resultant equal angles that are important.

first, you should be able to eliminate choices (b) and (d) fairly quickly, because it should be fairly clear that statement 2 is insufficient. this is the case because, if only statement 2 is true, then you have NO lengths to work with at all; the sides of the figure could be numbers in the millions, tiny decimals, or anything in between. this situation is clearly insufficient for determining the length of one of the sides.

TAKEAWAY #2: you can't determine lengths/sizes unless you know the length/size of at least one thing in the picture.

now examine statement one.

here are two different approaches:

1. ALGEBRAIC APPROACH: (this is the preferred approach - this algebra is straightforward, albeit a bit tricky, but none of it requires any creative thinking at all)
just take advantage of the fact that there are 180 degrees in a triangle, and also 180 degrees in a straight line.

TAKEAWAY #3: almost every problem mixing variables and angles comes down to exactly the same two things: 180° in a triangle, and 180° in a straight line.

look at angle ADB. since the angle adjacent to it is 2x°, and the two angles together make a straight line, angle ADB must be (180 - 2x)°.

now look at angle ABD. this is a bit trickier, but you use the fact that a triangle has 180 degrees. therefore, add the other two angles together, and then subtract that total from 180°.
ABD = 180° - (x° + (180 - 2x)°)
= 180° - (180° - x°)
= 180° - 180° + x°
= x°

TAKEAWAY #4: don't forget to distribute subtraction when you have quantities in parentheses.

therefore, triangle ABD is isosceles.
therefore, AD = BD. but we already know BD = BC, so AD = BD = BC = 6.
sufficient.

approach #2 follows.[/list]

2. PLUG IN NUMBERS:
just plug in two wildly different values for x, and see what happens. if you get the same result twice, you can be fairly confident that the same result will hold all the time.

TAKEAWAY #5: geometry diagrams don't change abruptly; they either change smoothly when you change the value of a variable, or they don't change at all. so, if a quantity remains the same when two wildly different values of an angle are plugged in, then you can be pretty sure that it stays the same no matter what.

let's try x = 1°
then angle BDC = 2°
so angle ADB = 178°
there are 180° in triangle ADB, so angle ABD must be 180 - (178 + 1) = 1°
omg, triangle ABD is isosceles
therefore AD = BD (= BC)

let's try x = 40° (which is rather different from 1°)
then angle BDC = 80°
so angle ADB = 100°
there are 180° in triangle ADB, so angle ABD must be 180 - (100 + 40) = 40°
omg, triangle ABD is isosceles
therefore AD = BD (= BC)

omg wow, it was isosceles both times
i think it's isosceles every time
so bc must be 6 every time
sufficient

good stuff

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pre-gmat: Master | Next Rank: 500 Posts; Posts: 105; Joined: Tue Jul 15, 2008 7:03 pm; Thanked: 5 times

by pre-gmat » Tue Sep 23, 2008 3:22 pm

I had another approach to this.

angle BDC is exterior angle for Angle DAB and Angle ABD

So since Angle BDC =2x Angle ABD is x.

this means side AD and Side BD are same (isosclees property)
Given BD=BC so AD=BD

Therefore A is correct answer.

B is not sufficient.

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 am; Thanked: 2256 times; Followed by:1535 members; GMAT Score:800

by lunarpower » Tue Sep 23, 2008 10:58 pm

pre-gmat wrote:I had another approach to this.

angle BDC is exterior angle for Angle DAB and Angle ABD

So since Angle BDC =2x Angle ABD is x.

this means side AD and Side BD are same (isosclees property)
Given BD=BC so AD=BD

Therefore A is correct answer.

B is not sufficient.

you can totally use the exterior angle approach, yes.

but the exterior angle approach doesn't discount the value of a GENERAL opener, such as the algebraic approach above. (i.e., the exterior angle approach is great when it's applicable, but it's applicable a lot less of the time than is the general approach outlined above.)

good stuff.

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pre-gmat: Master | Next Rank: 500 Posts; Posts: 105; Joined: Tue Jul 15, 2008 7:03 pm; Thanked: 5 times

by pre-gmat » Wed Sep 24, 2008 10:03 am

Thanks.

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