Triangle problem

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siddhartha123: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Mon Aug 01, 2011 6:15 am; Location: vijayawada; Thanked: 1 times

by siddhartha123 » Fri Aug 05, 2011 2:55 am

Hai every one ,

I have come across the question in one of practice test , here is the Q

A tringle ABC show below(not drawn to scale) ; find the value of AB , if AC=10,BC=X
i can remember only few options .

a.) 5
b.) 10
c.) 5 sqaureroot 2
d.) 5 sqaureroot 3

kindly help me out

thanks

Siddhartha.M

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Source: Beat The GMAT — Problem Solving | Fri Aug 05, 2011 2:55 am

gmatboost: Master | Next Rank: 500 Posts; Posts: 312; Joined: Tue Aug 02, 2011 3:16 pmFollowed by: 33 members; Location: New York City; Thanked: 130 times; GMAT Score:780

by gmatboost » Fri Aug 05, 2011 10:17 am

Hi Siddhartha,

You should recognize the 45 and 60 degree angles as belonging to common right triangles: 30-60-90 and 45-45-90.

It's not clear how this helps right away, but if you draw an altitude from C to AB (altitude means a line that is perpendicular to AB), you will suddenly have two common right triangles:
1. 30-60-90 on the left.
2. 45-45-90 on the right.

Let D be the point where the new line hits AB.
You can use the side ratios of those common right triangles to find AD, CD, BD and BC.

Give it a shot and let me know what you think.

If you need a hint, the side ratios are:
[spoiler]30-60-90 x : xroot(3) : 2x
45-45-90 x : x : root(2)[/spoiler]

Greg Michnikov, Founder of GMAT Boost

GMAT Boost offers 250+ challenging GMAT Math practice questions, each with a thorough video explanation, and 100+ GMAT Math video tips, each 90 seconds or less.
It's a total of 20+ hours of expert instruction for an introductory price of just $10.
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siddhartha123: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Mon Aug 01, 2011 6:15 am; Location: vijayawada; Thanked: 1 times

by siddhartha123 » Sat Aug 06, 2011 1:26 am

gmatboost wrote:Hi Siddhartha,

You should recognize the 45 and 60 degree angles as belonging to common right triangles: 30-60-90 and 45-45-90.

It's not clear how this helps right away, but if you draw an altitude from C to AB (altitude means a line that is perpendicular to AB), you will suddenly have two common right triangles:
1. 30-60-90 on the left.
2. 45-45-90 on the right.

Let D be the point where the new line hits AB.
You can use the side ratios of those common right triangles to find AD, CD, BD and BC.

Give it a shot and let me know what you think.

If you need a hint, the side ratios are:
[spoiler]30-60-90 x : xroot(3) : 2x
45-45-90 x : x : root(2)[/spoiler]

thanks that is really helpful ; it didnt strike me at time of test .

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