triangle ABC

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netigen: Legendary Member; Posts: 631; Joined: Mon Feb 18, 2008 11:57 pmFollowed by: 3 members; Thanked: 29 times

by netigen » Sat May 24, 2008 1:47 pm

In triangle ABC, AB has a length of 10 and D is the midpoint of AB. What is the length of line segment DC?

(1) Angle C= 90
(2) Angle B= 45

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Source: Beat The GMAT — Data Sufficiency | Sat May 24, 2008 1:47 pm

chidcguy: Master | Next Rank: 500 Posts; Posts: 438; Joined: Mon Feb 12, 2007 9:44 am; Thanked: 26 times

by chidcguy » Sat May 24, 2008 2:41 pm

OA C?

I will explain if my ans is correct. If you can post the answer with the spoiler, that would be great.

Thanks

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netigen: Legendary Member; Posts: 631; Joined: Mon Feb 18, 2008 11:57 pmFollowed by: 3 members; Thanked: 29 times

by netigen » Sat May 24, 2008 2:50 pm

OA is A

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chidcguy: Master | Next Rank: 500 Posts; Posts: 438; Joined: Mon Feb 12, 2007 9:44 am; Thanked: 26 times

by chidcguy » Sat May 24, 2008 4:48 pm

I was running and was thinking about the Q again and I realized that I blew it. Even though I selected C, I was doubtful. Arghhh

Here is what I think

Why A is the answer. We know that angle C =90 and that means AB is the hypotenuse. Also we know that D is the mid point of AB, Now the question is asking about the length of CD. Isn't CD the perpendicular bisector of AB? I am not sure. Just because a line segment divides another line segment into two halves does not make it perpendicular bisector. I dont think so. Can some one tell me this?

see this https://www.mathwarehouse.com/geometry/s ... heorem.php.

So that means AD/AC=BD/BC We know that AD=BD and hence AC=BC. we already know that C is 90, that means we have a right angled isosceles triangle ABC. Given that angles BCD and ACD are equal, they both have to be 45 forming another 2 45,45,90 triangles

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netigen: Legendary Member; Posts: 631; Joined: Mon Feb 18, 2008 11:57 pmFollowed by: 3 members; Thanked: 29 times

by netigen » Sat May 24, 2008 5:26 pm

CD may or may not be the perpendicular bisector.

Also, there is a formula which says that the median of the hypotenuse will be equidistant from the three vertices. You can reach the answer by similar triangles.

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chidcguy: Master | Next Rank: 500 Posts; Posts: 438; Joined: Mon Feb 12, 2007 9:44 am; Thanked: 26 times

by chidcguy » Sat May 24, 2008 5:58 pm

Good to know that, In a right triangle, the mid point on the hypotenuse is equidistant from the three vertices.

I don't think we can say that CD is the perpendicular bisector. If CD is the perpendicular bisector that angle is 90 and we are given in other case with B=45, which means that the triangle CDB is right angled isosceles triangle.

As the answer is not in sync with the perpendicular bisector thought, CD can't be the perpendicular bisector.

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kausis: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Mon May 26, 2008 6:20 am; Thanked: 1 times

by kausis » Tue May 27, 2008 5:31 am

Since angle subtended by a diameter within a semicircle is always that of 90 degrees, if we assume that AB is the diameter of a circle and it extends and angle of 90 degrees on any point of the circumference, then, D being the centre of the circle, AD = BD = CD = radius = 5.

At the same time, nothing much can be derived from the measure of angle B since the point C could be at any point on the ray defining an angle of 45 degrees extending from B.

Hence the answer would be [A]

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netigen: Legendary Member; Posts: 631; Joined: Mon Feb 18, 2008 11:57 pmFollowed by: 3 members; Thanked: 29 times

by netigen » Tue May 27, 2008 11:41 am

Dude you are assuming that this triangle is inscribed in a circle with diameter as AB. This will not hold true until and unless we know that DC = AD = BD or we know for sure from the question that triangle is inscribed in a circle with dia = AB

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