Question 20 - Geometry Practice questions

[jyothishvarma]

Given a triangle with sides 5 and 13.
What is the length of segment BC?
(1) Angle ABC is 90 degrees.
(2) The area of the triangle is 30.

1 is sufficient (for obvious reasons).
Why is 2 not sufficient? If area and 2 sides are given for a triangle, can we not find the 3rd side using the following formula (area of a scalene triangle - isn't it applicable to all kinds of triangle) ?
s = (a + b + c) / 2
area = sqrt(s * (s - a) * (s - b) * (s - c))

Thanks in advance!

[saurav.raaj]

Should be!

[JonathanSchneider]

You've quoted Heron's formula. See here: http://mathopenref.com/heronsformula.html

Be careful with this formula, though: you only want to use it when you know all three sides from the outset. Don't use it to find the sides. Why not? Because you're dealing with even exponents, which hide multiple solutions.

An easier approach to this problem:

If the triangle is in fact a right triangle, then we would have a 5-12-13 triangle. You know that the area is 30. You also know that area = .5bh. Rather than calling the base and height 5 and 12, you could call the base 13, and calculate the height to be 4.6~. Note, though, that you can make two such triangles, one where the length of five tilts "in," above the center of the base of 13, the other where the length of five tilts "out," away from the base.

[malikrulzz]

The answer should be C. For A either give the exact sides whter AB= 5 or whta

[esledge]

Yes, the answer should be (C). As Jonathan says above about (2) alone, there are several ways to draw the triangle (i.e. with the "base" being any of the 3 sides, really, and playing with the various angles or the "tilt" of the triangle.) You need (1) with (2) to clarify which triangle we have (i.e. 5-12-13).

Furthermore, it's a good test taking strategy to brainstorm a bit on paper for geometry or other picture-type DS problems. When looking at (2) alone, ask yourself "What if angle ABC isn't 90 degrees? Is that possible, and how?" By doing so, you prevent yourself from erroneously considering (1) with (2) unintentionally or making assumptions for the sake of convenience. Assumptions are common on triangle problems--people assume right triangles, simply because the many formulas that apply make them more fun!

[michael.a.gabe]

Shouldn't this answer actually be D?

From (1), we know that is is 90 degrees and therefore a 5-12-13 triangle. This is actually what the answer says it should be.

I believe (2) is also sufficient. The base can be drawn from angle B to line AC, which would split the triangle into 2 90 degree triangles. Knowing the area of the large triangle, this measure of the base can be calculated as (30*2)/13. With this information, we can calculate the line segments of the 2 split triangles.

Am I missing something? Thanks

[RonPurewal]

Shouldn't this answer actually be D?

From (1), we know that is is 90 degrees and therefore a 5-12-13 triangle. This is actually what the answer says it should be.

does the problem statement say which sides of the triangle (i.e., NAMES) are lengths 5 and 13?
if not, then we don't know which side is the hypotenuse (AC), and so we can't assume that 13 is the hypotenuse. we could have a triangle whose legs are 5 and 13, and so whose hypotenuse would be √(5^2 + 13^2), or √194.

if the two given sides are AB and AC (i.e., BC, which the problem is asking for, isn't one of them), and the 90 degree angle is at ABC, then we must have AC = hypotenuse = 13. (it's impossible for 5 to be the hypotenuse if 13 is a leg.) in this case, yes, you'll have a 5-12-13 triangle.

--

regarding statement (2):

As Jonathan says above about (2) alone, there are several ways to draw the triangle (i.e. with the "base" being any of the 3 sides, really, and playing with the various angles or the "tilt" of the triangle.)

well, there aren't "several" ways. there are only 2.
viz.:
one way is the aforementioned 5-12-13 triangle.
take this triangle and lay it down on the "13" side.
now erase the 12.
take the 5 and reflect it through an imaginary vertical line, while leaving the 13 alone. (this will create an obtuse angle; in fact, the angle will be exactly 180 minus the angle that used to be there.)
now re-connect the two corners of the triangle (which will need a segment longer than the original "12").
you have just created a new triangle whose base is still 13, and which still has the same height as the original triangle, and which still has a height of 5.
there you go.

by the way, there are only 2 ways to make such a triangle. if you rotate the triangle and/or use other sides as "base" (as suggested above), you will just create more triangles that are congruent to one or the other of the two already found.

I believe (2) is also sufficient. The base can be drawn from angle B to line AC, which would split the triangle into 2 90 degree triangles. Knowing the area of the large triangle, this measure of the base can be calculated as (30*2)/13. With this information, we can calculate the line segments of the 2 split triangles.

Am I missing something? Thanks[/quote]