MGMAT Question Back Geometry Q 22

[Saurabh Malpani]

What is the length of segment BC?

(1) Angle ABC is 90 degrees.

(2) The area of the triangle is 30.





I don't understand how B is not Correct.

My Apporach: Let's Drop a Perpendicular from B on AC and let's call it BD

Now the Area of the Triange is Given = 30. So, .5*AC*BD,

From this BD is Given = 12
I can calculate AC= 5.

Now I using Pythagoras in Triangle BDC. I Can Calculate BC.

So what's wrng with my solution? Where am I wrong?

[shaji]

Statement 2 is sufficient as it suggests that BC is 12.

What is the length of segment BC?

(1) Angle ABC is 90 degrees.

(2) The area of the triangle is 30.





I don't understand how B is not Correct.

My Apporach: Let's Drop a Perpendicular from B on AC and let's call it BD

Now the Area of the Triange is Given = 30. So, .5*AC*BD,

From this BD is Given = 12
I can calculate AC= 5.

Now I using Pythagoras in Triangle BDC. I Can Calculate BC.

So what's wrng with my solution? Where am I wrong?

[RonPurewal]

No sir, statement 2 isn't sufficient; there are two ways to draw the triangle so that its area is 30. To produce the other triangle, imagine drawing a vertical axis through A, and then reflecting AB over that axis to produce AB' (a new segment, also of length 5, that runs northwest/southeast-ish). Don't reflect C. Then the new triangle AB'C still has base 13, still has the same height as ABC (which happens to be 60/13, if AC is chosen as the base - but you can just look at the picture's symmetry and see that they're the same), and so its area is still 30.

Statement 1 is indeed sufficient, as it enables you to use the Pythagorean theorem (notice that there's no point in actually USING the theorem; this is data sufficiency, so you're done the minute you notice that you CAN use the theorem).

If my directions aren't clear enough, post your email address and I'll send you a hand-drawn picture of the diagram.

The OA is given as A, right?

[shaji]

U R correct. Angle BAC can be obtuse as well, resulting in a larger BC.

The correct answer is A indeed.

No sir, statement 2 isn't sufficient; there are two ways to draw the triangle so that its area is 30. To produce the other triangle, imagine drawing a vertical axis through A, and then reflecting AB over that axis to produce AB' (a new segment, also of length 5, that runs northwest/southeast-ish). Don't reflect C. Then the new triangle AB'C still has base 13, still has the same height as ABC (which happens to be 60/13, if AC is chosen as the base - but you can just look at the picture's symmetry and see that they're the same), and so its area is still 30.

Statement 1 is indeed sufficient, as it enables you to use the Pythagorean theorem (notice that there's no point in actually USING the theorem; this is data sufficiency, so you're done the minute you notice that you CAN use the theorem).

If my directions aren't clear enough, post your email address and I'll send you a hand-drawn picture of the diagram.

The OA is given as A, right?

[shaji]

The diagram of the triBAC suggests that BAC is accute, If suggestions are to be ignored!!!, if and only if, I say , can Statement 2 may be deemed insufficient. On the GMAT, I presume, there is a caution that diagrams are not drawn to scale. If the question setter thought implications in daigrams are to be considered as depicted and marked D as the correct answer, then its again a 'no sir' matter.

U R correct. Angle BAC can be obtuse as well, resulting in a larger BC.

The correct answer is A indeed.

No sir, statement 2 isn't sufficient; there are two ways to draw the triangle so that its area is 30. To produce the other triangle, imagine drawing a vertical axis through A, and then reflecting AB over that axis to produce AB' (a new segment, also of length 5, that runs northwest/southeast-ish). Don't reflect C. Then the new triangle AB'C still has base 13, still has the same height as ABC (which happens to be 60/13, if AC is chosen as the base - but you can just look at the picture's symmetry and see that they're the same), and so its area is still 30.

Statement 1 is indeed sufficient, as it enables you to use the Pythagorean theorem (notice that there's no point in actually USING the theorem; this is data sufficiency, so you're done the minute you notice that you CAN use the theorem).

If my directions aren't clear enough, post your email address and I'll send you a hand-drawn picture of the diagram.

The OA is given as A, right?

[RonPurewal]

Diagrams, generally, do not have to be "to scale". Therefore, you should NOT assume that things are as they look.

This should be rather obvious in the case of Data Sufficiency problems: if diagrams were to scale, then each diagram would have one and only one possible interpretation (i.e., whatever the scale measurements would be). If that were the case, then every diagram would be sufficient to determine ... just about everything!

[bongbong]

1 is definitely sufficient. but 2 we do not know that this is a right triangle. we just know that the area is 30. we cannot tell what the base is here.

[StaceyKoprince]

Just to clarify on the diagrams thing.

On problem solving questions, diagrams are drawn to scale UNLESS otherwise noted.

On data sufficiency, you can only assume that the information given is in the proper order (eg, if you had square ABCD, that's really the order of the vertices; it's not ACBD). You cannot assume that the picture is drawn to scale, or that the picture is drawn in the only way that it can be drawn.

[dhoomektu]

Also if you actually solve for BC using Heroe's Formula you will get two values 12 and 15.6025 both of which will have area of 30; hence 2 is insufficient.

[Guest]

The only reasoning that makes sense here is of Ron's. It would be nice if you could post your construction.

The argument that the angle could be obtuse/acute/right angle is flawed as it is possible there could be only ONE value of BC that could satisfy the area of 30, so the assumption that we are not given the angle has no bearing on the problem. For eg. if the angle was assumed to be obtuse and quick calculation with the area shows that the height required is 20 (just for the sake of the argument). This height would then be physically impossible and we would have to assume that the angle is acute or right.

[RonPurewal]

The only reasoning that makes sense here is of Ron's. It would be nice if you could post your construction.

check it out:
here or here.
(the first one shows up bigger, but the second doesn't show any adverts)

i showed the construction in two different ways - one using the '5' as the base and the other using the '13' as the base. either way, you can figure out that there are two different possibilities.