If side AB measures 3 and side BC measures 4, what is the length of li

Isn't it true that on official problem solving questions, if the figure doesn't say not drawn to scale, then it is?

We can measure the shape and eliminate A D and E. B is also likely too short.

It has to be C. Done in about 30 seconds.

I'd also like confirmation about that. Is that true?

OFFICIAL GUIDE:

Problem Solving
Figures: All figures accompanying problem solving questions are intended to provide information useful in solving the problems. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

Data Sufficiency:
Figures:
• Figures conform to the information given in the question, but will not necessarily conform to the additional information given in statements (1) and (2).
• Lines shown as straight are straight, and lines that appear jagged are also straight.
• The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero.
• All figures lie in a plane unless otherwise indicated.

Hope it helps.

Thanks Bunuel. This is the crazy part: "Figures are drawn as accurately as possible. Exceptions will be clearly noted.". I think you guys can make a bigger deal out of this. On some questions, like this one, knowing that is night and day. You can literally just measure your way to the answer if the answers are spread out.

VERITAS PREP OFFICIAL SOLUTION:

While many will rush into an abyss of Pythagorean Theorem, thinking out loud can show you a calm, proactive way to do this.

“Thinking out loud…I know that it’s a right triangle so if AB = 3 and BC = 4, it’s a 3-4-5 and side AC is 5. And as much as I want side AC to be cut in half by point D I don’t think I can do that. There are three different right triangles so I could go nuts with Pythagorean Theorem but that’s a lot of work. Thinking out loud, I also know that the perimeter is 3 + 4 + 5 and the area is 1/2(base)(height) so that’s 1/2 (3)(4) = 6. But what can I do with that?

Thinking out loud…the answer choices are all divided by 5…why do they all look like that? The only 5 in the problem so far is the 5 that’s side AC. Why would I multiply or divide by that?

Thinking out loud…BD is definitely going to be smaller than 4 because there’s no way it’s longer than side BC. So it can’t be E. But what else do I know about BD? It’s perpendicular to side AC, and AC is 5 and that’s that 5 in the denominator. Thinking out loud…what if I drew the triangle so that AC was on the bottom and not on the side? Then BD would be the height of triangle ABC and AC would be the base…but wait, I already know the area is 6, so that area 1/2 (side BD)(5) has to be 6, which means that side BD has to be 12/5, answer choice C.”

The takeaway here is that almost no one sees the area relationship with side BD right away, and that’s okay. The key to working on problems like these is staying loose and filling in unknowns. You can’t simply do math on paper and follow a set of steps…you need to do some thinking out loud and talk to yourself as you solve. For each of Drake, Nicki, and Wayne the phrase “thinking out loud” is followed by a wild description of how much money they have. Follow that “thinking out loud” philosophy and you’ll be on a similar pace with the help of an elite MBA.

Ha! I never would have guessed the official explanation would have gone that route. I would've expected a math heavy approach.

I still have to say, I don't think even this explanation makes a big enough deal out of the 'drawn to scale' rules. I bet if you did a poll, you'd discover most other students don't know that official 'drawn to scale' rule for PS that you shared.

why doesn't the orthocentre of right angle triangle work here i.e perpendicular to hypotenuse from B should bisect the hypotenuse and that is also equal to half the hypotenuse