Geometry question #20 from Geometry Question Bank

[sbrara]

If you look on the Geometry question bank (not HW bank) #20:

What is the length of segment BC?

(1) Angle ABC is 90 degrees.

(2) The area of the triangle is 30.



My answer was D, both sufficient, though the answer is A according to MGMAT. I understand the (1) portion, but looking at (2) independently:

We are given the height (60/13) because the area is 30. Can't we draw a line down from angle ABC perpendicular to AC, which gives us a 90 degree triangle. from there we have two sides, 5 and 60/13. can't we use the Pythagorean Theorem to determine what A to the intersection is. From that we can determine what the intersection to C is, which is 13 - (A to intersection distance). Now we have 2 angles again for a right triangle, from that we can use the Pythagorean theorem to determine the hypotenuse of Triangle ABC.

Please explain how we cannot do that??

[esledge]

We are given the height (60/13) because the area is 30. Can't we draw a line down from angle ABC perpendicular to AC, which gives us a 90 degree triangle. from there we have two sides, 5 and 60/13. can't we use the Pythagorean Theorem to determine what A to the intersection is. From that we can determine what the intersection to C is, which is 13 - (A to intersection distance). Now we have 2 angles again for a right triangle, from that we can use the Pythagorean theorem to determine the hypotenuse of Triangle ABC.

Please explain how we cannot do that??

I know this one well, because I answered D, too, when I first did this under timed conditions. It's not that you can't do that--that's exactly the right math. It's just that you've overlooked a completely different possible scenario.

When you say that the height must be 60/13, that uses side AC with length 13 as the base. But you are making an assumption: that the height (drawn perpendicular from the base, to the opposite corner B) is drawn within the triangle. The height can also be measured outside the triangle: imagine the apex at point B "leaning" to the left of base AC, rather than hovering somewhere above it. In such a case, the height is drawn perpendicular to an extension of AC up to B, outside the triangle. (There's a picture of both scenarios in the explanation.)

For either scenario, you'd use Pythagorean theorem as you describe.

For the height-inside-the-triangle scenario, you'd add a point D on AC such that BD is the height (i.e. BD is perpendicular to AC). Using BD = 60/13 (given by the area), by Pythagorean theorem, AD = 25/13, DC = 144/13, so BC = 12.

For the height-outside-the-triangle scenario, you'd add a point D on the extension of AC, to the left of A. By the same Pythagorean logic, AD would be 25/13. The next right triangle would be BCD, with the right angle at D. By Pythagorean theorem again: BC = squareroot[(60/13)^2 +(13+25/13)^2] = about 15.6.

[jssaggu.tico]

What about Heron's formula?
http://en.wikipedia.org/wiki/Heron%27s_formula

We know that Area = (s(s-a)(s-b)(s-c))^1/2 = 30..................1
where s= (a+b+c)/2
a= 13
b= 5
c = ?

Hence, we will get c by solving quadratic eqn 1.
That will make second statement sufficient.

Please suggest!

[tim]

Um, did you solve the quadratic equation you got from Heron’s formula? Emily told us there were two different possible triangles, and it sounds like we’re setting ourselves up to get the exact same result if we solve a quadratic and get two values..

[jp.jprasanna]

Um, did you solve the quadratic equation you got from Heron’s formula? Emily told us there were two different possible triangles, and it sounds like we’re setting ourselves up to get the exact same result if we solve a quadratic and get two values..

Dear instructors - Could you please share the the two diagrams please. I really don't see how!

Although I marked A, my reasonings were completely diff

5 and 13 cant be the base and height to give us an area 30

So one has to be the base other (unknown value must be the height) or vise=versa so Area = 1/2 * 5 * x = 30 give me a different value of x and 1/2 * 13 * x = 30 gives me a diff value of x hence Option B NS.

Am i OK?

Cheers
Jp

[jp.jprasanna]

Um, did you solve the quadratic equation you got from Heron’s formula? Emily told us there were two different possible triangles, and it sounds like we’re setting ourselves up to get the exact same result if we solve a quadratic and get two values..

Dear instructors - Could you please share the the two diagrams please. I really don't see how!

Although I marked A, my reasonings were completely diff

5 and 13 cant be the base and height to give us an area 30

So one has to be the base other (unknown value must be the height) or vise=versa so Area = 1/2 * 5 * x = 30 give me a different value of x and 1/2 * 13 * x = 30 gives me a diff value of x hence Option B NS.

Am i OK?

Cheers
Jp

Ok i think my reasoning is wrong... For any triangle sum of 2 sides my be greater than the 3 side

so 1/2 *13* X =30 give me x =4.6 approx say 5

5 + 5 is not greater than 13....!!!!!!!! :-(((

Please help me!

Cheers
Jp

[tim]

JP, none of what you're saying here makes much sense. In other words, as near as I can tell it's all wrong. In your first example, it appears you are calling two completely different things both x, which is a problem in itself, but neither of these is actually what you're trying to solve for. In your second attempt, it looks like you are again trying to solve for some x that has nothing to do with the answer to the problem..

Have you read the explanation for the problem? That's really the best way to approach things. Let us know if you have any questions after reading the explanation..