Manhattan Question Bank

The OA is correct.

stmt 1 is obvious. So I will skip it.

stmt 2: The figure lulls into a trap. Unless mentioned we cannot assume
that the figure is the Right triangle. Looking at sizes (5,13) and it looks like a right angle triangle... GMAT TRAP !!!.


Without assuming the right triangle, we cant find the the 3rd side. If u use
DSS: u will get the following inequality.

13-5 < side BC < 13+5

So side BC can be as small as 8 and as big as 18. So clearly insufficient.

<in previous post I had an incorrect statement. I have that stmt as follows>

A triangle with a Pythagorean triplet ({5,12,13}, {3,4,5} etc) as the dimensions of its sides, It implies that the triangle is Right triangle.
And the converse is true.


HT helps

vittalgmat wrote:
I have a conjecture. Someone pls let me know if this is right:
A triangle with a Pythagorean triplet ({5,12,13}, {3,4,5} etc) as the dimensions of its sides, does not imply that the triangle is Right triangle.



HT helps

Hi Vittalgmat,

Converse of Pythagorean theorem is also true... That is when there is a triangle with sides that satisfy the theorem such as (5,12,13), the triangle is right angled!

thanks!

Stmt 2 statement . I have a different approach.
We know the area of the triangle.

If we draw a perpendicular from B to AC at point D, We can find the height of the because we know the area of the triangle.
(0.5) * BD* 13 = 30.

BD = 60/13

The perpendicular divides the triangle in to 2 small triangles. Lets Say BAD and BCD (both of them are right triangles). Now we have 2 triangles and we know the 2 sides for both of those triangles.

Step1
Calculate AD = SQRT(AB^2 - BD^2)

Step 2
Calculate CD = 13 - AD

Step 3
Calculate BC = SQRT(BD^2 + BC^2)

Let me know if i missed a point.

i dont think wht aroon has said is rite..i havent read this anywhere,

This was a PM transaction I had with Ian Stewart which, in summary, means
pythagorean triplets <==> Right Triangle

Yes, it's one of the things you learn in trigonometry: the angles of a triangle determine the ratio of its sides, and vice versa. So there is only one kind of triangle with, say, the sides 5-12-13, and that triangle has a 90 degree angle. In case you're interested, you can test out any combination of sides and find the angles here:

www.sizes.com/numbers/triangles.htm

In brief, if the numbers a, b and c satisfy the equation a^2 + b^2 = c^2, then a triangle with sides a, b and c is a right triangle.

All the best,
Ian

vittalgmat wrote:

Hi Ian,
I had question in the context of this link

https://www.beatthegmat.com/manhattan-qu ... tml#110470

I am wondering whether the presence of a pythagorean triplet as the dimensions of a triange imply that the triangle is right angled triangle.

In other words

Does pythagorean triplet <----> right triangle??

thanks
-Vittal

pandeyvineet24 wrote:Stmt 2 statement . I have a different approach.
We know the area of the triangle.

If we draw a perpendicular from B to AC at point D, We can find the height of the because we know the area of the triangle.
(0.5) * BD* 13 = 30.

BD = 60/13

The perpendicular divides the triangle in to 2 small triangles. Lets Say BAD and BCD (both of them are right triangles). Now we have 2 triangles and we know the 2 sides for both of those triangles.

Step1
Calculate AD = SQRT(AB^2 - BD^2)

Step 2
Calculate CD = 13 - AD

Step 3
Calculate BC = SQRT(BD^2 + BC^2)

Let me know if i missed a point.

This should work, shouldn't it? I did the same thing, just slightly different method:

30 = (1/2)(base)(height)
30=(1/2)(13)(height)
height = 60/13

Now use this on the smaller triangle created by drawing a line straight down from B:

(60/13)^2 = 5^2 + (13-x)

solve for x (I called the segment DC 'x') and you have the base of the right handle triangle created by cutting the original one in half

now it's just height^2 = x^2 = (BC)^2


Shouldn't that all be solvable, at least in theory?

Thanks for the response. I'm not sure I can say I have my head around it 100%, but I do believe you're right. So the lesson is that unless marked in the picture, assume nothing about how it's drawn and try drawing it different ways to test?

Does anyone have any general advice/insight into how I could have seen that (2) was insufficient in the first place?


Thanks