Lesson: Triangles - Part II

Comment on Triangles - Part II

At 7:21, if we draw an altitude to the isosceles triangle with one of the equal sides as base, would the altitude still bisect it?
gmat-admin's picture

Good question.
The answer is no. In order for the diagonal to bisect the top angle, the non-equal side must be the base.
Try it with a 45-45-90 triangle, and you'll see what I mean.

Hi Brent,

A little confused with this. Suppose if the isosceles triangle is ABC with sides AB=AC, then will the altitude drawn from angle A bisect angle BAC in two equal parts? I thought it is only a bisector segment dividing side BC in two equal parts.

Thanks
gmat-admin's picture

Yes, the altitude drawn from point A WILL bisect angle BAC in two equal parts.

Think of it this way (using the diagram you describe with point D representing the point where the altitude intersects side BC)

We get two triangles: ∆ADB and ∆ADC

First, recognize that ∠ABD = ∠ACD, since those are the equal angles in ∆ABC

Second, recognize that ∠ADB = ∠ADC = 90°, since AD is the ALTITUDE

Third, since triangles ∆ADB and ∆ADC already have two angles in common, the third angles must also be equal. That is ∠BAD = ∠CAD

Does that help?

Cheers,
Brent

Amazing. Without an inkling of doubt.
Is a bisector drawn always perpendicular to the line it bisects or this is the property applicable only in case of Isosceles and Equilateral triangle?

Thanks.
gmat-admin's picture

The bisector is perpendicular ONLY IF the triangle is either isosceles and equilateral.

Sir is there any easy solution for the question given below
https://gmatclub.com/forum/in-the-figure-above-point-o-is-the-center-of-the-circle-and-107874.html
gmat-admin's picture

Have you seen Karishma's solution: https://gmatclub.com/forum/in-the-figure-above-point-o-is-the-center-of-...

If you have any questions about her solution, let me know and we can go from there.

sir, i am having doubt in this also,

https://gmatclub.com/forum/in-triangle-abc-point-x-is-the-midpoint-of-side-ac-and-82671.html
gmat-admin's picture

One of the statements from the DS question reads as follows
"(1) x^2 + y^2≠ z^2"

Does this statement translate into the triangle being an equilateral triangle?
gmat-admin's picture

Question link: https://gmatclub.com/forum/is-the-area-of-the-triangular-region-above-le...

IF it were the case that x² + y² = z², then we'd know that the triangle is an RIGHT TRIANGLE with z as the hypotenuse.

Since we're told that x² + y² ≠ z², then we know that z is not the hypotenuse of a right triangle.

ASIDE #1: This does not necessarily mean that the given triangle is not a right triangle. It could be the case that side x is the hypotenuse of the triangle, in which case, we'd get y² + z² = x²

ASIDE #2: It COULD also be the case that the given triangle is an equilateral triangle, but it could also not be an equilateral triangle.

Does that help?

Cheers,
Brent

Hi Brent

Helps perfectly thank you, I saw in a later video this topic is discussed in more depth.

Regards

Hi Brent,

Could you please present your answer on this?
https://gmatclub.com/forum/in-triangle-abc-point-x-is-the-midpoint-of-side-ac-and-82671.html

The question introduces properties of Midsegements, something i haven't come across in our course module.

Thanks.
gmat-admin's picture

You bet!

Here's my step-by-step solution: https://gmatclub.com/forum/in-triangle-abc-point-x-is-the-midpoint-of-si...

Cheers,
Brent

Question Link: https://gmatclub.com/forum/in-the-figure-above-point-o-is-the-center-of-the-circle-and-107874.html

Would you mind explaining the solution. Specifically I would like to know the approach on how the deduce or crack the given info. I seem to understand but it when encountered again, I can't get it right
gmat-admin's picture

Hi emailme,

Here's my step-by-step solution: https://gmatclub.com/forum/in-the-figure-above-point-o-is-the-center-of-...

Cheers,
Brent

Hi,

In the following question, how did we figure out the height of the triangle?:
https://gmatclub.com/forum/given-the-three-points-below-what-is-the-area-of-the-triangle-formed-238212.html

Thanks!
gmat-admin's picture

Hi aanchal890,

Link to my solution: https://gmatclub.com/forum/given-the-three-points-below-what-is-the-area...

First notice that the red base is on the line y = -3
We know this because the y-coordinates on the base are -3.

Next, notice that the y-coordinate of the top vertex is 3, since the coordinates are (0, 3)

So, the height = 3 - (-3) = 6

Does that help?

Cheers,
Brent

Yes! Thank you for the clarification

Sir, could you please solve this problem your way?

https://gmatclub.com/forum/in-the-figure-point-d-divides-side-bc-of-triangle-abc-into-segments-126934.html

Thank you.

gmat-admin's picture

Thank you, sir, for the lucid solution!

Hi Brent, need your support for the below question.

https://gmatclub.com/forum/what-is-the-area-of-a-triangle-created-by-the-intersections-158727.html

Hi Brent, the question below is testing different properties of Triangles. Are they part of the module somewhere?

https://gmatclub.com/forum/in-the-figure-point-d-divides-side-bc-of-triangle-abc-into-segments-126934.html
gmat-admin's picture

My mistake. The question largely tests one's knowledge of Special Right Triangles. So, I have moved the link to the video lesson that covers this concept: https://www.gmatprepnow.com/module/gmat-geometry/video/870

Cheers,
Brent

Is the reverse true for both Isosceles and Equilateral triangles?
I mean if the line bisects the base then does it also imply the line is perpendicular? or is it only the other way round.
gmat-admin's picture

That's correct.
If we have an isosceles or equilateral triangle, and a line drawn from a vertex bisects the opposite side, then line is also perpendicular to the opposite side.

ASIDE: Just to be clear, for isosceles triangles, the side that is bisected must be the side that is NOT one of the two equal sides.

Cheers,
Brent

Hi Brent,

In the question below, my answer is D. Here's how.

https://gmatclub.com/forum/in-the-triangle-shown-y-z-if-j-x-k-what-is-the-least-value-of-263317.html

The question is asking to calculate least possible value of k, which is greater than x.

Stmt 1: y=5 and since z>y, z could 6 and therefore x could be in the range of 1 to 11, making the least possible value of k as 1

Adopted a similar approach for stmt 2.

Where am I wrong here?
gmat-admin's picture

Question link: https://gmatclub.com/forum/in-the-triangle-shown-y-z-if-j-x-k-what-is-th...

That's a good idea, but since we don't know the value of z, we can't make any conclusions about x.
For example, if z = 1000, then x cannot equal 1.

The same applies to statement 2.

Does that help?

Cheers,
Brent

But isn't that in order to find the least possible value of 'x' we either maximize or minimize values of other variables. Isn't that allowed in GMAT because I have seen many explanations of Buennel where he has done so.
gmat-admin's picture

I should note that it's EXTREMELY rare to have a Data Sufficiency question featuring a max/min component. In fact, I'm not sure I've actually seen an official GMAT question with a similar construction.

True be told, I was initially going to argue that the question was vulnerable to ambiguity.

Here's the dilemma: When we're given the lengths of TWO sides of a triangle, we can easily determine the range of possible lengths for the 3rd (unknown) side. However, what conclusions can we draw when we know the length of just ONE side (as is the case with this DS question)?

Statement 1 tells us that y = 5, so it's possible that z = 5.00000000001, in which case, the length of the 3rd side must be greater than 0.00000000001.
So, k can actually be less than 1 (notice that we aren't told that any of the values are integers).

Ultimately, I think the question inherently flawed (and un-GMAT-like)

Cheers,
Brent

Hi Brent,

Need your help:
A triangle has side lengths of a, b, and c centimeters. Does each angle in the triangle measure less than 90 degrees?
1. The 3 semicircles whose diameters are the sides of the triangle have areas that are equal to 3 cm2, 4 cm2, and 6 cm2, respectively.
2. c < a + b < c + 2
gmat-admin's picture

Another killer question!

My solution: https://gmatclub.com/forum/a-triangle-has-side-lengths-of-a-b-and-c-cent...

Cheers,
Brent



A triangle has side lengths of a, b, and c centimeters. Does each angle in the triangle measure less than 90 degrees?
1. The 3 semicircles whose diameters are the sides of the triangle have areas that are equal to 3 cm2, 4 cm2, and 6 cm2, respectively.
2. c < a + b < c + 2
sir i did this question in different way
rephrasing the question is the triangle acute angled triangle
side1 side2 side3
so acute angle property (side1)^2 <(side2)^2 +(side3)^2
from statement 1 we can get sides and if they satisfy the above equation or not
hence a is sufficient.
is this approach correct?
gmat-admin's picture

Question link: https://gmatclub.com/forum/a-triangle-has-side-lengths-of-a-b-and-c-cent...

Yes, that approach is perfectly valid.

By the way, here's my (different) solution: https://gmatclub.com/forum/a-triangle-has-side-lengths-of-a-b-and-c-cent...

Cheers,
Brent

In part 5.25 of your video, how did you determine the height was 7.5? I would have assumed you would have required the length of the line that extends to your perpendicular line (which would have been 4+x) and using pythagorean theorem solved for height.
gmat-admin's picture

Sorry, the main purpose of that example was to show how to find the area of a triangle.
I just picked a random height to show how the formula would work.

For that particular triangle, finding the actual height would require some math that is not required for the GMAT.

Cheers,
Brent

https://gmatclub.com/forum/what-is-the-area-of-triangular-region-abc-above-143500.html

If the statement (1) had said the product of BD and DC was 20, would this have been enough to answer the area of ABC?

As there's a perpendicular line BD, I assume that it has created two congruent triangles and as such DC would equal AD. Thus, in this case the area of the entire triangle would be 20
gmat-admin's picture

Question link: https://gmatclub.com/forum/what-is-the-area-of-triangular-region-abc-abo...

If statement (1) had said the product of BD and DC = 20, we wouldn't have enough information to answer the area of ABC.

The presence of the perpendicular does not guarantee that AD = DC.

I drew some diagrams to reinforce this point: https://imgur.com/a/qYeN4pj

NOTE: the property you've described applies only to ISOSCELES triangles. Since we don't know whether triangle ABC is isosceles, we can't apply that property.

Cheers,
Brent

at minute 7 can't you just draw a line (the height) through the top 60 degree angle. This bisect the angle and divides the triangle into two. Then you have a 30, 60, 90 triangle and solve accordingly. That way it would just have to multiple 3 x 3srt3 which gives you 9 srt3.
gmat-admin's picture

That strategy works perfectly.
However, you can save some time by applying the area formula for equilateral triangles.

Hi!

It seems that for question https://gmatclub.com/forum/in-triangle-abc-point-x-is-the-midpoint-of-side-ac-and-82671.html , there is a new concept about midpoints and similar triangles.

Is there a general rule we can remember for such questions? Thanks.
gmat-admin's picture

Question link: https://gmatclub.com/forum/in-triangle-abc-point-x-is-the-midpoint-of-si...

Here are two takeaways:

Take triangle ABC
Add the midpoint of side AB and call it point X
Add the midpoint of side AC and call it point Y
Connect points X and Y

1) line segment XY will be parallel to side BC.
2) triangle AXY is similar to triangle ABC

Thanks!

Hi Brent,

I wanted to check with you on the approach I used for this question: https://gmatclub.com/forum/in-triangle-abc-above-if-ad-bc-what-is-the-value-of-x-347007.html.

S1: D is midpoint of side AC.
Let AD = P, so AC = 2P, and since BC = AD, then BC is also P. Since AC = 2P, BC = P, it will be a 30, 60, 90 right triangle. So angle ACB is 60 degrees. Within triangle DBC, angle ACB is 60 degrees, angle DBC = 90-x and given angle BDC is 2x, all add to 180. Solving for x, x=30.

S2. Same as S1 since it's been realized as a 30, 60, 90 right triangle.

Thank you!
gmat-admin's picture

That's a good idea, but having one side of a triangle be twice as long as one of the other sides doesn't necessarily mean the triangle is a 30-60-90 triangle.

Consider the following graphic: https://imgur.com/jrAJy7t

In the graphic, the red line is twice as long as the blue line. However, there are infinitely many triangles that can be created using those two sides. The first triangle is a 30-60-90 right triangle, but the other two are not.

So, even though it turns out that the triangle is, indeed, a 30-60-90 triangle, we need more information (in addition to one side being twice as long as another side) in order to conclude this.

I hope that helps.

Thank you Brent, that helps!

Brag up your Geometry Mastery

Now that you’ve mastered GMAT Geometry, it’s time to let the world know!

Change Playback Speed

You have the option of watching the videos at various speeds (25% faster, 50% faster, etc). To change the playback speed, click the settings icon on the right side of the video status bar.

Have a question about this video?

Post your question in the Comment section below, and a GMAT expert will answer it as fast as humanly possible.

Free “Question of the Day” emails!