Lesson: Special Right Triangles

Comment on Special Right Triangles

For the following question, can you please tell me why we are adding root 2 for the perimeter?

https://gmatclub.com/forum/figures-x-and-y-above-show-how-eight-identical-triangular-pieces-of-ca-207393.html

The hypotenuse of a right triangle is 10 cm. What is the perimeter, in centimeters, of the triangle?

(1) The area of the triangle is 25 square centimeters.
(2) The 2 legs of the triangle are of equal length.
gmat-admin's picture

Hi santhosh1989,

Here's my step-by-step solution to that question: https://gmatclub.com/forum/the-hypotenuse-of-a-right-triangle-is-10-cm-w...

Cheers,
Brent

Hi Brent.

Your solution to this problem makes sense, However, another solution has me puzzled by "Math Revolution"
https://gmatclub.com/forum/the-hypotenuse-of-a-right-triangle-is-10-cm-what-is-the-93911-20.html

His solution where he pulls (a +b)^2 just seems weird. I'd like to know how he got to the "root 200" and proceeds to solve for an answer. I know an answer is not needed but the continuation to a solution helps me see learn and solve other problems like this might they come up.

Thanks.


gmat-admin's picture

Yes, Math Revolution's solution (here https://gmatclub.com/forum/the-hypotenuse-of-a-right-triangle-is-10-cm-w...) is somewhat counter-intuitive.

Here's the basic idea:

The perimeter = a + b + c (where c is the right triangle's hypotenuse, and a and b are the lengths of the two legs)

Statement 1 tells us that the area is 25
This means ab/2 = 25, which means ab = 50.

We also know that c = 10

From here, Math Revolution decides to see what (a + b)² evaluates to be.

(a + b)² = a² + 2ab + b²
= (a² + b²) + 2ab

The Pythagorean Theorem tells us that a² + b² = c²
Since c = 10, we can write: a² + b² = 10² = 100

We also know that ab = 50

So, let's take (a² + b²) + 2ab and replace a² + b² with 100 AND replace ab with 50. We get...

(a² + b²) + 2ab = 100 + 2(50)
= 100 + 100
= 200

This means (a + b)² = 200, which means a + b = √200 = 10√2

At this point, we can determine the perimeter.

a + b + c = (a + b) + c
= 10√2 + 10
DONE!

Does that help?

Cheers,
Brent

Thanks Brent !

From here, Math Revolution decides to see what (a + b)² evaluates to be.
(a + b)² = a² + 2ab + b²
= (a² + b²) + 2ab

I understand that we have the "Pythagorean formula and the Area formula" and because of that we now have a system of two equations, so therefore it is solvable, but I'm still confused as to how one would see, or know to take (a + b)² and "square" it? It seems so "Non-Sequitur" and out there for lack of a better term or description. What am I missing, I still don't or wouldn't ever get to do that.
gmat-admin's picture

I wouldn't be concerned about not seeing that particular approach; most people will miss it.

The idea here is that we know that the Pythagorean Theorem tells us that a² + b² = c²

We also know (from statement 1) that ab = 50, which means 2ab = 100

So, we have one piece of information about a² + b² AND we have information about 2ab.

Notice that (a + b)² = a² + 2ab + b²

Since the given information seems similar to the expansion of (a + b)², Math Revolution decided to see where it might take him/her.

Cheers,
Brent

Thanks Brent,

Maybe later it will click with me, but for now I will just take your advice and forget about it... however, I most likely won't, Lol!
gmat-admin's picture

Ha - I'm the same way!

In the below problem, how do one deduce the 30-60-90 triangle from the given info. What property am I missing. I could assume it will form a right angle, say 90 and then deduce 30 (180 - 90-60= 30).
https://gmatclub.com/forum/a-ladder-of-a-fire-truck-is-elevated-to-an-angle-of-60-and-134073.html

Thanks in advance

gmat-admin's picture

Good question!

First, we're told that the ladder forms a 60-degree angle with the ground.

Second, the height of any triangle is the length of the line from the top of the triangle to the base, SUCH THAT this line is perpendicular to the base (see Jeff's diagram here: https://gmatclub.com/forum/a-ladder-of-a-fire-truck-is-elevated-to-an-an...)

Does that help?

Cheers,
Brent

Hi Brent,

Could you please help me understand the approach adopted to solve the following question:

https://gmatclub.com/forum/figures-x-and-y-above-show-how-eight-identical-triangular-pieces-of-ca-207393.html

Thanks!
gmat-admin's picture

You bet.

Here's my step-by-step solution: https://gmatclub.com/forum/figures-x-and-y-above-show-how-eight-identica...

Please let me know if you have any questions.

Cheers,
Brent

Hi Brent,

In the below question (https://gmatclub.com/forum/figures-x-and-y-above-show-how-eight-identical-triangular-pieces-of-ca-207393.html), I solved to get 2:3. I did not understand why the outer side of the square becomes the hypotenuse? For me, there are two right angle triangles and the outer sides of the square are legs of the 2 triangles. Therefore, the perimeter of X equals to 4 and that of Y equals to 6. Ratio: 2:3
gmat-admin's picture

Question link: https://gmatclub.com/forum/figures-x-and-y-above-show-how-eight-identica...

It's not easy to see how the sides in each shape correspond.

Take a look at the last diagram in my solution at https://gmatclub.com/forum/figures-x-and-y-above-show-how-eight-identica...

Does that help?

Cheers,
Brent

Hi Brent,
Grateful if you could advise on how to solve the below-mentioned problem.
Thanks a lot
Fatima-Zahra

https://gmatclub.com/forum/in-the-figure-shown-the-length-of-line-segment-qs-is-117577.html

When adding the altitude to an equilateral triangle we get two 30-60-90 triangles as your video mentions. The line formed is a perpendicular bisector as is the case when we draw an altitude to calculate the height.

It is me or is this the same as saying that a perpendicular bisector splits the angle (in this 60 degrees) into two equal angles as well as the line?
gmat-admin's picture

Those two sentences mean the same thing.

https://gmatclub.com/forum/figures-x-and-y-above-show-how-eight-identical-triangular-pieces-of-ca-207393.html

Why are not able to conceptualise the square as being made of two large isosceles triangles instead of 4 smaller triangles? I know it doesn't work as it will yield the same results as the perimeter of a square: perimeter of rectangle (4:6) but why?
gmat-admin's picture

Question link: https://gmatclub.com/forum/figures-x-and-y-above-show-how-eight-identica...

The question specifically tells us to consider the square and rectangle each consisting of four identical triangular pieces of cardboard.

If we want to treat the square as consisting of TWO right triangles, how do you then use those same TWO triangles to create the rectangle?

Here's my solution: https://gmatclub.com/forum/figures-x-and-y-above-show-how-eight-identica...

Cheers,
Brent

Hi Brent,

Could you please explain how you would approach this question.

https://gmatclub.com/forum/a-certain-right-triangle-has-sides-of-length-x-y-and-z-wh-168727.html

I'm still not following. You got y to equal to 10 with setting x as 0.5. But the answer E means y < sqr root of 3 / 4. How is 10 < sqr root 3 / 4?
gmat-admin's picture

My bad.
I meant to say the correct answer is A.
B, C, D and E all provide a MAXIMUM value of y.
I've edited my response accordingly.

Cheers,
Brent

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

I can't seem to arrive at the solution on this one. What's the best approach in this type of question?
gmat-admin's picture

That's a very tough question!

Here's my solution: https://gmatclub.com/forum/in-the-figure-point-d-divides-side-bc-of-tria...

Cheers,
Brent

Hi Brent,

I have posted my doubts to the question. Request you to help with the same - https://gmatclub.com/forum/in-the-figure-point-d-divides-side-bc-of-triangle-abc-into-segments-126934-20.html#p2380414

Warm Regards,
Pritish
gmat-admin's picture

My solution is here: https://gmatclub.com/forum/in-the-figure-point-d-divides-side-bc-of-tria...

Your questions:

1) When you drew the segment EB, how did you determine that it will split angle B (i.e. 45) into 30 and 15? Why could it not be any other split like 20 and 25 or 10 and 35 etc?
From an earlier step, we learned that ED = 1 (from analyzing the 30-60-90 right triangle)
We were also told that BD = 1 (from the given info)
This means ∆EBD is an isosceles triangle, which means ∠ECB = ∠EBC = 30°

2) In triangle AEC we know that angle O is 90, but then how did you conclude that the other 2 angles are 45 each and not any other split?
I found that angle in two steps (different from Bunuel)
You can see my solution for that part.

Cheers,
Brent

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