Suppose you have a triangle with sides {6,7,8} — how do you find the height?

This is a question some GMAT test takers ask. They know they would need the height to find the area, so they worry: how would I find that height. The short answer is: fuhgeddaboudit!

## Which height?

First of all, the “height” of a triangle is its altitude. Any triangle has three altitudes, and therefore has three heights. You see, any side can be a base. From any one vertex, you can draw a line that is perpendicular to the opposite base — that’s the altitude to this base. Any triangle has three altitudes and three bases. You can use any one altitude-base pair to find the area of the triangle, via the formula A = (1/2)bh.

In each of those diagrams, the triangle ABC is the same. The green line is the altitude, the “height”, and the side with the red perpendicular square on it is the “base.” All three sides of the triangle get a turn.

## Finding a height

Given the lengths of three sides of a triangle, the only way one would be able to find a height and the area from the sides alone would involve **trigonometry**, which is well beyond the scope of the GMAT. You are 100% NOT responsible for knowing how to perform these calculations. This is several levels of advanced stuff beyond the math you need to know. Don’t worry about that stuff.

In practice, if the GMAT problem wants you to calculate the area of a triangle, they would have to *give* you the height. The only exception would be a right triangle — in a right triangle, if one of the legs is the base, the other leg is the altitude, the height, so it’s particularly easy to find the area of right triangles.

## Some “more than you need to know” caveats

*If you don’t want to know anything about this topic that you don’t absolutely need for the GMAT, skip this section*!

a. Technically, if you know the three sides of a triangle, you could find the area from something called Heron’s formula, but that’s also more than the GMAT will expect you to know. More than you needed to know!

b. If one of the angles of the triangle is obtuse, then the altitudes to either base adjacent to this obtuse angle are outside of the triangle. Super-technically, an altitude is not a segment through a vertex perpendicular to the opposite base, but instead, a segment through a vertex perpendicular to *the line containing* the opposite base.

In the diagram above, in triangle DEF, one of the three altitudes is DG, which goes from vertex D to the infinite straight line that contains side EF. That’s a technicality the GMAT will not test or expect you to know. Again, more than you needed to know!

c. If the three sides of a triangle are all nice pretty positive integers, then in all likelihood, the actual mathematical value of the altitudes will be ugly decimals. Many GMAT prep sources and teachers in general will gloss over that, and for the purposes of easy problem-solving, give you a nice pretty positive integer for the altitude also. For example, the real value of the altitude from C to AB in the 6-7-8 triangle at the top is:

Not only are you 100% NOT expected to know how to find that number, but also most GMAT practice question writers will spare you the ugly details and just tell you, for example, altitude = 5. That makes it very easy to calculate the area. Yes, technically, it’s a white lie, but one that spares the poor students a bunch of ugly decimal math with which they needn’t concern themselves. Actually, math teachers of all levels do this all the time — little white mathematical lies, to spare students details they don’t need to know.

So far as I can tell, the folks who write the GMAT itself are sticklers for truth of all kinds, and do not even do this “simplify things for the student” kind of white lying. They are more likely to circumvent the entire issue, for example, by making all the relevant lengths variables or something like that. Yet again, more than you needed to know!

## What you need to know

You need to know basic geometry. Yes, there is tons of math beyond this, and tons more you could know about triangles and their properties, but you are not responsible for any of that. You just need to know the basic geometry of triangles, including the formula A = (1/2)b*h. If the triangle is not a right triangle, you have absolute no responsibility for knowing how to find the height — it will always be given if you need it. Here’s a free practice question for you.

1) http://gmat.magoosh.com/questions/81

I am in Algebra 1, and I am being asked this question on a pop quiz- like homework, and I need to find the height of an isosceles triangle with two sides of 7 units and the base is 8 units.

PLEASE HELP!

We don’t really offer homework or quiz answers here. And as Mike says, the GMAT doesn’t really do triangle height questions– the height will generally be given. I’ve still decided to approve and reply to your comment Nathan, because you’ve touched on a concept that IS important on the GMAT: The Pythagorean theorem, as it applies to right triangles.

It’s important to remember that any isosceles triangle can be bisected into two right triangles by drawing a straight line from the angle between the two equal sides down to the middle of the base. The vertical line of the right angle on a triangle is considered the triangle’s height as well as its side. And the Pythagorean theorem states that there where

aandbare the sides of a right angle on a right triangle andcis the side that is opposite of the right angle on the right triangle,a^2 +b^2 =c^2. Armed with thisasquared plusbsquared pluscsquared formula, you can calculate the missing height of any right triangle as long as you have the length of the other side of the right angle (b^2) and the length of the side opposite the right angle (c^2).This is important on the GMAT because some exam problems that look like they could be dealing with the unknown height of an isosceles triangle are really asking you to calculate the length of one side of a right triangle, which doubles as the height of an isosceles triangle. And Nathan, this means you can also calculate the height of an isosceles triangle with the Pythagorean theorem by cutting the isosceles in half and treating its height as one side of a right triangle. In the isosceles triangle you described, half the base squared plus 7 squared equals the height!

What is the formula for finding the height of a trapezoid for instance, b1 is 4 b2 is 2.5 and area is 3.9 what is the height?

Hi there!

The area of a trapezoid is [(a+b)/2]h so if we want to rearrange that to solve for h, we can say that the height is:

Area/[(a+b)/2]

In the case of the numbers you give, that means:

Height = 3.9/[(4+2.5)/2] = 3.9/[(6.5)/2] = 3.9/[3.25] = 1.2

I hope that helps! 🙂

If I have a triangle with a base of 12 units and two side lengths of 10 units how long will the altitude be?

This is an isosceles triangle which can be cut into 2 right triangles with base as 6 and hypotenuse as 10. So, the 3rd side of the right triangle which also is the altitude of the isosceles triangle is 8.

If I was given the lengths of a triangle 10, 12, and 15 cm and the area 60 cm how would I find the altitude

Good day mr. mike.

Can you please prove me wrong that (as i can only verify these in smaller numbers)

If n^2+n is m, and if m is divisible by 3, m is the average of twin primes juxtaposed at m.

Dear Walter,

I’m happy to respond. 🙂 First of all, rest assured that absolutely no simply algebraic procedure will always generate prime numbers. This has been proven impossible. I would recommend looking at this blog:

http://magoosh.com/gmat/2013/gmat-quant-must-be-true-problems/

under the section “Patterns of Prime Numbers.”

For this, if n = 9, then m = 90, and 91 = 7*13.

If n = 11, then m = 132, and 133 = 7*19

If n = 12, then m = 156, and 155 = 5*31

You see, n^2 + n = n(n + 1) is always even, and if it’s divisible by 3 also, then it has 2 and 3 and 6 as factors, which leaves fewer possible factors for the numbers adjacent to it. That’s why primes are relatively common adjacent to this number, and why we have to go up to n = 18, m = 342, to get to an example that has two non-prime neighbors:

341 = 11*31

343 = 7^3

Does all this make sense?

Mike 🙂

Mike, i have been solving this problem for weeks and eureka i have solve it fast. this problem is very useful in surveying especially before the advent of high tech electronic gadgets because you have to walk around the perimeter to measure the lenghts so you can get the area bounded. in geometry its called ASA or side angle side and until now, the only way to get altitude is by getting the lenghts of all segments which is way way back to heros time. and i must guess that most of our knowledge of the postulates in geometry eg. concurrence of altitudes of bisectors etc. stems from tinkering with how to find the altitude of the triangle. i was thinking of writing a book about this because the my formula only needs one side and the two angles.

Dear Walter,

I’m glad you solved your problem. I point out that walking out the perimeter to spec the three sides would determine the triangle via SSS. It’s true that either SSS or ASA or AAS or SAS completely determine a triangle, so any of those should determine the area completely. Frankly, I believe the concurrence of angle bisectors and of medians and etc. was known to Euclid, who didn’t seem particularly concerned with finding altitudes. Nevertheless, the details of geometry are always fascinating to investigate. Best of luck to you.

Mike 🙂