### Related Articles:

### Geometric Mean

- Geometric Mean: The geometric mean between two positive numbers a and b is x, where
- Similar Triangles: If the altitude is drawn from the vertex of the right angle of a right triangle to its hypotenuse, then the two triangles formed are similar to the given triangle and to each other
- The measures of the altitude drawn from the vertex of the right angle of a right triangle to its hypotenuse is the geometric mean between the measures of the two segments of the hypotenuse

### Geometric Mean

Right triangle ABC, ―BD ⊥―AC , write all the pairs of similar triangles.

Right triangle ABC, if ―BD ⊥―AC , then ―BD is the geometric mean between ―AC and ―AD .

Right triangle ABC, if ―BD ⊥―AC , then [AC/AB] = [AB/AD].

Right triangle ABC, ―BD ⊥―AC , AD = 3, CD = 9, find BD.

- [AD/BD] = [BD/CD]
- [3/BD] = [BD/9]

find the values of x and y.

- [3/x] = [x/5]
- x = √{15}
- [3/y] = [y/(3 + 5)]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Geometric Mean

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Geometric Mean 0:04
- Geometric Mean & Example
- Similar Triangles 4:32
- Similar Triangles
- Geometric Mean-Altitude 11:10
- Geometric Mean-Altitude & Example
- Geometric Mean-Leg 14:47
- Geometric Mean-Leg & Example
- Extra Example 1: Geometric Mean Between Each Pair of Numbers 20:10
- Extra Example 2: Similar Triangles 23:46
- Extra Example 3: Geometric Mean of Triangles 28:30
- Extra Example 4: Geometric Mean of Triangles 36:58

### Geometry Online Course

### Transcription: Geometric Mean

*Welcome back to Educator.com.*0000

*We are going to go over the geometric mean for our next lesson.*0002

*For the geometric mean, here, we have a/x, which is equal to x/b.*0006

*Now, if you remember, a few lessons ago, we went over ratios and proportions.*0019

*When we have a proportion, we have two ratios that are equal to each other.*0025

*Here, these numbers right here are called the means, and these numbers here are called extremes.*0032

*So, when we talk about the geometric mean, we are talking about this number and this number.*0049

*The geometric mean between two positive numbers, a and b, is x.*0060

*If we are given two numbers, a and b, those numbers would be considered the extremes.*0067

*And then, x, which is the geometric mean, is going to go here and here.*0073

*Now, when we solve proportions, remember: we cross-multiply; so we are going to do this number times this number, is equal to a times b.*0082

*If we are asked to find the geometric mean between two positive numbers--let's say 3 and 8--find the geometric mean between 3 and 8...*0094

*Now, actually, before we continue with that problem, let's go back to this.*0114

*If we cross-multiply, then we are going to get x ^{2} = a times b; then how would you find x?*0121

*x is going to be the square root of ab.*0134

*The geometric mean x is going to be the square root of this number times that number.*0140

*This is how you would find the geometric mean between the two numbers.*0148

*Now, back to this: 3 and 8--find the geometric mean between 3 and 8.*0152

*What you can do is go ahead and solve it like this: plug it into here and find the square root of 3 times 8.*0159

*Or you can just first set it up as a proportion; and this actually makes it a lot easier to see, especially if you are given*0168

*different versions of this problem, like if you are given the geometric mean and asked to find one of these numbers.*0179

*It is always easiest to put it into this proportion first, because you know that these two numbers here have to be the geometric mean.*0186

*If they are asking you to find the geometric mean, that means that that would be this number down here, and that number up there, for your proportion.*0196

*And then, 3 and 8 are going to go here and here; so this is how you would set it up.*0209

*Find the geometric mean--that means that these numbers right here are what you are looking for.*0216

*To solve it out, cross-multiply: x ^{2} is equal to 3 times 8, which is 24, so x is equal to √24.*0220

*And that is going to be 2√6, or you can use your calculator to change that to a decimal.*0235

*But this would be the answer, right here: this is the geometric mean between 3 and 8.*0247

*Again, set it up as a proportion; put the two numbers up here and down there.*0254

*Make sure that the geometric mean goes here and here, because those are the means; and then, just cross-multiply to solve.*0262

*Similar triangles: now, here we have a right triangle, and remember: an altitude is a segment*0274

*starting from a vertex and going to the side opposite that vertex so that it is perpendicular to it.*0286

*If we draw an altitude from this vertex (and it has to be drawn from that right angle)--*0299

*so then here is our right angle, and you are going to draw an altitude so that it is perpendicular to the side opposite--*0307

*in this case, it is the hypotenuse, because it is coming from that right angle--then the two triangles formed*0318

*are similar to the given triangle and to each other.*0325

*Before this altitude was drawn, we only had one triangle.*0330

*Now, after the altitude, we have three triangles: we have the big one; we have this one right here; and we have this one right here--three triangles.*0335

*Now, this theorem is saying that, once that altitude is drawn from that right angle to the hypotenuse, all three triangles are now similar.*0345

*Remember: similar triangles are triangles that have congruent angles, but proportional sides.*0356

*"Similar" just means that they have the same shape, but a different size; that is "similar" or "similarity."*0368

*So, here we know that we have three triangles--the same shape, but slightly different sizes.*0377

*To state all three triangles so that the corresponding parts are in order, so that we can say that they are similar,*0386

*I can name all three; it doesn't matter which order, so if I want to say the big one first:*0397

*the big one is going to be triangle ABC; it is similar to...and then we are going to just name another triangle.*0406

*Now, remember: it has to be corresponding to the order; so AB in the big triangle*0419

*(and that is kind of hard to see)...the easiest way to see their corresponding parts is to look at it from long leg*0429

*to short leg, and then hypotenuse; those are the three parts that make up a right triangle.*0445

*It is always easiest to determine whatever you are stating--which side it would be considered from that triangle.*0453

*And then, you can just look at the corresponding part of the other triangle.*0464

*So, AB from the big triangle is considered the long leg; this would be AB.*0468

*AB is the long leg; BC is the short leg; and then, AC is the hypotenuse.*0479

*If AB is the long leg, then I have to mention that one first.*0487

*And remember: B is the right angle; so then, the right angle is going to go in the middle of the next triangle that I am stating.*0491

*Then, let's say I am going to name this triangle--not the very small one, but the medium-sized one, this one right here.*0499

*I am going to label this D; so remember, I have to state the long first.*0510

*The long leg from this triangle would be AD, because here is the right angle; AD is the long leg.*0516

*Now, does it matter if I say AD or DA?*0525

*Well, I know that, since B is my right angle for the first triangle, I have to state the right angle in that same order.*0528

*So, it is going to be AD, triangle AD, because AD is my long leg, and D is my right angle; and then, it would be AD, and then B.*0540

*So again, from the big triangle, it is triangle ABC, with AB as a long leg, BC as a short leg, and AC as the hypotenuse.*0556

*For my second triangle that I am naming, AD is my long leg; BD is my short leg; and AB is my hypotenuse.*0571

*And then, see how D is the right angle; the middle angle that I am listing is the right angle.*0583

*So then, the third triangle has to be in the same order, so from this one right here,*0593

*which one is my long leg? Which one is my short leg? And which one is my hypotenuse?*0600

*My long leg from here...this is the right angle...would be BD or DB.*0606

*Now, from this triangle, where is my right angle?--this one, D.*0613

*I know that that letter D is going to go in the middle.*0618

*So then, BD is my long leg; D is my right angle; C is my missing vertex.*0622

*BD is my long leg; DC is my short leg; and BC is my hypotenuse.*0634

*So now, I have listed all three triangles within their corresponding parts.*0642

*Just keep in mind that, when the altitude is drawn from the right angle, you are actually splitting this up into three similar triangles--not congruent, but similar.*0653

*Here, we have a right triangle, again, with the altitude drawn from the right angle of that triangle down to the hypotenuse.*0672

*When you have this diagram, the measure of the altitude, BD, is going to be the geometric mean between AD and DC.*0685

*Remember how, when we went over geometric mean, we found the geometric mean between two numbers.*0710

*So then, the two numbers that are given are a and b, and then, this is a geometric mean and a geometric mean.*0719

*This right here, BD, would be the geometric mean between the two parts of the hypotenuse.*0731

*Just to read this to you: the measures of the altitude drawn from the vertex of the right angle of the right triangle*0739

*to its hypotenuse is the geometric mean between the measures of the two segments of the hypotenuse.*0748

*So, because this right here--see how it splits up the hypotenuse into two parts, AD and DC--those two parts,*0756

*then it is as if AD is a here; DC is b; so then, those are the two numbers; and BD, the altitude, would be the geometric mean.*0765

*So then, if you were to write this out, this would be like BD is the geometric mean between AD and DC, these two parts.*0778

*Then, AD would go up here, and DC would go right there; that is like a and b--those are the two numbers.*0788

*And then, the geometric mean, which is the altitude, BD, is going to go here and here.*0798

*So, whenever they say that something is the geometric mean, that something goes in here and here,*0808

*because remember, again: these two would be the means.*0816

*Let's say that this is x, and we want to find the geometric mean between...let's say AD is 20 and DC is 5.*0823

*So then, we are trying to find the altitude.*0838

*Now, we know, because of this theorem, that the altitude, BD, is the geometric mean between AD and DC, these two parts.*0840

*And this would be the geometric mean; so to solve it out, I am going to make it into my proportion.*0855

*The geometric mean is BD; that is going to go here and here.*0863

*My two numbers, a and b, are going to go here and here; so when I solve it out, it is going to become x ^{2} = 100, so x is 10.*0867

*So then, this right here...the measure of the altitude is 10.*0879

*Now, one more: here, for this one, if the altitude is drawn to the hypotenuse of a right triangle,*0888

*then the measure of a leg of the triangle (we have two legs: it is AB and BC, the two legs of the big triangle)*0897

*is the geometric mean between the measures of the hypotenuse*0909

*(this whole thing, AC) and the segment of the hypotenuse adjacent to that leg.*0917

*"Adjacent" means "next to"; so then, to find the geometric mean, we need two numbers.*0925

*The two numbers would be the hypotenuse and the segment of the hypotenuse adjacent to that leg--*0936

*the part of the hypotenuse that is close to that leg that we are trying to find.*0944

*Here, for example, we are looking for this leg; this is the geometric mean between...the two numbers would be*0950

*the whole thing, the hypotenuse, AC, and the part of the hypotenuse (remember, this altitude*0961

*can divide this hypotenuse into two parts, AD and BC) that is adjacent (meaning closest to, next) to that leg that we are trying to find.*0971

*It would be right here--that part; that means that we make our proportion...*0983

*Again, the geometric mean...BC goes here and here; or you can write BC, BC.*0992

*And then, let's write that so that you know what we are looking for: BC is going to go here and here.*1000

*And then, the two numbers that we are going to use are AC, the whole hypotenuse, and then this part right here; that is DC.*1012

*OK, and then you would solve it that way.*1028

*Now, let's say that we are looking for this leg.*1030

*If this is the leg that we are looking for--let's say y--then it is going to be the geometric mean*1033

*between, again, the whole thing, AC, and the part of the hypotenuse adjacent to that leg that we are looking for.*1039

*In this case, if we are looking for y, this leg, we are not going to be using DC, because that is not the one adjacent to that.*1048

*Then, we are going to be using AC and AD, because that is the part of the hypotenuse that is adjacent to this leg that we are looking for.*1056

*Again, y, y...or you can write AB, AB...that is going to be the whole thing, AC, and AD.*1068

*It is the geometric mean between those two numbers.*1083

*Let's say that this whole thing is 20, and DC is 5.*1086

*Now, let's look for BC; BC is the geometric mean between 20 and 5; we put 20 up here, x, and 5.*1098

*It is going to be x ^{2} = 100; x = 10; so this right here is 10.*1114

*To find y, if this is 5, and this whole thing is 20, I know that AD has to be 15, because 15 + 5 makes up this whole thing, which is 20.*1124

*So, AC is 20, over y, equals y over this part, which is 15.*1140

*y ^{2} (I am going to cross-multiply) equals...this would be 300, so y is going to be √300, which is 10√3.*1154

*That is how you would find the geometric mean.*1181

*Now, the one before (remember, the altitude one)--the altitude is the geometric mean between part and part of the hypotenuse.*1185

*The leg is the geometric mean between the whole and this part close to it.*1197

*Again, this leg is the geometric mean between the whole and the part close to it.*1203

*Those are the two theorems, there; now let's go over our examples.*1210

*Find the geometric mean between each pair of numbers.*1215

*Find the geometric mean: to use our proportion, the geometric mean goes here and here;*1220

*and then, the two numbers that we are going to use, that are given to us--those are the extremes, 8 and 11.*1231

*Make sure that whatever the geometric mean is...it has to go here and here.*1241

*Then, cross-multiply; I get x ^{2} = 88, so x = √88, and then that would simplify to...let's see...2√22.*1247

*You can simplify it, if your teacher wants you to round it to several decimal places, or whatever it is.*1287

*And you would have to use your calculator to figure that out in decimals.*1293

*But this is how it would simplify.*1299

*The next one: 5 and 2/3--again, 5/x = x/(2/3).*1303

*Now, I am doing it this way, just so that it is easier to set up.*1320

*But you also know that x is going to equal the square root of AB; so you can just use that if you want to.*1325

*That is where you just take the two numbers, multiply it, and then take the square root of it to find the geometric mean.*1333

*Or you can just set it up so that this is easier to understand, because you know that these two numbers make up the means.*1341

*So, you know that whatever it says the mean is, you write that here.*1350

*In this case, we are looking for it; that is why we have x's there.*1354

*And then, the two numbers that are given would go there.*1360

*Then, x ^{2} =...here, this becomes...5 times 2/3 is going to be 10/3, so x is the square root of 10/3.*1365

*And then, just use your calculator for that.*1380

*If you have to leave it in radical form, it would just be √10/√3.*1383

*And then, I would have to rationalize this denominator, so it is going to be...see how √3/√3 is just equal to 1.*1391

*So, this would be √30/3, because √3 times √3 equals 3.*1400

*Now again, if you don't understand this, just go ahead and use your calculator; just do 10/3, and then you can just take the square root of that.*1408

*Or if your calculator will allow you, just do the square root of 10/3.*1416

*Name the three similar triangles: again, we have the altitude from the right angle of this big triangle.*1428

*The altitude is from the right angle to the hypotenuse of the big triangle.*1440

*We are going to name the three triangles that are similar.*1447

*Now, again, we can start with whatever triangle we want.*1451

*If you do this on your own, then your triangles will probably be different; it is probably going to be listed differently than how I list it.*1458

*But that is fine, just as long as whatever you wrote, the three triangles, are corresponding with each other in the parts.*1466

*The first triangle that I want to name is the big one.*1475

*I want to name the big one first, and I am going to say that triangle...let's do the hypotenuse first:*1478

*MOP...that P, my right angle, would be last; so MO, and then P is my right angle.*1486

*And that is going to be similar to triangle...make sure that...MO is my hypotenuse, and then OP would be*1505

*the short side; so triangle...and then, let's see, let's do this one...MPN--let's try that,*1524

*because again, P is my right angle; so then, N, which is the right angle for this triangle, has to be listed last.*1544

*So, is MP my hypotenuse?--yes, so that is right.*1551

*How about PN--is PN my short side?--yes, so this is correct.*1555

*And then, it is similar to...what is my other one?--the small triangle right here, right?*1571

*So then, this is the right angle; that is going to go last; and the triangle's hypotenuse would be...*1581

*let's see, OPN...OP is my hypotenuse; N is my right angle; and now, let's look at PN.*1590

*For all of these other ones, OP was my short leg; PN was also my short leg of this triangle.*1604

*Is PN the short side of this small triangle? It is actually not, so this is wrong.*1615

*I know that OP is my hypotenuse, but then, instead of saying OPN, I would have to say PON.*1629

*N is in the correct position, because that is the right angle, and that has to go last.*1641

*I know that, instead of OP, like how we had it, it would have to go PO.*1645

*PON: that way, PO is my hypotenuse, and then ON is my short side.*1654

*Those are my three triangles that are similar.*1667

*Now, again, if you had your first triangle listed out differently, and if you used another triangle, that is fine;*1670

*just make sure that the other two triangles that you list out are similar to that triangle.*1680

*It is always easiest to just maybe write some symbols like this, like how I did it.*1688

*And then, just make sure that you know that your hypotenuse has to go hypotenuse, short leg, long leg, right angle...*1694

*use those to help you list it in the corresponding order.*1704

*The next example: we are going to find the values of x and y.*1712

*We are actually going to do a few of these for the next example, too.*1714

*For the first one, let's see: it is a right angle's altitude; that means that I know that this altitude is the geometric mean between these parts of the hypotenuse.*1720

*So, that means that the altitude is the geometric mean, x and x, between 4 and 7.*1743

*Solve this out; this is going to be x ^{2} = 28, so x is √28; does that simplify?*1757

*Let's see: it is going to be 2√7.*1770

*Again, if you want to change it to decimals, then use your calculator.*1779

*2√7 is this value right here.*1784

*And then, to find y, you can do two things: you can use that second theorem*1787

*that says that the leg of the big triangle is the geometric mean between the whole hypotenuse*1796

*and the part that is adjacent to it, and solve it out this way; or you can just, using this right triangle,*1807

*now that you know what this side is, and you have this side as 7, use the Pythagorean theorem.*1819

*This would be a ^{2} + b^{2} = c^{2} (or y^{2}).*1825

*Either way, it does not matter; you will still get the same answer.*1835

*Let's just go ahead and solve it this way, with the geometric mean.*1840

*So, we are going to write y and y there; this is the geometric mean between the whole thing--*1847

*what is the whole thing?--4 + 7 is the whole thing; the whole hypotenuse would be 11;*1856

*and then, the part of the hypotenuse that is close to this leg that we are looking for is 7.*1864

*So, y ^{2} = 77; y = √77; and that is it; so we have here and here.*1871

*The next one: Let's see, this altitude is the geometric mean between this part and this part.*1886

*Now, this one is given to me; I am not looking for the altitude.*1896

*But the theorem does not change; the theorem stays the same.*1901

*It is still saying that the altitude is a geometric mean between this part and this part.*1904

*So, when I write it in my proportion, I still have to keep this altitude, whatever it is,*1910

*whether it is x or whether it is given--I still have to write it as my mean, there, and then between what?-- between this, which is x, and 5x.*1917

*So, it still stays the same, just like this; whatever this is right here, that is going to go here and here.*1934

*In the same way, whatever the altitude is, the altitude is the geometric mean; so then, that is going to go here and here.*1943

*When you solve it out, it is going to be 25 ^{2} = 5x^{2}; you are just multiplying through.*1950

*And then, 25 ^{2} is 625; that is equal to 5x^{2}.*1964

*If you divide the 5, that is going to give you 125 = x ^{2}, so √125 is going to be equal to x, which is 5√5.*1975

*And that is a 5 right there; so 5√5 is x, so this is going to be 5√5.*2009

*This is going to be 5 times x, which is 5 times 5√5, is 25√5.*2020

*And then, y: again, you can use the Pythagorean theorem: 25 ^{2} + (5√5)^{2} = y^{2}.*2032

*Or you can just use this: y is a geometric mean between the whole thing--*2042

*the whole thing would be 25√5 + 5√5; that is going to be 30√5--*2058

*and this part that is close to the leg, 5√5.*2068

*y ^{2} is equal to...this is going to be 30 times 5, is going to be 150; and then, √5 times √5 becomes 5.*2076

*So then, it is 150 times 5; so again, √5 times √5 is (√5) ^{2}.*2104

*Remember: that square root and the square cancel each other out, so it just becomes 5.*2114

*And then, this becomes 750; and then, y becomes √750, and to simplify that out, it becomes...*2119

*let's see...25...OK, well, here, y equals 5√30; I believe that is correct.*2158

*Or you can just, again, change it to decimals with your calculator.*2191

*It is kind of a large number, so just go ahead and use your calculator.*2196

*And that is it for this problem; here is x, and here is y, for this.*2205

*Now, we are going to go on to the next problem; and for the next one,*2214

*we are going to go over a couple more, just so you can get more familiar with these types of problems.*2217

*Again, here we have the geometric mean, the altitude being the geometric mean between this part and this part.*2223

*So then, the geometric mean is going to go here and here, between these two numbers, x and 8.*2236

*8x = 9; x is equal to 9/8; and then, for the y, again, you can use the same concept,*2250

*the theorem that says that this is the geometric mean between the whole thing and then this part.*2268

*Or we can use the Pythagorean theorem; let's go ahead and just use the Pythagorean theorem this time.*2274

*So then, a ^{2}, the leg squared, plus the other leg squared, is equal to the hypotenuse squared.*2280

*3 ^{2} + 8^{2} = y^{2}.*2288

*3 ^{2} is 9, plus 64, is equal to y^{2}; 73 = y^{2}; y = √73.*2294

*And you can just leave it like that; that does not simplify.*2311

*The next one: the same thing: this is the geometric mean between this part and that part.*2320

*So then, 12...12...the geometric between 8 and x...8x = 144; divide the 8; x is equal to 18.*2331

*And then, here, to find the y, the same thing happens.*2359

*We are going to say that this...we know that, since this is 18, we can use the Pythagorean theorem.*2378

*12 ^{2} + 18^{2} is equal to y^{2}.*2385

*Or we can use the theorem; let's just use the theorem: y...y; that is the geometric mean between the whole thing...*2390

*18 + 8 is 26; that is the whole thing; this whole thing would be 26...and then this part that is close to it is 18.*2400

*Those are my extremes, and then my mains would be this: y ^{2} is equal to--*2418

*and you can just use your calculator for this part...it is 26 times 18, which is 468;*2427

*and then, y is equal to the square root of 468, and again, you can just use your calculator to simplify that out.*2440

*That is it for this lesson; thank you for watching Educator.com.*2456

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