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Explore the Back-StoryLet's learn how to calculate the geometric mean. This is a concept video useful for Mathematics Olympiad and ISI and CMI Entrance.

Suppose a and b are positive numbers then their geometric mean is defined as square root of a times b. This is the formula of geometric mean.

We will construct the geometric mean of a and b geometrically. So, here are the steps:

- Let's start by drawing the diameter of a circle.
- Suppose the two endpoints are X and Y of the diameter.
- Choose any point M on this particular diameter
- Now, let's construct a perpendicular at M which hits the circle at N.
- M divides the diameter into two parts XM and YM. Suppose, the length of XM and YM be a and b respectively.
- Now, we want to show that the length of MN is the square root of a times b, i.e., the geometric mean of a and b.

7. Now, join XN and YN, and then look at the triangles YMN and XMN and try to show that these two triangles are actually similar. Now, see the following proof:

8. Since, we know that similar triangles have proportional sides. So, we can say,

Hence, this is proved.

Hope you liked it!

- AM-GM Inequality - Math Olympiad Concepts Video
- Triangular Number Sequence - Explanation with Application

Let's learn how to calculate the geometric mean. This is a concept video useful for Mathematics Olympiad and ISI and CMI Entrance.

Suppose a and b are positive numbers then their geometric mean is defined as square root of a times b. This is the formula of geometric mean.

We will construct the geometric mean of a and b geometrically. So, here are the steps:

- Let's start by drawing the diameter of a circle.
- Suppose the two endpoints are X and Y of the diameter.
- Choose any point M on this particular diameter
- Now, let's construct a perpendicular at M which hits the circle at N.
- M divides the diameter into two parts XM and YM. Suppose, the length of XM and YM be a and b respectively.
- Now, we want to show that the length of MN is the square root of a times b, i.e., the geometric mean of a and b.

7. Now, join XN and YN, and then look at the triangles YMN and XMN and try to show that these two triangles are actually similar. Now, see the following proof:

8. Since, we know that similar triangles have proportional sides. So, we can say,

Hence, this is proved.

Hope you liked it!

- AM-GM Inequality - Math Olympiad Concepts Video
- Triangular Number Sequence - Explanation with Application

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