How to find the area of an isosceles triangle

Sometimes the question of how to find the area of an isosceles triangle rises not only before students or students, but also in real, practical life. For example, during construction, it becomes necessary to finish the facade part, which is under the roof. How can I calculate the amount of material I need?

Often with similar tasks, craftsmen who work with cloth or leather face. After all, many details that are to be found to the master, have just the shape of an isosceles triangle.

So, there are several ways to help find the area of an isosceles triangle. The first is the calculation of its base and height.

For the solution, we need to construct for clarity the triangle MNP with the base MN and the height PO. Now we will finish something in the drawing: from point P draw a line parallel to the base, and from point M - a line parallel to the height. The intersection point is called Q. To find out how to find the area of an isosceles triangle, we need to consider the resulting quadrilateral MOPQ, in which the side of the given triangle MP is already its diagonal.

We first prove that this is a rectangle. Since we built it ourselves, we know that the sides of MO and OQ are parallel. And the sides of QM and OP are also parallel. The angle POM is straight, so the angle OPQ is also a straight line. Consequently, the resulting quadrilateral is a rectangle. Find its area is not difficult, it is equal to the product of PO on OM. OM is half the base of this triangle MPN. It follows that the area of the rectangle we constructed is equal to the half-product of the height of a right-angled triangle on its base.

The second stage of the problem before us, how to determine the area of a triangle, is the proof of the fact that the rectangle we obtained corresponds to a given isosceles triangle, that is, that the area of the triangle is also equal to the half-product of the base and height.

Let us compare the triangle PON and PMQ for the beginning. They are both rectangular, since the right angle in one of them is formed by the height, and the right angle in the other is the angle of the rectangle. Hypotenuses in them are sides of an isosceles triangle, hence, are also equal. The PO and QM cateches are also equal as the parallel sides of the rectangle. Hence, both the area of the triangle PON, and the triangle PMQ are equal to each other.

The area of the rectangle QPOM is equal to the areas of the triangles PQM and MOP in the sum. Replacing the superimposed triangle QPM with the triangle PON, we obtain in sum the triangle given to us for the derivation of the theorem. Now we know how to find the area of an isosceles triangle on the basis and height - to calculate their half-product.

But you can learn how to find the area of an isosceles triangle on the base and side. Here, too, there are two options: the theorem of Geron and Pythagoras. We consider the solution using the Pythagorean theorem. For example, let's take the same isosceles triangle PMN with the height PO.

In the rectangular triangle, POM MP is the hypotenuse. Its square is equal to the sum of the squares of PO and OM. And since OM is half the base, which we know, we can easily find OM and raise the number square. By subtracting the obtained number from the square of the hypotenuse, we find out what the square of the other leg, which in the isosceles triangle is the height, is equal to. Finding the square root of the difference and recognizing the height of a right triangle, you can give an answer to the task assigned to us.

You just need to multiply the height by the bottom and divide the resulting result in half. Why this should be done, we explained in the first version of the proof.

It happens that you need to make calculations on the side and the corner. Then we find the height and the base, using the formula with sines and cosines, and, again, multiply them and divide the result in half.