Areas of polygons

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Here are some problems related to the concept of area of ​​a polygon. We give an axiomatic definition of the area of ​​a polygon, listing those properties that we need.

A1. The area of ​​a polygon is a positive number.
A2. The areas of equal (congruent) polygons are equal.
A3. If a polygon is cut into several parts, then its area is equal to the sum of the areas of these parts.
A4. The area of ​​a triangle is equal to half the product of its base and height. (We consider the unit of scale to be given; it is easy to show that this product does not depend on which side and the corresponding height we take.)
The letter A was chosen because the word “Axiom” begins with it. From these four properties of the axioms you can derive all the theorems about the areas of polygons that you study in school.

It is much more difficult to prove that, indeed, it is possible to associate with each polygon $M$ in the plane the area number $S(M)$, its area, so that properties A1–A4 are satisfied. This is true, but it is usually not proven in a school course and is considered very dynamite. Instead of A4, we can consider the main property from which A4 is derived:

A4’. The area of ​​a square with side 1 is 1.

For example:

a) The area of ​​a parallelogram is equal to the product of its base and its height;
b) The area of ​​a trapezoid is equal to the product of half the sum of the bases and the height.
Think about how to prove these statements (that is, derive from axioms A1-A4).
We’ll start by proving another formula for the area of ​​a trapezoid.

1. The area of ​​a trapezoid is equal to the product of one of the sides and the perpendicular dropped onto it from the middle of the other side.

Proof. Let $ABCD$ be the given trapezoid $(AD \parallel BC)$, $K$ the midpoint of side $CD$, $KH$ the perpendicular dropped from the point $K$ to the line $AB$. Let us draw through the point $K$ a line parallel to the line $AB$. Let $M$ and $P$ be its points

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7. Prove that the area of ​​the right one is the largest and the smallest of its diagonals.

Note. Take a look at figure4.

8. Prove that the area of ​​a polygon circumscribed about a circle is equal to the product of the radius of this circle and the perimeter of the polygon.

In these problems, you can bypass the “cut and fold” technique. Note that, in principle, this technique can always be used when you need to prove the equality of the areas of two polygons; The following theorem is true: if two polygons are equal in size, then one of them can be cut into parts from which the other polygon is composed.

This theorem is not difficult to prove *). But practically, in order to establish equality of areas or find the ratio of areas, it is not at all necessary to “cut and add”. For example, you can very quickly compare the areas of two triangles using the formula we called A4. It immediately follows, in particular, also correspondences for areas:

if the vertex of a triangle is moved along a straight line parallel to its base, then the area does not change (Fig. 5; triangles ABC

*) See the book by V. G. Boltyansky “Equal-sized and equally spaced figures”, series “Popular lectures on mathematics”, vol. 22, Gostekhidat, M., 1956.

and ADB have a common base AB and equal heights);

if one of the sides of a triangle adjacent to a given angle is increased by k times, then its area will also increase by k times (see Fig. 6; triangles ABC and ABD have a common height AH, therefore the ratio of their areas is equal to the ratio of the bases:

$$S_{\triangle ABC} / S_{\triangle ABD} = BC / BD.$$

This implies a more general correspondence:

If two triangles have a common angle, then their area is proportional to the product of the sides enclosing the angle.

(see Fig. 7; $$S_{\triangle ABC} / S_{\triangle ABD} = BC / BD,$$
$$S_{\triangle ABD} / S_{\triangle EBD} = AB / EB,$$
$$S_{\triangle ABC} / S_{\triangle EBD} = (AB \cdot BC) / (EB \cdot BD)).$$

In particular, if two triangles are similar and the sides of one are k times larger than the corresponding sides, then the area of ​​the first is $k^2$ times the area of ​​the other.

In the following problems, the solution is based precisely on these considerations: on comparing the areas of the triangle

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nicknames that either have the same bases, or the same heights, or the same angles. Try to find the simplest and most beautiful solution!

9. O is the intersection point of the segments AC and BD (Fig. 8). In order for the areas of triangles AOB and DOC to be equal, it is necessary and sufficient that lines BC and AD are parallel. Prove it!

To solve this problem, you need to prove two statements: (1) if the areas of triangles AOB and DOC are equal, then lines BC and AD are parallel; (2) if lines BC and AD are parallel, then the areas of triangles AOB and DOC are equal.

10. Prove that a convex quadrilateral is a moed and polko moed parallel when each of its diagonals bisects its area adb.

And here, as in the previous problem, you need to prove two theorems: direct and inverse.

11. In triangle ABC, the line passing through vertex A and dividing the median BM in the ratio 1:2, counting from the vertex, we intersect side BC at point K. Find the ratio of the areas of triangles ABK and ABC.

Note: Draw a line parallel to AK through point M and use it to find the ratio of segments BK and BC.

12. On the continuation of the convex side BC of the quadrilateral ABCD, find a point O such that the area adb of the quadrilateral ABCD is equal to the area of ​​the triangle ABO.

Note: Draw a line through point D parallel to diagonal AC.

Does problem 12 always have a solution? Is it always the only one?

This problem allows you to turn any convex polygon into an equal-sized one with fewer sides.

13. Two straight lines are drawn through the middle of the altitude of an isosceles triangle, connecting it with the vertices of the base (Fig. 9). What part of the area of ​​the triangle is made up of the parts on which the triangle is still cut by straight lines?

14. Given triangle ABC. We extend its side AB beyond vertex B with the segment BP=AB, side AC beyond vertex A with the segment AM=CA, and side BC beyond vertex C with the segment KC=BC. How many times is the area adb of triangle PKM greater than the area of ​​triangle ABC?

Note: Contain points M and B, P and C, A and K.

15. Form an analocine problem for a quadrilateral and solve it. Derive the result of problem 6 from it.

16. On the sides of the convex quadrilateral ABCD, points M, P, K, H mak are taken, such that AM : MB = 3 : 5; BP: PC=1: 3; CK: KD=4: 5; DH : HA = 1 : 8. Find the ratio of the area of ​​the quadrilateral MBPKDH to the area of ​​the quadrilateral ABCD. Think, given any relationship between AM and MB, BP and PC and mak, can this problem be solved?

Note: Draw a diagonal and use A3.

17. In a convex quadrilateral, the midpoints of adjacent hundred-