Areas | Problems 17-27 | Problem Solving | Volchkevich | Geometry Lessons Grades 7-8
17. The vertex of a parallelogram and the midpoints of the opposite
sides form a triangle. What fraction of its area is its area of the entire parallelogram?
18. How can a square be cut into three equal parts by two lines passing through its vertex?
19. A point inside a parallelogram was connected to all of its vertices. The resulting segments divided it into four triangles. The areas of three of them, taken in order, are 2, 4, and 5. Find the area of the fourth.
20. Two parallelograms are arranged as shown in the figure. Prove that their areas are equal.
21. Two points on the sides of a parallelogram were connected to its three vertices as shown in the figure. Prove that the area of one of the shaded parts in the figure is equal to the sum of the areas of the others.
22. A parallelogram was cut into four smaller parallelograms. Two of them, shaded in the figure, have equal
areas. Prove that their common vertex lies on the diagonal
of the large parallelogram.
23. An arbitrary point M is chosen inside parallelogram ABCD. Line BM intersects AD at point E. Prove that the areas
of triangles AMD and CME are equal.
24. (Kite Lemma.) The diagonals divide a quadrilateral into four triangles. Prove that the product of the areas of two triangles adjacent to its opposite sides is equal to the product of the areas of the other two
triangles.
25. The diagonals of a trapezoid divide it into four triangles.
The areas of two of these triangles, adjacent to the bases, are equal to 1 and 4.
Find the area of the trapezoid.
26. A diagonal is drawn in a parallelogram, and a line is drawn through a vertex not lying on it. They divided the parallelogram into
three triangles and a quadrilateral. The areas of the two triangles in the figure are 1 and 3. Find the area of the quadrilateral.
27. (Theorem on the ratio of areas of triangles with an equal angle.) Two triangles have an equal angle. Prove
that their areas are related as the products of the sides enclosing
that angle.