Triangles with a common base and equal heights have equal areas (equal-area transformations). This property is often used in relationships between the areas of parallelograms and triangles.
Parallelograms and Triangles
Area of parallelogram ABCD = 2 × area of △ABC (bisected by a diagonal)
Example ①
When the area of parallelogram ABCD is 24 cm², find the area of △ABC.
Solution: △ABC = 24 ÷ 2 = 12 cm²
When the area of parallelogram ABCD is 24 cm², find the area of △ABC.
Solution: △ABC = 24 ÷ 2 = 12 cm²
Example ② (Equal-Area Transformation)
Points A and B lie on parallel lines l and m, respectively, and point C is taken on l. How can you construct another triangle that has the same area as △ABC and shares vertex B?
Solution: Take another point D on l so that, using BC as the base, the height (the distance between l and m) is the same. Then the area of △DBC = the area of △ABC.
Points A and B lie on parallel lines l and m, respectively, and point C is taken on l. How can you construct another triangle that has the same area as △ABC and shares vertex B?
Solution: Take another point D on l so that, using BC as the base, the height (the distance between l and m) is the same. Then the area of △DBC = the area of △ABC.
Key Point
- Common base + equal heights → equal areas
- Area of a parallelogram = 2 × the area of a triangle formed by a diagonal
Practice Problem
- Find the area of a parallelogram with a base of 8 cm and a height of 5 cm, and the area of the triangle formed by its diagonal.
Answer and Explanation
- Answer: Parallelogram: 40 cm²; triangle: 20 cm².