A ray is drawn from vertex A of parallelogram ABCD, which intersects side BC at point P
A ray is drawn from vertex A of parallelogram ABCD, which intersects side BC at point P and diagonal BD at point M. Find the area of triangle BMP if it is known that the area of triangle ABCD is equal to 14 and the area of parallelogram ABCD is 84
The diagonal of the parallelogram divides it into two equal triangles. Savd = Svsd = Savs / 2 = 84/2 = 42 cm. Since the area of the triangle ABM, by convention, is equal to 14 cm2, then Samd = Savd – Savm = 42 – 14 = 28 cm2.
In triangles ABM and AMD, the height drawn from the vertex A is common, therefore, the ratio of the areas of these triangles is equal to the ratio of their bases.
Samd / Sawm = DM / BM = 28/14 = 2/1.
Triangles BMP and AMD are similar in two angles, the angle BMP = AMD as vertical angles at the intersection of straight lines AR and VD, angle BPA = PAD as criss-crossing angles at the intersection of parallel straight lines BP and BC of the secant AR.
The coefficient of similarity of triangles is: K = DK / BM = 2 (determined earlier).
The ratio of the areas of similar triangles is equal to the square of the similarity coefficient.
K2 = Samd / Svmr.
4 = 28 / Svmr.
Svmr = 28/4 = 7 cm2.
Answer: The area of the BMP triangle is 7 cm2.
