In a triangle ABC, DE is parallel to BC; AD = a, DB = a + 4, AE = 2a + 3, EC = 7a. What is the value of 'a’ if a > 0 ?

5 6 3 4

3

By using the concept , we know that \(\frac{AD}{DB}\) = \(\frac{AE}{EC}\) \(\frac{a}{a + 4}\) = \(\frac{2a + 3}{7a}\) 7a² = 2a²  + 3a + 8a + 12 5a² - 11a - 12 = 0 5a² - 15a + 4a - 12 = 0 5a ( a  - 3 ) + 4 ( a - 3 ) = 0 5a + 4 = 0   is not possible So, a - 3 = 0 a = 3