Question

In ∆ABC, DE ∥ BC so that AD = (7x − 4) cm, AE = (5x − 2) cm, DB = (3x + 4) cm and EC = 3x cm. Then, we have:


(a) x = 3
(b) x = 5
(c) x = 4
(d) x = 2.5

Answer 1

(c) x = 4

It is given that $DE\parallel BC$.
Applying Thales' theorem, we get:

$$\begin{gathered} \frac{AD}{BD}=\frac{AE}{EC} \\ \Rightarrow \frac{7x-4}{3x+4}=\frac{5x-2}{3x} \\ \Rightarrow 3x\left(7x-4\right)=\left(5x-2\right)\left(3x+4\right) \\ \Rightarrow 21x^2-12x=15x^2+20x-6x-8 \\ \Rightarrow 21x^2-12x=15x^2+14x-8 \\ \Rightarrow 21x^2-12x-15x^2-14x+8=0 \\ \Rightarrow 6x^2-26x+8=0 \\ \Rightarrow 2\left(3x^2-13x+4\right)=0 \\ \Rightarrow 3x^2-13x+4=0 \\ \Rightarrow 3x^2-12x-x+4=0 \\ \Rightarrow 3x\left(x-4\right)-1\left(x-4\right)=0 \\ \Rightarrow \left(x-4\right)\left(3x-1\right)=0 \\ \Rightarrow x-4=0\mathrm{or}3x-1=0 \\ \Rightarrow x=4\mathrm{or}x=\frac{1}{3} \\ \mathrm{If}x=\frac{1}{3},7x-4=-\frac{5}{3}<0;\mathrm{it}\mathrm{is}\mathrm{not}\mathrm{possible}. \\ \mathrm{Therefore},x=4 \end{gathered}$$