Question : In $\triangle ABC$, $D$ and $E$ are points on sides $AB$ and $AC$, such that $DE$ II $BC$. If $AD=x+3$, $DB =2 x-3$, $A E=x+1$ and $EC=2 x-2$, then the value of $x$ is:
Option 1: $\frac{4}{5}$
Option 2: $\frac{1}{2}$
Option 3: $\frac{3}{5}$
Option 4: $\frac{1}{5}$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
Correct Answer: $\frac{3}{5}$
Solution :
As $DE \parallel BC$, and $\angle A$ is common in both $\triangle ABC$ and $\triangle ADE$.
⇒ $\triangle ABC \sim \triangle ADE$
⇒ $\frac{AB}{AD} = \frac{AC}{AE}$
⇒ $\frac{AD+DB}{AD} = \frac{AE+EC}{AE}$
⇒ $\frac{ (x + 3 + 2x - 3)}{(x + 3)} = \frac{(x + 1 + 2x-2)}{(x + 1)}$
⇒ $\frac{(3x)}{(x+3)}=\frac{(3x-1)}{(x + 1)}$
⇒ $3x^2+3x= 3x^2 - x + 9x - 3$
⇒ $3x = 8x - 3$
⇒ $5x = 3$
$\therefore x= \frac{3}{5}$
Hence, the correct answer is $\frac{3}{5}$.
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Get Updates Brochure
Your Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.
