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In $\triangle ABC, DE || BC$. If $AE = (2x + 1)$ cm, $EC= 4$ cm, $AD= (x + 1)$ cm and $DB = 3$ cm, then value of $x$ is:
In $\angle ABC, DE || BC$. If $AE = (2x + 1)$ cm, $EC= 4$ cm, $AD= (x + 1)$ cm and $DB = 3$ cm, then value of $x$ is:
a) 1
b) 1/2
c) -1
d) 1/3
b) 1/2
c) -1
d) 1/3
This Question has 1 answers.
Given that $ \displaystyle DE \parallel BC $, by the **Basic Proportionality Theorem (Thales' Theorem)**, we have:
$ \displaystyle \frac{AD}{DB} = \frac{AE}{EC} $
Substituting the given values:
$ \displaystyle \frac{x + 1}{3} = \frac{2x + 1}{4} $
Cross multiplying:
$ \displaystyle (x + 1) \times 4 = (2x + 1) \times 3 $
$ \displaystyle 4x + 4 = 6x + 3 $
Rearranging:
$ \displaystyle 4 + 1 = 6x - 4x $
$ \displaystyle 5 = 2x $
$ \displaystyle x = \frac{5}{2} $
Thus, the value of $ \displaystyle x $ is **$ \displaystyle \frac{5}{2} $**.
$ \displaystyle \frac{AD}{DB} = \frac{AE}{EC} $
Substituting the given values:
$ \displaystyle \frac{x + 1}{3} = \frac{2x + 1}{4} $
Cross multiplying:
$ \displaystyle (x + 1) \times 4 = (2x + 1) \times 3 $
$ \displaystyle 4x + 4 = 6x + 3 $
Rearranging:
$ \displaystyle 4 + 1 = 6x - 4x $
$ \displaystyle 5 = 2x $
$ \displaystyle x = \frac{5}{2} $
Thus, the value of $ \displaystyle x $ is **$ \displaystyle \frac{5}{2} $**.
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