Question

In ABC, DE , find the value of x. IF AD=x DB=x+1 AE=3cm EC=5cm​

Answer 1

Answer:

i hope this is your answer hope it help..

Answer 2

Note: there is an error in the question as in place of just DE it should be DE || BC.

Given: In triangle ABC $AD=x, DB=x+1, AE=3 cm, EC=5 cm$.

To find: the value of x.

Solution:

Draw the required figure.

Understand that according to thales theorem, If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio.

Find the value of x.

$\[\frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}\]$

$\[ \Rightarrow \frac{x}{{x + 1}} = \frac{3}{5}\]$

$\[\begin{array}{l} \Rightarrow 5x = 3\left( {x + 1} \right)\\ \Rightarrow 5x = 3x + 3\\ \Rightarrow 5x - 3x = 3\end{array}\]$

$\[\begin{array}{l} \Rightarrow 2x = 3\\ \Rightarrow x = \frac{3}{2}\end{array}\]$

Hence, the value of x is $\[\frac{3}{2}\]$.