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In the following figures, $ \Delta \mathrm{ADE} $ and $ \Delta \mathrm{ABC} $ are similar triangles. Find the values of ' $ \mathrm{x} $ ' and measure of $ \mathrm{AB} $.

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Given:


$\Delta$ADE and $\Delta$ABC are similar.


To do:


We have to find the values of x and measures of AB.

Solution:

a)  Triangle ADE and ABC are similar.

So, the ratio of their sides is equal.

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

$\frac{x}{x-2} = \frac{x - 3}{x - 4}$

Cross multiply,

$x (x - 4) = (x - 3)(x - 2)$

$x^2- 4 x  =  x^2 - 3x - 2x + 6$

$x^2- 4x  = x^2 - 5 x + 6$ 

$x^2 - 4 x  - ( x^2- 5 x + 6) = 0$

$x^2 - 4 x  -  x^2 + 5 x - 6 = 0$

$-4 x + 5 x - 6 = 0$

$x - 6 = 0$


$x  =  6$ cm

AB  =  AD $+$ DB

AB  =  $x + x -2$

AB  =  $6 + 6 -2  =  12 - 2  =  10$


AB = 10 cm.

 b) Triangle ADE and ABC are similar.

So, the ratio of their sides is equal.

$\frac{AD}{DC} = \frac{AE}{EB}$

$\frac{x}{x - 2} = \frac{x + 2}{x - 1}$

Cross multiply,

$x (x - 1)  =  (x + 2) (x - 2)$

$x^2-  x  =  x^2+2 x - 2 x - 4$

$x^2-  x  =  x^2 - 4$

 

$x^2 -  x  - ( x^2- 4 ) = 0$

$x^2 -  x  -  x^2+ 4 = 0$

$- x  + 4 = 0$

$- x  =- 4$


$x  =  4$ cm.

AB  = AE  $+$ EB

AB  = $x + 2 + x -1$

AB  = $4 + 2 + 4 - 1  =  10 - 1 = 9$ 


AB = 9 cm.


Updated on: 10-Oct-2022

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