Question

What value(s) of x will makr DE || AB, in the given figure ?

AD = 8x + 9,
CD = x + 3,
BE = 3x + 4,
CE = x.​

Answer 1

Answer :-

• The value x = 2 will make DE // AB.

Step - By - Step Explanation :-

Given :-

• In ∆ABC, DE // AB
• AD = 8x + 9
• CD = x + 3
• BE = 3x + 4
• CE = x

By Basic proportionality theorem,

• If DE // AB then we should have

CD/DA = CE/EB

.°. x+3/8x+9 = x/3x+4

=> (x + 3) (3x + 4) = x (8x + 9)

=> x (3x + 4) + 3 (3x + 5) = 8x² + 9x

=> 3x² + 4x + 9x + 12 = 8x² + 9x

=> 8x² + 9x - 3x² - 4x - 9x - 12 = 0

=> 5x² - 4x - 12 = 0

=> 5x² - 10x + 6x - 12 = 0

=> 5x (x - 2) + 6 (x - 2) = 0

=> (5x + 6) (x - 2) = 0

=> 5x + 6 = 0 or x - 2 = 0

=> x = -6/2 or x = 2

Here x cannot be negative.

.°. The value x = 2 will make DE // AB .

Answer 2

Given, DE//AB

$\frac{AD}{DC} = \frac{BE}{EC} \\ \\ \frac{8x + 9}{x + 3} = \frac{3x + 4}{x} \\ \\ {8x}^{2} +9x= {3x}^{2} +13x+12 \\ \\ {5x}^{2} −4x−12=0 \\ \\ {5x}^{2} −10x+6x−12=0 \\ \\ 5x(x−2)+6(x−2)=0 \\ \\ (5x+6)(x−2)=0 \\ \\ x = \frac{ - 6}{5} \: and \: 2$

Therefore x is 2.