Question

2. If in AABC, for what value of x, DE || AB, where !
CE = x,
Value of X, DE || AB, where AD = 8x + 9, CD = x+3, BE = 3x+4and CE=x​

Answer 1

Step-by-step explanation:

The value of x is 2

Step 1: Given Data DE IS Parallel to AB, AD=8X+9, CD=X+3.BE=3X+4, CE=X

Step 2: (Therefore CD/DA is equal to CE/EB)

Step 3: (Cross Multiplication on both sides)

Step 4: To calculate the value of x

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

Step 5:

 (Multiply – sign on both the sides)

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

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

Step 6: x-2=0  (Assign x-2 is equal to zero)

x=2

Hope you all understand the answer dear......

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.