Question

1. In a triangle ABC, DE || BC and AD = 4x - 3, DB = 3x - 1 , AE = 8x - 7, EC = 5x - 3 find the value of x.

Class 10

Content Quality Solution Required

No Spamming

Answer 1

Hey!!!!

Good Morning

Difficulty Level : Easy

Chances of being asked in Board : 50%

I'm sorry I'm out of country so couldn't arrange paper, kindly consider the drawing

_____________________

Here DE II BC

Thus by BPT(Basic Proportionality Theorem or Thales Theorem), we know

=> AD/DB = AE/EC

Thus replacing values

=> (4x - 3)/(3x - 1) = (8x - 7)/(5x - 3)

=> (4x - 3)(5x - 3) = (8x - 7)(3x - 1)

=> 20x² - 12x - 15x + 9 = 24x² - 8x - 21x + 7

=> 20x² - 27x + 9 = 24x² - 29x + 7

=> 4x² - 2x - 2 = 0

=> 4x² - 4x + 2x - 2 = 0

=> 4x(x - 1) + 2(x - 1) = 0

=> (4x + 2)(x - 1) = 0

Thus x = 1 or x = - 1/2 <<<<<< Answer

________________

Hope this helps ✌️

Good morning :-)

Answer 2

HELLO !!!!....

GOOD MORNING :))

__________________________


●GIVEN :-

AD = 4x - 3,
DB = 3x - 1 ,
AE = 8x - 7,
EC = 5x - 3

Since DE || BC

So, by basic proportionality theorem ,

AD/DB = AE/EC

$\frac{4x - 3}{3x - 1} = \frac{8 x- 7}{5x - 3} \\ \\ = > 20 {x}^{2} - 27x + 9 = 24 {x}^{2} - 29x + 7 \\ \\ = > 4 {x}^{2} - 2x - 1 = 0 \\ \\ = > 2 {x}^{2} - x - 1 = 0 \\ \\ = > 2 {x}^{2} - 2x + x - 1 = 0 \\ \\ = > 2x(x - 1) + 1(x - 1) = 0 \\ \\ = > (2x + 1) \: \: \: (x - 1) = 0 \\ \\ = > x = 1 \: \: \: or \: \: \: - \frac{1}{2}$


__________________________


HOPE THIS HELPS U ☺☺

CHEERS! ! !

# NIKKY ✌ ✌