In ∆ABC, D and E are points on the side AB and
AC respectively such that DE || BC. AD = 4x-3,
AE = 8x - 7, BD = 3x - 1 and CE = 5X - 3, find the
value of x.[CBSE 2002, SP 2010] { Answer :- x=1}
Answers
Step-by-step explanation:
The basic proportionality theorem states that if a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.
Given : Δ ABC & DE || BC , AD = (8x -7), DB = 5x – 3, AE= 4x – 3 and EC = 3x -1
So, AD/DB=AE/ EC
[By using basic proportionality Theorem]
Then,( 8x–7)/(5x–3) = (4x–3)/(3x–1)
(8x – 7)(3x – 1) = (5x – 3)(4x – 3)
24x² – 8x - 21x + 7 = 20x² - 15x -12x + 9
24x² – 29x + 7 = 20x² - 27x + 9
24x² - 20x² – 29x + 27x + 7 - 9= 0
4x² – 2x – 2 = 0
2(2x² – x – 1) = 0
2x² – x – 1 = 0
2x² – 2x + x – 1 = 0
[By Middle term splitting]
2x(x – 1) + 1(x – 1) = 0
(x – 1)(2x + 1) = 0
x = 1 or x = -1/2
[Since the side of triangle can never be negative]
Therefore, x = 1.
Hence,the value of x is 1 cm.
11) Given : Δ ABC & DE || BC ,
It is given that AD = 4x – 3, BD = 3x – 1, AE = 8x – 7 and EC = 5x – 3
So, AD/DB = AE/ EC
[By using basic proportionality Theorem]
Then, (4x–3)/(3x–1) =(8x–7)/(5x–3)
(4x – 3)(5x – 3) = (3x – 1)(8x – 7)
20x² – 12x – 15x + 9 = 24x² – 21x - 8x + 7
20x² - 27x + 9 = 24x² - 29x + 7
20x² - 24x² - 27 x + 29 x + 9 - 7= 0
- 4x² + 2x + 2 = 0
4x² – 2x – 2 = 0
4x² – 4x + 2x – 2 = 0
[By Middle term splitting]
4x(x -1) + 2(x - 1) = 0
(4x + 2)(x - 1) = 0
x = -2/4 or x = 1
[side of triangle can never be negative]
Therefore, x = 1
Hence,the value of x is 1 cm.
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hope it helps you.
Answer:
Step-by-step explanation:
