In triangle ABC, D and E are point on the side a b and ac respectively such that De parallel BC if ad is equal to 1 by 3 BD is equal to 4.5 cm find AC?
Answers
[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.
12) Given : Δ ABC & DE || BC , AD = 2.5 cm, AE = 3.75 cm and BD = 3 cm
So, AD/DB=AE/ EC
[By using basic proportionality Theorem]
Then, 2.5/ 3 = 3.75/CE
2.5CE = 3.75 x 3
CE = (3.75×3)/2.5
CE=11.25/2.5
CE = 4.5
Now, AC = AE + EC
AC = 3.75 + 4.5
AC = 8.25 cm.
Hence, the length of AC is 8.25 cm.