Math, asked by sonusahani7, 10 months ago

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

Answered by dhruvsingla222
1

[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.

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