In a triangle DE//BC, 2 EX=CX & AD=BD. Then prove AE:EX=3:1
Answers
Answer:
1. In a Δ ABC, D and E are points on the sides AB and AC respectively such that DE || BC.
i) If AD = 6 cm, DB = 9 cm and AE = 8 cm, Find AC.
Solution:
Given: Δ ABC, DE ∥ BC, AD = 6 cm, DB = 9 cm and AE = 8 cm.
Required to find AC.
By using Thales Theorem, [As DE ∥ BC]
AD/BD = AE/CE
Let CE = x.
So then,
6/9 = 8/x
6x = 72 cm
x = 72/6 cm
x = 12 cm
∴ AC = AE + CE = 12 + 8 = 20.
ii) If AD/DB = 3/4 and AC = 15 cm, Find AE.
Solution:
Given: AD/BD = 3/4 and AC = 15 cm [As DE ∥ BC]
Required to find AE.
By using Thales Theorem, [As DE ∥ BC]
AD/BD = AE/CE
Let, AE = x, then CE = 15-x.
⇒ 3/4 = x/ (15–x)
45 – 3x = 4x
-3x – 4x = – 45
7x = 45
x = 45/7
x = 6.43 cm
∴ AE= 6.43cm
iii) If AD/DB = 2/3 and AC = 18 cm, Find AE.
Solution:
Given: AD/BD = 2/3 and AC = 18 cm
Required to find AE.
By using Thales Theorem, [As DE ∥ BC]
AD/BD = AE/CE
Let, AE = x and CE = 18 – x
⇒ 23 = x/ (18–x)
3x = 36 – 2x
5x = 36 cm
x = 36/5 cm
x = 7.2 cm
∴ AE = 7.2 cm
iv) If AD = 4 cm, AE = 8 cm, DB = x – 4 cm and EC = 3x – 19, find x.
Solution:
Given: AD = 4 cm, AE = 8 cm, DB = x – 4 and EC = 3x – 19
Required to find x.
By using Thales Theorem, [As DE ∥ BC]
AD/BD = AE/CE
Then, 4/ (x – 4) = 8/ (3x – 19)
4(3x – 19) = 8(x – 4)
12x – 76 = 8(x – 4)
12x – 8x = – 32 + 76
4x = 44 cm
x = 11 cm
v) If AD = 8 cm, AB = 12 cm and AE = 12 cm, find CE.
Solution:
Given: AD = 8 cm, AB = 12 cm, and AE = 12 cm.
Required to find CE,
By using Thales Theorem, [As DE ∥ BC]
AD/BD = AE/CE
8/4 = 12/CE
8 x CE = 4 x 12 cm
CE = (4 x 12)/8 cm
CE = 48/8 cm
∴ CE = 6 cm
vi) If AD = 4 cm, DB = 4.5 cm and AE = 8 cm, find AC.
Solution:
Given: AD = 4 cm, DB = 4.5 cm, AE = 8 cm
Required to find AC.
By using Thales Theorem, [As DE ∥ BC]
AD/BD = AE/CE
4/4.5 = 8/AC
AC = (4.5 × 8)/4 cm
∴AC = 9 cm
vii) If AD = 2 cm, AB = 6 cm and AC = 9 cm, find AE.
Solution:
Given: AD = 2 cm, AB = 6 cm and AC = 9 cm
Required to find AE.
DB = AB – AD = 6 – 2 = 4 cm
By using Thales Theorem, [As DE ∥ BC]
AD/BD = AE/CE
2/4 = x/ (9–x)
4x = 18 – 2x
6x = 18
x = 3 cm
∴ AE= 3cm
viii) If AD/BD = 4/5 and EC = 2.5 cm, Find AE.
Solution:
Given: AD/BD = 4/5 and EC = 2.5 cm
Required to find AE.
By using Thales Theorem, [As DE ∥ BC]
AD/BD = AE/CE
Then, 4/5 = AE/2.5
∴ AE = 4 × 2.55 = 2 cm
ix) If AD = x cm, DB = x – 2 cm, AE = x + 2 cm, and EC = x – 1 cm, find the value of x.
Solution:
Given: AD = x, DB = x – 2, AE = x + 2 and EC = x – 1
Required to find the value of x.
By using Thales Theorem, [As DE ∥ BC]
AD/BD = AE/CE
So, x/ (x–2) = (x+2)/ (x–1)
x(x – 1) = (x – 2)(x + 2)
x2 – x – x2 + 4 = 0
x = 4
x) If AD = 8x – 7 cm, DB = 5x – 3 cm, AE = 4x – 3 cm, and EC = (3x – 1) cm, Find the value of x.
Solution:
Given: AD = 8x – 7, DB = 5x – 3, AER = 4x – 3 and EC = 3x -1
Required to find x.
By using Thales Theorem, [As DE ∥ BC]
AD/BD = AE/CE
(8x–7)/ (5x–3) = (4x–3)/ (3x–1)
(8x – 7)(3x – 1) = (5x – 3)(4x – 3)
24x2 – 29x + 7 = 20x2 – 27x + 9
4x2 – 2x – 2 = 0
2(2x2 – x – 1) = 0
2x2 – x – 1 = 0
2x2 – 2x + x – 1 = 0
2x(x – 1) + 1(x – 1) = 0
(x – 1)(2x + 1) = 0
⇒ x = 1 or x = -1/2