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In a ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 6 cm, DB = 9 cm and AE = 8 cm, Find AC.
Given: Δ ABC where Length of side AD = 6 cm, DB = 9 cm, AE = 8 cm Also, DE ∥ BC, To find : Length of side AC By using Thales Theorem we get, (As DE ∥ BC) AD/BD = AE/CE – equation 1 Let CE = x. So then putting values in equation 1, we get 6/9 = 8/x 6x = 72 cm x = 72/6 cm x = 12 cm ∴ AC = AE + CE =Read more
Given: Δ ABC where
Length of side AD = 6 cm, DB = 9 cm, AE = 8 cm
Also, DE ∥ BC,
To find : Length of side AC
By using Thales Theorem we get, (As DE ∥ BC)
AD/BD = AE/CE – equation 1
Let CE = x.
So then putting values in equation 1, we get
6/9 = 8/x
6x = 72 cm
x = 72/6 cm
x = 12 cm
∴ AC = AE + CE = 12 + 8 = 20.
Therefore, Length of side AC is 20 cm.
See lessIn a ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD/DB = 3/4 and AC = 15 cm, Find AE.
Given: Length of side AD/BD = 3/4 and AC = 15 cm To find: Length of side AE By using Thales Theorem, we get AD/BD = AE/CE – equation 1 Let, AE = x Then CE = 15 – x. Now, putting values in equation 1, ⇒ 3/4 = x/ (15 – x) 45 – 3x = 4x -3x – 4x = – 45 7x = 45 x = 45/7 x = 6.43 cm ∴ AE= 6.4Read more
Given:
Length of side AD/BD = 3/4 and AC = 15 cm
To find: Length of side AE
By using Thales Theorem, we get
AD/BD = AE/CE – equation 1
Let, AE = x
Then CE = 15 – x.
Now, putting values in equation 1,
⇒ 3/4 = x/ (15 – x)
45 – 3x = 4x
-3x – 4x = – 45
7x = 45
x = 45/7
x = 6.43 cm
∴ AE= 6.43cm
Therefore, Length of side AE is 6.43 cm
See lessIn a ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD/DB = 2/3 and AC = 18 cm, Find AE.
Given: Length of side AD/BD = 2/3 and AC = 18 cm To find: Length of side AE. By using Thales Theorem, we get AD/BD = AE/CE – equation 1 Let, AE = x Then CE = 18 – x Now, putting values in equation 1, ⇒ 23 = x/ (18 – x) 3x = 36 – 2x 5x = 36 cm x = 36/5 cm x = 7.2 cm ∴ AE = 7.2 cmRead more
Given:
Length of side AD/BD = 2/3 and AC = 18 cm
To find: Length of side AE.
By using Thales Theorem, we get
AD/BD = AE/CE – equation 1
Let, AE = x
Then CE = 18 – x
Now, putting values in equation 1,
⇒ 23 = x/ (18 – x)
3x = 36 – 2x
5x = 36 cm
x = 36/5 cm
x = 7.2 cm
∴ AE = 7.2 cm
Therefore, Length of side AE is 7.2 cm
See lessIn a ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC.If AD = 4 cm, AE = 8 cm, DB = x – 4 cm and EC = 3x – 19, find x.
Given: Length of side AD = 4 cm, AE = 8 cm, DB = x – 4 and EC = 3x – 19 To Find: length of side x. By using Thales Theorem, we get, AD/BD = AE/CE – equation 1 Now, putting values in equation 1, 4/ (x – 4) = 8/ (3x – 19) 4(3x – 19) = 8(x – 4) 12x – 76 = 8(x – 4) 12x – 8xRead more
Given:
Length of side AD = 4 cm, AE = 8 cm, DB = x – 4 and EC = 3x – 19
To Find: length of side x.
By using Thales Theorem, we get,
AD/BD = AE/CE – equation 1
Now, putting values in equation 1,
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
Therefore, Length of side x is 11 cm
See lessIn a ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC.If AD = 8 cm, AB = 12 cm and AE = 12 cm, find CE.
Given: Length of side AD = 8 cm, AB = 12 cm, and AE = 12 cm. To find: Length of side CE. By using Thales Theorem, we get AD/BD = AE/CE – equation 1 Now, putting values in equation 1, 8/4 = 12/CE 8 x CE = 4 × 12 cm CE = (4 × 12)/8 cm CE = 48/8 cm ∴ CE = 6 cm Therefore, LRead more
Given:
Length of side AD = 8 cm, AB = 12 cm, and AE = 12 cm.
To find: Length of side CE.
By using Thales Theorem, we get
AD/BD = AE/CE – equation 1
Now, putting values in equation 1,
8/4 = 12/CE
8 x CE = 4 × 12 cm
CE = (4 × 12)/8 cm
CE = 48/8 cm
∴ CE = 6 cm
Therefore, Length of side x is 6 cm
See lessIn a ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC.If AD = 4 cm, DB = 4.5 cm and AE = 8 cm, find AC.
Given: Length of side AD = 4 cm, DB = 4.5 cm, AE = 8 cm To find: Length of side AC. By using Thales Theorem, we get AD/BD = AE/CE – equation 1 Now, putting values in equation 1, 4/4.5 = 8/AC AC = (4.5 × 8)/4 cm ∴AC = 9 cm Therefore, Length of side AC is 9 cm
Given:
Length of side AD = 4 cm, DB = 4.5 cm, AE = 8 cm
To find: Length of side AC.
By using Thales Theorem, we get
AD/BD = AE/CE – equation 1
Now, putting values in equation 1,
4/4.5 = 8/AC
AC = (4.5 × 8)/4 cm
∴AC = 9 cm
Therefore, Length of side AC is 9 cm
See lessIn a ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 2 cm, AB = 6 cm and AC = 9 cm, find AE.
Given: Length of side AD = 2 cm, AB = 6 cm and AC = 9 cm To find: Length of side AC. Length of DB = AB – AD = 6 – 2 = 4 cm By using Thales Theorem, we get AD/BD = AE/CE – equation 1 Now, putting values in equation 1, 2/4 = x/(9 – x) 4x = 18 – 2x 6x = 18 x = 3Read more
Given:
Length of side AD = 2 cm, AB = 6 cm and AC = 9 cm
To find: Length of side AC.
Length of DB = AB – AD = 6 – 2 = 4 cm
By using Thales Theorem, we get
AD/BD = AE/CE – equation 1
Now, putting values in equation 1,
2/4 = x/(9 – x)
4x = 18 – 2x
6x = 18
x = 3 cm
∴ AE = 3cm
Therefore, Length of side AE is 3 cm
See lessIn a ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC.If AD/BD = 4/5 and EC = 2.5 cm, Find AE.
Given: Length of side AD/BD = 4/5 and EC = 2.5 cm To find: Length of side AC. By using Thales Theorem, we get AD/BD = AE/CE – equation 1 Now, putting values in equation 1, Then, 4/5 = AE/2.5 ∴ AE = 4 × 2.55 = 2 cm Therefore, Length of side AE is 2 cm
Given:
Length of side AD/BD = 4/5 and EC = 2.5 cm
To find: Length of side AC.
By using Thales Theorem, we get
AD/BD = AE/CE – equation 1
Now, putting values in equation 1,
Then, 4/5 = AE/2.5
∴ AE = 4 × 2.55 = 2 cm
Therefore, Length of side AE is 2 cm
See lessIn a ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = x cm, DB = x – 2 cm, AE = x + 2 cm, and EC = x – 1 cm, find the value of x.
Given: Length of side AD = x, DB = x – 2, AE = x + 2 and EC = x – 1 To find: Value of x. By using Thales Theorem, we get AD/BD = AE/CE – equation 1 Now, putting values in equation 1, x/(x – 2) = (x + 2)/(x – 1) x(x – 1) = (x – 2)(x + 2) x 2– x – x2Read more
Given:
Length of side AD = x, DB = x – 2, AE = x + 2 and EC = x – 1
To find: Value of x.
By using Thales Theorem, we get
AD/BD = AE/CE – equation 1
Now, putting values in equation 1,
x/(x – 2) = (x + 2)/(x – 1)
x(x – 1) = (x – 2)(x + 2)
x 2– x – x2 + 4 = 0
∴ x = 4
Therefore, the value of x is 4
See lessIn a ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC.If AD = 8x – 7 cm, DB = 5x – 3 cm, AE = 4x – 3 cm, and EC = (3x – 1) cm, Find the value of x.
Given: Length of side AD = 8x – 7, DB = 5x – 3, AER = 4x – 3 and EC = 3x -1 To find : Value of x By using Thales Theorem, we get AD/BD = AE/CE – equation 1 Now, putting values in equation 1, (8x – 7)/(5x – 3) = (4x–3)/ (3x–1) (8x – 7)(3x – 1) = (5x – 3)(4x – 3)Read more
Given:
Length of side AD = 8x – 7, DB = 5x – 3, AER = 4x – 3 and EC = 3x -1
To find : Value of x
By using Thales Theorem, we get
AD/BD = AE/CE – equation 1
Now, putting values in equation 1,
(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
Since, we know that the side of triangle is always positive.
Therefore, we take the positive value.
∴ x = 1.
Therefore, the value of x is 1.
See less