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In a Δ ABC, AD is the bisector of ∠A, meeting side BC at D.
(i) If BD = 2.5 cm, AB = 5 cm, and AC = 4.2 cm, find DC.
(ii) If BD = 2 cm, AB = 5 cm, and DC = 3 cm, find AC.
(iii) If AB = 3.5 cm, AC = 4.2 cm, and DC = 2.8 cm, find BD.
(iv) If AB = 10 cm, AC = 14 cm, and BC = 6 cm, find BD and DC.
(v) If AC = 4.2 cm, DC = 6 cm, and BC = 10 cm, find AB.
(vi) If AB = 5.6 cm, BC = 6 cm, and DC = 3 cm, find BC.
(vii) If AB = 5.6 cm, BC = 6 cm, and BD = 3.2 cm, find AC.
(viii) If AB = 10 cm, AC = 6 cm, and BC = 12 cm, find BD and DC.
(i) It is given that BD = 2.5 cm, AB = 5 cm, and AC = 4.2 cm.
In Δ ABC, AD is the bisector of ∠A, meeting side BC at D.
We need to find DC,
Since, AD is ∠A bisector,
5DC = 4.2 x 2.5
DC = (4.2 x 2.5)/5
DC = 2.1
(ii) It is given that BD = 2 cm, AB = 5 cm, and DC = 3 cm
In Δ ABC, AD is the bisector of ∠A, meeting side BC at D
We need to find AC.
Since, AD is ∠A bisector.
Therefore,
(since AD is the bisector of ∠A and side BC)
2AC = 5 × 3
AC = 15/2
AC = 7.5 cm
(iii) It is given that AB = 3.5 cm, AC = 4.2 cm, and DC = 2.8 cm
We need to find BD.
Since, AD is ∠A bisector
(since, AD is the bisector of ∠A and side BC)
BD = (3.5 × 2.8)/4.2
BD = 7/3
BD = 2.3 cm
(iv) It is given that AB = 10 cm, AC = 14 cm, and BC = 6 cm
In Δ ABC, AD is the bisector of ∠A meeting side BC at D
We need to find BD and DC.
Since, AD is bisector of ∠A
(AD is bisector of ∠ A and side BC)
14x = 60 – 6x
20x = 60
x = 60/20
BD = 3 cm and DC = 3 cm.
(v) It is given that AC = 4.2 cm, DC = 6 cm, and BC = 10 cm.
We need to find out AB,
Since, AD is the bisector of ∠A
6AB = 4.2 x 4
AB = (4.2 x 4)/6
AB = 16.8/6
AB = 2.8 cm
(vi) It is given that AB = 5.6 cm, BC = 6 cm, and DC = 3 cm
We need to find BC,
Since, AD is the ∠A bisector
DC = 2.8 cm
And, BC = 2.8 + 3
BC = 5.8 cm
(vii) It is given that AB = 5.6 cm, BC = 6 cm, and BD = 3.2 cm
In Δ ABC, AD is the bisector of ∠A , meeting side BC at D
AC = (5.6 x 2.8)/3.2
AC = 4.9 cm
(viii) It is given that AB = 10 cm, AC = 6 cm, and BC = 12 cm
In Δ ABC, AD is the ∠A bisector, meeting side BC at D.
We need to find BD and DC
Let BD = x cm
Then,
6x = 120 – 10x
16x = 120
x = 120/16
x = 7.5
Now, DC = 12 – BD
DC = 12 – 7.5
DC = 4.5
BD = 7.5 cm and DC = 4.5 cm.
AE is the bisector of the exterior ∠CAD meeting BC produced in E. If AB = 10 cm, AC = 6 cm, and BC = 12 cm, Find CE.
It is given that AE is the bisector of the exterior ∠CAD
Meeting BC produced E and AB = 10 cm, AC = 6 cm, and BC = 12 cm.
Since AE is the bisector of the exterior ∠CAD.
72 + 6x = 10x
4x = 72
x = 18
CE = 18 cm
Δ ABC is a triangle such that AB/AC = BD/DC, ∠B = 70, ∠C = 50, find ∠BAD.
It is given that in Δ ABC, ABAC = BDDC AB/AC = BD/DC, ∠B = 70 and ∠C = 50
We need to find ∠BAD
In Δ ABC,
∠A = 180 - (70 + 50)
= 180 - 120
= 60
Since, AB/AC = BD/DC
Therefore, AD is the bisector of ∠A
Hence, ∠BAD = 60/2 = 30
Check whether AD is the bisector of ∠A of ΔABC in each of the following:
(i) AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm
(ii) AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm
(iii) AB = 8 cm, AC = 24 cm, BD = 6 cm and BC = 24 cm
(iv) AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 2 cm
(v) AB = 5 cm, AC = 12 cm, BD = 2.5 cm and BC = 9 cm
(i) It is given that AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm
We have to check whether AD is bisector of ∠A
First we will check proportional ratio between sides.
Now,
Hence, AD is not the bisector of ∠A.
(ii) It is given that AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm.
We have to check whether AD is the bisector of ∠A
Hence, AD is the bisector of ∠A.
(iii) It is given that AB = 8 cm, AC = 24 cm, BD = 6 cm and BC = 24 cm.
DC = BC – BD
DC = 24 – 6
DC = 18
(iv) It is given that AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 2 cm.
First, we will check proportional ratio between sides.
(v) It is given that AB = 5 cm, AC = 12 cm, BD = 2.5 cm and BC = 9 cm.
In fig. AD bisects ∠A, AB = 12 cm, AC = 20 cm, and BD = 5 cm, determine CD.
It is given that AD bisects ∠A
AB = 12 cm, AC = 20 cm, and BD = 5 cm.
We need to find CD.
Since AD is the bisector of ∠A
12 × DC = 20 × 5
DC = 100/12
DC = 8.33 cm
∴ CD = 8.33 cm.
In ΔABC, if ∠1 = ∠2, prove that,
We need to prove that,
In ΔABC,
∠1 = ∠2
So, AD is the bisector of ∠A
D and E are the points on sides BC, CA and AB respectively. of a ΔABC such that AD bisects ∠A, BE bisects ∠B and CF bisects ∠C. If AB = 5 cm, BC = 8 cm, and CA = 4 cm, determine AF, CE, and BD.
It is given that AB = 5 cm, BC = 8 cm and CA = 4 cm.
We need to find out, AF, CE and BD.
40 – 5B D = 4 BD
9 BD = 40
So, BD = 40/9
Since, BE is the bisector of ∠B
5CE = 32 – 8CE
5CE + 8CE = 32
13CE = 32
So, CE = 32/13
Now, since, CF is the bisector of ∠C
8AF = 20 – 4AF
12AF = 20
So, 3AF = 5
AF = 5/3 cm, CE = 32/12 cm
and BD = 40/9 cm
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