If a ∆ 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, AC = 6 cm and DC = 3 cm, find BC.
(vii) If AD = 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 $B D=2.5 \mathrm{~cm}, A B=5 \mathrm{~cm}$ and $A C=4.2 \mathrm{~cm}$.
In $\triangle A B C, A D$ is the bisector of $\angle A$, meeting side $B C$ at $D$.
We have to find $D C$.
Since $A D$ is $\angle A$ bisector
Then $\frac{A B}{A C}=\frac{2.5}{D C}$
$\frac{5}{4.2}=\frac{2.5}{D C}$
$5 D C=4.2 \times 2.5$
$D C=\frac{4.2 \times 2.5}{5}$
$=2.1$
Hence $D C=2.1 \mathrm{~cm}$
(ii) It is given that $B D=2 \mathrm{~cm}, A B=5 \mathrm{~cm}$ and $D C=3 \mathrm{~cm}$.
In $\triangle A B C, A D$ is the bisector of $\angle A$, meeting side $B C$ at $D$.
We have to find $A C$.
Since $A D$ is $\angle A$ bisector
So $\frac{A B}{A C}=\frac{B D}{D C}(A D$ is bisector of $\angle A$ and side $B C)$
Then
$\frac{5}{A C}=\frac{2}{3}$
$\Rightarrow 2 A C=5 \times 3$
$\Rightarrow A C=\frac{15}{2}$
$=7.5$
Hence $A C=7.5 \mathrm{~cm}$
(iii) It is given that $A B=3.5 \mathrm{~cm}, A C=4.2 \mathrm{~cm}$ and $D C=2.8 \mathrm{~cm}$.
In $\triangle A B C, A D$ is the bisector of $\angle A$, meeting side $B C$ at $D$
We have to find $B D$.
Since $A D$ is $\angle A$ bisector
So $\frac{A B}{A C}=\frac{B D}{D C}(A D$ is bisector of $\angle A$ and side $B C)$
Then
$\frac{3.5}{4.2}=\frac{B D}{2.8}$
$\Rightarrow B D=\frac{3.5 \times 2.8}{42}$
$\Rightarrow B D=\frac{7}{3}$
$=2.3$
Hence $B D=2.3 \mathrm{~cm}$
(iv) It is given that $A B=10 \mathrm{~cm}, A C=14 \mathrm{~cm}$ and $B C=6 \mathrm{~cm}$.
In $\triangle A B C, A D$ is the bisector of $\angle A$, meeting side $B C$ at $D$.
We have to find $B D$ and $D C$.
Since $A D$ is $\angle A$ bisector
So $\frac{A B}{A C}=\frac{B D}{D C}\left(A D^{\text {is bisector of }} \angle A\right.$ and side $\left.B C\right)$
Let $\mathrm{BD}=x \mathrm{~cm}$. Then $\mathrm{CD}=(6-x) \mathrm{cm}$
Then,
1014=x6-x⇒14x=60-10x⇒24x=60⇒x=6024=2.5
Hence, BD = 2.5 cm and DC = 6 − 2.5 = 3.5 cm.
(v) It is given that $A C=4.2 \mathrm{~cm}, D C=6 \mathrm{~cm}$ and $B C=10 \mathrm{~cm}$.
In $\triangle A B C, A D$ is the bisector of $\angle A$, meeting side $B C$ at $D$.
We have to find $A B$.
Since $A D$ is $\angle A$ bisector
$S O \frac{A C}{A B}=\frac{D C}{B D}$
Then
$\frac{4.2}{A B}=\frac{6}{4}$
$\Rightarrow 6 A B=4.2 \times 4$
$\Rightarrow A B=\frac{4.2 \times 4}{6}$
$=\frac{16.8}{6}$
Hence $A B=2.8 \mathrm{~cm}$
(vi) It is given that $A B=5.6 \mathrm{~cm}, B C=6 \mathrm{~cm}$ and $D C=3 \mathrm{~cm}$.
In $\triangle A B C, A D$ is the bisector of $\angle A$, meeting side $B C$ at $D$.
We have to find $B C$.
Since $A D$ is $\angle A$ bisector
So $\frac{A C}{A B}=\frac{B D}{D C}$
Then
$\frac{6}{5.6}=\frac{3}{D C}$
$\Rightarrow D C=2.8$
So
$B C=2.8+3$
$=5.8$
Hence $B C=5.8 \mathrm{~cm}$
(vii) If it is given that $A B=5.6 \mathrm{~cm}, B C=6 \mathrm{~cm}$ and $B D=3.2 \mathrm{~cm}$.
In $\triangle A B C, A D$ is the bisector of $\angle A$, meeting side $B C$ at $D$
$\therefore \frac{A B}{A C}=\frac{B D}{D C}$
$\frac{5.6 \mathrm{~cm}}{A C}=\frac{3.2 \mathrm{~cm}}{2.8 \mathrm{~cm}} \quad[D C=B C-B D]$
$A C=\frac{5.6 \times 2.8}{3.2} \mathrm{~cm}=4.9 \mathrm{~cm}$
(viii) It is given that $A B=10 \mathrm{~cm}, A C=6 \mathrm{~cm}$ and $B C=12 \mathrm{~cm}$.
In $\triangle A B C, A D$ is the bisector of $\angle A$, meeting side $B C$ at $D$.
We have to find $B D$ and $D C$.
Since $A D$ is $\angle A$ bisector
So $\frac{A C}{A B}=\frac{D C}{B D}$
Let BD = x cm
Then
$\frac{6}{10}=\frac{12-x}{x}$
$\Rightarrow 6 x=120-10 x$
$\Rightarrow 16 x=120$
$\Rightarrow x=\frac{120}{16}$
$\Rightarrow x=7.5$
Now
$D C=12-B D$
$=12-7.5$
$=4.5$
Hence $B D=7.5 \mathrm{~cm}$ and $D C=4.5 \mathrm{~cm}$
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