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Standard IX

Mathematics

Application of Similarity

In ABC, the...

Question

In $△ A B C$ , the internal bisector $A D$ and $∠ A$ meets the side $B C$ at $D$ . if $B D = 2.5 c m, A B = 5 c m$ and $A C = 4.2 c m$ , then find $D C$

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Solution

In $△ A B C, A D$ is the internal bisector of $∠ A$ .
$∴ \frac{B D}{D C} = \frac{A B}{A C}$ (angle bisector theorem)
$\Rightarrow D C = \frac{B D \times A C}{A B}$
Thus, $D C = \frac{2.5 \times 4.2}{5} = 2.1 c m$ .

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Similar questions

Q. In a

Δ A B C, A D

is the bisector of

∠ A,

meeting side

B C

D

.
If

B D = 2.5 c m, A B = 5 c m

. and

A C = 4.2 c m

,
Find

D C

Q. In a Δ ABC, AD is the bisector of ∠A, meeting side BC at D. If BD = 2.5 cm, AB = 5 cm, and AC = 4.2 cm, find DC.

Q. In

△ A B C

, the internal bisector

A D

∠ A

meets the side

B C

D

. If

B D = 2.5

cm,

A B = 5

cm and

A C = 4.2

cm, then find

D C

In a $Δ A B C$ , 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 AD= 5.6 cm, AC = 6 cm and DC = 3 cm, find BC.
(vii) If AB= 3.6 cm, BC e 6 cm and BD = 32 cm, find AC.
(viii) If AB =10 cm, AC = 6 cm and BC = 12 cm, find BD and DC

In triangle ABC, AD is the bisector of angle A, meeting BC at D, if BD = 2.5 cm, AB = 5cm and AC = 4.2 cm then DC

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In △ABC, the internal bisector AD and ∠A meets the side BC at D. if BD=2.5 cm,AB=5 cm and AC=4.2 cm, then find DC

In △ABC,AD is the internal bisector of ∠A.∴BDDC=ABAC (angle bisector theorem)⇒DC=BD×ACABThus, DC=2.5×4.25=2.1 cm.

In $△ A B C$ , the internal bisector $A D$ and $∠ A$ meets the side $B C$ at $D$ . if $B D = 2.5 c m, A B = 5 c m$ and $A C = 4.2 c m$ , then find $D C$

In $△ A B C, A D$ is the internal bisector of $∠ A$ .
$∴ \frac{B D}{D C} = \frac{A B}{A C}$ (angle bisector theorem)
$\Rightarrow D C = \frac{B D \times A C}{A B}$
Thus, $D C = \frac{2.5 \times 4.2}{5} = 2.1 c m$ .