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Standard X

Mathematics

Internal Angle Bisector Theorem

In a ABC, AD ...

Question

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.

3.2 cm

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2.5 cm

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5 cm

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2.1 cm

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Solution

The correct option is D 2.1 cm
Since,
AD is the bisector of ∠A meeting side BC at D in Δ ABC
Hence, Δ ABD ~ Δ ACD

Now,
⇒ $\frac{A B}{A C} = \frac{B D}{C D}$

⇒ $\frac{5}{4.2} = \frac{2.5}{C D}$

⇒ $5 \times C D = 2.5 \times 4.2$

⇒ $C D = 2.1 c m$

$∴$ DC = 2.1 cm.

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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

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

A D

is the internal bisector of

∠ A

of triangle

A B C

and meets side

B C

D

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B D = 5 c m, B C = 7.5 c m

then

A B : A C

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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.

The correct option is D 2.1 cmSince, AD is the bisector of ∠A meeting side BC at D in Δ ABC Hence, Δ ABD ~ Δ ACD Now, ⇒ABAC=BDCD ⇒54.2=2.5CD ⇒5×CD=2.5×4.2 ⇒CD=2.1 cm ∴ DC = 2.1 cm.

The correct option is D 2.1 cm
Since,
AD is the bisector of ∠A meeting side BC at D in Δ ABC
Hence, Δ ABD ~ Δ ACD

Now,
⇒ $\frac{A B}{A C} = \frac{B D}{C D}$

⇒ $\frac{5}{4.2} = \frac{2.5}{C D}$

⇒ $5 \times C D = 2.5 \times 4.2$

⇒ $C D = 2.1 c m$

$∴$ DC = 2.1 cm.