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Standard IX

Mathematics

Application of Similarity

In ABC, the...

Question

In $△ A B C$ , the internal bisector $A D$ of $∠ A$ meets the side $B C$ at $D$ . If $B D = 2.5$ cm, $A B = 5$ cm and $A C = 4.2$ cm, then find $D C$

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Solution

By Angle Bisector Theorem,
Let $D C = x$
$\Rightarrow \frac{A B}{A C} = \frac{B D}{D C}$
$\Rightarrow \frac{5}{4.2} = \frac{2.5}{x}$
$\Rightarrow 5 x = 2.5 \times 4.2$
$\Rightarrow 5 x = 10.5$
$\Rightarrow x = 2.1$ cm
Therefore, $D C = 2.1$ cm

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Similar questions

Q. In

△ A B C

, the internal bisector

A D

and

∠ A

meets the side

B C

D

. if

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

and

A C = 4.2 c m

, then 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.

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

Q. In 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 cmand 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.

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

By Angle Bisector Theorem,Let DC=x⇒ABAC=BDDC⇒54.2=2.5x⇒5x=2.5×4.2⇒5x=10.5⇒x=2.1 cmTherefore, DC=2.1 cm

In $△ A B C$ , the internal bisector $A D$ of $∠ A$ meets the side $B C$ at $D$ . If $B D = 2.5$ cm, $A B = 5$ cm and $A C = 4.2$ cm, then find $D C$

By Angle Bisector Theorem,
Let $D C = x$
$\Rightarrow \frac{A B}{A C} = \frac{B D}{D C}$
$\Rightarrow \frac{5}{4.2} = \frac{2.5}{x}$
$\Rightarrow 5 x = 2.5 \times 4.2$
$\Rightarrow 5 x = 10.5$
$\Rightarrow x = 2.1$ cm
Therefore, $D C = 2.1$ cm