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1. In a Δ ABC, AD is the bisector of ∠ A, meeting side BC at D.(iii) if \[\mathbf{AB}\text{ }=\text{ }\mathbf{3}.\mathbf{5}\text{ }\mathbf{cm},\] \[\mathbf{AC}\text{ }=\text{ }\mathbf{4}.\mathbf{2}\text{ }\mathbf{cm},\] and \[\mathbf{DC}\text{ }=\text{ }\mathbf{2}.\mathbf{8}\text{ }\mathbf{cm},\] find BD.(iv) if \[\mathbf{AB}\text{ }=\text{ }\mathbf{10}\text{ }\mathbf{cm},\] \[\mathbf{AC}\text{ }=\text{ }\mathbf{14}\text{ }\mathbf{cm},\]and \[\mathbf{BC}\text{ }=\text{ }\mathbf{6}\text{ }\mathbf{cm},\] find BD and DC.
1. In a Δ ABC, AD is the bisector of ∠ A, meeting side BC at D.(iii) if \[\mathbf{AB}\text{ }=\text{ }\mathbf{3}.\mathbf{5}\text{ }\mathbf{cm},\] \[\mathbf{AC}\text{ }=\text{ }\mathbf{4}.\mathbf{2}\text{ }\mathbf{cm},\] and \[\mathbf{DC}\text{ }=\text{ }\mathbf{2}.\mathbf{8}\text{ }\mathbf{cm},\] find BD.(iv) if \[\mathbf{AB}\text{ }=\text{ }\mathbf{10}\text{ }\mathbf{cm},\] \[\mathbf{AC}\text{ }=\text{ }\mathbf{14}\text{ }\mathbf{cm},\]and \[\mathbf{BC}\text{ }=\text{ }\mathbf{6}\text{ }\mathbf{cm},\] find BD and DC.
CBSE | Class 10 | Exercise 4.3 | Maths | RD Sharma | Triangles
1. In a Δ ABC, AD is the bisector of ∠ A, meeting side BC at D.(iii) if \[\mathbf{AB}\text{ }=\text{ }\mathbf{3}.\mathbf{5}\text{ }\mathbf{cm},\] \[\mathbf{AC}\text{ }=\text{ }\mathbf{4}.\mathbf{2}\text{ }\mathbf{cm},\] and \[\mathbf{DC}\text{ }=\text{ }\mathbf{2}.\mathbf{8}\text{ }\mathbf{cm},\] find BD.(iv) if \[\mathbf{AB}\text{ }=\text{ }\mathbf{10}\text{ }\mathbf{cm},\] \[\mathbf{AC}\text{ }=\text{ }\mathbf{14}\text{ }\mathbf{cm},\]and \[\mathbf{BC}\text{ }=\text{ }\mathbf{6}\text{ }\mathbf{cm},\] find BD and DC.

Given:

 Δ ABC and AD bisects ∠A, meeting side BC at D.

\[~AB\text{ }=\text{ }3.5\text{ }cm\], \[AC\text{ }=\text{ }4.2\text{ }cm,\] and \[DC\text{ }=\text{ }2.8\text{ }cm\].

Required to find: BD

AD is the bisector of ∠ A meeting side BC at D in Δ ABC

⇒ $\frac{AB}{AC}=\frac{BD}{DC}$

$\frac{3.5}{4.2}=\frac{BD}{2.8}$

$4.2\times BD=3.5\times 2.8$

BD = 7/3$BD=\frac{7}{3}$

\[\therefore BD\text{ }=\text{ }2.3\text{ }cm\]

Given:

 In Δ ABC, AD is the bisector of ∠A meeting side BC at D.

  \[AB\text{ }=\text{ }10\text{ }cm,\] \[AC\text{ }=\text{ }14\text{ }cm,\] and \[BC\text{ }=\text{ }6\text{ }cm\]

Required to find: BD and DC.

 AD is bisector of ∠A

AB/AC = BD/DC (AD is bisector of ∠ A and side BC)

$\frac{10}{14}=\frac{x}{\left( 6-x \right)}$

\[14x\text{ }=\text{ }60\text{ }\text{ }6x\]

\[20x\text{ }=\text{ }60\]

\[x\text{ }=\frac{60}{20}\]

\[\therefore \] \[BD\text{ }=\text{ }3\text{ }cm\] and \[DC\text{ }=\text{ }\left( 6\text{ }\text{ }3 \right)\text{ }=\text{ }3\text{ }cm\]         

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