Class 10: Triangles – Exercise 7.3

$\displaystyle \textbf{Question 1: }\text{In a }\triangle ABC,\ AD\text{ is the bisector of }\angle A,\text{ meeting side }BC\text{ at }D.$
$\displaystyle \text{(i) If }BD=2.5\text{ cm},\ AB=5\text{ cm and }AC=4.2\text{ cm, find }DC.$
$\displaystyle \text{(ii) If }BD=2\text{ cm},\ AB=5\text{ cm and }DC=3\text{ cm, find }AC.\hfill\text{[CBSE 2001]}$
$\displaystyle \text{(iii) If }AB=10\text{ cm},\ AC=6\text{ cm and }BC=12\text{ cm, find }BD\text{ and }DC. \hfill\text{[CBSE 2001]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }AD\text{ is the bisector of }\angle A,\text{ by the Angle Bisector Theorem,}$
$\displaystyle \frac{BD}{DC}=\frac{AB}{AC}.$

$\displaystyle \text{(i) }\frac{2.5}{DC}=\frac{5}{4.2}$
$\displaystyle DC=\frac{2.5\times4.2}{5}=2.1\text{ cm}$
$\displaystyle \therefore DC=2.1\text{ cm.}$

$\displaystyle \text{(ii) }\frac{2}{3}=\frac{5}{AC}$
$\displaystyle 2AC=15$
$\displaystyle AC=7.5\text{ cm}$
$\displaystyle \therefore AC=7.5\text{ cm.}$

$\displaystyle \text{(iii) }\frac{BD}{DC}=\frac{10}{6}=\frac{5}{3}$
$\displaystyle \therefore BD:DC=5:3$
$\displaystyle \text{Also, }BD+DC=BC=12\text{ cm}$
$\displaystyle BD=\frac{5}{5+3}\times12=\frac{15}{2}=7.5\text{ cm}$
$\displaystyle DC=\frac{3}{5+3}\times12=\frac{9}{2}=4.5\text{ cm}$
$\displaystyle \therefore BD=7.5\text{ cm and }DC=4.5\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{In the adjoining figure, }AE\text{ is the bisector of the exterior }\angle CAD$
$\displaystyle \text{ meeting }BC \ \text{produced in }E.\text{ If }AB=10\text{ cm},\ AC=6\text{ cm and }BC=12\text{ cm, find }x.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Since }AE\text{ is the exterior angle bisector of }\angle CAD,$
$\displaystyle \frac{BE}{CE}=\frac{AB}{AC}\qquad\text{[By the External Angle Bisector Theorem]}$
$\displaystyle \text{Here, }BE=BC+CE=12+x\text{ and }CE=x.$
$\displaystyle \therefore \frac{12+x}{x}=\frac{10}{6}=\frac{5}{3}$
$\displaystyle 3(12+x)=5x$
$\displaystyle 36+3x=5x$
$\displaystyle 2x=36$
$\displaystyle x=18$
$\displaystyle \therefore x=18\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{In the adjoining figure, }\triangle ABC\text{ is a triangle such that }$
$\displaystyle \frac{AB}{AC}=\frac{BD}{DC}, \ \angle B=70^\circ,\ \angle C=50^\circ.\text{ Find }\angle BAD.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{AB}{AC}=\frac{BD}{DC}$
$\displaystyle \therefore AD\text{ is the bisector of }\angle A\qquad\text{[By the converse of Angle Bisector Theorem]}$
$\displaystyle \angle A=180^\circ-(\angle B+\angle C)$
$\displaystyle =180^\circ-(70^\circ+50^\circ)=60^\circ$
$\displaystyle \therefore \angle BAD=\frac{1}{2}\angle A=\frac{1}{2}\times60^\circ=30^\circ$
$\displaystyle \therefore \angle BAD=30^\circ.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{In the adjoining figure, check whether }AD\text{ is the bisector of }\angle A\text{ of }$
$\displaystyle \triangle ABC \ \text{in each of the following:}$
$\displaystyle \text{(i) }AB=4\text{ cm},\ AC=6\text{ cm},\ BD=1.6\text{ cm and }CD=2.4\text{ cm}$
$\displaystyle \text{(ii) }AB=8\text{ cm},\ AC=24\text{ cm},\ BD=6\text{ cm and }BC=24\text{ cm}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{By the converse of the Angle Bisector Theorem, if }\frac{BD}{DC}=\frac{AB}{AC},$
$\displaystyle \text{then }AD\text{ is the bisector of }\angle A.$
$\displaystyle \text{(i) }\frac{AB}{AC}=\frac{4}{6}=\frac{2}{3}$
$\displaystyle \frac{BD}{DC}=\frac{1.6}{2.4}=\frac{2}{3}$
$\displaystyle \therefore \frac{AB}{AC}=\frac{BD}{DC}$
$\displaystyle \therefore AD\text{ is the bisector of }\angle A.$

$\displaystyle \text{(ii) }DC=BC-BD=24-6=18\text{ cm}$
$\displaystyle \frac{AB}{AC}=\frac{8}{24}=\frac{1}{3}$
$\displaystyle \frac{BD}{DC}=\frac{6}{18}=\frac{1}{3}$
$\displaystyle \therefore \frac{AB}{AC}=\frac{BD}{DC}$
$\displaystyle \therefore AD\text{ is the bisector of }\angle A.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }D,\ E\text{ and }F\text{ are the points on sides }BC,\ CA\text{ and }AB\text{ respectively of }$
$\displaystyle \triangle ABC \ \text{such that }AD\text{ bisects }\angle A,\ BE\text{ bisects }\angle B\text{ and }CF\text{ bisects }\angle C.$
$\displaystyle \text{If }AB=5\text{ cm},\ BC=8\text{ cm and }CA=4\text{ cm, determine }AF,\ CE\text{ and }BD.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }CF\text{ bisects }\angle C,\text{ by the Angle Bisector Theorem,}$
$\displaystyle \frac{AF}{FB}=\frac{AC}{CB}=\frac{4}{8}=\frac{1}{2}$
$\displaystyle \therefore AF:FB=1:2$
$\displaystyle \text{Since }AF+FB=AB=5\text{ cm},$
$\displaystyle AF=\frac{1}{1+2}\times5=\frac{5}{3}\text{ cm}$
$\displaystyle \text{Since }BE\text{ bisects }\angle B, \\ \text{by the Angle Bisector Theorem,}$
$\displaystyle \frac{AE}{EC}=\frac{AB}{BC}=\frac{5}{8}$
$\displaystyle \therefore AE:EC=5:8$
$\displaystyle \text{Since }AE+EC=AC=4\text{ cm},$
$\displaystyle CE=\frac{8}{5+8}\times4=\frac{32}{13}\text{ cm}$
$\displaystyle \text{Since }AD\text{ bisects }\angle A,\text{ by the Angle Bisector Theorem,}$
$\displaystyle \frac{BD}{DC}=\frac{AB}{AC}=\frac{5}{4}$
$\displaystyle \therefore BD:DC=5:4$
$\displaystyle \text{Since }BD+DC=BC=8\text{ cm},$
$\displaystyle BD=\frac{5}{5+4}\times8=\frac{40}{9}\text{ cm}$
$\displaystyle \therefore AF=\frac{5}{3}\text{ cm},\ CE=\frac{32}{13}\text{ cm and }BD=\frac{40}{9}\text{ cm.}$
$\displaystyle \\$