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Question

In a $∆ABC,AD$ is the bisector or $\angle A.$ (i) If AB = 6.4 cm, AC = 8 cm and BD = 5.6 cm, find DC. (ii) If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC. (iii) If AB = 5.6 cm, BD = 3.8 cm and BC = 6 cm, find AC. (iv) If AB = 5.6 cm, AC = 4 cm and DC = 3 cm, find BC.

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Solution

(i) $\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{AD}\mathrm{bisects}\angle \mathrm{A}.\phantom{\rule{0ex}{0ex}}\mathrm{Applying}\mathrm{angle}-\mathrm{bisector}\mathrm{theorem}\mathrm{in}△\mathrm{ABC},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}\phantom{\rule{0ex}{0ex}}⇒\frac{5.6}{\mathrm{DC}}=\frac{6.4}{8}\phantom{\rule{0ex}{0ex}}⇒\mathrm{DC}=\frac{8×5.6}{6.4}=7\mathrm{cm}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ (ii) $\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{AD}\mathrm{bisects}\angle \mathrm{A}.\phantom{\rule{0ex}{0ex}}\mathrm{Applying}\mathrm{angle}-\mathrm{bisector}\mathrm{theorem}\mathrm{in}△\mathrm{ABC},\mathrm{we}\mathrm{get}:$ $\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}\phantom{\rule{0ex}{0ex}}\mathrm{Let}\mathrm{BD}\mathrm{be}x\mathrm{cm}.\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{DC}=\left(6-x\right)\mathrm{cm}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{6-x}=\frac{10}{14}\phantom{\rule{0ex}{0ex}}⇒14x=60-10x\phantom{\rule{0ex}{0ex}}⇒24x=60\phantom{\rule{0ex}{0ex}}⇒x=2.5\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{BD}=2.5\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{DC}=6-2.5=3.5\mathrm{cm}\phantom{\rule{0ex}{0ex}}$ (iii) $\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{AD}\mathrm{bisects}\angle \mathrm{A}.\phantom{\rule{0ex}{0ex}}\mathrm{Applying}\mathrm{angle}-\mathrm{bisector}\mathrm{theorem}\mathrm{in}△\mathrm{ABC},\mathrm{we}\mathrm{get}:$ $\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}\phantom{\rule{0ex}{0ex}}\mathrm{BD}=3.2\mathrm{cm},\mathrm{BC}=6\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{DC}=6-3.2=2.8\mathrm{cm}\phantom{\rule{0ex}{0ex}}⇒\frac{3.2}{2.8}=\frac{5.6}{\mathrm{AC}}$ $⇒AC=\frac{5.6×2.8}{3.2}=4.9\mathrm{cm}$ (iv) $\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{AD}\mathrm{bisects}\angle \mathrm{A}.\phantom{\rule{0ex}{0ex}}\mathrm{Applying}\mathrm{angle}-\mathrm{bisector}\mathrm{theorem}\mathrm{in}△\mathrm{ABC},\mathrm{we}\mathrm{get}:$ $\frac{BD}{DC}=\frac{AB}{AC}\phantom{\rule{0ex}{0ex}}⇒\frac{BD}{3}=\frac{5.6}{4}\phantom{\rule{0ex}{0ex}}⇒BD=\frac{5.6×3}{4}\phantom{\rule{0ex}{0ex}}⇒BD=4.2\mathrm{cm}$ Hence, BC = 3 + 4.2 = 7.2 cm

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