Question

For a circle which is inscribed in a $\Delta$ ABC having sides 8 cm, 10 cm and 12 cm (figure), match the columns.

• A. $A⟶1B⟶4C⟶3D⟶2$
• B. $A⟶2B⟶3C⟶4D⟶1$
• C. $A⟶1B⟶2C⟶3D⟶4$
• D. $A⟶4B⟶2C⟶3D⟶1$

Answer 1

The correct option is B $A⟶2B⟶3C⟶4D⟶1$

$Applytheconceptthatthelengthofthetwotangents,drawnfromapointtoacircle,areequal.$

$Given-ABCisa\Delta anditsincircletouchesthesidesAB,BC\&ACatD,E\&Frespectively.AB=12cm,BC=8cm\&AC=10cm.Tofindout-thevaluesofAD,BE,CFand\frac{AD}{AF}=?Solution-Hereweusetheconceptthatthelengthofthetwotangents,drawnfromapointtoacircle,areequal......\left(i\right).LetAD=xcm.........\left(ii\right).thenBD=AB-AD=12-x........\left(iii\right).So,by\left(i\right)BE=BD=12-x.......\left(iv\right).Also,by\left(i\right)AF=AD=x.......\left(v\right).\therefore CF=AC-AF=10-x.......\left(vi\right).i.eby\left(i\right)CE=CF=10-x.SoBE=BC-CE=8-(10-x)=x-\mathrm{2......}\left(vii\right).Again,by\left(i\right)BD=BE=x-\mathrm{2.......}\left(viii\right).Now,bycomparing\left(iii\right)\&\left(viii\right),wehave12-x=x-2⟹x=7.Substitutingx=7in\left(ii\right),\left(iv\right)\&\left(vi\right)wehaveAD=x=7cm,BE=12-x=(12-7)cm=5cmandCF=10-x=(10-7)cm=3cmandfrom\left(v\right)\frac{AD}{AF}=\frac{x}{x}=1.Sothereqiredmatchingsare-A⟶2B⟶3C⟶4D⟶1Ans-OptionB$