Question

In $\Delta ABC,ifO\text{is the centroid of}\Delta ABCandar\left(\Delta OAB\right)=36c{m}^2,thenar\left(\Delta ABC\right)is$

• A.
  $112c{m}^2$
• B.
  $96c{m}^2$
• C.
  $84c{m}^2$
• D.
  $108c{m}^2$

Answer 1

The correct option is D

$108c{m}^2$


Let AF, BG and EC be the medians of $\Delta ABC.$ Given that O is the centroid of $\Delta ABC.$

Let $AM\perp BC.$

Here, $BF=FC,AE=EBandAG=GC.Now,Ar\left(\Delta ABF\right)=\frac{1}{2}\times Base\times Altitudeof\Delta ABF=\frac{1}{2}\times BF\times AM=\frac{1}{2}\times FC\times AM=\frac{1}{2}\times FC\times Altitudeof\Delta AFC=Ar\left(\Delta AFC\right)$

Thus, median divides a triangle into two equal halves.

Similarly, $Ar\left(\Delta BOE\right)=Ar\left(\Delta AOE\right)\dots \left(i\right)Ar\left(\Delta AOG\right)=Ar\left(\Delta OGC\right)\dots \left(ii\right)andAr\left(\Delta BOF\right)=Ar\left(\Delta OFC\right)\dots \left(iii\right)Now,Ar\left(\Delta ABF\right)=Ar\left(\Delta AFC\right)\Rightarrow Ar\left(\Delta AEO\right)+Ar\left(\Delta EOB\right)+Ar\left(\Delta OBF\right)=Ar\left(\Delta AOG\right)+Ar\left(\Delta GOC\right)+Ar\left(\Delta OFC\right)\Rightarrow 2Ar\left(\Delta AEO\right)=2Ar\left(\Delta AOG\right)\text{[From (i), (ii) and (iii)]}\Rightarrow Ar\left(\Delta AEO\right)=Ar\left(\Delta AOG\right)\dots \left(iv\right)Also,Ar\left(\Delta ABG\right)=Ar\left(\Delta BGC\right)\Rightarrow Ar\left(\Delta AOG\right)+Ar\left(\Delta AOE\right)+Ar\left(\Delta BEO\right)=Ar\left(\Delta GOC\right)+Ar\left(\Delta OFC\right)+Ar\left(\Delta BOF\right)\Rightarrow 2Ar\left(\Delta BEO\right)=2Ar\left(\delta BOF\right)\Rightarrow Ar\left(\Delta BEO\right)=Ar\left(\Delta BOF\right)..\left(v\right)\text{From (i), (ii), (iii), (iv) and (v), we get}Ar\left(\Delta BOE\right)=Ar\left(\Delta AOE\right)=Ar\left(\Delta AOG\right)=Ar\left(\Delta OGC\right)=Ar\left(\Delta BOF\right)=Ar\left(\Delta OFC\right)Thus,ar\left(\Delta ABC\right)=3\times Ar\left(\Delta OAB\right)=3\times 36=108c{m}^2$

Hence, the correct answer is option (d).