Question

It is proposed to add to a square lawn with each side 50 m, two circular ends, the centre of each circle being the point of intersection of the diagonals of the square. Find the area of the whole lawn.( Take $\pi =3.14$)

• A. $4153.5m^2$
• B. $6212.5m^2$
• C. $3212.5m^2$
• D. $4412.5m^2$

Answer 1

The correct option is C $3212.5m^2$

$Given-AlawnconsistsofasquareABCDofside50m.ItisflankedbytwosegmetsAED\&BFC.ThecetresofthesegmentslieonthepointofintersectionOofthediagonalsAC\&BDofthesquare.Tofindout-ar.ABFCDE=?Solution-ar.ABFCDE=arADE+arABCD+arBCF=2ar.ADE+arABCD..........\left(i\right).(sincearADE=arBCF)NowABCDisasquareofside50m.\therefore ar.ABCD={side}^2={50}^2{cm}^2=2500{cm}^2.........\left(ii\right)andthediagonalAC=side\times \sqrt{2}=50\times \sqrt{2}m.Sincediagonalofasquarebisecteachotheratrightangles,wehave\angle BOC=\angle AOD={90}^{o}andOA=OC=\frac{1}{2}\times AC=\frac{1}{2}\times 50\times \sqrt{2}m=25\times \sqrt{2}m\therefore Bythegivenconditiontheradiusofthesegment=r=25\times \sqrt{2}mandthecentralangle=\theta ={90}^o..............\left(iii\right)Weknowthatar.segment=[\frac{\theta }{{360}^o}\times \pi -sin\frac{\theta }{2}cos\frac{\theta }{2}]\times r^{2}when\theta istheanglesubtendedbythearcofthesegmenttothecentreofthecircle\left(orcentralangle\right),r=radiusofthecircle.So2\times ar.ADE=2\times [\frac{{90}^o}{{360}^o}\times \pi -sin\frac{{90}^o}{2}cos\frac{{90}^o}{2}]\times {25\times \sqrt{2}}^2{m}^2\left(fromiii\right)=2\times [\frac{1}{4}\times 3.14-\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}]\times 1250m^2=[1.57-1]\times 1250m^2=712.5m^2.......\left(iv\right)So,by\left(i\right)ar.ABFCDE=(2500+712.5)m^2(usingii+iv)=3212.5m^2.Ans-OptionC.$