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Question Number 67852 by mr W last updated on 01/Sep/19

Commented by Prithwish sen last updated on 01/Sep/19

  A=(0,a), B = (((−4)/a),0),C=(((20)/a),−9a),D=((2/a),0)  E=(0,((−3a)/2))   Equation of line passing through A and D  AC⇒a^2 x+2y−2a=0....(i)  Equation of line passing through B and E  BC⇒3a^2 x+8y+12a=....(ii)  Solving (i) and (ii) we get C  ∴ The required area = △ODC+△OEC                                               = (1/2)×(2/a)×9a+(1/2)×((3a)/2)×((20)/a)                                      = 9+15 = 24

A=(0,a),B=(4a,0),C=(20a,9a),D=(2a,0)E=(0,3a2)EquationoflinepassingthroughAandDACa2x+2y2a=0....(i)EquationoflinepassingthroughBandEBC3a2x+8y+12a=....(ii)Solving(i)and(ii)wegetCTherequiredarea=ODC+OEC=12×2a×9a+12×3a2×20a=9+15=24

Commented by Prithwish sen last updated on 01/Sep/19

Commented by mr W last updated on 01/Sep/19

correct! thanks sir!

correct!thankssir!

Commented by Prithwish sen last updated on 02/Sep/19

welcome sir.

welcomesir.

Answered by MJS last updated on 01/Sep/19

((op)/2)=1 ⇒p=(2/o)  ((pq)/2)=2 ⇒ q=2o  ((qr)/2)=3 ⇒ r=(3/o)  A= ((o),(0) )  B= ((0),((2/o)) )  C= (((−2o)),(0) )  D= ((0),((−(3/o))) )  AB: y=−(2/o^2 )x+(2/o)  CD: y=−(3/(2o^2 ))x−(3/o)  AB∩CD=E= (((10o)),((−((18)/o))) )  ∣OA∣=o  ∣OD∣=(3/o)  ∣AD∣=((√(o^4 +9))/o)  ∣AE∣=((9(√(o^4 +4)))/o)  ∣DE∣=((5(√(4o^4 +9)))/o)  area OAED =area OAD +area ADE =  =(3/2)+((45)/2)=24

op2=1p=2opq2=2q=2oqr2=3r=3oA=(o0)B=(02o)C=(2o0)D=(03o)AB:y=2o2x+2oCD:y=32o2x3oABCD=E=(10o18o)OA∣=oOD∣=3oAD∣=o4+9oAE∣=9o4+4oDE∣=54o4+9oareaOAED=areaOAD+areaADE==32+452=24

Commented by mr W last updated on 01/Sep/19

thanks sir!

thankssir!

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