239
Question
In the following diagram ABCD is a square and E, F, G and H are mid points of the sides
In the following diagram ABCD is a square and E, F, G and H are mid points of the sides
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Solution
The correct options are
A point G is (3, 7)
B point E is (5, 1)
D Area of ΔHME is 5 sq. units
Given−A(2,0),B(x,y),C(p,q)&D(0,6)aretheverticesofthesquareABCD.Tofindout−(i)ar.ABCD=?((ii)thecoordinatesofE,G&F=?Solution−SinceABCDisasquare,AB=BC=CD=DAandthediagonalsBD=AC=√32×AD.Weusethedistanceformula,d=√(x1−x2)2+(y1−y2)2,togetAD.SoAD=√(2−0)2+(0−6)2units=√40units=AB=BC=CD............(i)∴ar.ABCD=AD2=(√40)2sq.units=40sq.unitsAlsoBD=AD2+AB2=√80units=AC(diagonalsofasquarebisecteachotheratrightangles)............(ii)Now,sinceABCDisasquare,ThelinesGME&HMFwilldividear.ABCDinto4equalsquares.(eachofthemjoinsthemidpointsoftheoppositesides).∴ar.AHME=14ar.ABCD⟹ar.AHME=14×40sq.units=10sq.units.AgainthelineHEisadiagonalofthesquareAHME.Soar.ΔHME=12×ar.AHME⟹ar.ΔHME=12×10sq.units=5sq.units.LetthecoordinatesofBbe(x,y)∴dAB=√(x−2)2+(y−0)2=√40(fromi)⟹x2+y2−4x−36=0.........(iii)AlsodBD=√(x−0)2+(y−6)2=√80(fromii)⟹x2+y2−12y−44=0.........(iv).Subtracting(iv)from(iii),12y−4x+8=0⟹y=x−23......(v).Substitutingy=x−23in(iiii),x2+(x−23)2−4x+4−36x−324=0⟹x2−4x−32=0⟹(x−8)(x+4)=0⟹x=(8,−4).Werejectx=−4sinceABCDisinthefirstquadrant.Soputtingx=2in(v,y=8−23=2.i.e(x,y)=(8,2).Similarly,letthepointCbeC(p,q).ThenAC2=(p−6)2+(q−0)2=80⟹p2+q2−4p−76=0..........(vi)andBC=DC⟹(p−8)2+(q−2)2=(p−0)2+(q−6)2⟹2p−q−4=0⟹q=2p−4.......(vii).Puttingq=2p−4in(vi),p2+(2p−4)2−4p−76=0⟹p2−4p−12=0⟹(p−6)(p+2)=0⟹p=(6,−2).WerejectthenegativevalueasABCDisinthefirstquadrant.Soputtingp=6in(vi)q=2×6−4=8.∴C(p,q)=(6+8).∴Usingmidpointformula,ThecoordinatesofEare(8+22,2+02)=(5,1),ThecoordinatesofFare(8+62,2+82)=(7,5)andThecoordinatesofGare(6+02,8+62)=(3,7).Alsoar.HME=5sq.units.Ans−OptionsA,B&D.
A point G is (3, 7)
B point E is (5, 1)
D Area of ΔHME is 5 sq. units
Given−A(2,0),B(x,y),C(p,q)&D(0,6)aretheverticesofthesquareABCD.Tofindout−(i)ar.ABCD=?((ii)thecoordinatesofE,G&F=?Solution−SinceABCDisasquare,AB=BC=CD=DAandthediagonalsBD=AC=√32×AD.Weusethedistanceformula,d=√(x1−x2)2+(y1−y2)2,togetAD.SoAD=√(2−0)2+(0−6)2units=√40units=AB=BC=CD............(i)∴ar.ABCD=AD2=(√40)2sq.units=40sq.unitsAlsoBD=AD2+AB2=√80units=AC(diagonalsofasquarebisecteachotheratrightangles)............(ii)Now,sinceABCDisasquare,ThelinesGME&HMFwilldividear.ABCDinto4equalsquares.(eachofthemjoinsthemidpointsoftheoppositesides).∴ar.AHME=14ar.ABCD⟹ar.AHME=14×40sq.units=10sq.units.AgainthelineHEisadiagonalofthesquareAHME.Soar.ΔHME=12×ar.AHME⟹ar.ΔHME=12×10sq.units=5sq.units.LetthecoordinatesofBbe(x,y)∴dAB=√(x−2)2+(y−0)2=√40(fromi)⟹x2+y2−4x−36=0.........(iii)AlsodBD=√(x−0)2+(y−6)2=√80(fromii)⟹x2+y2−12y−44=0.........(iv).Subtracting(iv)from(iii),12y−4x+8=0⟹y=x−23......(v).Substitutingy=x−23in(iiii),x2+(x−23)2−4x+4−36x−324=0⟹x2−4x−32=0⟹(x−8)(x+4)=0⟹x=(8,−4).Werejectx=−4sinceABCDisinthefirstquadrant.Soputtingx=2in(v,y=8−23=2.i.e(x,y)=(8,2).Similarly,letthepointCbeC(p,q).ThenAC2=(p−6)2+(q−0)2=80⟹p2+q2−4p−76=0..........(vi)andBC=DC⟹(p−8)2+(q−2)2=(p−0)2+(q−6)2⟹2p−q−4=0⟹q=2p−4.......(vii).Puttingq=2p−4in(vi),p2+(2p−4)2−4p−76=0⟹p2−4p−12=0⟹(p−6)(p+2)=0⟹p=(6,−2).WerejectthenegativevalueasABCDisinthefirstquadrant.Soputtingp=6in(vi)q=2×6−4=8.∴C(p,q)=(6+8).∴Usingmidpointformula,ThecoordinatesofEare(8+22,2+02)=(5,1),ThecoordinatesofFare(8+62,2+82)=(7,5)andThecoordinatesofGare(6+02,8+62)=(3,7).Alsoar.HME=5sq.units.Ans−OptionsA,B&D.
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