Please solve q 91?

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2 Answers
May 19, 2018

#|ABC|=30 " unit"^2#

Explanation:

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Let #|ABC|# denote area of #DeltaABC#
Given #BD:CD=2:1, => |GBD|=2*|GDC|=2xx4=8#
let #|GDE|=a, => |CDE|=7-a#,
as #|EBD|:|EDC|=2:1, => (8+a):(7-a)=2:1, => a=2#
as #|GDB|:|GDE|=8:a=8:2=4:1#,
#=> BG:GE=4:1#
#=> |ABG|:|AGE|=4:1#,
let #|AGE|=b, => |ABG|=4b#
as #BD:DC=2:1, => |ABD|:|ADC|=2:1#
#=> (8+4b):(7+b)=2:1#
#=> 8+4b=2*(7+b)#
#=> b=3, 4b=12#
#=> |ABC|=8+4b+7+b=8+12+7+3=30 " units"^2#

May 19, 2018

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Given

  • #(BD)/(CD)=2/1#
  • #DeltaGEC=3#
  • #DeltaGCD=4#

The ratio of areas two triangles of same height is equal to the ratio of their bases.
So we can write

#(DeltaBGD)/(DeltaGCD)=(BC)/(CD)=(2CD)/(CD)=2#

#=>w/2=4=>w=8.....[1]#

For similar reason

#(x+y)/z=(BG)/(GE)=(w+4)/3=(8+4)/3=4#

#=>x+y-4Z=0.....[2]#

Similarly

#(x+y)/w=(z+3)/4#

#=>(x+y)/8=(z+3)/4#

#=>x+y-2z=6.....[3]#

By [2] and [3] we get #z=3#

So #x+y=12....[4]#

Again

#x/(w+4)=y/(z+3)#

#x/(8+4)=y/(3+3)#

#=>x/12=y/6#

#=>x=2y.....[5]#

By [4] and [5] we get #3y=12=>y=4#
and #x=2y=8#

So area of #Delta ABC=w+x+y+z+3+4#

#=>Delta ABC=8+8+4+3+3+4=30#sq unit