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mydreammba
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Area of a triangle 10 Jan 2012, 03:33
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Can anyone please help with this.....i tried to solving but i was unable to arrive at a proper solution



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BDSunDevil
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Area of a triangle 10 Jan 2012, 03:45
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LN= 2\sqrt{2}
LM= 2\sqrt{5}
MN= 2\sqrt{5}

therefore LMN is and isosceles triangle with the height of 3\sqrt{2}
area=1/2*2\sqrt{2}*3\sqrt{2}
=6
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Area of a triangle 10 Jan 2012, 06:01
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Length of LM is root20
Length of MN is root20
Length of NP is root8

it is isosceles triangle
its area is c/4*sqroot of 4asquare-csquare ( its formula of isoceles triangle)
we will get 6 as answer ( substiute c as sqrt8 and a as sqrt20 )
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Area of a triangle 10 Jan 2012, 06:38
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BDSunDevil
LN= 2\sqrt{2}
LM= 2\sqrt{5}
MN= 2\sqrt{5}

therefore LMN is and isosceles triangle with the height of 3\sqrt{2}
area=1/2*2\sqrt{2}*3\sqrt{2}
=6
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How did you get height as 3/Squareroot(2)----->I am getting height as Squareroot(15)????
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Area of a triangle 10 Jan 2012, 07:06
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draw bisector of angle M and its property of isocceles triangle that bisector of angle LMN is perpendicluar to base BC and and it bisects it
assume prep is MK so will get right angle trianlge LKM with base sqrt2 and hyp 2sqrt5 u will get height 3sqt2
otherwise simlply remember formula which i mentioned above i think it is fast
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Area of a triangle 10 Jan 2012, 07:11
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if P is the midpoint of LN,
LP=NP=root2 (since, LN=LP+NP=2root2)
MP=height= root(LM^2- (root2)^2)
=root(20-2)=root(18)=3root2
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Area of a triangle 10 Jan 2012, 07:19
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Thanks a lot...Got it......initially i messed up



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