Please solve q 42 ?

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2 Answers
May 11, 2018

(MC)/(AM)=3

Explanation:

enter image source here
Given CL=AL=2BL,
let CL=AL=2a, => BL=a
DeltaCLA is a right triangle,
=> AC^2=CL^2+AL^2=(2a)^2+(2a)^2=8a^2,
=> AC=2asqrt2
DeltaALB is also a right triangle,
AB^2=AL^2+LB^2=(2a)^2+a^2=5a^2
=> AB=sqrt5a
Let |ABC| be area of DeltaABC,
=> |ABC|=1/2*CB*AL=1/2*3a*2a=3a^2
=> |ABC|=1/2*AB*MC=3a^2
=> 1/2*sqrt5a*MC=3a^2
=> MC=(2*(3a^2))/(sqrt5a)=(6a)/sqrt5
Similarly, AM^2=AC^2-MC^2=(2asqrt2)^2-((6a)/sqrt5)^2=(4a^2)/5
=> AM=sqrt((4a^2)/5)=(2a)/sqrt5

=> (MC)/(AM)=((6a)/sqrt5)/((2a)/sqrt5)=3

May 11, 2018

#(MC)/(AM)=3

Explanation:

solution 2
enter image source here
given CL=AL=2BL,
let CL=AL=2a, => BL=a
DeltaALB is a right triangle
AB^2=AL^2+LB^2=(2a)^2+a^2=5a^2
=> AB=sqrt5a,
Area of DeltaABC=1/2*CB*AL=1/2*AB*MC
=> MC=(CB*AL)/(AB)=(3a*2a)/(sqrt5a)=(6a)/sqrt5
DeltaCMB and DeltaALB are similar
=> (CM)/(MB)=(AL)/(AB)=2/1
=> MB=1/2*CM=(6a)/(2sqrt5)=(3a)/sqrt5
=> AM=AB-MB=sqrt5a-(3a)/sqrt5
=(5a-3a)/sqrt5=(2a)/sqrt5

=> (MC)/(AM)=((6a)/sqrt5)/((2a)/sqrt5)=3