Question

Please solve q 42 ?

Answer 1

`(MC)/(AM)=3`

Explanation:

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`

Answer 2

#(MC)/(AM)=3

Explanation:

solution 2

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`