Delta ABC ~ Delta LMN. In Delta ABC, AB ...

`Delta ABC ~ Delta LMN`. In `Delta ABC, AB = 5.5 cm`, BC = 6 cm, CA = 4.5 cm. If MN = 4.8 cm then construct `Delta ABC` and `Delta LMN`

Text Solution

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For `triangleABC` , the lengths of three sides are known
`thereforetriangleABC` can be constructed
`triangleABC~triangleLMN`
`therefore(AB)/(LM)=(BC)/(MN)=(AC)/(LN)" "...("Corresponding sides of similar triangles are in proportion")`
`therefore(5.5)/(LM)=6/(MN)=(4.5)/(LN)=5/4`
`therefore(5.5)/(LM)=5/4`
`thereforeLM=(5.5xx4)/5`
`thereforeLM=1.1xx4 `
`thereforeLM=4.4cm`
`6/(MN)=5/4`
`thereforeMN=(6xx4)/5`
`thereforeMN=(6xx4)/5`
`thereforeMN=(24)/5`
`thereforeMN=4.8 cm`
`(4.5)/(LN)=5/4`
`thereforeLN=(4.5xx4)/5`
`thereforeLN=(18)/5`
`thereforeLN=3.6cm`
For `triangleLMN` the lengths of three sides are known
`therefore triangleLMN` can vbe constructed
Construction :

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