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DeltaABC~DeltaDEF, then (AB)/(DE)=(……….)...

`DeltaABC~DeltaDEF`, then `(AB)/(DE)=(……….)/(EF)`

A

AC

B

DF

C

BC

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
NA
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Delta ABC=Delta DEF, Prove that (AB)/(DE)=(AB+BC+AC)/(DE+EF+DF)

DeltaABC~DeltaDEF, BC=4.8cm EF=7.2cm . Complete the following activity to find A(DeltaABC):A(DeltaDEF) . Activity: DeltaABC~DeltaDEF (A(DeltaABC))/(A(DeltaDEF))=(BC^(2))/(square^(2)) .....Theorem on _________ :.(A(DeltaABC))/(A(DeltaDEF))=(4.8^(2))/((square)^(2)) :.(A(DeltaABC))/(A(DeltaDEF))=4/(square) A(DeltaABC):A(DeltaDEF)=4:square

Knowledge Check

  • If in DeltaABC and DeltaDEF , (AB)/(DE)=(BC)/(FD) , then they will be similar, when

    A
    `angleB=angleE`
    B
    `angleA=angleD`
    C
    `angleB=angleD`
    D
    `angleA=angleF`

  • If DeltaABC and DeltaDEF are so related the (AB)/(FD)=(BC)/(DE)=(CA)/(EF) , then which of the following is true?

    A
    `angleA=angleF and angleB=angleD`
    B
    `angleC=angleL and angleA=angleD`
    C
    `angleB=angleF and angleC=angleD`
    D
    `angleA=angleE and angleB=angleD`

  • If DeltaABC ~ DeltaDEF such that A(DeltaABC) = 4A(DeltaDEF) and AC = 6 cm, then DF =

    A
    2 cm
    B
    3 cm
    C
    6 cm
    D
    9 cm

    • Similar Questions

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    If DeltaABC ~ DeltaDEF and /_A=45^(@) , /_E=87^(@) , then /_C= …… .

    If DeltaABC~DeltaDEF , AB=4cm, DE=6,EF=9 cm and FD=12 cm, then find the perimeter of DeltaABC .

    DeltaABC ~ DeltaDEF such that AB = 9.1 cm and DE =6.5 cm. If the perimeter of DeltaDEF is 25 cm, what is the perimeter of DeltaABC ?

    In DeltaABC and DeltaDEF , it is given that AB = DE and BC = EF. In order that DeltaABC~=DeltaDEF , we must have

    DeltaABC and DeltaDEF are similar such that (AB)/(DE)=(BC)/(EF) . Area of the two triangles are respectively 16cm^(2) and 49cm^(2) . If BC=2sqrt(2)cm , then what is length of EF ?

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