Question

If in two triangles DEF and XYZ, DE/YZ=EF/ZX=FD/XY then

(I) traingle XYZ is similar to traingle FDE
(II)traingle XYZ is similar to traingle DEF
(III)traingle FED is similar to traingle XYZ
(IV)traingle EFD is similar to traingle XYZ
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Answer 1

Answer:

option no. (ii) is correct

Answer 2

Solution :-

given that, in two triangles DEF and XYZ ,

→ DE/YZ = EF/ZX = FD/XY -------- Eqn.(1)

• Corresponding sides are in same ratio .

we know that,

• When corresponding sides of two ∆'s are in same ratio , both ∆'s are similar .

Now, checking given options we get,

(I) traingle XYZ is similar to traingle FDE

→ ∆XYZ ~ ∆FDE

then,

→ XY/FD = YZ/DE = XZ/FE

or,

→ FD/XY = DE/YZ = FE/XZ

or,

→ DE/YZ = EF/ZX = FD/XY

since it is equal to Eqn.(1) , given ∆'s are similar .

(II)traingle XYZ is similar to traingle DEF

→ ∆XYZ ~ ∆DEF

then,

→ XY/DE = YZ/EF = XZ/DF

or,

→ DE/XY = EF/YZ = DF/XZ

since it is not equal to Eqn.(1) , given ∆'s are not similar .

(III)traingle FED is similar to traingle XYZ

→ ∆FED ~ ∆XYZ

then,

→ FE/XY = ED/YZ = FD/XZ

since it is not equal to Eqn.(1) , given ∆'s are not similar .

(IV) traingle EFD is similar to traingle XYZ

→ ∆EFD ~ ∆XYZ

then,

→ EF/XY = FD/YZ = ED/XZ

since it is not equal to Eqn.(1) , given ∆'s are not similar .

Hence, Option (I) traingle XYZ is similar to traingle FDE is correct answer .

Learn more :-

In the figure along side, BP and CP are the angular bisectors of the exterior angles BCD and CBE of triangle ABC. Prove ∠BOC = 90° - (1/2)∠A .

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