X and Y are two similar triangular rooftops.
So, the ratio of their sides is $\frac{A B}{E F} = \frac{A C}{E G} = \frac{B C}{F G} = 2$

False

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True

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Solution

The correct option is A False

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Similar questions

\frac{A B}{E F} = \frac{A C}{E G} = \frac{B C}{F G} = 2

△ A B C,

a, b

c

A, B

C

a = 7, b = 3, c = 5

\frac{a b}{(tan \frac{B}{2} + tan \frac{C}{2}) (tan \frac{A}{2} + tan \frac{C}{2})} + \frac{b c}{(tan \frac{A}{2} + tan \frac{C}{2}) (tan \frac{A}{2} + tan \frac{B}{2})} + \frac{a c}{(tan \frac{B}{2} + tan \frac{C}{2}) (tan \frac{A}{2} + tan \frac{B}{2})}

k,

\sqrt{k} + \frac{1}{2}

In the given figure, ∠DFE $= 90^{\circ}$ , $F G ⊥ E D$ , If $G D = 8$ , $F G = 12$ , find (i) $E G$ (ii) $F D$ and (iii) $E F$

In the given figure , line DE || line GF ray EG and ray FG are bisectors of ∠DEF and ∠DFM respectively. Prove that, (i) ∠DEG=1/2∠EDF
(ii) EF=FG

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