Question

Q.4) In the figure, ABCD is a square. ∆BCE on side

BC & ∆ACF on diagonal AC are similar. Complete
the following activity to show that A(∆BCE) =½ A(∆ACF).

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Answer 1

Answer:

ABCD is a square. △BCE is described on side BC is similar to △ACF desctibed on diagonal AC.

Since ABCD is a square. Therefore,

AB=BC=CD=DA and AC=

2

BC [∵Diagonal=

2

(side)]

Now, △BCE∼△ACF

⇒

Area(△ACF)

Area(△BCE)

=

AC

2

BC

2

⇒

Area(△ACF)

Area(△BCE)

=

(

2

BC)

2

BC

2

=

2

1

⇒ Area(△BCE)=

2

1

Area(△ACF) [Hence proved]

Answered By

toppr