Question

In the given figure, ABCD is a square.
∆BCE on side BC and ∆ACF on the diagonal AC are similar to each other.
then show that A(∆BCE)= ½ A(∆ACF)
​

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]

Step-by-step explanation:

Hope it helps

#BayBlader

Answer 2

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)