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)
Answers
Answered by
8
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
Answered by
6
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)
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