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State true or false.

In the following figure, $A B C$ is a right triangle right angled at $A$ is a right triangle right angled at , $B C E D, A C F G$ , and $A B M N$ and are squares on the sides $B C, C A$ are squares on the sides and $A B$ and respectively. Line segment $A X ⊥ D E$ respectively. Line segment meets $B C$ meets at $Y$ at . Then $a r e a C Y X E = 2 a r e a Δ F C B$ . Then .

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True

False

answer is A.

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Detailed Solution

Given that a figure.
Question Image

We have to show that

a r e a C Y X E = 2 a r e a Δ F C B

.
In

Δ FCB, Δ ACE

∠ F C A = ∠ B C E

= 90 °

[since it is a square]
Adding

∠ A C B

on both sides,

∠ F C A + ∠ A C B = ∠ B C E + ∠ A C B

∠ F C B = ∠ A C E

Since the sides of a square are equal.

F C = A C

[Sides of a square ACFG]

C B = C E

[Sides of a square BDEC]
By SAS congruence rule,

Δ FCB ≅ Δ ACE

And, so,

Δ F C B = Δ A C E

. ……(1)

Δ A C E

and parallelogram

C Y X E

are on the same base

C E

and between the same parallels

C E

and

A X

.
Hence,

2 a r e a (Δ A C E) = a r e a (C Y X E)

……(2)
By combining (1), (2),

a r e a C Y X E = 2 a r e a Δ F C B

.
Hence, the statement is true.
Therefore, option 1 is correct.

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