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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 $a r (B Y X D) = a r (A B M N)$ . Then .

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True

False

answer is A.

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Detailed Solution

Given that a figure.
Question Image

Consider,

Δ M B C, Δ A B D

.
Since the sides

B C = B D

of a square BCED are equal.
And, MB = AB are equal since they are sides of the square ABMN.
Now,

∠ M B C = ∠ A B D

since they both are of measure

90 ° + ∠ A B C

each.
Hence, by using SAS congruence theorem,

Δ M B C ≅ Δ A B D

.
And,

a r e a (M B C) = a r e a (A B D)

. ……(1)
So,

Δ A B D

and square BYXD have the same base BD and are between the same parallels BD and AX.
Thus,

a r e a (Δ A B D) = 12 × a r e a (B Y X D) ⇒ a r e a (B Y X D) = 2 × a r e a (Δ A B D)

By (1),

⇒ a r (B Y X D) = 2 a r (Δ M B C)

……(2)
Square ABMN and

Δ M B C

have the same base and are between the same parallels MB and NC.

2 a r e a (Δ M B C) = a r e a (s q u a r e ABMN)

……(3)
By combining (2), (3),

⇒ a r e a (B Y X D) = a r e a (A B M N)

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

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