Question

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Answer 1

Hey !! ^_^


Here is your answer

⬇️⬇️⬇️⬇️⬇️⬇️⬇️


1) In ∆ MBC and ABD ..

Given
.
BC = BD ...( SiDe of the square BCED )


MB = AB .....( Side of the square ABMN) .

∠MBC = ∠ABD ...( Both are 90 )

So ,

∆ MBC Congruent to ∆ ABD

......................................................

2) ∆ ABD and □ BYXD have the same base BD and between the parallel BD and AX ...

So ,

area(ABD) = 1/2. area ( BYXD) ..

and

∆ MBC Congruent to ∆ ABD.... (proved earlier)

Therefore ..

ar( MBC ) = ar ( ABD ) = 1/2 ar ( BYXD) .

or ..

ar(BYXD) = 2 ar( MBC) ..

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3) Square ABMN and ∆ MBC have the same MB and Between the same parallel MB and NAC

Therefore .

ar(MBC) = 1/2ar(ABMN)

ar(AMBN) = 2ar(MBC)

=> 2ar( MBC) = ar(BYXD)......(proved earlier)

=> ar(ABMN) = ar( BYXD) ..


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4) .In ∆ ACE and ∆ BCF

CE = BC......... ( side of the square) ..

AC = CF.....( " )

∠ACE = ∠BCF ..( BOTh are 90 )

So ,
∆ ACE are Congruent to ∆ BCF ..


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5) .∆ ACE and square CYXE have the same base CE and between the same parallel CE and AXY ..


So ,

ar(ACE) = 1/2 ar ( CYXE )

ar(FCB) = 1/2 ar ( CYXE)

or ..

ar(CYXE) = 2ar ( ACE )

__________________________

6) ∆BCF and Square ACFG have the Same base and between the Same parallel ..CF and BAG ..

ar(BCF) = 1/2ar (ACFG) .

or ..

ar(BCF) = 1/2ar(CYXE) ..

So ,

ar(ACFG) = ar(CYXE)

_________________________`_

7 ) from 3 and 6 ..

ar(BYXD) = ar(ABMN)

and ar(CYXE) = ar(ACFG)

and

ar(BCED) = ar(ABMN) + ar(ACFG) ..


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Answer 2

Hey mate ^_^

=======
Answer:
=======

Given:

ABC is a right angled triangle in which ∠A= 90°. BCED,ACFG,ABMN are squares.

Proof:

_______________________

(i)

We know that each angle of a square is 90°.
Hence, ∠ABM = ∠DBC = 90º
∠ABM + ∠ABC = ∠DBC + ∠ABC
∠MBC = ∠ABD
In ΔMBC and ΔABD,
∠MBC = ∠ABD (Proved above)
MB = AB (Sides of square ABMN)
BC = BD (Sides of square BCED)
∴ ΔMBC ≅ ΔABD (SAS congruence rule)

_______________________

(ii)

We have
ΔMBC ≅ ΔABD
ar (ΔMBC) = ar (ΔABD) ... (1)
It is given that AX ⊥ DE and BD ⊥ DE (Adjacent sides of square BDEC)
BD || AX (Two lines perpendicular to same line are parallel to each other)

ΔABD & rectangle BYXD are on the same base BD and between the same parallels BD and AX.

ar(∆ABD)= ½ ar(BYXD)
ar(BYXD)=2ar(∆ABD)
ar(BYXD) = 2ar (ΔMBC)..........(2)
[Using equation (1)]

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(iii)

ΔMBC & square ABMN are lying on the same base MB and between same parallels MB and NC.

ar(∆MBC)=1/2ar(ABMN)

2 ar (ΔMBC) = ar (ABMN)

ar (BYXD) = ar (ABMN) [Using equation (2)] ... (3)

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(iv)

We know that each angle of a square is 90°.
∴ ∠FCA = ∠BCE = 90º
∠FCA + ∠ACB = ∠BCE + ∠ACB
∠FCB = ∠ACE

In ΔFCB and ΔACE,
∠FCB = ∠ACE
FC = AC (Sides of square ACFG)
CB = CE (Sides of square BCED)
ΔFCB ≅ ΔACE (SAS congruence rule)

_______________________

(v)

It is given that AX ⊥ DE and CE ⊥ DE (Adjacent sides of square BDEC)

Hence, CE || AX (Two lines perpendicular to the same line are parallel to each other)

ΔACE & rectangle CYXE are on the same base CE and between the same parallels CE and AX.

Ar(∆ACE)= 1/2ar(CYXE)

ar (CYXE) = 2 ar (ΔACE) ... (4)

We had proved that
∴ ΔFCB ≅ ΔACE
ar (ΔFCB) ≅ ar (ΔACE) ... (5)

On comparing equations (4) and (5),

ar (CYXE) = 2 ar (ΔFCB) ... (6)

_______________________

(vi)

ΔFCB & square ACFG are lying on the same base CF and between the same parallels CF and BG.

Ar(∆FCB)= 1/2ar(ACFG)

ar (ACFG) = 2 ar (ΔFCB)

ar (ACFG) = ar (CYXE) [Using equation (6)] ... (7)

_______________________

(vii)

From the figure, it is evident that
ar (BCED) = ar (BYXD) + ar (CYXE)

ar (BCED) = ar (ABMN) + ar (ACFG) [Using equations (3) and (7)]

_______________________

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