State true or false.In the following figure,

ABC
is a right triangle right angled at

A
is a right triangle right angled at ,

BCED,ACFG
, and

ABMN
and are squares on the sides

BC,CA
are squares on the sides and

AB
and respectively. Line segment

AX⊥DE
respectively. Line segment meets

BC
meets at

Y
at . Then

area(CYXE) = area(ACFG)
. Then .

Question

State true or false.In the following figure,

ABC
  is a right triangle right angled at

A
  is a right triangle right angled at ,

BCED,ACFG
 ,  and

ABMN
  and  are squares on the sides

BC,CA
  are squares on the sides and

AB
 and  respectively. Line segment

AX⊥DE
  respectively. Line segment meets

BC
 meets  at

Y
  at . Then

area(CYXE) = area(ACFG)
 . Then .

Accepted Answer

Given that a figure.We have to show that

area(CYXE) = area(ACFG)
 .In

Δ FCB, ΔACE
 ,

∠FCA=∠BCE

=90°
 [since it is a square]Adding

∠ACB
  on both sides,

∠FCA+∠ACB=∠BCE+∠ACB

∠FCB=∠ACE
 Since the sides of a square are equal.

FC=AC
   [Sides of a square ACFG]

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

Δ FCB ≅ ΔACE
 .And, so,

ΔFCB=ΔACE
 .                              ……(1)

ΔACE
  and parallelogram

CYXE
  are on the same base

CE
  and between the same parallels

CE
  and

AX
 .Hence,

2area(ΔACE)=area(CYXE)
            ……(2)By combining (1), (2),

area
    
     CYXE
   =2area
    
     ΔFCB
   
 .                      ……(3)

ΔFCB
  and rectangle

ACFG
  are on the same base

FC
 .Also, they lie between the same parallels

FC,BG
 .Hence,

2area(ΔFCB)=
 area of rectangle

ACFG
        ……(4)By (3), (4), we get,

area(CYXE)=area(ACFG)
 .Hence, the statement is true.Therefore, option 1 is correct.

Detailed Solution