Question -
Answer -
(i) We know that eachangle of a square is 90°. Hence, ∠ABM = ∠DBC = 90º
∴∠ABM+∠ABC = ∠DBC+∠ABC
∴∠MBC = ∠ABD
In ∆MBC and ∆ABD,
∠MBC = ∠ABD (Proved above)
MB = AB (Sides ofsquare ABMN)
BC = BD (Sides ofsquare BCED)
∴ ∆MBC ≅ ∆ABD (SAS congruency)
(ii) We have
∆MBC ≅ ∆ABD
∴ar (∆MBC) = ar (∆ABD)… (i)
It is given that AX ⊥ DE and BD ⊥ DE (Adjacent sides of square BDEC)
∴ BD || AX (Two linesperpendicular to same line are parallel to each other)
∆ABD and parallelogramBYXD are on the same base BD and between the same parallels BD and AX.
Area (∆YXD) = 2 Area(∆MBC) [From equation (i)] … (ii)
(iii) ∆MBC andparallelogram ABMN are lying on the same base MB and between same parallels MBand NC.
2 ar (∆MBC) = ar(ABMN)
ar (∆YXD) = ar (ABMN)[From equation (ii)] … (iii)
(iv) We know that eachangle of a square is 90°.
∴∠FCA = ∠BCE = 90º
∴∠FCA+∠ACB = ∠BCE+∠ACB
∴∠FCB = ∠ACE
In ∆FCB and ∆ACE,
∠FCB = ∠ACE
FC = AC (Sides ofsquare ACFG)
CB = CE (Sides ofsquare BCED)
∆FCB ≅ ∆ACE (SAS congruency)
(v) AX ⊥ DE and CE ⊥ DE (Adjacent sides of square BDEC) [given]
Hence,
CE || AX (Two linesperpendicular to the same line are parallel to each other)Consider BACE andparallelogram CYXE
BACE and parallelogramCYXE are on the same base CE and between the same parallels CE and AX.
∴ar (∆YXE) = 2ar (∆ACE)… (iv)
We had proved that
∴ ∆FCB ≅ ∆ACE
ar (∆FCB) ≅ ar (∆ACE) … (v)
From equations (iv)and (v), we get
ar (CYXE) = 2 ar(∆FCB) … (vi)
(vi) Consider BFCB andparallelogram ACFG
BFCB and parallelogramACFG are lying on the same base CF and between the same parallels CF and BG.
∴ar (ACFG) = 2 ar(∆FCB)
∴ar (ACFG) = ar (CYXE)[From equation (vi)] … (vii)
(vii) From the figure,we can observe that
ar (∆CED) = ar(∆YXD)+ar (CYXE)
∴ar (∆CED) = ar(ABMN)+ar (ACFG) [From equations (iii) and (vii)].