Question

Pls solve this... I hve my xams :(
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Answer 1

ABCD is a square⇒AB=BC=CD=DA=21 cm
Ar. semi. having diameter AD=πr²/2 sq.units
=[(22/7)×(21/2)×(21/2)]/2=90.75 cm²
Ar. semi. having diameter BC=(22/7)×(21/2)²/2=90.75 cm²
Ar.ΔAOB=Ar.ΔCOD=Ar.ΔBOC=Ar.ΔAOD(Diagonals of a square divides a square into four equal Δs)....(1)
Ar.ABCD=21×21=441 cm²
∴Ar.BOC=1/4(Ar.ABCD)=1/4(441)=110.25 cm²
Also,Ar.AOD=Ar.AOB=110.25 cm²......[from (1)]
Area of the shaded region=(Ar. semi. having diameter AD+Ar. semi. having diameter BC+Ar.BOC+Ar.AOD)
=(90.75+90.75+110.25+110.25)cm²=402 cm²(Ans.)

Answer 2

Answer:

this is not there in ICSE syllabus