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Question

If A,A1,A2,A3 are the areas of the inscribed and escribed of a ABC, then:

A
A1+A2+A3=π(r1+r2+r2)
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B
1A1+1A2+1A3=1A
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C
1A1+1A2+1A3=s2πr1r2r3
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D
A1+A2+A3=π(4R+r)
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Solution

The correct options are
A A1+A2+A3=π(r1+r2+r2)
B 1A1+1A2+1A3=1A
C 1A1+1A2+1A3=s2πr1r2r3
D A1+A2+A3=π(4R+r)
A=πr2,A1=πr21,A2=πr22 and A3=πr23
Option(a)
A1+A2+A3=π(r1+r2+r3)
Option(b)
1A1+1A2+1A3=1π(1r1+1r2+1r3)
=1π(1r)
=1πr2=1A
Option(c)
s2πr1r2r3=s2π3(sa)(sb)(sc)
=s2(sa)(sb)(sc)π3
=sπ where =s(sa)(sb)(sc)
=1rπ
=1πr2
=1A
=1A1+1A2+1A3
Option(d)
A1+A2+A3=π(r1+r2+r3)
=π(4R+r)

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