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Q.

If (a,b),(c,d),(e,f) are the vertices of a triangle such that a , c,e are in G.P. with common ratio r and b,d,f are in G.P. with  common ratio s then the area of the triangle is

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a

ab2(r+1)(s+2)(s+r)

b

ab2(r−1)(s−1)(s−r)

c

ab2(r−1)(s+1)(s−r)

d

(r+1)(s+1)(s−r)

answer is B.

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Detailed Solution

Here c=ar,e=ar2,d=bs,f=bs2The area of triangle,A=12ab1cd1cf1=12[(cf−ed)+(be−af)+(ad−ac)]=12abrs2−abr2 s+abr2−abs2+abs−bar=12abrs(s−r)+r2−s2+(s−r)=12ab(s−r)[rs−s−r+1]=12ab(s−r)(s−1)(r−1)
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