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Question

If 1,2 be the areas of two triangles with vertices (b,c),(c,a),(a,b), and (acb2,abc2),(bac2,bca2),(cba2,cab2), then 12=(a+b+c)2

A
True
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B
False
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Solution

The correct option is A True
T1=(b,c) (c,a) (a,b)=(x1,y1) (x2,y2) (x3,y3)
T2=[(acb2),(abc2)],[(bac2),(bca2)],[(cba2),(cab2)]
T1=12[c(ca)+a(ab)+b(bc)]
=12[c2ac+a2ab+b2bc]
T2=12[(abc2)[bac2cb+a2]+(bca2)[cba2ac+b2]+(cab2)[acb2ba+c2)]
=12[[(abc2)(ac)(a+b+c)]+(bca2)[(ba)(a+b+c)]+(cab2)[(cb)(a+b+c)]]
=(a+b+c)2[a2babcac2+c3+b2cabca2b+c2aabcb2c+b3]
=(a+b+c)2(a3+b3+c33abc)
=(a+b+c2)[a+b+c][a2+b2+c2abbcca]
Δ1Δ2=(a+b+c)2[a2+b2+c2bcbaca](a2+b2+c2bcbaca)
=(a+b+c)2

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