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Q.

The vertices of a triangle are A(–10, 8) B(14, 8) and C(–10, 26). Let G,I,O,S be the centroid in center orthocenter ,circum center respectively of ABC

Column - IColumn - I
a.The inradius r isp.15
b.The circum radius R isq.12
c. The area of ΔIGo is r.6
d. The area of ΔSGI is s.3

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a

a-q; b-r; c-p; d-s

b

a-r; b-p; c-r; d-s

c

a-s; b-q; c-r; d-p

answer is A.

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Detailed Solution

S=18+24+302=30

 Area ABC=216

a) r=Δs=21636=6

b) R=302=15, sinθ triangle ABC is right angled 

c) I=(4,14);G=(2,14);O=A=(10,8)ΔIGO=6

d) S=(2,17)ΔSGI=3

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The vertices of a triangle are A(–10, 8) B(14, 8) and C(–10, 26). Let G,I,O,S be the centroid in center orthocenter ,circum center respectively of ∆ABCColumn - IColumn - Ia.The inradius r isp.15b.The circum radius R isq.12c. The area of ΔIGo is r.6d. The area of ΔSGI is s.3