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Question
SANGUES
8. In Fig 6.34.0
6.54.O is a point in the interior of a triangle
COD IBC, OE LAC and OF LAB.Show that
A2+OB:+OC-OD--OF-OF=AF+BD+CE.
f) AF+BD + CE-=AE+CD+BF:
6 Aladder 10 m long reaches a window 8 m above the
ground. Find the distance of the foot of the ladder
from base of the wall.
In attached to a vertical pole of height 18 m
tached to the other
Fig. 6.54
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Q1
In Fig., O is a point in the interior of a triangle ABC,OD⊥BC,OE⊥AC and OF⊥AB. Show that (i)OA2+OB2+OC2−OD2−OE2−OF2=AF2+BD2+CE2, (ii) AF2+BD2+CE2=AE2+CD2+BF2.
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Q2
In the above figure, O is a point in the interior of a triangle ABC,OD⊥BC,OE⊥AC and OF⊥AB. Show that: (i) OA2+OB2+OC2−OD2−OE2−OF2=AF2+BD2+CE2 (ii) AF2+BD2+CE2=AE2+CD2+BF2
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