Question

In


$\red {\tt \huge{ΔABC, \: 'O' \: is \: Circum \: Center, ' H ' \: is \: a \: Ortho \: Center}}$
Then Show That


$\sf{ \huge{ \pink{(I) \overline{OA+OB+OC=OH}}}}$

$\huge \bf {\color{cyan}{(II) \overline{ HA+HB+HC=2HO}}}$
​

Answer 1

Answer:

tex] \red {\tt \huge{ΔABC, \: 'O' \: is \: Circum \: Center, ' H ' \: is \: a \: Ortho \: Center}}[/tex]

Then Show That

$\sf{ \huge{ \pink{(I) \overline{OA+OB+OC=OH}}}}$

$\huge \bf {\color{cyan}{(II) \overline{ HA+HB+HC=2HO}}}$

Answer 2

Answer:

Correct option is

D

4cm

CQ and CP are the tangents to a circle

So, CQ=CP (Tangents to a circle from a common point (C) are equal in length)

CP=CQ=11

Also, BC+BQ=11

⇒BC+7=11

⇒BQ=4

So, BR=BQ=4. (Tangents to a circle from a common point (B) are equal in length)