Question

1.
EXERCISE 7.5 (Optional)*
ABC is a triangle. Locate a point in the interior of A ABC which is equidistant from
the vertices of A ABC.
distant from all the sides of​

Answer 1

Answer:

Let PQ and RS intersect at point O.

Join OA, OB and OC.

Now in AOM and  BOM,

AM = MB [By construction]

AMO = BMO = [By construction]

OM = OM [Common]

AOM BOM [By SAS congruency]

OA = OB [By C.P.C.T.] …..(i)

Similarly, BON CON

OB = OC [By C.P.C.T.] …..(ii)

From eq. (i) and (ii),

OA = OB = OC

Hence O, the point of intersection of perpendicular bisectors of any two sides of ABC equidistant from its vertices.

HOPE THIS HELPS U

PLZ MARK AS BRAINLYEST

Answer 2

Answer:

SEE THE ATTACHMENT FOR SOLUTION.