1) Let two circles having their centres as O and O' intersect each other at point A and B respectively. Let us join OO'.
In triangle (AOO') and triangle(BOO'),
OA = OB (Radius of first circle)
O'A = O'B (Radius of second circle)
OO' = OO' (Common)
Hence, triangle (AOO') is congruent to triangle(BOO'), (By SSS)
?OAO' = ? OBO' (By CPCT)
Therefore, line of centres of two intersecting circles subtends equal angles at the two points of intersection.
2. Let ABCD be the square with side s and diagonal AC = 10 cm
Since triangle ABC is a right angled triangle, therefore by pythagoras theorem we have
AB2 + BC2 = AC2
s2 + s2 = 102
2s2 = 100
s2 = 50
Area of a square = s2 = 50 cm2