Answer:
Let P be the external point. O be the origin. join O and P get OP and nearest point on the circle from P be A.
Let Q be the point onthe circle in which, tangent make 90° with radius at Q.
PQ = 8 and OQ = 6
we get a right angled triangle PQO right angled at Q.
so, OP^2 = OQ^2 + PQ^2= 8^2 + 6^2 = 64 + 36 =1==
therefore OP =10cm
we need nearest point from P, which is PA
PA = OP - OA= 10 -6=4cm
3 sides of square is 3*12=36 cm
perimeter for the full circle is 2 * pi * radius
which is =2 * 22/7 * 12 =75.43 cm
so 360° =75.43 cm
for 150° = 75.43/360 *150 =31.43
all together =36 +31.43 =67.43 cm
You do what I’d 9-3 , 8-0 , then combined it and you will get the right answer
Sqrt -13 = sqrt13 i
sqrt -26 = sqrt2 * sqrt13i
sqrt-13 * sqrt -26 = sqrt2 * 13* i^2 = -13 sqrt2 (because i^2 = -1)
answer is -13 sqrt2
