Question

what is the solution of ax2+by3+cz​

Answer 1

Answer:

A and B are the centres of the circles with radii 5cm and 3cm respectively.

C is the mid-point of AB.

Extend AB upto O point on circumference of outer circle.

AB=AO−BO=5−3=2cm (since AO and BO are radii of larger and smaller circles)

AC=

2

AB

=

2

2

=1cm

now in right angled triangle AMP

AC=1cm,AP=5cm

by pythagoras thm.

AP

2

=PC

2

+AC

2

PC

2

=

AP

2

−AC

2

PC

2

=

5

2

−1

2

Therefore PQ=2PC=2.

24

=4.

6

cm [CP=CQ]

So , option A is the answer.