Question

help me with easy solution ​

Answer 1

Answer:

hope this helps

Step-by-step explanation:

It'll help to draw a circle if you want.

Join OP

PQ=1/2PR=1/2×30=15cm

perpendicular from the center of circle to chord bisects the latter. In triangle OPQ, use Pythagorean theorem:

OP=$\sqrt{OQ^{2} +PQ^{2} } =\sqrt{15^{2}+8^{2} } cm=17$

$\sqrt{225+64} =17\\\\\\225+64=289\\\\\\\sqrt{289} =17$

Answer 2

Answer:

so the answer of this question is 17 cm

Step-by-step explanation:

Given AB=30cm

⇒AM+MB=30

⇒2AM=30(∵AM=MB)

⇒AM=15cm

In Δle OMB(OMB=90∘)

O→ Center of circle

AB→ chord

∴OM^2+MB^2=OB^2

⇒8^2+15^2=R^2

⇒R^2=289

⇒R=17cm

image for solution is attached with it.