Please solve fast it's urgent.

1.	Please solve fast it's urgent.
Answer» *Hey mate* *here* is *your* *answer*: LET the point at a distance of 4 cm be PO Given: PA=8 cm and PO=4 cm, <POA=90°[Line joining centre of circle to a chord are perpendicular BISECTOR] APPLY PYTHAGORAS THEOREM: $(8) {}^{2} = (4) {}^{2} + ao$ $= = > 64 - 16 = (ao) {}^{2}$ $ao = \sqrt{48}$ $ao = 4 \sqrt{3cm}$ Now, area of SEMI circle**= $\frac{1}{2} \times \pi \times r {}^{2}$ $\frac{1}{2} \times 3.14 \times (4 \sqrt{3} ) {}^{2} = area$ $area = 75cm^2$

Answer»

Hey mate here is your answer:

LET the point at a distance of 4 cm be PO