1 the shape of the top of a table in a restaurant is that of a sector of a - Maths - Areas Related to Circles

Question

1 the shape of the top of a table in a restaurant is that of a
sector of a circle with centre O an dang;e bod=90 if bo=od=60cm,
find the area of the table and the perimeter of the table top
2 triangle ABC is right angled at A semicircles are drawn on AB,
AC and BC as diameters find the area of the shaded region
3 OPQR is a rhombus , three of whose vertices lie on a circle
with centre O if the area of the rhombus is 32root3 cm sq , find
the radius of the circle
4 O is the centre of the biggre circle an dAC is the diameter
another circle with AB as diameter is drawn if AC=54cm and BC=10cm,
find the area of the shaded region

Gursheen kaur. · Accepted Answer

SOLUTION 1)

OB=OD = 60 cm and âˆ BOD=90Â° (i) Area of top of table =Area of circle -Area of sector BOD $= (π r^{2}) - (\frac{θ}{360^{0}} \times π r^{2}) = [π {(60)}^{2} - \frac{90^{0}}{360^{0}} \times π {(60)}^{2}] {c m}^{2} = [π {(60)}^{2} - \frac{1}{4} \times π {(60)}^{2}] {c m}^{2} = π {(60)}^{2} [1 - \frac{1}{4}] {c m}^{2} = \frac{22}{7} {\times (60)}^{2} \times \frac{3}{4} {c m}^{2} = 8485 71 {c m}^{2} (i i) P e r i m e t e r o f t o p o f t a b l e = O B + O D + L e n g t h o f m a j o r a r c B D = 60 + 60 + \frac{270}{360} \times 2 \times \frac{22}{7} \times 60 = 120 + 282 85 = 402 85 c m$ SOLUTION 3)