5. The plaine of the ray and one quarter of the periphery, is the content of the semicircle.
As here thou seest: For the product of 7, the halfe of the diameter, multiplyed by 11, the quarter of the periphery, doth make 77, for the content of the semicircle.
This may also be done by taking of the halfe of the circle now measured.
And
6. The plaine made of the ray and halfe the base, is the content of the Sector.
Here are three sectours, ae the base of 12 foote: And ei in like manner of 12 foote. The other or remainder ia of 7 f. and 3/7 of one foote. The diameter is 10 foote. Multiply therefore 5, halfe of the diameter, by 6 halfe of the base, and the product 30, shall be the content of the first sector. The same shall also be for the second sectour. Againe multiply the same ray or semidiameters 5, by 3.5/7, the halfe of 7.3/7, the product of 18.4/7 shall be the content of the third sector. Lastly, 30 + 30 + 18.4/7 are 78.4/7, the content of the whole circle.
And