(2.) By Note I., the perpendicular from the centre on

Sum

=4800, the square of the slant height.

..the slant height=4800=69-282032.

Hence

240×69-282032

=8313-8438, slant surface.

2

2.598076 × 40 × 40 =4156.9216, area of the base. .. 12470-7654- whole surface. 4156-9216, area of base.

Mult. by 60 =

20

83138-432, solidity.

(3.) By Note I., the perpendicular from the centre on

1.720477 × 22 = 6·881908, area of the base.

26-881908, whole surface.

Perpendicular from the centre on the side =

x2 cot. 36°

=cot, 36°.

.. Log. perp.

Log.cot. 36°-10-138739.

2

(3.) √(12)2+52=√169 =13, the slant height.

13

3-1416 x 24 x 1
•7854×24x24x5

=490·0896, convex surf.

=753.984, solidity.

-30, the diameter of the base.

✅✅(36)2+(15)2=√1521=39, the slant height.

the difference of the perpendiculars from the centres of the ends upon the sides, the square of which increased by the square of the altitude, will be the square of the slant height.

...√/(12)2+3=/147-12-124356- the slant height.

3(15+9)

Hence

x12.124356-436-476816, surf. of sides.

2

=132.501978, area of ends.

568-978794, whole surf.

{(15)2+92} ×·433013

{(15)2+15×9+92} × ×·433013-763-834932,

solidity.

(2.) √√(12)2+(13)2=12′093382, slant height.

×12-093382=169.307348, slant. surf.

4(5+2)

2

the perpendiculars from the centres of the ends upon the

28 × 2-598076 × 3=218.238384, solidity.

(5+3)×3-1416 × 4·123106=51.812599, slant. surf.

(6.) ×3-1416×(20+10)×13=612-612, slant surf.

PROBLEM VI.

(1.) (35×2+55) × 17·145 × 15=5357-8125, solid in.

(2) (9 × 2411) × 10 × g=1933 cubic inches.

(3.) (26x2+15) x 28 x 1-5628 solid inches.

(1.) The height =7912 × sin. 23° 28′.

The circumference =7912 × 31416, and the product of these two will be the surface sought, which is therefore (7912)2 x 3.1416x sin. 23° 28'; hence