In a right circular cone, using S for lateral area, T for total area, s for slant height, r for radius, c for circumference, and h for altitude, solve the following:

5. Given r=4 in., s=6 in.; find S and T. 6. Given c=25 in., s=8 in.; find S and T. 7. Given s=8 ft. 6 in., c=3 ft. 7 in.; find S. 8. Given s=7.2 in., r=5.3 in.; find S and T. 9. Given h=12 in., r=5 in.; find S and T. 10. Given c=30 in., s=8 in.; find h and S. 11. Given S=200 sq. in., r=6 in.; find s. 12. Given S=256 sq. cm., s= = 16 cm.; find r. 13. Given T=500 sq. cm., r = 10 cm.; find s.

14. The circumference of the base of a conical church steeple is 35 ft. and the altitude is 73 ft. Find the lateral area. Ans. 1281-sq. ft. 15. Find the lateral edge and the lateral area of a regular pyramid each side of whose triangular base is 10 ft., and whose altitude is 18 ft. Ans. 18.90+ ft., 273.45

16. A tower with a regular hexagonal base has dimensions as shown in the figure. The pitch of the pyramdial roof is 1, which means that OV=11⁄2 QR. Find the lateral area of the tower.

17. Cut out of heavy paper a sector of a circle, with a radius of 3 in. and the central angle 120°. Bring the edges together and paste them. Find the lateral area and the total area of the cone thus formed.

18. The cone formed by revolving an equilateral triangle about one of its altitudes has a lateral area equal to twice the area of the base.

sq. ft.

V

19. Find the area of the surface of the solid formed by revolving an equilateral triangle, having a side of 12 in., about one of its sides.

20. Find the area of the surface of the solid formed by revolving a right triangle, having a base of 8 in. and an altitude of 6 in., about its hypotenuse.

AREA OF FRUSTUM OF PYRAMID OR CONE

721. Truncated Pyramid, or Cone. The portion of a pyramid, or cone, included between the base and a section not parallel to the base is called a truncated pyramid, or cone.

722. Frustum. The portion of a pyramid, or cone, included between the base and a section parallel to the base is called a frustum of the pyramid, or cone.

D Ni

ADA A

723. The base and the parallel section are called the bases of the frustum. The perpendicular distance between the bases is the altitude of the frustum.

724. The slant height of a frustum of a regular pyramid, or of a cone, is that portion of the slant height of the pyramid, or cone, included between the bases of the frustum.

Thus AB and CD are slant heights, and QP and MN are altitudes.

725. Midsection. If a solid has parallel bases, the section parallel to the bases and half way between them is called the midsection.

In the figure, NR is the midsection.

726. Prove the following facts concerning frustums:

(1).The lateral faces of a frustum of a regular pyramid are congruent isosceles

trapezoids.

D

M

R

A

B

(2) The lateral edges of a frustum of a regular pyramid are equal; and the slant height is the same for all the faces.

727. Theorem. The lateral area of a frustum of a regular pyramid is equal to half the product of the slant height and the sum of the perimeters of the bases. S=1⁄2s(P+p).

Given the frustum of a regular pyramid with bases AC and EG.

To prove that S=s(P+p), where S denotes lateral area, s slant height, and P and p the perimeter of the lower and upper base respectively.

Proof.

Area AFs(AB+EF).

Area BG=s(BC+FG).

etc.

E

H

G

S=s[(AB+BC+···)+(EF+FG+· · ·)].

F

C

B

§ 370

Hence

§ 105

But

AB+BC+=P, and EF+FG+··· = p.

Why?

728. Theorem. The lateral area of a frustum of a right circular cone is equal to half the product of the slant height and the sum of the circumferences of the bases.

Given the frustum of a right circular cone. To prove ST(R+r)s, where S denotes lateral area, s slant height, and R and r the radius of the lower and upper base respectively.

Proof. Let m be the slant height of the cone and n the slant height of the part above the frustum.

S=mπR-nπг=π(mR-nr).

Rr=mn.

mr-nR=0.

2

R

Then

Why?

But

§ 428

And therefore

§398

AREA OF FRUSTUM OF PYRAMID OR CONE35

729. Theorem. The lateral area of a frustum of a right circular cone is equal to the product of the altitude and the circumference of a circle whose radius is the perpendicular at the midpoint of an element, and terminated by the axis.

Outline of proof. Let a be the length of EF, the perpendicular at the midpoint of an element and terminated by the axis. Let r' be the radius of the midsection, and s and h the slant height and altitude respectively.

Show that ADEF AABC.

Then sha : r', and sr' = ah.

2011

Ea

hD

A B

Show that R+r=2r', where R and r are the radii of the bases.
Then, substituting in the formula of § 728, S=2πah.

EXERCISES

1. Prove the theorem of § 728 by considering the lateral area of a frustum of a right circular cone as the limit of the area of a frustum of a pyramid.

2. Prove the theorem of § 727 by using a method similar to that of § 728 for the frustum of a cone.

3. Show that the lateral area of a frustum of a right circular cone is equal to the slant height times the perimeter of the midsection.

4. Find the lateral area of the frustum of a regular quadrangular pyramid, if an edge of the lower base is 16 in., an edge of the upper base 12 in., and the slant height 18 in.

5. Find the total area of the frustum of a right circular cone, if the radii of the bases are 6 ft. and 8 ft. respectively, and the slant height is 7 ft. 6. A portion of the roof of a tower is a frustum of a right circular The radii of the bases are 10 ft. and 6 ft. respectively, and the altitude is 8 ft. Find the number of square feet in this part of the roof.

cone.

7. The altitude of a right circular cone is h. How far from the vertex must a plane be passed parallel to the base so that the lateral area of the cone cut off shall equal the lateral area of the frustum. Ans. h√2.

8. The diameter of the bottom of a pail is 10 in. and that of the top 12 in. Find the number of square inches of tin in the pail if the slant height is 11 in. Ans. 458.67.

9. Determine the diameter of the blank to make a pressed basin of the form shown in the figure. The depth is 3 in., the bottom has a diameter of 6 in., the top an inside diameter of 7 in., and the rim is in. wide.

Ans.

11.4+ in. SUGGESTION. Consider the rim as a ring between two concentric circles. The blank must have an area equal to the total area of the basin.

Find

10. A pie tin has a diameter of 7 in. at the bottom and 9 in. at the top inside. It is 1 in. deep and has a flange in. wide around the edge. the diameter of the blank required to make the tin.

11. Find the area generated by revolving a square a inches on a side about a diagonal.

12. A tower whose cross section is a regular octagon 6 ft. on a side has a roof that is full pitch. Find the area of the roof.

By full pitch is meant that the vertex of the roof is the same distance above the plates as the width of the tower from face to face.

13. The conical roof over a water tank is half pitch. the diameter of the tank is 12 ft. and the roof projects 1

Ans.

Find its area if ft. at the eaves. 221.5 sq. ft.

By half pitch is meant that the vertex of the roof is one-half the diameter of the tower above the top of the tower.

14. Find the total area of the solid generated by revolving an isosceles trapezoid about a line connecting the middle points of its bases, if the bases are 8 in. and 12 in. respectively, and the angles at one base are each 60°.

16. Determine how to draw a pattern for the lower part of the funnel shown in the figure of the previous exercise.