VOLUME OF PYRAMIDS & CONES BACK NEXTThe formula for thevolume of pyramids and conestells youhow much space is inside each object.For these two solid shapes, the volume formula is the same: it's one third of the area of the base times the height.Volume of Pyramids or Cones = BaseheightWhy? Here it is in a nutshell. The volume of three pyramids is equal to the volume of one prism with the same base and height. Similarly, the volume of three cones is equal to the volume of one cylinder with the same circular base and height.

The volume of each cone is equal to Bh= (28.3 10) = 94 cm3. All three cones combined equals 283 cm3. The volume of the cylinder is equal toBh= 28.3 10 = 283 cm3, ta da!

The volume of each pyramid is equal to Bh= (18 8) = 48 cm3. All three pyramids combined equals 144 cm3. The volume of the prism is equal toBh= 18 8 = 144 cm3.To find the volume of a sphere, follow this simple formula (which took a brilliant ancient Greek mathematician named Archimedes years to derive):Volume of a Sphere = 4/3 x radius cubed= 4/3r3

Surface area of a right pyramidThe faces bounding a right pyramid consist of a number of triangles together with the base. To find the surface area, we find the area of each face and add them together. Depending on the information given, it may be necessary to use Pythagoras Theorem to calculate the height of each triangular face. If the base of the pyramid is a regular polygon, then the triangular faces will be congruent to each other.

EXAMPLEVABCDis a square-based pyramid with vertexVandbaseABCD, withVvertically above the centre of thesquare base. The height of the pyramid is 4 cm andthe side length of the base is 6 cm, find the surfacearea of the pyramid.SOLUTIONWe need to find the heightVEof triangleVBC,using Pythagoras Theorem.VE2=VO2+OE2

= 42+ 32

= 25

HenceVE= 5 cm.Area ofVCB= CB VE

=65

= 15 cm2

Area of base= 66

= 36 cm2

Surface area= 415 + 36

= 96 cm2

The surface area of the pyramid is 96 cm2.

click for screencastEXERCISE 1When it was built, the Great Pyramid of Cheops in Egypt had a height of 145.m and its base was a square of side length 229 m. Find its surface area in square metres, correct to three significant figures.Volume of a pyramidHere is a method for determining the formula for the volume of a square-based pyramid.Consider a cube of side length 2x. If we draw the four long diagonals as shown, then we obtain six square-based pyramids, one of which is shaded in the diagram.Each of these pyramids has base area 2x 2xand heightx.Now the volume of the cube is 8x3.So the volume of each pyramid is8x3=x3. Since the base area of each pyramid is 4x2it makes sense to write the volume asVolume =4x2 x=area of the baseheight.We can extend this result to any pyramid by using a geometric argument, giving the following important result.Volume of a pyramid =area of the baseheight.See the appendix on the pyramid for details.

click for screencastEXERCISE 2Find the volume of the Great Pyramid of Cheops whose height is 145.75 m and whose base is a square of side length 229 m. Give answer in cubic metres correct to two significant figures.

click for screencastEXERCISE 3Find the volume of the diamond, with height 24 cm and side length 10 cm as shown.

return to topCONESTo create aconewe take a circle and a point, called thevertex, which lies above or below the circle. We then join the vertex to each point on the circle to form a solid.

If the vertex is directly above or below the centre of the circular base, we call the cone aright cone. In this section only right cones are considered.If we drop a perpendicular from the vertex of the cone to the circular base, then the length of this perpendicular is called theheighthof the cone.The length of any of the straight lines joining the vertex to the circle is called theslant heightof the cone. Clearlyl2=r2+h2, whereris the radius of the base.

Surface area of a coneSuppose the cone has radiusr, and slant heightl, then the circumference of the base of the cone is 2r.To find the area of the curved surface of a cone, we cut and open up the curved surface to form a sector with radiusl, as shown below.

In the figure to the right below the ratio of the area of the shaded sector to the area of the circle is the same as the ratio of the length of the arc of the sector to the circumference of the circle.

Thus the fraction of the area of the whole circle taken up by the sector is=.Hence, the area of the sector isl2=rl.In conclusion, the area of the curved surface of the cone isrlAdding this to the base, we haveSurface area of a cone =rl+r2.

click for screencastEXERCISE 4Find the surface area of the solid with dimensions shown.

Volume of a coneWhen developing the formula for the volume of a cylinder in the moduleArea Volume and Surface Area,we approximated the cylinder using inscribed polygonal prisms. By taking more and more sides in the polygon, we obtained closer and closer approximations to the volume of the cylinder. From this, we deduced that the volume of the cylinder was equal to the area of the base multiplied by the height.We can use a similar approach to develop the formula for the volume of a cone.Given a cone with base radiusrand heighth, we construct a polygon inside the circular base of the cone and join the vertex of the cone to each of the vertices of the polygon, producing a polygonal pyramid. By increasing the number of sides of the polygon, we obtain closer and closer approximations to the cone. Hence,Volume of a cone =area of the baseheight=r2h

click for screencastEXERCISE 5Find the volume of the solid described in the previous exercise.return to topOBLIQUE PRISMS, CYLINDERS AND CONESWe have seen that the volume of a right rectangular prism is area of the base multiplied by the height. What happens if the base of the prism is not directly below the top?Cavalieris first principlestates that if the cross-sections of two solids, taken at the same distance above the base, have the same area, then the solids have the same volume.

We will not give a proof of Cavalieris principle here. To present a rigorous proof requires integration and slicing ideas.It allows us to say that the volume of any rectangular prism, right or oblique, is given by the area of the base multiplied by the height.The same applies to oblique cylinders and cones.More on conesThe portion of a right cone remaining after a smaller cone is cut off is called afrustum. Suppose the top and bottom of a frustum are circles of radiusRandr, respectively, and that the height of the frustum ish, while the height of the original cone isH. The volume of the frustum is by the difference of the volumes of he two cones and is given byVolume of a frustum =[H(R2r2) +r2h].

Using similar triangles, we can eliminateHand the formula can be rewritten asVolume of a frustum =h(R2+r2+rR).EXERCISE 8Derive these results.Similarly, it can be shown that the surface area of the frustum of a cone with base radiir and Rand slant heights, is given bySurface Area of a frustum = (r2+R2) + (r+R)s.

Note that whenr=lwe obtain the surface area of a cylinder.EXERCISE 9Derive this formula.The concepts and insights developed in finding the formulas for areas and volumes are used in Physics and Engineering to find such quantities as thecentre of massand themoment of inertiaof a solid body. Thus the development of volume formulas are important for students, as is the careful memorizing of the key formulas, such as the volume of a sphere.APPENDIX ON THE PYRAMIDIn the sectionPyramids, we showed that a pyramid whose base is a square of sides 2x 2xand heightxhas volume 2x 2x xwhich isbase height.

In this appendix, we will show how to extend this result to any rectangular based pyramid.It can also be shown that the volume of any pyramid is given by area of base perpendicular height.Cavalieris first principlestates that given two solids of the same height, whose cross-sections, taken at the same distance above the base, are of equal area, then the solids have the same volume.Cavalieris second principlestates that if the cross-sections of two solids taken at the same distance above the base have areas in the ratioa:b, then the solids have their volumes in the ratioa:b.We will use the second principle to show that the formula holds for any square pyramid of heighth.Take two square pyramids of heighth, one with base square length 2hand one with base square length 2l. From our earlier discussion, we know that the volume of the first pyramid isV1= 4h2h. LetV2be the volume of the second pyramid.Take a slices atBandB1, distancedfrom the vertices, as shown in the diagram,so thatAB=A1B1=d.

LetBC=a,B1C1=b. Referring to the diagrams,ABCis similar toADE.Hence=impliesa=d.Hence the area of the cross-section for the first pyramid is given by 4a2but this equals 4d2.Now in the second pyramid,A1B1C1is similar toA1D1E1.Hence=impliesb=Thus the area of the cross-section is 4b2but this equals 4l2d2. Now the ratios of the areas is 4a2:4b2= 4d2:=h2:l2.This ratio is independent of d and so from Cavalieris second principle, the ratio of the volumes is alsoh2:l2. This gives us=HenceV2== 4l2hwhich is base perpendicular height.This tells us that the formula base perpendicular height is valid for a square pyramid of heighth.Suppose we have two pyramids, one with a square base side length 2xand heighthand one whose base is a rectangle with sides 2y 2z, also of heighth. The volume of the first pyramid is 4h2hby our discussion above.We can choosexsuch that 4x2= 4yz, so that the area of the rectangle equals the area of the square.

Take slices at a distancedfrom the vertex in each pyramid. Using similar triangles, we can easily show that so=anda=. Thus the area of the cross-section in the first pyramid isA1= 4a2=.In the second pyramid, we can again use similar triangles, as we did above, to show thaty1=. Similarly, ifz1is half the width of the rectangular slice, thenz1=giving the area of the cross-section to be 4y1z1=.Sincezy=x2, the cross-sections are equal in area and since the pyramids have the same height, their volumes are equal by Cavalieris first principle. Thus the volume of the rectangular pyramid is 2y 2zh, or base areaxheight.It remains to say that a similar method can be used to progress from a pyramid with rectangular base to one whose base is a regular polygon, although the technical details are more complicated and will not be given here.return to topREFERENCESA History of Mathematics: An Introduction, 3rd Edition, Victor J. Katz, Addison-Wesley, (2008)History of Mathematics, D. E. Smith, Dover publications New York, (1958)return to topANSWERS TO EXERCISESEXERCISE 1137 000 m2EXERCISE 22 500 000 m3EXERCISE 3800 cm3EXERCISE 4726m2EXERCISE 5(1800 + 36)m3EXERCISE 6The ratios of the areas of the cross-sections taken at the same heights is 4x2: cdFrom Cavalieris second principle:Volume of pyramid =x3=cdxEXERCISE 7Surface area =(3+ 66)cm2Volume = 105cm3EXERCISE 8Volume of frustrum = volume of large cone Volume of small cone=R2Hr2(Hh)=((R2r2)H+r2h)By similarity,H=. Substituting in the above formula gives the result.EXERCISE 9Surface area of the frustrum =(r2+R2) +Rlr(ls)By similar triangles,Rs=Rllr(r2+R2) +Rlr(ls) =(r2 +R2) +(r+R)sEXERCISE 102r 2r= 4r2EXERCISE 11aThe square of the radius of the cross-section =r2h2(By Pythagoras theorem)Area of cross-section=(r2h2)bArea of cross-section =r2h2cVolume of cylinder volume of cone =r3r3=r3EXERCISE 12Surface area = 2r 2R= 42rRSolid Mensuration Problem SetThis is just the first batch of problems. Enjoy Solving.

1. Each side of the cube is increased by 1%. By what percent is the volume of the cube increased? Ans. 3.03%2.Given a sphere of diameter d, what is the percentage increase in its diameter when the surface area increases by 21% Ans. 10%3. How many times does the volume of the sphere increase if the radius is doubled? Ans. 8 times4. A circular cone having an altitude of 9 m is divided into two segments having the same vertex. If the smaller altitude is 6 m, find the ratio of the volume of the small cone to the big cone. Ans. 0.296. Ans. 13503.4 cu. cm.5. Find the volume of the cone to be constructed from a sector having a diameter of 72 cm and a central angle of 2106. What is the height of a circular cone having a slant height of (10x ) and a base diameter of 2x? Ans. 3x7. The ratio of the volume to the lateral area of a right circular cone is 2:1. If the altitude is 15 cm, what is the ratio of the slant height to the radius? Ans. 5:28. A regular triangular pyramid has an altitude of 9m and a volume of 187.06 cu. m. What is the base edge in meters? Ans. 12m9. The volume of the frustum of a regular triangular pyramid is 135 cu.m. The lower base is an equilateral triangle with an edge of 9 cm. The upper base is 8 m above the lower base. What is the upper base edge in meters? Ans. 3m10. What is the frustum of a cone whose upper base is 15 cm in diameter and lower base 10 cm. in diameter with an altitude of 25 cm? Ans. 3108.87 cm3