mensuration

7.5 A Summary of Useful Mensuration Formulas

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This section lists a number of formulas relating to measuring geometric figures. (The word “mensuration” is one of those traditional, old-fashioned words that means the branch of mathematics that deals with measuring geometric figures.) Mastering the list, including the ideas behind the derivations of the formulas (where possible; some require calculus, and so their derivations will be postponed until later in the program), will be excellent preparation for college or university mathematics.Plane FiguresRectanglesA rectangle is a quadrilateral (a four-sided figure) that has opposite sides equal and parallel, and all four internal angles are right angles.The area of a rectangle is its length times its width:

A=LWA=LW

The perimeter of a rectangle is the sum of the lengths of its four sides:

P=2L+2WP=2L+2W

A square is a special case of a rectangle for which the length and width are the same. If we label the length of each side of a square by xx, then the area of a square is A=x2A=x2, and the perimeter of a square is P=4xP=4x.TrianglesThe figure shows a triangle embedded within a rectangle. It is traditional to label the sides of the containing rectangle as bb (for base) and hh (for height); thus, the corresponding side of the triangle is called its base, and the distance from the vertex opposite the base to the base is called the height of the triangle. (The dashed line in the diagram is perpendicular to the base of the triangle.) Of course, if the triangle is oriented differently, then its base and height will be different; thus, the base and height of a triangle depend on its orientation.

By carefully examining the diagram, can you argue that the area of the triangle is half of the area of the containing rectangle? That is, the area of the triangle is

A=12bhA=12bh

TrapezoidsA trapezoid is a quadrilateral with two opposite sides parallel; the other two opposite sides may or may not be parallel. Thus, the class of trapezoids includes squares, rectangles, rhombuses, and parallelograms as special cases.The height of a trapezoid is defined to be the perpendicular distance between the parallel sides. The parallel sides are known as the bases of the trapezoid.By separating a trapezoid into two triangles with a diagonal line segment, it follows from the formula for the area of a triangle that the area of a trapezoid is

AA=12ah+12bh=12(a+b)hA=12ah+12bhA=12(a+b)h

In words, the area of a trapezoid is the average base times the height.The same area formula also applies to rhombuses and parallelograms, because they are special cases of trapezoids. A parallelogram is a quadrilateral in which each pair of opposite sides is congruent and parallel, although adjacent sides may have different lengths. A rhombus is a parallelogram in which all four sides have the same length.PolygonsThe sum of the angles in a triangle is 180∘180∘, as shown in Chapter 6. A quadrilateral can be subdivided into two triangles by a diagonal line segment, and so it follows that the sum of the internal angles in a quadrilateral is 2×180∘=360∘2×180∘=360∘. A pentagon can be subdivided into three triangles by two diagonals, which means that the sum of the internal angles in a pentagon is 3×180∘=540∘3×180∘=540∘. Similarly, a polygon with nn sides can be separated into (n–2)(n–2) triangles by (n–3)(n–3) diagonals, and so the sum of the internal angles in a polygon with nn sides is (n–3)×180∘=360∘(n–3)×180∘=360∘.CirclesFor a circle that has radius rr, the circumference is

C=2πrC=2πr

and the area is

A=πr2A=πr2

Both of these formulas can be proved using calculus, and we shall do so in Chapter 5 of our forthcoming Calculus, to appear on this site.The area of a sector of a circle is proportional to the size of the central angle of the sector: A=12θr2A=12θr2, where the central angle θθ is measured in radians. (If you are

unfamiliar with radian measure, then an alternative formula is A=12srA=12sr, where ss is the distance around the arc of the sector.)Three-Dimensional FiguresCubeEach of the six faces of a cube is a square. If each side length is xx, then the volume of the cube is V=x3V=x3, and the surface area of the cube is A=6x2A=6x2.Rectangular ParallelepipedEach of the six faces of a rectangular parallelepiped is a rectangle; another name for this kind of figure is a rectangular box. (Think of a shoe-box as an example.) If the side lengths of the rectangular parallelepiped are LL, WW, and HH, then the volume of the rectangular parallelepiped is V=LWHV=LWH, and its surface area is A=2LW+2LH+2WHA=2LW+2LH+2WH.CylinderA right circular cylinder has circular ends, where the radii of the circular ends are perpendicular to the axis (the line segment joining the centres of the circular ends). The volume of a right circular cylinder is the area of the base times the height hh: V=πr2hV=πr2h. The surface area is the sum of the areas of the two bases plus the area of the lateral part of the cylinder; the area of the lateral part can be determined by cutting the cylinder in a straight line parallel to the axis, and unwrapping the cylinder to produce a rectangle of height hh and width equal to the circumference of the circular bases, which is 2πr2πr. Thus, the surface area of a cylinder is A=2πr2+2πrhA=2πr2+2πrh.For a general cylinder, the bases need not be circles and the axis need not be perpendicular to the bases, although the bases should be congruent and the planes that contain the bases should be parallel. The volume of a general cylinder is the area of one base times the length of the lateral edge (i.e., the distance between corresponding points on the bases), and the lateral surface area is the perimeter of one base times the length of the lateral edge.A prism is a type of general cylinder where the bases are congruent polygons. A general parallelepiped is a type of a general cylinder, where the bases are parallelograms. A rectangular parallelepiped is a type of general cylinder (in fact, a right cylinder), where the bases are rectangles.ConeOne can use calculus to show that the volume of a right circular cone with height hh and radius of base rr is V=13πr2hV=13πr2h. That is, the volume of the right circular cone is one-third the volume of the smallest coaxial right circular cylinder that can contain the cone.The surface area of a right circular cone can be obtained by cutting the base out of the cone, and then cutting the curved part of the cone with a straight cut from a point on the circular base to the apex, and then opening up the curved part and laying it out flat to obtain a sector of a circle of radius ss. (In fact, the reverse procedure is a common way to create a cone from a flat piece of paper.) It follows that the surface area of a cone is

AAA=πr2+12θs2=πr2+12(2πr2πs⋅2π)s2=πr2+πrsA=πr2+12θs2A=πr2+12(2πr2πs⋅2π)s2A=π

r2+πrs

In a right circular cone the line segment joining the centre of the base to the apex is perpendicular to the base. In general, the base of a cone need not be a circle, and even if the base is a circle the apex need not be directly over the centre of the base. A pyramid is an example of a general cone for which the base is a polygon.For a general cone, the volume is one-third the area of the base times the height (i.e., the perpendicular distance from the apex to the plane containing the base).SphereA sphere is a three-dimensional figure for which each point is the same distance rr from a fixed point, the centre of the sphere. The distance rr is called the radius of the sphere. It can be proved using calculus that the volume of a sphere with radius rr is V=43πr3V=43πr3, and its surface area is A=4πr2A=4πr2.*In reviewing the formulas of this section, you’ll notice that perimeter and circumference formulas are linear in terms of the relevant lengths, areas and surface areas are quadratic in terms of the relevant lengths, and volumes are cubic in terms of the relevant lengths. This makes sense in terms of units, doesn’t it? For example, if all relevant lengths are measured in cm, then perimeters and circumferences are also measured in cm, but areas are measured in cm22, and volumes are measured in cm33.

EXERCISES1. Determine the area of each plane figure.a rectangle with dimensions 77 cm and 33 cmShow Answera triangle with base 44 cm and height 55 cmShow Answera triangle with sides 33 m, 44 m, and 55 mShow Answera trapezoid with bases of 88 cm and 99 cm, and height 1111 cmShow Answera circle with radius 2.302.30 kmShow Answera circular sector with central angle π/3π/3 radians and radius 5.015.01 cmShow Answer2. Determine the perimeter/circumference of each plane figure.a rectangle with dimensions 77 cm and 33 cmShow Answera triangle with base 44 cm and height 55 cmShow Answera triangle with sides 33 m, 44 m, and 55 mShow Answera trapezoid with bases of 88 cm and 99 cm, and height 1111 cmShow Answera circle with radius 2.302.30 kmShow Answera rhombus with one side of length 3.93.9 cm

Show Answer3. Determine the volume of each three-dimensional figure.a cube with side length 2.502.50 cmShow Answera rectangular parallelepiped (box) with sides of length 22 cm, 55 cm, and 99 cmShow Answera right circular cylinder, with height 5.005.00 cm and radius of base 4.224.22 cmShow Answera cone with height 11.211.2 cm and radius of base 9.179.17 cmShow Answera sphere of radius 6.5806.580 cmShow Answer4. Determine the surface area of each three-dimensional figure.a cube with side length 2.502.50 cmShow Answera rectangular parallelepiped (box) with sides of length 2.002.00 cm, 5.005.00 cm, and 9.009.00 cmShow Answera right circular cylinder, with height 5.005.00 cm and radius of base 4.224.22 cmShow Answera cone with height 11.211.2 cm and radius of base 9.179.17 cmShow Answera sphere of radius 6.5806.580 cmShow Answer5. Determine the area of a square that has perimeter 2828 cm.Show Answer6. Determine the circumference of a circle that has area 9π9π m22.Show Answer7. Determine the volume of a cube that has surface area 2424 m33.Show Answer8. Determine the surface area of a sphere that has volume 17.0017.00 cm33.Show AnswerCHALLENGE PROBLEMHeron’s formula for the area of a triangleExercises require only the application of techniques discussed in this section; problems may require deeper thought and some insight.Show that the area of a triangle with sides of lengths aa, bb, and cc is A=s(s–a)(s–b)(s–c)−−−−−−−−−−−−−−√A=s(s–a)(s–b)(s–c)where s=12(a+b+c)s=12(a+b+c). (This is known as Heron’s formula.)

CHALLENGE PROBLEMA formula for the volume of a truncated coneExercises require only the application of techniques discussed in this section; problems may require deeper thought and some insight.A cone is cut by a plane perpendicular to its base, and the top (the part with the apex) is removed, leaving a three-dimensional figure with two circular bases and a curved lateral side, called a truncated cone. The circular bases have radii r1r1 and r2r2, and the distance between the centres of the two circular bases isthe truncated cone.Show AnswerHISTORYThe marvelous mechanical arguments of ArchimedesArchimedes (c. 287 BC–c. 212 BC) used ingenious mechanical arguments to determine many geometric properties, such as formulas for the volume and surface area of a sphere, volume of a cone, and many others. His arguments involved imagining that each geometric figure were made of a material with constant density, and then balancing each figure against a figure with known properties. Using his “law of the lever” (nowadays this is expressed in terms of torque, that the net torque on a system in equilibrium is zero), Archimedes proved that the volume of a sphere is 2/32/3 the volume of the smallest cylinder that can contain the sphere, and similarly the surface area of a sphere issurface area of the smallest cylinder that can contain the sphere.Archmedes considered the determination of the volume and surface area of a sphere to be his greatest achievement, and he had a sculpted sphere and cylinder placed on his tomb.

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