Draw any two rectangles M and N. We then have : Given M and N rectangles, with altitudes a and a', and bases Proof. 1. Let R be a rect. with altitude a and base b'. 2. M and R have tively; find ratio M to 3. R and N have = = altitudes; and bases b and b', respecR by § 275. bases; and altitudes a and a', respec tively; find ratio R to N by § 276. 4. Multiplying results of steps (2) and (3), gives (1). DEFINITIONS 278. The area of a surface is its ratio to another surface, called the unit of surface, adopted arbitrarily as the unit of measure (§ 180). The usual unit of surface is a square whose side is some linear unit; for example, a square inch or a square foot. 279. We call two surfaces equivalent (~) when their areas are equal. The dimensions of a rectangle are its base and altitude. 280. The area of a rectangle is equal to the product of its base and altitude. In all propositions relating to areas, the unit of surface (§ 278) is understood to be a square whose side is the linear unit. Draw any rectangle M, and a square N whose side is 1. We then have: Given a the altitude and b the base, of rect. M; and N the unit of surface, a square whose side is the linear unit, To Prove that, if N is the unit of surface, Note. The statement of Prop. III is an abbreviation of the following: If the unit of surface is a square whose side is the linear unit, the number which expresses the area of a rectangle is equal to the product of the numbers which express the lengths of its sides. An interpretation of this form is always understood in every proposition relating to areas. 281. By § 280, the area of a square equals the square of its side. Ex. 1. Find the ratio of the area of a square to the product of its diagonals. Ex. 2. The area of a rectangle whose base is 24 is 432. Find the diagonal of the rectangle. Ex. 3. How many shingles will it take to cover a roof 36 feet by 18 feet (one side) with shingles that average 4 inches wide, 16 inches long, and are laid 6 inches to the weather? PROP. IV. THEOREM 282. The area of a parallelogram is equal to the product of its base and altitude. Draw ABCD, AD (b) being the base; draw DF (a) the altitude meeting BC at F. We then have : Given b the base, and a the altitude, of ☐ ABCD. To Prove area ABCD = a X b. (1) Proof. 1. Draw line AE || DF, meeting CB prolonged at E. 2. Then, rt. A ABE and DCF are equal. (§§ 61, 104) 3. If from figure ADCE we take ▲ ABE, there remains AC; if we take ▲ DCF, there remains rect. AF. 4. Then, area ABCD = area AEFD; whence equation (1). 283. It follows from § 282 that: 1. Two parallelograms having equal bases and equal altitudes are equivalent (§ 279). 2. Two parallelograms having equal altitudes are to each other as their bases. 3. Two parallelograms having equal bases are to each other as their altitudes. 4. Any two parallelograms are to each other as the products of their bases by their altitudes. Ex. 4. The area of a parallelogram is 288, the base is twice the altitude. Find the dimensions. Construct the parallelogram. Can more than one such parallelogram be drawn? How many and what parts are necessary for a definite figure? Why? Ex. 5. Find the ratio of the area of a rhombus to the product of its diagonals. PROP. V. THEOREM 284. The area of a triangle is equal to one-half the product of its base and altitude. Draw A ABC having base BC (b); draw AE1 BC, meeting BC, or BC extended, at E. We then have : Given b the base, and a the altitude of ▲ ABC. (Draw line AD || BC, and line CD || AB. divides ABCD into two equal A.) 285. It follows from § 284 that: By § 105 AC 1. Two triangles having equal bases and equal altitudes are equivalent. 2. Two triangles having equal altitudes are to each other as their bases. 3. Two triangles having equal bases are to each other as their altitudes. 4. Any two triangles are to each other as the products of their bases by their altitudes. 5. The area of any triangle is one-half that of a parallelogram having the same base and altitude. Ex. 6. Draw a line dividing a given right triangle into two equivalent isosceles triangles. Ex. 7. Prove that each of its medians divides a triangle into two equivalent parts. Ex. 8. The base of a triangle is 37 feet, the altitude is 32 feet. How many square yards in its area? Ex. 9. The area of a triangle is 216, the altitude is 12; find the base. Ex. 10. If E and F are the middle points of sides AB and CD, respectively, of parallelogram ABCD, the lines AF, EF, and CE divide the parallelogram into four equal triangles. A E B F Ex. 11. If the middle point of any side of a parallelogram be joined to the opposite vertices, the triangle included by these lines and the opposite side is equivalent to one-half the parallelogram. Ex. 12. If the middle point of a diagonal of any quadrilateral be joined to the opposite vertices, the quadrilateral is divided into two pairs of equivalent triangles, and into two equivalent parts. 286. The area of a trapezoid is equal to one-half the sum of its bases multiplied by its altitude. Draw trapezoid ABCD, AB (b) and DC (b') being || sides. From D draw altitude DE (a) meeting AB at E. We then have : Given AB (b) and DC (b') the bases, and DE (a) the altitude, of trapezoid ABCD. To Prove area ABCD = a × 1 (b + b'). (Draw diagonal BD. The trapezoid is composed of two A whose altitude is a, and bases b and b', respectively.) 287. Since the line joining the middle points of the nonparallel sides of a trapezoid is equal to one-half the sum of the bases (§ 132), it follows that The area of a trapezoid is equal to the product of its altitude by the line joining the middle points of its non-parallel sides. Note. The area of any polygon may be obtained by finding the sum of the areas of the triangles into which the polygon may be divided by drawing diagonals from any one of its vertices. But in practice it is better to draw the longest diagonal, and draw perpendiculars to it from the remaining vertices of the polygon. The polygon will then be divided into right tri angles and trapezoids; and by measuring the lengths of the perpendiculars, and of the portions of the diagonal which they intercept, the areas of the figures may be found by §§ 284 and 286. PROP. VII. THEOREM 288. Two similar triangles are to each other as the squares of their homologous sides. Draw ▲ ABC; and construct ▲ A'B'C' similar to ▲ ABC, AB (c) and A'B' (c') being homologous sides. We then have : Given c and c' homologous sides of AABC and A'B'C', respectively. To Prove ABC c2 A'B'C' c'2 (1) Proof. 1. Draw altitudes CD(h) and C'D'(h') 1 AB and A'B', respectively. 2. Find ratio ABC to A'B'C' in terms of h and c, and h' and c'. 3. Express ratio h to h' in terms of c and c'. 4. Substituting in result of step 2, gives (1). (§ 285, 4) (§ 245) 289. Note. Two similar triangles are to each other as the squares of any two homologous lines. Ex. 13. A field has two sides parallel, the lengths of these sides being 160 yards and 302 yards, respectively. The distance between the parallel sides is 48 rods. How many acres has the field? Draw a diagram. Ex. 14. A field has two parallel sides whose lengths are 322 and 168 yards, respectively. The area of the field is 12 acres. Find the distance between the parallel sides and make a diagram of the field. Ex. 15. A square field contains 3 acres 12 square rods. Find its perimeter in rods, correct to three decimal places. |