(2) By formula (a) we find S=b• h=4X1.4=5.6. 2. Verify formula (a) for b=3.8, h=5; for b=6, h=2.6; for b=3, h=4.2. 3. Verify formula (a) for b=4.6, h=2.4 (see Fig. 253). ABCD (1)+(2)+(3)+(4) =8+2(1%)+4(10)+(1%) (1%) sq. cm. 4. Verify formula (a) for b=1.4, h=0.32. 232. Exercises 1, 2, 3, and 4, § 229, show that formula (a) holds when b and h are integers (whole numbers). Exercises 1, 2, 3, and 4, § 231, show that formula (a) holds also when b and h are decimal fractions. 233. Now let b and h be irrational numbers, such as If the theorem holds in this case, we should find: S=bXh=Vi2XV 27=V22 •3•33=i8. (?) Taking for b and h the values given in I, II, III, and IV below, we find the corresponding values of S by multiplying. By taking b and h to a sufficiently large number of decimal places, the difference between S and 18 may be made as small as may be wished. To prove from these facts that our theorem holds true when b and h are irrational numbers would require a fuller discussion of such numbers than is desirable here. Sufficient has been shown, however, to make us feel that this conclusion is correct, for the inexactness of the result seems to come from our inability to express exactly, without the root sign, the dimensions of the rectangle, and not because the theorem is untrue. 1. Make a similar discussion for b=√/5 and h=√7; for b=1/2 and h=1/3. EXERCISES 1. Prove that any two parallelograms are to each other as the product of their bases and altitudes. 2. Prove that the areas of parallelograms having equal bases are to each other as their altitudes. 3. Prove that the areas of parallelograms having equal altitudes, are to each other as their bases. 4. Prove that the area of a parallelogram is equal to the product of the base and altitude. (Use Exercise 2.) 5. Prove that the area of a triangle is one-half the product of the base and altitude. (Use Exercise 4 and Fig. 254.) Thus the area of a triangle can be computed if the base and the altitude are known. 6. Show that the area of a triangle can be expressed in terms of any two sides, and the sine of the included angle. In Fig. 254 let a=side D B, b=side D A; then (§ 137, p. 104) show that h = a sin D. .. Area A B D = b · h = b · a sin D. 7. Prove that triangles having equal bases and equal altitudes are equal. 8. Prove that the areas of triangles are to each other as the products of the bases and altitudes. 9. Prove that the areas of triangles having equal bases are to each other as the altitudes. 10. Prove that the areas of triangles having equal altitudes are to each other as the bases. II. To bisect a triangle by a line drawn from any vertex to the opposite side. See Exercise 7. 12. To divide a triangle into three equal triangles (trisect) by lines from any vertex to the opposite side. 234. The area of a triangle can be expressed in terms of the perimeter of the triangle and the radius of the inscribed circle, by dividing it into three triangles whose common altitude is the radius,. and whose bases are the sides. 1. Prove that the area of a triangle is equal to one-half the perimeter times the radius of the inscribed circle (see Fig. 255). The area of a triangle may be expressed in terms of the sides of the triangle and the radius of the circumscribed circle. 2. Prove that the area of a triangle is equal to the product of the three sides divided by four times the radius of the circumscribed circle. (See § 189.) The area of a triangle can be expressed in terms of the sides alone, but in the proof of the formula, the following theorem is needed: PROPOSITION III 235. Theorem: The square on the hypotenuse of a right triangle is equal to the sum of the squares on the sides including the right angle. Hypothesis: A B C (Fig. 256) is a triangle having a right angle C; S1, S2, and S are the squares on the sides a, b, and c, respectively; Conclusion: S=S1+S2. Proof: Draw CD AB, dividing S into rectangles R1 and R. Draw A E and CF. Show that triangle E BA and S, have equal bases and altitudes. But A B EFB C. (1) Similarly, prove that triangle F B C=R1. (2) (3) For EB=BC (?) AB=BF (?) ZABE=ZFBC From (1), (2), (3) it follows that †S,=}R,. (?) Therefore S1 =R1. (4) Similarly, draw B G and C H, and prove S2 = R2. Therefore SS, +S2. Q.E.D. |