Proof. 1. ВС⊥AB. Construction 2... AB is tangent to circle with diameter BC. § 163 3. .·. AE × AF = AB2. § 203 4. But AE = AM and EF=BC (diam.)= AB. By substituting AN-AB for AE and AN for AF in step 3, then applying Ax. V and § 118, (5), let the student com AB AN plete the proof that ANNB Write the construction and proof in full. EXERCISES 1. Draw a line-segment 1 in. long and one 4 in. long. Construct the mean proportional between them. Measure the resulting mean proportional. Is the construction accurate? 2. Divide a line-segment 4 in. long internally in extreme and mean ratio. Measure the parts to the nearest sixteenth of an inch. Is the construction accurate? 3. If a and b are two given line-segments, construct a line-segment n such that n = Vab. 4. Compute the parts of a line-segment 20 in. long when divided in extreme and mean ratio. 5. Experience has shown that a book, photograph, or other rectangular object is most pleasing to the eye when its length and width are obtained by dividing the half-perimeter in extreme and mean ratio. Find to the nearest integer the width of such a book whose length is 8 in. 6. Measure the lengths and widths of several books, picture frames, or windows, and find whether or not the Golden Section is apparently used in their construction. 7. Given the longer part of a line-segment divided in extreme and mean ratio, construct the line-segment. 8. From a given external point O, draw a secant OB to intersect a given circle so that AB = OA × ОВ. 2 that 9. Find a point P in the arc AB so 10. Find a point such that tangents from it to two given circles shall be equal. 11. Through one of the intersections of two given circles draw a line such that the two chords intercepted on it by the circles shall be in the ratio of 3 to 4. M 0 A C B 12. Draw a circle which shall pass through two given points and be tangent to a given line. 14. Draw a circle through a given point and tangent to two given lines. SUGGESTION. - Draw a line through the given point and perpendicular to the bisector of the angle formed by the given lines. On this line locate a second point at the same distance from the bisector as that of the given point. Then apply Ex. 12. CHAPTER X AREAS OF POLYGONS 208. Area of a surface. - The area of a surface is the numerical measure of the surface, the unit of measure being the surface of a square whose side is some linear unit. Thus, in the adjoining figure, if the surface of one of the small squares is taken as the unit of measure, the area of the rectangle is 32. In the theorems and construc tions involving areas of polygons, by the words "rectangle," "parallelogram," etc., are meant the areas of the figures mentioned. 209. Equal polygons. - Two polygons are said to be equal when they have equal areas. It is evident that two congruent polygons are equal, but that two equal polygons are not necessarily congruent. Thus, in the adjoining figures, the square and isosceles triangle are equal, for the isosceles triangle is formed by divid ing the square into two triangles, A and B, and placing them in A B BA new positions. But the square and isosceles triangle are not congruent. 210. The area of a rectangle. - If the base AB of a rectangle ABCD is 7 linear units, and the altitude AD is 5 linear units, the rectangle may be divided D into 5 rows of squares with 7 squares in each row. Hence the total number of squares is 5 × 7 or 35. In general, if the altitude of a rectangle is a linear units and the A base is b of those units, the num C B ber of unit squares of surface, or the area of the rectangle, is ab. a b This relation may be proved to be equally true when the altitude and base have no common unit of measure, or when their numerical measures can be expressed only approximately, although the proof is not attempted in this course. That is, The area of a rectangle is equal to the product of its altitude and base. 211. Corollary 1. — The area of a square is equal to the square of its side. For, if the side is a, the area equals a × a or a2, by § 210. 212. Corollary 2. - Two rectangles with equal altitudes are to each other as their bases. For, if a and b are the altitude and base, respectively, of rectangle M, and a and care the altitude and base, respectively, of rectangle N, then by § 210, 213. Corollary 3. - Two rectangles with equal bases are to each other as their altitudes. The proof is left to the student. EXERCISES 1. Find the area of a rectangle whose base is 8.45 in. and altitude 5.38 in. 2. Find the area of a square whose side is 6 in. 3. The area of a square is 160 sq. rd. Find the side of the square to hundredths of a rod. 4. All lots of a city block are 120 ft. long. If lot A in this block is 50 ft. wide and lot B is 60 ft. wide, compare the areas of the two lots (find their ratio in lowest terms). 5. Prove that any two rectangles are to each other as the products of their bases and altitudes. 6. If one field is 80 rd. long and 40 rd. wide, and another field is 60 rd. square, find the ratio of their areas without computing them. 7. A rectangle is 50 in. wide and 200 in. long. Compare its perimeter with that of an equal square. 8. The perimeter of a rectangular lot is 220 yd. and its area 2400 sq. yd. Find the dimensions of the lot. SUGGESTION. - Form an equation. 9. Find the approximate areas of the parallelogram and triangle below, using one of the small squares as the unit of area. In counting the squares, include a square in the figure if one half of it or more lies within the figure; otherwise do not include it. |