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(CHAPTER XVIII) The proof of lines in proportion from similar triangles has two parts: 1. Proving triangles similar by one of the summary, Art. 267; 2. The corresponding sides of similar triangles are in proportion, Art. 259. The proof of the product of two lines equals the product of two other lines, has three parts: 1. Proving triangles similar; 2. Corresponding sides in proportion; 3. Product of the means equals product of the extremes, Art. 242.

269. Method is 259. The same thing can be proved of medians and angle-bisectors.

271. Method is 242. Join the ends of two chords. 272. Method is 242. Join the alternate intersections. 273. Method is 242. Join the intersection points.

274. The last three theorems can be grouped under one statement: If two chords intersect (internally or externally), the product of the parts of one chord equals the product of the parts of the other chord.

276. Method is: 1. Art. 265; 2. Art. 259; 3. Art. 259.
277. Method is 276, 2. Make a right triangle.
278. Method is 277 (or 276, 3).

(CHAPTER XIX] This chapter is given up entirely to the Pythagorean theorem and its applications. The method of proof for the theorem is Art. 276, 3. Draw a perpendicular from the right angle to the hypotenuse.

(CHAPTER XX] Similar polygons follow the definition, Art. 253. They depend usually upon similar triangles. Circles follow regular polygons.

285. Method is Art. 253. Use similar triangles.
286. Method is 267b. Use similar polygons, Art. 253.
287. Method is Art. 285 (or 253).
288. Method is Art. 253.

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See Proportion, 8. b d

b + d + f 290. Method is Arts. 253 and 289. 291. Method is Arts. 290 and 269. Draw radii and apothem. 293. The theorem of limits is to be treated informally.

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There are two parts to the condition: 1. The variables are always equal. 2. Each approaches a limit. Under these conditions, their limits are equal.

294. Measurement of a circle is proved from the measurement of a regular polygon by means of 293. Method is 291 and 293. Inscribe regular polygons with the same number of sides.

C C1 295. Method is Art. 294:

D Di С 296. 7, whence C = 2 * R. An important formula.

D 297. Method is Art. 230. The value of a cannot be found exactly. Its commonly accepted value is 3.1416–, or nearly 34. Compared to our number system, a is incommensurable.

[CHAPTER XXI]

“The area of a rectangle equals the product of the base and altitude,” Art. 300, is put first as the basis. All other theorems of area depend either directly or indirectly upon this. The circle follows the regular polygon.

300. This may be taken as a definition, or its commensurable case may be treated informally by dividing it into unit squares.

301. Method is 300.

302. Method is 300. Erect perpendiculars at the ends of the base. Show that the triangles formed are congruent; then the parallelogram equals the rectangle.

303. Method is 302. Make a parallelogram from the triangle. 304. Method is 303. Draw a diagonal making triangles.

As the mid-line of a trapezoid equals one-half the sum of the bases (Art. 117), Art. 304 can be read, “The area of a trapezoid equals the product of the altitude and the mid-line."

305. The area of any polygon can be found by dividing into triangles, and measuring the triangles.

306. Method is 303. Draw all the radii.

307. Method is 306 and 293. Circumscribe a regular polygon about the circle. 308. As area of a circle 12C X R, and C 2 + R,

Area of a circle = a R2. An important formula.

309. A sector is to its circle its central angle is to four right angles.

The area of a segment of a circle is found by subtracting the area of the triangle formed by the chord and the radii from the area of the sector. This can be done if the radius is known and the central angle is 30°, 45°, 60°, 90°, and their supplements; also 36° and 72o.

(CHAPTER XXII]

Equality of areas can be developed from congruent triangles. Another method is to show that their area formulas are equal. This is easily done in the first four theorems and Art. 322. The other theorems can depend upon the first four.

312. Method is 302. Another proof is by Art. 52. See 302. 313. Method is 303. 314. Method is 304. See note. (Or by Art. 52.) 315. Method is 302 and 303. 316. This is the Pythagorean theorem again, but proved here by

The previous proof was algebraic. The method is Arts. 52 and 315.

317. Method is 316. 318. Method is 316. 319. Method is 317. 320. Method is 313.

321. The algebraic expression a2 means in geometry the area of a square whose side is a. The algebraic expression ab means in geometry the area of a parallelogram whose base is b and altitude a.

322. Method is 277 and 321.

areas.

(CHAPTER XXIII]

The theorems of this chapter depend upon the corresponding theorems of area, except 331, 332, 333, and 334. These four depend either directly or indirectly upon 328.

325. Method is 302.
326. Method is 302 (or 325).
327. Method is 302 (or 325).
328. Method is 303.
329. Method is 303 (or 328).
330. Method is 303 (or 328).

331. Method is 328. Draw corresponding altitudes.

332. Method is 328. Draw corresponding altitudes. Another method is Art. 331.

333. Method is Art. 332 and 289. Draw corresponding diagonals. 334. Method is Art. 333 and 269. 335. Method is Art. 308 (or 334 and 293).

[CHAPTER XXIV] The Pythagorean theorem, Art. 282, is the basis. 339. Method is Art. 282. 340. Method is Art. 282.

341. Arts. 282, 339, and 340 can be grouped under one statement: The square of one side of a triangle equals the sum of the squares of the other two sides plus twice the product of one of those sides and the external projection of the other upon it.

Thus if we draw an obtuse triangle, we have Art. 340. If the vertex of the triangle be moved to the right or left, the external projection becomes smaller. When this external projection becomes zero, we have Art. 282. Continuing this movement the external projection becomes less than zero or negative. We then have Art. 339.

342. Method is 339 and 340. 348. Method is 270. 344. Method is 270. Circumscribe a circle about the triangle. 345. Method is 303. Draw the altitude. 346. Method is 282. 347. Method is 346.

(CHAPTER XXV]

Extreme and Mean Ratio; Square Ratio; Incommensurable Cases. 351. Method is Art. 273.

352. Method is Art. 351. Angles of 36°, 60°, and 90°, their halves, their sums and differences, are practically the only angles that can be constructed by straight edge and compass.

353. Method is Art. 352.
354. Method is 326 and 322.
355. Method is 354 and 287.
356. Method is 320 and 247.
358. Method is 177 and 293.
359. Method is 237 and 293.
360. Method is 300 and 293.

SUMMARIES OF METHODS

ARRANGED ALPHABETICALLY IN Two GROUPS, GENERAL AND

GEOMETRICAL

GENERAL

235. Analysis of Construction Problems:

Construction problems usually depend upon the drawing of locus lines or of a triangle. One and only one triangle can be constructed under the conditions of congruent triangles. When the construction is not clear, it is often advisable to:

1. Draw the figure as if the construction was completed. 2. Mark all known lines and angles.

3. From these known parts, construct some part of the figure. (Usually a triangle.)

4. Around this figure, complete the construction.

Analysis of Numerical Exercises:
1. Draw the figure.
2. Mark all known parts with their numbers.

3. Recall the theorem that connects the known parts with the desired part by an equation.

4. Write and solve this equation.

156. Analysis of Theorems:

1. Read the statement carefully, separating the given conditions from the thing to be proved.

If the statement is in the form of a conditional sentence, the condition is given and the conclusion is to be proved.

Thus, If two sides of a triangle are equal, | the opposite angles are equal.

If the statement is in the form of a declarative sentence, the subject and its modifiers form the given conditions; the rest of the statement

to be proved.

Thus, Two straight lines perpendicular to a third straight line | are parallel.

2. Having determined the thing to be proved, recall the methods of proving that thing.

3. Select the method most suited to the given conditions.

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