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Geometry of the 1950's (continued)

Many math teachers claim that geometry has changed over the last half-century. In fact, given the advent of integrated math and Common Core Standards, some (including myself) feel that the meat of geometry has been replaced by a utilitarian skeleton that does little to satisfy.

As a a means to reflect on whether or not geometry has changed, consider the following AMSCO Exam from January 1954. Can you (or your students) successfully pass it?

Note: I am not saying that this exam represents the geometry I believe should be taught (As it doesn't)...but it does provide a contrast from the geometries taught separated by a 60-year period.


Part II (Part I appeared last week)
Answer three questions from Part II. Values in square brackets are credit values.
  • Prove: Two right triangles are congruent if the hypotenuse and a leg of one are equal to the hypotenuse and a leg of the other. [10]
  • Diagonal AC of parallelogram ABCD is extended through A to point E and through C to point F, making CF equal to AE. Lines ED, DF, FB, and BE are drawn. prove that EDFB is a parallelogram. [10]
  • Prove: If two chords intersect within a circle, the product of the segments of one is equal to the product of the segments of the other. [10]
  • In triangle ABC, sides AB and AC are equal. A line through B intersects AC at D. BD is extended through D to point E, and CE is drawn. prove that BE is greater than CE. [10]
  • Paralell lines r and s are d distance apart and point P is any point between the two lines.
    1. What is the length of the radius of a circle that is tangent to both r and s? [1]
    2. What is the locus of the center of a circle that is tangent to both r and s? [2]
    3. What is the locus of the center of a circle whose radius is d/2 and which passes through P? [2]
    4. On your answer paper draw the figure at the right. [Construction of r parallel to s is not required.] Now construct a circle tangent to the two parallel lines and passing through the given point P. [4]
    5. How many different circles are there that satisfy the requirements specified in part 3? [1]
Part III

Answer two questions from Part III.

  • In the accompanying figure, BCD is a semicircle and AB and ED are equal quadrants. (A quadrant is a quarter of a circle.) AE is 42 inches and the diameter of the semicircle is 28 inches. Find the area of the figure. [Use π = 22/7] [10]
  • The longer base of an isosceles trapezoid is 40, one leg is 20 and one of the base angles is 63o.
    1. Find to the nearest integer: (a) the altitude of the trapezoid [3] and (b) the shorter base of the trapezoid [4]
    2. Using the results found in answer to part 1, find the area of the trapezoid. [3]
  • Quadrilateral ABCDE is inscribed in a circle and arcs AB, BC, CD, and DA are in the ratio 3:3:4:2.
    1. Find the number of degrees in each of the fouyr arcs. [2]
    2. If AC is drawn, find the number of degrees in angles CAB and ACD. [2]
    3. If chord AD is 10, find the (a) circumference of the circle, and (b) area of triangle ABC. [3]
  • For each statement listed, the numerical value, correct to the enarest tenth, is given in the list below (a)-(g). List the numbers 1-5 on your answer paper and after each number write the letter indicating the numerical value of the corresponding statement. [10]
    1. The side of a square whose diagonal is 10
    2. The area of a rhombus whose diagonals are 4.8 and 4.
    3. The hypotenuse of a right triangle whose legs are are 6 and 14.
    4. The area of a regular hexagon whose side is 2.
    5. The length of an arc of 80o in a circle whose circumference is 37 1/2.
(a) 7.1 (b) 8.3 (c) 9.6 (d) 10.4 (e) 13.4 (f) 15.2 (g) 15.3


Note: Something odd is that this ARCO exam supposedly is from January 1954, yet the copyright of the text providing the exam has a copyright of 1950!

 

Source: I. Dressler, Reviewing Plane Geometry. AMSCO, 1950


Hint: None provided, except think like a student in 1954!

 


Solution Commentary: None provided....for each question, you can search out a reasonable answer...even now in the year 2014!