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Parallelograms, Rectangles and Squares.
1. With the parallel rulers, or by other means, draw a pair of parallel lines AB, CD, and also another pair of parallel lines EF, GH, inclined to the former pair at any angle.
The figure KLMN is called a parallelogram, i.e., a parallelogram is a foursided figure whose opposite sides are parallel.
With the dividers compare the lengths of KL and NM; also the lengths of KN and LM. With the bevel compare the magnitudes of the angles NKL and NML; also the magnitudes of the angles KLM and KNM.
Construct two or more parallelograms with sides of different lengths, and angles of different magnitudes; and in the case of each compare the magnitudes of the opposite sides and angles.
The result of such observations will be that the opposite sides and angles of parallelograms are equal.
We may also prove this as follows:
Draw KM, the
diameter or diagonal, as it is called, of the paralleloWe have in the two triangles NKM, LMK,—
The side KM common to both,
the alternate angles NKM, LMK equal,
Hence (Ch. II. 5) these triangles are equal in all respects, i.e.,
KN = ML,
Of course the triangle KNM, if cut out, can be fitted on the triangle MLK, and is equal to it, i.e., the diagonal of a parallelogram bisects it.
2. Draw a pair of parallel lines, AB and CD. In AB take any length KL, and, adjusting the dividers to it, in CD mark off an equal length MN. Join K, M and L, N.
Using the dividers, what do you note with reference to the lengths of KM and LN? Using the bevel or parallel rulers, what do you note with reference to the position of KM and LN with respect to one another?
Draw other parallel lines, mark off on them equal lengths, join the extremities of these equal lengths, and repeat the examination as to the lengths and relative position of the joining lines.
The result of such observations will be that the straight lines joining the extremities of equal and parallel straight lines are themselves equal and parallel.
In the triangles
We may prove this as follows: LKN, MNK, the sides LK, KN are equal to the sides MN, NK; and the angles LKN, MNK are equal. Hence
these triangles are equal in all respects. Therefore LN and MK are equal. Also the alternate angles LNK and MKN are equal, and therefore LN and MK are parallel.
3. With the power of drawing parallel lines we have another means of bisecting a line, indeed of dividing a line into any number of
Let AB be the line to be bi
sected. Draw through A any A other line AC, and with the
dividers mark off on it equal lengths AD, DE. Join BE, and with the parallel rulers draw DF parallel to BE. F is the bisection of AB. For, drawing FG parallel to AC, the triangles ADF, FGB are evidently equal, and AF is equal to FB.
In employing this method of bisecting a line, we may avoid altogether drawing the lines AC, &c. For, place the edge of the ruler against A, and, close to the edge of the ruler, with the sharp points of the dividers, mark the points D and E (the distances AD, DE being equal). Then place the edge of the parallel rulers against the points B and E, and move the edge, parallel to itself, back to D. The point in which the edge cuts AB is its middle point, and can be marked with a point of the dividers.
It is well to so place AC and the points D and E, that the lines DF, EB cut AB at nearly 90°. The point F is thus located with most definiteness.
We leave to the pupil to discover for himself, following the suggestion here given, a means of dividing a straight line into any number of equal parts.
4. Having drawn two parallels, adjust the points of the dividers to a distance of, say, 1 inch from one another. Place one point at A, and let the other
point of the dividers meet the other parallel at B. Then inch from A gives the middle point of AB. Draw a number of lines through C, and terminated by the parallels. Using the dividers, C will be found to be the middle point of all these lines.
5. If the angles of a parallelogram are right angles, it is called a rectangle.
If the adjacent sides, and therefore all the sides, of a parallelogram are equal, it is called a rhombus.
If the angles of the rhombus are right angles, the figure is called a square, i.e., a square is a foursided figure with all its sides equal, and all its angles right angles.
Construct the following parallelo
Sides 50 and 80 millimetres, and included angle 45°. Sides 40 and 110 millimetres, and included angle 110°. Sides 2 and 3 inches, and included angle 58°.
In all cases test the equality of the opposite sides and angles.
Construct the following rhombuses:
Sides 70 millimetres, and one angle 60°.
Sides 3 inches, and one angle 15°.
whose side is 4
square whose side is 50 millimetres; inches; whose side is 70 millimetres;
1. With two equal triangles, cut out of paper, form a parallelogram.
2. Draw a number of straight lines of various lengths, and, by the method of § 3, bisect them, using points only in your construction. With the dividers test the accuracy of your construction.
3. Draw a number of straight lines of various lengths, and, by the method of § 3, trisect them, using points only in your construction. With the dividers test the accuracy of your construction.
4. Draw both diagonals in a number of parallelograms, and examine how the point in which the diagonals intersect divides them. Give proof.
5. Draw two lines whose intersection bisects both, and show by using parallel rulers that the lines joining the extremities of the bisected lines are parallel in pairs. Give proof.
6. Two equal and parallel lines are joined towards opposite parts. How do the joining lines divide each other?
7. With compasses and ruler only, construct a four-sided figure with opposite sides equal. How are opposite sides placed with respect to each other? Apply tests. Give proof.
This exercise explains the principle of the construction of parallel rulers. 8. With protractor and ruler, construct a four-sided figure with one pair of opposite angles equal and each 75°, and the other pair of opposite angles equal and each 105°. What is the figure? Apply