EXERCISES.

ON THE CONSTRUCTION OF QUADRILATERALS.

Draw a rhombus each of whose sides is equal to a given straight line PQ, which is also to be one diagonal of the figure.

Ascertain (without measurement) the number of degrees in each angle, giving a reason for your answer.

2. Draw a square on a side of 2.5 inches. Prove theoretically that its diagonals are equal; and by measuring the diagonals to the nearest hundredth of an inch test the correctness of your drawing.

3. Construct a square on a diagonal of 3·0", and measure the lengths of each side. Obtain the average of your results.

4. Draw a parallelogram ABCD, having given that one side AB 5.5 cm., and the diagonals AC, BD are 8 cm., and 6 cm. respectively. Measure AD.

5. The diagonals of a certain quadrilateral are equal, (each, 6·0 cm. ), and they bisect one another at an angle of 60°. Shew that five independent data are here given.

Construct the quadrilateral. Name its species; and give a formal proof of your answer. Measure the perimeter. If the angle between the diagonals were increased to 90°, by how much per cent. would the perimeter be increased?

6. In a quadrilateral ABCD,

AB 5'6 cm., BC=2.5 cm., CD=40 cm., and DA=33 cm. Shew that the shape of the quadrilateral is not settled by these data. Draw the quadrilateral when (i) A=30° (ii) A=60°. Why does the construction fail when A=100° ?

Determine graphically the least value of A for which the construction fails.

7. Shew how to construct a quadrilateral, having given the lengths of the four sides and of one diagonal. What conditions must hold among the data in order that the problem may be possible?

Illustrate your method by constructing a quadrilateral ABCD, when (i) AB=3′0′′, BC=17′′, CD=2.5′′, DA=2·8′′, and the diagonal BD=2.6". Measure AC.

(ii) AB=3.6 cm., BC=7·7 cm., CD=6.8 cm., DA=5·1 cm., and the diagonal AC 8.5 cm. Measure the angles at B and D.

LOCI.

DEFINITION. The locus of a point is the path traced out by it when it moves in accordance with some given law.

Example 1.

Suppose the point P to move so that its distance from a fixed point O is constant (say 1 centimetre).

Then the locus of P is evidently the circumference of a circle whose centre is O and radius 1 cm.

Example 2. Suppose the point P moves at a constant distance (say 1 cm.) from a fixed straight line AB.

Then the locus of P is one or other of two straight lines parallel to AB, on either side, and at a distance of 1 cm. from it.

A

P

B

Thus the locus of a point, moving under some given condition, consists of the line or lines to which the point is thereby restricted; provided that the condition is satisfied by every point on such line or lines, and by no other.

When we find a series of points which satisfy the given law, and through which therefore the moving point must pass, we are said to plot the locus of the point.

PROBLEM 14.

To find the locus of a point P which moves so that its distances from two fixed points A and B are always equal to one another.

Here the point P moves through all positions in which PA = PB; .. one position of the moving point is at O the middle point of AB.

Suppose P to be any other position of the moving point; that is, let PA

PB.

That is, every point P which is equidistant from A and B lies on the straight line bisecting AB at right angles.

Likewise it may be proved that every point on the perpendicular through O is equidistant from A and B.

This line is therefore the required locus.

PROBLEM 15.

To find the locus of a point P which moves so that its perpen dicular distances from two given straight lines AB, CD are equal to one another.

the

Then in the As PMO, PNO,

PMO, PNO are right angles,

because the hypotenuse OP is common,

and one side PM =

=

one side PN;

.. the triangles are equal in all respects; Theor. 18. so that the POM = the PON.

Hence, if P lies within the

of that angle;

BOD, it must be on the bisector

and, if P is within the AOD, it must be on the bisector of that angle.

It follows that the required locus is the pair of lines which bisect the angles between AB and CD.

INTERSECTION OF LOCI.

The method of Loci may be used to find the position of a point which is subject to two conditions. For corresponding to each condition there will be a locus on which the required point must lie. Hence all points which are common to these two loci, that is, all the points of intersection of the loci, will satisfy both the given conditions.

EXAMPLE 1. To find a point equidistant from three given points A, B, C, which are not in the same straight line.

(i) The locus of points equidistant from A and B is the straight line PQ, which bisects AB at right angles.

(ii) Similarly, the locus of points equidistant from B and C is the straight line RS which bisects BC at right angles.

Hence the point common to PQ and RS must satisfy both conditions: that is to say, X the point of intersection of PQ and RS will be equidistant from A, B, and C.

A

R

P

X

EXAMPLE 2. To construct a triangle, having given the base, the altitude, and the length of the median which bisects the base.

Let AB be the given base, and P and

Q the lengths of the altitude and median respectively.

Then the triangle is known if its vertex is known.

(i) Draw a straight line CD parallel to AB, and at a distance from it equal to P: then the required vertex must lie on CD.

(ii) Again, from O the middle point of AB as centre, with radius equal to Q, describe a circle:

CE

A

then the required vertex must lie on this circle.

Hence any points which are common to CD and the circle, satisfy both the given conditions: that is to say, if CD intersect the circle in E, F, each of the points of intersection might be the vertex of the required triangle. This supposes the length of the median Q to be greater than the altitude.

It may happen that the data of the problem are so related to one another that the resulting loci do not intersect. In this case the problem is impossible.