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781. Through a given point situated between the sides of an angle, to draw a line terminating at the sides of the angle, and in such a manner as to be bisected at the point.

SUG.-Conceive the point as situated in the third side of a triangle of which the two given lines are the other two.

P

FIG. 419.

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782. To draw a line parallel to the base of a triangle so as to divide the triangle into two equivalent parts. 43PBP8. i be

SUG. PB' = DB2 AB'. See (344, 362).

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783. To construct a square when the difference between the diagonal and a side is given.

SUG.-Consider the angles.

FIG. 420.

784. To determine the point in the circumference of a circle from which chords drawn to two given points shall have a given

ratio.

SUG.-Draw a chord dividing the chord joining the given points in the required ratio, and bisecting one of the subtended arcs.

785. To bisect a given triangle by a line drawn from one of its angles.

786. To bisect a given triangle by a line drawn from a given point in one of its sides.

FIG. 421

787. In the base of a triangle find the point from which lines extending to the sides, and parallel to them, will be equal.

788. To construct a parallelogram having the diagonals and one side given.

789. To construct a triangle when the three altitudes are given.

SUG.-What is the relation of the perpendiculars to the sides upon which they fall? If a triangle can be formed with the perpendiculars as sides, how will it compare with the first triangle? How proceed when the perpendiculars will not form a triangle?

790. What is the area of the sector whose arc is 50°, and whose radius is 10 inches? 103

Leg

791. To construct a square equivalent to the sum, or to the difference of two given squares.

792. To divide a given straight line in the ratio of the areas of two given squares.

793. To construct a triangle, when the altitude, the line bisecting the vertical angle, and the line from the vertex to the middle of the base are given.

SUG.-The centre of the circle circumscribing the required triangle is in the perpendicular to the base at its middle point; and the intersection of this perpendicular and the bisectrix is a point in this circumference.

Show that the bisector always lies between the perpendicular and the medial line.

794. Through a given point, draw a line such that the parts of it, between the given point and perpendiculars let fall on it from two other given points, shall be equal.

What would be the result, if the first point were in the straight line joining the other two?

795. From a point without two given lines, to draw a line such that the part intercepted between the given lines shall be equal to the part between the given point and the nearest line.

SUG.-Produce the lines till they meet, if necessary. Draw a line through the given point parallel to one of the lines, and produce it till it meets the other.

796. Given one angle, a side adjacent to it, and the difference of the other two sides, to construct the triangle.

QUERIES.-How if b> a? How if B is obtuse?

797. To pass a circumference through two given points, having its centre in a given line.

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A

FIG. 422

798. To draw a line parallel to a given line and tangent to a given circumference.

SUG.-Draw a diameter perpendicular to the given line.

FIG. 423.

799. To draw a common tangent to two given circles.

SUG.-1ST METHOD.-There are two sets of tangents, AC, BD, and A'D', B'C'. For the first, observe that if PE = AP- OC, OE is parallel to AC, etc.

FIG. 424.

2D METHOD. PO, p'O' being parallel to each other, pp'T gives the intersection of the tangent with the line passing through the centres, since

po p'O' OT: O'T, or pO -p'O' : p'O' :: 00': O'T.

Also, POP'O' : P'O' : : 00' : O'T.

Hence O'T is constant for all positions of the parallel radii. Prove that if the parallel radii are on different sides of the line joining the centres, T'is the point where the internal tangent cuts OO'.

QUERIES.-How many tangents can be drawn-1st. When the circles are external one to the other; 2d. When they are tangent externally; 3d. When they intersect each other; 4th. When tangent internally; 5th. When one lies within the other?

FIG. 425.

800. To describe a circle tangent to a given circumference and also to a given line at a given point.

SUG's. There may be two cases1st. When the given circle is exterior to the one sought; and 2d. When it is interior. In either case the centre of the required circle is in the perpendicular AO'. In the former case, O, the centre of the required circle, is at r+r' from C; and in the latter O' is

atr' from C. AOr, and CD AB = AB' r'.

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801. To construct a trapezoid when the

four sides are given.

SUG. Knowing the difference between the two parallel sides, we may construct the triangle AEC, and hence the trapezoid.

FIG. 426.

802. On a given line, to construct a polygon similar to a given polygon.

SUG's. One method may be learned from (90). Ex. 8, page 152, furnishes another method. The following is an elegant method: To construct on A' homologous with A, a polygon similar to P. Place A' parallel to A, and the figure will suggest the construction.

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FIG. 427.

803. To pass a plane through a given line and tangent to a given sphere.

SUG's.-Pass a plane through the centre of the sphere and perpendicular to the given line. Through the point of intersection and in this secant plane draw tangents to the great circle in which the secant plane intersects the surface of the sphere. The points of tangency will be the points of tangency of the required planes (?), of which there are thus seen to be two.

804. DEF.-A Tangent Plane to a cylindrical or conical surface is a plane which contains an element of the surface, but does not cut the surface. The element which is common to the surface and the plane is called the Element of Contact.

805. To pass a plane through a given point and tangent to a given cylinder of revolution.

SUG's.-1st. When the point is in the surface of the cylinder. Through the point draw an element of the cylinder, by passing a line parallel to the axis, or to any given element. Through the same point pass a plane perpendicular to this element, making a right section (a circle). To this circle draw a tangent. The plane of the element and tangent is the tangent plane required. [The student should prove that any point in the plane affirmed to be tangent, not in the element passing through the given point, is without the cylinder.]. 2d. When the given point is without the cylinder. Pass a plane through the given point perpendicular to the axis of the cylinder, thus making a right section of the cylinder (a circle). In this secant plane draw tangents to the section. Through the points of contact of these tangents draw elements of the cylinder. These elements are the elements of contact of the tangent planes. Hence planes passing through them and the given point are the tan

gent planes required. [The student should remember that this is but an out line, and be careful to fill it up, giving the proof.]

806. To pass a plane through a given point and tangent to a conical surface of revolution.

807. To find, with the compasses and ruler, the radius of a material sphere whose centre is inaccessible.

B

FIG. 428.

SUG's. With one point of the compasses at any point in the surface, as A, trace a circle of the sphere, as acb. The chord Aa is measured by the distance between the compass points. In like manner measure three other chords, as ac, ab, and be. Draw a plane triangle having these chords for its sides, and circumscribe a circle about it. Thus aD is found. Knowing aƐ, and aƊ, and remembering that AaB is right angled at a, the triangle AaB can be drawn in a plane (?), whence AO becomes known.

SECTION III.

APPLICATIONS OF ALGEBRA TO GEOMETRY.

808. The mathematical method which is called technically Applications of Algebra to Geometry consists in finding, by means of equations, the numerical values of the unknown parts of a geometrical figure, when a sufficient number of the parts are given numerically.

809. By reference to the COMPLETE SCHOOL ALGEBRA, page 238, it will be seen that the algebraic solution of a problem consists of two parts: 1st. The Statement, which is the expressing by one or more equations of the conditions of the problem, i. e., the relations between the known and unknown quantities (parts of the figure) to be compared; and 2d. The Solution of these equations, so as to find the values of the unknown quantities in known ones.

810. In applying the equation for the solution of such problems as are now proposed, we have to depend upon our previously acquired knowledge of the properties of geometrical figures for the relations between the known and unknown quantities, which will enable us to form the necessary equations, i. e., to make the State

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