PROBLEM IX. To construct a square equal to the sum of two given squares; also a square equal to the difference of two given squares. 1o. Let A and B be the sides of the given squares, and let A be the greater. Construct a right angle CDE; make DE equal to A, and DC equal to B; draw CE, and on it construct a square: this square D A B E will be equal to the sum of the given squares (P. XI.). 2. Construct a right angle CDE. Lay off DC equal to B; with C as a centre, and CE, equal to A, as a radius, describe an arc cutting DE at E; draw CE, and on DE construct a square: this square will be equal to the difference of the given squares (P. XI., C. 1). Scholium. A polygon may be constructed similar to either of two given polygons, and equal to their sum or difference. For, let A and B be homologous sides of the given polygons. Find a square equal to the sum or difference of the squares on A and B; and let X be a side of that square. On X as a side, homologous to A or B, construct a polygon similar to the given polygons, and it will be equal to their sum or difference (P. XXVII., C. 2). GEOMETRY. EXERCISES. 1. The altitude of an isosceles triangle is 3 feet, each of the equal sides is 5 feet; find the area. 2. The parallel sides of a trapezoid are 8 and 10 feet, and the altitude is 6 feet; what is the area? 3. The sides of a triangle are 60, 80, and 100 feet, the diameter of the inscribed circle is 40 feet; find the area. 4. Construct a square equal to the sum of the squares whose sides are respectively 16, 12, 8, 4, and 2 units in length. 22. 5. Show that the sum of the three perpendiculars drawn from any point within an equilateral triangle to the three sides is equal to the altitude of the triangle. 6. Show that the sum of the squares of two lines, drawn from any point in the circumference of a circle to two points on the diameter of the circle equidistant from the centre, will be always the same. 7. The distance of a chord, 8 feet centre of a circle is 3 feet; what is the circle? 10 8. Construct a triangle, having given the vertical angle, the line bisecting the base, and the angle which the bisecting line makes with the base. 9. Show that if a line bisects the exterior vertical angle of a triangle, the dis long, from the diameter of the tances of the point in which it meets the base produced, from the extremities of the base, are proportional to the other two sides of the triangle. 10. The segments made by a perpendicular, drawn from a point on the circumference of a circle to a diameter, are 16 feet and 4 feet; find the length of the perpendicular. 11. Two similar triangles, ABC and DEF, have the homologous sides AC and DF equal respectively to 4 feet and 6 feet, and the area of DEF is 9 square feet; find the area of ABC. - 12. Two chords of a circle intersect; the segments of one are respectively 6 feet and 8 feet, and one segment of the other is 12 feet; find the remaining segment. 13. Two circles, whose radii are 6 feet and 10 feet, intersect, and the line joining their points of intersection is 8 feet; find the distance between their centres. 13.63+ 14. Find the area of a triangle whose sides are respectively 31, 28, and 20 feet. 15. Show that the area of an equilateral triangle is equal to one fourth the square of one side multiplied by V3; or to the square of one side multiplied by .433. 16. From a point, O, in an equilateral triangle, ABC, the distances to the vertices were measured and found to be: OB = 20, OA 28, OC 31; find the area of the triangle and the length of each side. [AD is made equal to OA, CD to OB, CF to OC, BF to OA, BE to OB, AE to OC.] E B 31 3 201 A BOOK V. REGULAR POLYGONS.-AREA OF THE CIRCLE. DEFINITION. 1. A REGULAR POLYGON is a polygon which is both equilateral and equiangular. PROPOSITION I. THEOREM. Regular polygons of the same number of sides are similar. Let ABCDEF and abcdef be regular polygons of the same number of sides: then they are similar. For, the corresponding angles in each are equal, because any angle in either E polygon is equal to twice F as many right angles as the polygon has sides, less four right angles, divided α B by the number of angles (B. I., P. XXVI., C. 4); and further, the corresponding sides are proportional, because all the sides of either polygon are equal. (D. 1): hence, the polygons are similar (B. IV., D. 1); which was to be proved. PROPOSITION II. THEOREM. The circumference of a circle may be circumscribed about any regular polygon; a circle may also be inscribed in it. 1o. Let ABCF be a regular polygon: then can the circumference of a circle be circumscribed about it. For, through three consecutive vertices A, B, C, describe the circumference of a circle (B. III., Problem XIII., S.). Its centre O lies on PO, drawn perpendicular to BC, at its BC, at its point P; draw OA and OD. Let the quadrilateral middle H OPCD be turned about the line OP, until PC falls on PB; then, because the angle A O: B C is equal to B, the side CD will take the direction BA: and because CD is equal to BA, the vertex D, will fall upon the vertex A; and consequently, the line OD will coincide with OA, and is, therefore, equal to it: hence, the circumference which passes through A, B, and C, passes. through D. In like manner, it may be shown that it passes through each of the other vertices: hence, it is circumscribed about the polygon; which was to be proved. 2°. A circle may be inscribed in the polygon. For, the sides AB, BC, &c., being equal chords of the circumscribed circle, are equidistant from the centre 0; hence, a circle described from O as a centre, with OP as a radius, is tangent to each of the sides of the polygon, and consequently, is inscribed in it; which was to be proved. D |