A B Let ABCD be a trapezoid of which AB and DC are the parallel sides. Then will its area be equal to half the product of the sum of AB and DC into its altitude, namely, the perpendicular distance between AB and DC. Draw the diagonal AC. Now, the area of the triangle ABC (Cor. 2, Theo. XXIV) is equal to half the product of its base AB into its altitude, which is the same as the altitude of the trapezoid; again, the area of the triangle ADC is equal to half the product of its base DC into its altitude, which is also the same as the altitude of the trapezoid. Therefore, adding together the area of the two triangles, we have the area of the whole figure equal to half the product of the sum of AB and DC into the altitude. Hence, the area of a trapezoid, etc. THEOREM XXVI. The square described on the hypotenuse of a right-angled triangle is equivalent to the sum of the squares described on the other two sides. A B E K H Let ABC be a triangle rightangled at B. It is to be proved that the square AEDC is equivalent to the sum of the squares ABGF and BHIC. Join FC, BE, and draw BK parallel to AE. Also observe that, since ABC and ABG are right angles, BC and BG form one straight line. Now, in the triangles EAB, CAF, the side EA is equal to the side CA, since they are sides of the same square, and, for the same reason, the side AB is equal to the side AF. But the included angles EAB, CAF are also equal; for each of them is composed of a right angle and the angle CAB. Therefore, the two triangles are equal (Theo. XII). But, since the triangle CAF and the square BAFG have the same base, AF, and the same altitude, namely, the perpendicular distance between the parallels AF, CG, the square must be double the triangle (Cors. 1 and 2, Theo. XXIV). For like reason, the parallelogram AEKL is double the triangle EAB. But the doubles of equals are equals; therefore, the parallelogram AEKL is equivalent to the square BAFG. In the same manner (by joining AI and BD), it may be shown that the parallelogram CDKL is equivalent to the square BCIH. Hence, the whole square ACDE is equivalent to the sum of the squares BAFG, BCIH. Therefore, the square described, etc. Cor. 1.-The hypotenuse is equal to the square root of the sum of the squares of the other two sides. Cor. 2. The square of either of the sides containing the right angle is equivalent to the square of the hypotenuse diminished by the square of the other side. By taking the square root of the remainder, the side itself will be found. Cor. 3.-If two right-angled triangles have the hypotenuse and one side of the one respectively equal to the hypotenuse and one side of the other, the third sides will also be equal. For, if H2 B2 = C2, A2 = A = B2 A2, and H2 = C. C2, and right-angled triangles have each a side equal to AB, and each a side equal to AC, make the pair equal to AB coincide; then, if the other equal pair do not coincide, we shall have two equal oblique lines from a point to a line, on one side of the perpendicular. Can this be? EXERCISES. 1. If the side of a square is 36 inches, what is its area in square inches? What in square feet? 2. If the base of a parallelogram is 3 feet, and its altitude 4 feet and 6 inches, what is its area? 3. If the base of a triangle is 50 yards, and its altitude 20 yards, what is its area? 4. If the parallel sides of a trapezoid are 12 rods and 16 rods, and its altitude 8 rods, what is its area? 5. If the sides containing the right angle af a rightangled triangle are 3 and 4, what is the length of the hypotenuse? 6. If the hypotenuse is 10, and one of the sides 8, what is the length of the other side? 7. Prove that if a parallelogram has one right angle it is a rectangle. 8. Prove that the diagonals of a rectangle are equal to each other. (Cor. 1, Theo. XXVI.) 9. Prove that a perpendicular is the shortest line that can be drawn to a straight line from a point without it. A B E S Let AB be a perpendicular to the line ES, and AC any oblique line. AB and AC with the segment BC will form a right-angled triangle. Ang. ACB < ang. ABC (Theo. X). Why can not AB be equal to AC (Theo. XIII)? Why not greater (Thep. XVI)? If AB is neither equal to nor greater than AC, what follows? Cor.-Prolong AB till BO is equal to AB. It is readily proved that CO AC. Hence AO, which is equal to twice AB, is less than AC + CO, or twice AC. Therefore, a straight line is the shortest distance between two points. Sch. 1. This corollary is sometimes given as an axiom, and sometimes as the definition of a straight line. Sch. 2.—Notice carefully the difference between this discussion and that proving Theo. VII. 10. Prove that if, in the last theorem, two oblique lines be drawn, the one cutting the given straight line the furthest from the foot of the perpendicular will be the longest. (Consult Theorems IX and XI). NOTE.-It is thought best here to illustrate by examples the application of the equation to geometry. The problems will be understood readily even by those pupils who have not studied algebra. point C to the line AB, as has been already done in No. 9. Using the letters attached to the several lines as general symbols for the numerical value of the lines, and applying Theo. XXVI, As d and e are any oblique lines, we prove the perpendicular less than any other line. (1) d2a2+ b2 and d2 (2) e2 = a2 + b2 + 2bs + s2. Sub. (1) from (2), e2 d2 .. e2> d2 and e> d. 2bs + s2; We see that of any number of oblique lines, the long est is the one which cuts the given line furthest from the foot of the perpendicular. Prove o = d. 12. To find the altitude of any equilateral triangle. The known sides of the triangle 2 S Hence, the altitude is equal to the half side multiplied by the square root of 3. 13. What is the altitude of an equilateral triangle whose sides are each 12 feet? 14. What is the side of an equilateral triangle whose altitude is 12 feet? What is its area? A B C 15. The two triangles, formed by drawing lines from a point within a parallelogram to the ends of two opposite sides, are, together, equal to one half the parallelogram. SECTION VII.-POLYGONS. DEFINITIONS. 1. A polygon of five sides is a pentagon; one of six sides, a hexagon; of seven sides, a heptagon; of eight, an octagon; of ten, a decagon; of twelve, a dodecagon. 2. A regular polygon is both equilateral and equiangular. Squares and equilateral triangles are regular polygons. |