Sch.-The converse of Prop. 41 is,—If a parallelogram is double of a triangle, and they have the same base, or equal bases upon the same straight line and towards the same parts; they shall be between the same parallels.

APP.-The general method for finding the area of a triangle, or of any figure that may be resolved into triangles, is founded on this proposition. The area of a parallelogram is measured by the product of the base and altitude ; and a triangle on the same, or on an equal base, being half the parallelogram, the area of a triangle is equal to the product of the altitude and half the base; or what is the same thing, equal to half the product of the base and altitude. The area of polygons is the sum of the areas of the triangles into which the polygons may be divided.

Dividing & circle by an infinite number of triangles having their common vertex in the centre, the area of a circle equals the product of the radius and of the semi-circumference.

PROP, 42.-PROB. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

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Sch.--To describe a triangle equal to a given parallelogram,FECG and having an angle equal to a given angle D. As in the last figure, produce the parallels GF and CE, and make EB equal to EC ; at B construct an angle equal to the given angle D, and produce BA to meet the parallel GA in A, and join AC: then the triangle ABC thus constructed will, on drawing BH parallel to FE, be found by Prop. 41 equal to the parallelogram FEOG,

PROP. 43.-THEOR, The complements of the parallelograms which are about the diameter of any parallelogram, are equal to one another.

DEM.—P. 34, Ax. 2, Ax. 8.
E. 1 Hyp. 1. | Let ABCD be a O , and AC

its diameter,
, 2. EH and GF Os about AC,
3. and BK and KD the o s , E

complements of the figure: 2 Conc. | then the complement BK =

the complement KD. D. 1 by Hyp.1l ::: ABCD is a D , and AC,

its diameter ; 2 P. 34. 1.:. the A ABC = the A ADC: 3 Hyp. 2. And • : AEKH is a O , and AK its diameter, 4 P. 32. 1.:. the A AEK = the A AHK : 5 P. 34. Also the A KGC = the A KFC. 6 D. 4 & 5. Then, ::: A AEK = A AHK,

and A KGC=A KFC; 71 Ax. 2.1 ... the two AS AEK and KGC=the two As AHK & KFC: 8l D. 2. But the whole A ABC = the whole A ADC ; 9| Ax. 3. therefore the rem. complement BK = the rem. complement

| KD. 10 Rec. Wherefore, the complements of the parallelograms, &c. Q.E.D.

APP.-If any parallelogram, as KD in the last figure, be given, another parallelogram, KB, may be found, equal to it, and having one side EK, equal to a given line.

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Geometrically we say, a given rectangle BC contains 12 square feet,-a given line BD, or its equal EH, 2 lineal feet; how can we apply another rectangle to BD or EH measuring 2 lineal feet, so that the area of the required rectangle may equal the area of the given rectangle? Produce indefinitely the parallels AC and BE, A AB and CE;

loodaherbourhome from B set off BD equal to the representative

value of 2 feet, and draw the diag. DE; produce DE until it cuts AC produced in F; the

diagonal DF will so divide or intersect AC produced, that the part CF shall equal the other side of the rectangle, of which the given side BD is 2 lineal feet.

For, complete the rectangle ADGF; EH equals BD, and EI equals CF; and the complement EG is equal to the complement EA: and if we divide AE into equal squares, and EG into equal squares, we find that in the first there are 12, and in the second 2 X 6, or 12 also.

PROP. 45.-PROB. . To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle.

Sol.-Pst. 1, P. 42, P. 44.-DEM.---Ax. 1, Ax. 2, P. 29, P. 14, P.30, Def. A.

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2. By principles already established, and by the problems for the conversion of rectilineal figures into parallelograms of equal area, we may change any right-lined figure, as ABCDE, first into a triangle, and then into a rectangle of equal area.

Join DA and DB to divide the given figure into triangles, and produce AB indefinitely. Through E

DM draw EH parallel to DA, and through C, CF parallel to DB; join DH and DF; then the triangle DHF is equal in area to the given figure ABCDE.

Next, through D draw DN parallel to HF; bisect HF, a side of the triangle DHF, in L; at L raise the perpendicular LM, and through F draw FN a parallel to LM; the fig. LMNF is a rectangle of the same altitude as the triangle, and on half its base, and is therefore equal in area to the triangle DHF.


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Cor. 1.—The squares on equal lines are equal; and if the squares are equal, the lines are equal.

2. Every parallelogram having onert. angle, has all its angles rt, angles.

APP -- The Geometrical Square is an instrument by means of which, and of the property of similar triangles that the sides about the equal angles are proportional, the height of an inaccessible object can be ascertained, provided a measurement to the perpendicular from the object can be made.-See Euclid Practically Applied, Prop. 46, book 1.

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