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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 a 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.

SOL.-P. 10, Pst. 1, P. 23, P. 31, Def. A.-DEM.-P. 38, P. 41, Ax. 6.

E. 1 Dat.

Given the ▲ ABC and the angle D;

2 Quas. 1 to describe a

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2 and having an angle

- the ▲ ABC,

the angle D.

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3 C.

4 P. 41.

5 Ax. 6.

6 C. 2.

Rec.

E

base EC, and BC || AG ; ▲ ABEA AEC, and ▲ ABC=

D

twice A AEC:

But the FECG and the A AEC are both on EC, and between the same ||s, EC, AG;

the parallelogram FECG = twice the A AEC; and.. the FECG = the ▲ ABC;

And CEF of FECG = the D.

Wherefore, there has been described a parallelogram, &c.

Q.E.F.

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.

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EH and GF

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3.

2 Conc.

78 about AC,

and BK and KD thes, E
complements of the figure:

then the complement BK =
the complement KD.

D. 1by Hyp.1. ABCD is a ☐, and AC

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10 Rec.

its diameter;

=

.. the AABC the▲ ADC:
And ·.· AEKH is a

.. the A AEK

Also the A KGC

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B G

, and AK its diameter,

the A AHK:

the ▲ KFC.

Then, '.' ▲ AEK = ▲ AHK,
and A KGC = A KFC ;

.. the two As AEK and KGC-the two AS AHK & KFC:
But the whole ▲ ABC

the whole ▲ ADC;

the rem. complement

therefore the rem. complement BK

KD.

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.

PROP. 44.-PROB.

To a given line to apply a parallelogram which shall be equal to a given triangle, and have one of its angles equal to a given angle.

SOL.-P 42, P. 31, Pst. 1, Pst. 2.-DEM.-P. 29, Ax. 9, Ax. 12, P. 43, Ax. 1, P. 15

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C. 6 P.31,Pst. 2 through K draw KL || AE or HF, and produce GB and

6

7 Sol.

HA to M and L

then the BL =

D. 1 by C.3&4..

2 P. 29.

3 Ax. 9.

4 Ax. 12.

5 Conc.

6 Pst. 2.

7 P. 31.

8 Pst. 2.

9 Conc.

10 P. 43.

11 C. 1 & Ax.1
12 P. 15 & C. 1
13 Ax. 1.
14 Rec.

the ▲ C,
D.

and has its ang. ABM =

HF falls on the parallels AH and EF, .. the s AHF and HFE together

two rt. angles;

two rt.

ands BHF and HFE together are two rt. angles;
but when int. angles on the same side are
angles, their sides meet, if produced;

HB and FE, being produced, will meet :

Let them be produced and meet in the point K;
through K draw KL || EA or FH;

and produce HA to L, and GB to M;

then FKLH is a 7, HK the diameter, AG and ME Os about HK, and LB and FB the complements;

and.. compl. LB = compl. BF:

But

And

.. also

BFA C, .. — LB = A C.

GBE ABM, and

GBE= =

ABM LD.

LD,

Therefore, to the st. line AB the parallelogram, &c. Q.E.D.

APP.-This proposition contains a kind of Geometrical Division.

Arithmetically we can say, a given triangle contains, for example, 20 square feet,-a given line 4 lineal feet; how can we apply a parallelogram to the 4 lineal feet, so that the area of a parallelogram shall equal that of the triangle? We divide 20 by 4, and the quotient 5 is the lineal measurement of the other side of the parallelogram.

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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

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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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4 Ax. 2.

5 P. 29.

6 Ax. 1.

7 D. 6.

8 P. 14. 9 C.

10 P. 29. 11 Add.

12 Ax. 2.

13 P. 29.

14 Ax. 1.

15 P. 14.

16 C. 2 & 3 17 P. 30.

18 C. 2 & 3

▲ ADB, and having an

FHKE;

FH the

B

H M

GM the A DBC, and

GHM=

E ;

then the fig. FKML is the

required.

to the st. line GH apply the
having an

< E = / FKH, and also = ▲ GHM;

FKH GHM:

To each of the equal angles add the

KHG;

then thes FKH + KHG = <s GHM + KHG;
Buts FKH and KHG : two rt. angles;

Zs GHM and KHG = two rt. angles :

Thus at the H in the st. line HG, the adjacents GHM
+ GHK
two rt. angles;

KH is in the same st. line with HM.

Again, *.* HG meets the parallels KM and FG,

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19 Def. A... fig. KFLM is a parallelogram.

20 C. 2 & 3 And. A ABD

21 Ax. 2.

22 Rec.

HF, and ▲ BDC = GM ;

therefore the whole rectil. fig. ABCD the whole
KFLM.

Wherefore, the parallelogram KFLM has been described, &c.

Q.E.F.

COR.--Hence a parallelogram equal to a given rectil. fig. can be applied to a given right line and in a given angle, by applying to the given st. line a parallelogram equal to the first triangle. ABD p. 44, and having an angle the given angle.

APP.--1. By this and the preceding Problem we may measure the superficial content of any rectilineal figure whatever. by first reducing it to triangles, and then making a rt. angled parallelogram equal to the sum of the triangles. We may also make a rt. angled parallelogram on a given line, and which shall be equal in area to several irregular figures. Also if we have several figures, we may make another equal to their difference.

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 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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PROP. 46.-PROB.

To describe a square on a given st. line.

SOL.--P. 11, P. 3, P. 31, Def. A.--DEM.--P. 34, Ax. 1, P. 29, Ax. 3, Def. 30.

E. 1 Dat..

2 Quæs.

C. 1 by P. 11.

2 P. 3.

3 P. 31.

4 Def. A.
5 Sol.

D. 1 by C. 4.
2 P. 34 & C. 2
3 Ax. 1.

4 D. 1 & 2.

5 C. 3.

6 P. 29.

7 C. 1.

8 Ax. 3.

9 P. 34.

10 D. 7 & 6.

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and

fig. ABDE is a parallelogram,
AB -

A

C

DE, AD BE, and AB = AD;
AB= AD DE EB;

the ABDE is equilateral.

Also, AD meets the parallels AB and DE,
.. the int. s BAD + ADE=two right angles :
but the BAD is a rt. angle.

.. ADE is also a rt. angle.

Now the opposite angles ofs are equal;
ABE opposite to ADE is a rt. angle,
and BED opposite to BAD is a rt. angle;

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11 D. 6, 7, & 9... the figure ADEB is a rectangle ;

12 D. 3.

13 Def. 30.

it is also equilateral;

therefore ABED is a square on AB.

Q.E.F.

E

COR. 1.-The squares on equal lines are equal; and if the squares are equal, the lines are equal.

2. Every parallelogram having one rt. 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 he object can be made.-See Euclid Practically Applied, Prop, 46, book 1.

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