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ANOTHER DEMONSTRATION.

Let A C and A' C' be two rectangles of equal altitudes. P. C

CPI

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Let b and b', S and S stand for the bases and areas of these rectangles respectively.

Take A D, DE, EF ....m in number and all equal, and A'D', D' E', E' F', F'G' ....n in number and all equal.

Complete the rectangles as in the figure.

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Now we can prove by superposition, that if A F be > A' G', rect. A P will be > rect. A' P'; and if equal, equal; and if less, less.

That is, if mb be > nb, m S is > nS'; and if equal, equal; and if less, less.

Hence, 6:8:;S: s', Euclid's Def., § 272

Q. E. D.

PROPOSITION II. THEOREM. 315. Two rectangles are to each other as the product of their bases by their altitudes.

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Let R and R' be two rectangles, having for their bases 6 and b', and for their altitudes a and a'.

R a Xb
We are to prove

R = a' Construct the rectangle S, with its base the same as that of R and its altitude the same as that of R'.

R a
Then

$ 314

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and

(rectangles having the same base are to each other as their altitudes); S b

§ 313 RO'' (rectangles having the same altitude are to each other as their bases). By multiplying these two equalities together

ва хъ
Ra X 6"

Q. E. D. 316. Def. The Area of a surface is the ratio of that surface to another surface assumed as the unit of measure.

317. DEF. The Unit of measure (except the acre) is a square a side of which is some linear unit; as a square inch, etc.

318. DEF. Equivalent figures are figures which have equal areas.

REM. In comparing the areas of equivalent figures the symbol (=) is to be read “equal in area.”

PROPOSITION III. THEOREM. 319. The area of a rectangle is equal to the product of its base and altitude.

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Let R be the rectangle, b the base, and a the alti

tude; and let U be a square whose side is the
linear unit.
We are to prove the area of R= a X 6.
ва хъ

§ 315 Ū = 1Xi' (two rectangles are to each other as the product of their bases and altitudes). R is the area of R,

$ 316

But

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320. ScHolium. When the base and altitude are exactly divisible by the linear unit, this proposition is rendered evident by dividing the figure into squares, each equal to the unit of

measure. Thus, if the base contain seven linear units, and the altitude four, the figure may be divided into twenty-eight squares, each equal to the unit of measure; and the area of the figure equals 7 X 4.

PROPOSITION IV. THEOREM. 321. The area of a parallelogram is equal to the product of its base and altitude. BE

C F B

C E

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Let A EFD be a parallelogram, A D its base, and C D

its altitude.
We are to prove the area of the D A E FD= ADXC D.
From A draw A B Il to DC to meet F E produced.

Then the figure A B C D will be a rectangle, with the same
base and altitude as the O AEFD.
In the rt. A ABE and C DF, .
AB=CD.

§ 126 (being opposite sides of a rectangle). and A E= DF,

§ 134 (being opposite sides of a 0); .. A ABE=ACDF,

$ 109 (two rt. A are equal, when the hypotenuse and a side of the one are equal

respectively to the hypoten use and a side of the other). Take away the ACD F and we have left the rect. A B C D. Take away the A ABE and we have left the D A E FD.

.. rect. A B C D= 0 A EFD. Ax. 3 But the area of the rect. A B C DE ADXCD, $ 319 (the area of a rectangle equals the product of its base and altitude). .. the area of the D A EFD= ADXC D. Ax. 1

Q. E. D. 322. COROLLARY 1. Parallelograms having equal bases and equal altitudes are equivalent.

323. Cor. 2. Parallelograms having equal bases are to each other as their altitudes ; parallelograms having equal altitudes are to each other as their bases; and any two parallelograms are to each other as the products of their bases by their altitudes.

PROPOSITION V. THEOREM.

• 324. The area of a triangle is equal to one-half of the product of its base by its altitude.

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Let ABC be a triangle, A B its base, and C D its

altitude.
We are to prove the area of the A ABC= } A B XC D.
From C draw C H l to A B.
From A draw A H Il to BC.

The figure A B C H is a parallelogram, § 136
(having its opposite sides parallel),

and A C is its diagonal. ..A ABC=A AHC,

§ 133 (the diagonal of a divides it into two equal o ). The area of the D A B C H is equal to the product of its base by its altitude.

§ 321 .. the area of one-half the O, or the A ABC, is equal to one-half the product of its base by its altitude,

ŽA B XCD.

or,

Q. E. D.

325. COROLLARY 1. Triangles having equal bases and equal altitudes are equivalent.

326. COR. 2. Triangles having equal bases are to each other as their altitudes ; triangles having equal altitudes are to each other as their bases; any two triangles are to each other as the product of their bases by their altitudes.

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