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(C.) “ Triangles upon the same base and between the same parallels
are equal to one another" (in area).-Euc. I., 37. Ex. ABC, DBC
NOTE. - It can be readily seen that ABC is a half of the parallelogram EBCA, and DBC is a half of the parallelogram DBCP. The parallelograms, standing on the same base BC, are equal, therefore the triangles are equal, as “the halves of equal things are equal,”—Euc. I., Ax. 7.
(D.) “ Triangles upon equal bases and between the same parallels are
equal to one another” (in area).—Euc. I., 38. Ec. ABC, DEF
NOTE.-It can be readily seen that ABC is a half of the parallelogram GBCA, and DEF is a half of the parallelogram DEFH. The parallelograms, standing on equal bases BC and EF, are equal, therefore the triangles are equal, as “the halves of equal things are equal.” —Euc. I., As. 7."
From the foregoing notes the truth of the following theorem will be readily seen :(E.) “If a parallelogram and a triangle be upon the same base and
between the same parallels, the parallelogram shall be double of the triangle" (in area).—Euc. I., 41. Ex. ABCD and EBC
(F.) “Triangles of the same altitude are one to the other (in area) as
their bases.”—Euc. VI., 1. EX. ACE, ABC
NOTE.—The base CE being double of the base BC, the area of
ACE is double of that of the triangle ABC. (G.) Triangles on the same base have to one another the ratio that
their altitudes have.-Ex. EAB, CAB
NOTE.—The altitude EA being triple of the altitude CA, the area of EAB is triple of that of the triangle CAB,
Problem 157. To construct a rectangle on a given line AB equal in area to a given square C.
1. From A in AB, draw a straight line AD equal to a side
of the given square, and making any angle with AB. 2. Join BD, and from A, with radius AD, describe an arc
DE, meeting AB in E.
3. Through E, draw EF parallel to BD (Pr. 9). Then AF
is a third proportional less between AB and a side of the square (Pr. 143), and AF is equal to the second side
of the required rectangle. 4. From A, raise a perpendicular AG (Pr. 2) equal to AF,
and complete the required rectangle AGHB, which will be equal in area to the given square C.
To construct a parallelogram on a given base AB equal in area to a given parallelogram CDEF.
1. Find AG the fourth proportional less to AB, CD, CF
2. At A, raise the perpendicular AH equal to AG, and
complete the parallelogram AHKB. Then the parallelogram AHKB is equal to the given parallelogram CDEF.
To construct a rectangle equal in area to a given triangle ABC, on one side BC of the given triangle.
1. Draw AD, the altitude of the given triangle (Pr. 21),
and bisect it in E. Then ED is equal to the other side of the required rectangle, BC being one.
2. At B, raise a perpendicular BF (Pr. 2) to meet the line
of bisection through E in F. 3. Make FG equal to BC, and join CG. Then BCGF is
the required rectangle equal to the given triangle ABC, and constructed on one of its sides BC.
To construct a rectangle on a given line AB which shall be equal in area to a given rectangle CDET.
1. Produce the side De beyond E, and make EG equal to
AB. 2. Draw a line from G through F, to meet DC produced in
II. Then CH is equal to the other side of the rectangle, of which EG is one.
3. From E in EF or EF produced, cut off EK equal to
А — В 4. Make KL equal to EG, and GL equal to EK, and join
KL and GÌ. Then EGLK is the rectangle required, and it is constructed on EG, which is equal to the given ling AB.
Problem 161. To construct a rectangle having a given side AB, equal in area to a given triangle CDE.
1. Find CF the altitude of the triangle CDE (Pr. 21).
AB 2. Bisect CF in G (Pr. 1), and produce the line of bisection
over DE, and erect the perpendiculars DH, EK. Then the rectangle DEKH is equal to the triangle CDE (Pr. 159).