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3. Produce DE beyond E, and DH beyond H; and make

EL equal to AB. 4. Draw a line from L through K, to meet DH produced

in M. Then HM is the second side of the required rectangle ELNO, and it is constructed on EL, which is equal to the given line AB.

NOTE.ELNO is equal to DEKH by Problem 160, and is therefore equal to the given triangle CDĚ.

Problem 162.

To draw a straight line equal to half the circumference of a given circle A.

1. Draw a diameter BC, and from B draw BD at right

angles to BC (Pr. 2) and equal to three times the radius of the circle.

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2. From C, draw a line CE at right angles to BC.
3. With the radius of the circle, cut off arc CF and bisect it.
4. From the centre of the circle, draw AE through the

point of bisection, meeting in E.
5. Join ED. Then ED will be the required straight line

equal to half the circumference of the given circle A.

Problem 163.

To construct a rectangle equal in area to a given circle A.

1. By the last problem it will be seen that a rectangle

FGHK can be constructed equal in area to a given

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circle by making two of its sides, FG and KH, equal to the length of half its circumference as ED, and the

other two sides FK, GH, equal to the radius AC. NOTE. -A triangle also can be constructed of the same area as a circle, ly making its base equal to half the circumference of the circle, and its altitude equal to twice its radius (=its diameter).

Problem 164.

To construct a parallelogram equal to any given triangle ABC both in area and perimeter.

1. Biscc: BC in D (Pr. 1).
2. Produce BA beyond A, and make AE equal to AC, and

bisect BE in F.
3. Find the altitude AG of the given triangle (Pr. 21), and

through A draw AH parallel to BC (Pr. 9).

4. From B, with BF as radius, describe an arc FK, cutting

AH in K; and from D, with the same radius, cut HA produced in the point L.

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5. Join BK, DL. Then the parallelogram BDLK is equal

both in area and perimeter to the given triangle ABC.

Problem 165.

To divide any parallelogram ABCD into two parts, proportionate in area to a given divided line EF, from a point ő in one side.

A

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IK

M

F

E 1. Divide

BC into the same proportions as the given divided line EF (Pr. 16) in the point G.

K

2. Through G, draw GH parallel to CD (Pr. 9), and bisect

GH in K (Pr. 1). 3. From the given point 0, draw a line through K, and

produce it to meet AD in L. Then OL will divide the given parallelogram ABCD into two parts proportionate in area to the given line EF.

NOTE.—To divide the parallelogram into two equal parts from any point, say L, measure off BM equal to DL, and join LM; then LM will divide the parallelogram into two equal parts.

Problem 166.

To make an irregular polygon equal to a given irregular polygon ABCDE.

1. Draw a line FG equal to AB.
2. With centre F, and radius A E, describe an arc, and with

centre G, and radiuş BE, cut it in H.

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3. With centre H, and radius ED, describe an arc, and with

centre G, radius BD, cut it in K. 4. With centre K, and radius DC, describe an arc, and with

centre G, radius BC, cut it in L. 5. Join FH, HK, KL, and LG. Then FHKLG will be the

required irregular polygon.

Problem 167.

To construct a square equal in area to a given rectangle ABCD.

1. Produce DC indefinitely beyond C, and make CE equal

to CB.

2. Bisect DE in F (Pr. 1), and on DE describe a semicircle.

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3. Produce the perpendicular CB to meet the semicircle in

G. Then CG is a mean proportional between the two adjacent sides DC, CB (Pr. 140); and CG is one side of

the required square. 4. On CG, complete the required square CGHK (Pr. 34),

which will be equal in area to the given rectangle ABCD.

Problem 168.

To construct a square that shall have an area of two square inches (or any number of square. inches).

1. Let ABCD be a rectangle having an area of two square

inches, its side AB being two inches (linear), and BC

one linear inch. 2. Find BE a mean proportional to the lines AB, BC (Pr.

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