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Problem 190.

To construct a triangle equal in area to any given circle A.

1. Draw any radius AB, and from B draw BC perpen

dicular to AB (Pr. 2), and equal in length to the circumference of the circle.

A

B

2. Join AC, and the triangle ABC is equal in area to the

given circle A.

Problem 191.

To construct any regular polygon (say a hexagon), equal in area to any given triangle ABC.

1. Divide a side BA of the given triangle into as many

equal parts as the required polygon has sides (six) (Pr. 15)

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2. Produce CB indefinitely beyond B, and through point 1

draw a line parallel to BC (Pr. 9). 3. Construct a hexagon D (Pr. 64), and draw lines from the

centre to two of the angles, as Da, Db. 4. At B in CB make angle CBE equal to aDb (Pr. 10), the

line BE cutting the parallel to BC in F. 5. On CB produced make BG equal to BF.

6. Find a mean proportional BH to the two segments CB,

BG (Pr. 140), which is equal to the radius of the circlé described from B.

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7. Within this circle inscribe the required hexagon

EKLMNO, which is equal in area to the given triangle
ABC.

Problem 192.
To construct a circle two-thirds the area of a given circle A.

1. Draw a radius AB, and divide it into three equal parts

(Pr. 15).

B

2. Prɔduce it to C, making AC equal to two of the equal 3. Find AD the mean proportional between AB and AC

parts.

(Pr. 140), which is the radius of the required circle. 4. From centre A, with radius AD, describe the inner circle,

whose area is two-thirds of that of the given circle A.

NOTE.-In the same manner, if the required circle is to be one-third of the given circle, mark off AE equal to one-third of A B. Then find the mean proportional between AB and AE, which will be the radius of the required circle.

Problem 193.

To construct any rectilineal figure, whose area shall have a given proportion to any other rectilineal figure of the same kind (say one-third).

(A) To construct a triangle one-third of a given triangle ABC.

1. Draw DE equal to the side BC. As one-third the area

is required, produce DE, making Ef equal to one-third

of DE. 2. Find EG a mean proportional to DE, EF (Pr. 140).

Then EG is equal to a side of the required triangle.

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3. Make BH equal to EG, and draw HK parallel to AC.

Then BHK will be the required triangle, and it is onethird of the given triangle ABC.

(B) To construct a parallelogram one-third of a given parallel

ogram ABCD.

1. As in case A, find EG the mean proportional (Pr. 140). 2. On AB, mark off AF equal to EG, and join AC.

3. Draw FH, HK, parallel to BC, CD. Then AFHK will

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H

B

A

be the required parallelogram, and it is one-third of the given parallelogram ABCD.

(C) To construct an irregular rectilineal figure one-third of a given irregular rectilineal figure ABCDHK.

1. As in case A, find EG the mean proportional (Pr. 140). 2. On AB, mark off AF equal to EG, and from A, draw

AC, AD, AH,

K,

D

A

F

B

3. From F, commence drawing a series of lines parallel to

the sides of the given figure, and the smaller rectilineal figure will be constructed.

NOTE.-As in the case of the circle, if the figures required be any other proportion of the given figures, e.g., three-fifths ; make EF three-fifths of DE, and find the mean proportional as before. That will be equal to a side of the figure required.

SECTION XII.

MISCELLANEOUS PROBLEMS.

Problem 194.

Through a given point A, to draw a line which would, if produced, pass through the angular point towards which the two given lines BC, DE converge.

1. Draw any convenient line FG, and join FA, GA.
2. Draw any line HK parallel to FG (Pr. 9).

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F
B H

K

E

3. Through H and K, draw HL and KL parallel to FA and

GA (Pr. 9), meeting each other in .. 4. Through A and L, draw AL, which produced is the con

vergent line required.

Problem 195.

To construct an isosceles triangle, having given its altitude AB, and CD the length of its equal sides.

1. Through B, draw EF of unlimited length, and at right

angles to AB (Pr. 2).

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