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PROPOSITION XXV. PROBLEM.

357. To construct a parallelogram equivalent to a given square, and having the sum of its base and altitude equal to a given line.

IL-*----------N

Let R be the given square, and let the sum of the

base and altitude of the required parallelogram be equal to the given line M N.

It is required to construct a O =R, and having the sum of its base and altitude = M N.

Upon M N as a diameter, describe a semicircle. At M erect a I M P, equal to a side of the given square R. Draw P Q Il to MN, cutting the circumference at S.

Draw SC I to M N. Any O having C M for its altitude and C N for its base, is equivalent to R. For SC is || to PM,

§ 65 (two straight lines I to the same straight line are II). So=PM.

$ 135 (Ils comprehended between Ils are equal).

..SC = PM = R. But MC:SC :: SC: CN,

$ 307 (a I let fall from any point in a circumference to the diameter is a mean

proportional between the segments of the diameter).
Then
SC= MCXCN,

$ 259 (the product of the means is equal to the product of the extremes).

Q. E. F. 358. SCHOLIUM. The problem is impossible when the side of the square is greater than one-half the line M N.

PROPOSITION XXVI. PROBLEM.

359. To construct a parallelogram equivalent to a given square, and having the difference of its base and altitude equal to a given line.

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Let R be the given square, and let the difference of

the base and altitude of the required parallelogram be equal to the given line MN.

It is required to construct a D = R, with the difference of the base and altitude = M N.

Upon the given line M N as a diameter, describe a circle.

From M draw MS, tangent to the O, and equal to a side of the given square R.

Through the centre of the O, draw S B intersecting the circumference at C and B.

Then any O, as R', having S B for. its base and SC for its altitude, is equivalent to R. For

SB:SM :: SM : SC, (if from a point without a 0, a secant and a tangent be drawn, the tangent is a mean proportional between the whole secant and the part without the O). Then S M = SBXSC;

§ 259 and the difference between S B and SC is the diameter of the O, that is, MN.

Q. E. F.

§ 292

PROPOSITION XXVII. PROBLEM.

360. Given x = V2, to construct x.

--

----------------0-B

Let m represent the unit of length.

It is required to find a line which shall represent the square root of 2.

On the indefinite line A B, take A C=m, and C D = 2 m.

On A D as a diameter describe a semi-circumference.

At C erect a I to A B, intersecting the circumference at E.

Then CE is the line required.
For AC : CE :: CE : CD,

§ 307 (the I let fall from any point in the circumference to the diameter, is a mean

proportional between the segments of the diameter);

$ 259

.:.C Ēo = A C X C D, ..CE=VAC XCD,

=V1 X 2 = V2.

Q. E. F.

Ex. 1. Given x = V5, y = 17, 2= 2 V3 ; to construct x, y, and z.

2. Given 2 : x :: 2 : 3; to construct x.
3. Construct a square equivalent to a given hexagon.

PROPOSITION XXVIII. PROBLEM.

361. To construct a polygon similar to a given polygon P, and equivalent to a given polygon Q.

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Let P and Q be two given polygons, and AB a side

of polygon P.

It is required to construct a polygon similar to P and equivalent to Q. Find a square equivalent to P,

§ 356

and let m be equal to one of its sides.

$ 356

Find a square equivalent to Q,

and let n be equal to one of its sides.

§ 304

Find a fourth proportional to m, n, and A B.
Let this fourth proportional be A' B'.

Upon A' B', homologous to A B, construct the polygon P' similar to the given polygon P.

Then P' is the polygon required.

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(similar polygons are to each other as the squares on their homologous sides);

Ax. 1

.. P' is equivalent to Q, and is similar to P by construction.

Q. E. F.

Ex. 1. Construct a square equivalent to the sum of three given squares whose sides are respectively 2, 3, and 5.

2. Construct a square equivalent to the difference of two given squares whose sides are respectively 7 and 3.

3. Construct a square equivalent to the sum of a given triangle and a given parallelogram.

4. Construct a rectangle having the difference of its base and altitude equal to a given line, and its area equivalent to the sum of a given triangle and a given pentagon.

5. Given a hexagon ; to construct a similar hexagon whose area shall be to that of the given hexagon as 3 to 2.

6. Construct a pentagon similar to a given pentagon and equivalent to a given trapezoid.

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