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PROPOSITION XV. PROBLEM.

385. To construct a polygon similar to two given similar polygons and equivalent to their difference.

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Let R and R' be two similar polygons, and AB and A'B' two homologous sides.

To construct a similar polygon equivalent to R' — R.

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From O as a centre, with a radius equal to A'B', describe an arc cutting PX at H, and join OH. Take A"B"

Proof.

=

PH, and on A"B", homologous to AB, construct R" similar to R.

Then R" is the polygon required.

R': R= A'B': AB,

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(similar polygons are to each other as the squares of their homologous sides).

By division, R'R: RA'B' — AB2 : AB2,

OH2 -- OP2 : OP2,

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PROPOSITION XVI. FROBLEM.

386. To construct a triangle equivalent to a given polygon.

D

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Let ABCDHE be the given polygon.

To construct a triangle equivalent to the given polygon.
Construction.

From D draw DE,

and from H draw HF to DE.

Produce AE to meet HF at F, and draw DF.

Again, draw CF, and draw DK | to CF to meet AF produced at K, and draw CK.

In like manner continue to reduce the number of sides of the polygon until we obtain the ▲ CIK.

Proof. The polygon ABCDF has one side less than the polygon ABCDHE, but the two are equivalent.

For the part ABCDE is common,

and the A DEFA DEH,

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(for the base DE is common, and their vertices F and H are in the line

FH to the base).

The polygon ABCK has one side less than the polygon ABCDF, but the two are equivalent.

For the part ABCF is common,

and the ACFKA CFD,

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(for the base CF is common, and their vertices K and D are in the line

KD to the base).

In like manner the ▲ CIK ABCF.

Q. E. F.

PROPOSITION XVII. PROBLEM.

387. To construct a square which shall have a given ratio to a given square.

D

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n

Let R be the given square, and the given ratio.

m

To construct a square which shall be to R as n is to m. Construction. Take AB equal to a side of R, and draw Ay, making any acute angle with AB.

On Ay take AE=m, EF=n, and join EB.

Draw FC to EB to meet AB produced at C.

On AC as a diameter describe a semicircle.

At B erect the 1 BD, meeting the semicircumference at D. Then BD is a side of the square required.

Proof. Denote AB by a, BC by b, and BD by x.

Now a:x= x:b; that is, x2 = ab.

Hence, a2 will have the same ratio to a2 and to ab.

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(a straight line drawn through two sides of a ▲, parallel to the third side, divides those sides proportionally).

Therefore a2: x2=m: n.

By inversion, x2: a2 =n: m.

Hence the square on BD will have the same ratio to R as n has to m.

Q. E. F.

PROPOSITION XVIII. PROBLEM.

388. To construct a polygon similar to a given polygon and having a given ratio to it.

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Let R be the given polygon and the given ratio.

m

To construct a polygon similar to R, which shall be to R as

n is to m.

Construction. Find a line A'B', such that the square constructed upon it shall be to the square constructed upon AB as n is to m.

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Upon A'B' as a side homologous to AB, construct the polygon S similar to R.

Proof.

Then S is the polygon required.

S: R= A'B': AB,

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(similar polygons are to each other as the squares of their homologous sides).

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Ex. 304. Find the area of a right triangle if the length of the hypotenuse is 17 feet, and the length of one leg is 8 feet.

Ex. 305. Compare the altitudes of two equivalent triangles, if the base of one is three times that of the other.

Ex. 306. The bases of a trapezoid are 8 feet and 10 feet, and the altitude is 6 feet. Find the base of an equivalent rectangle having an equal altitude.

PROPOSITION XIX. PROBLEM.

389. To construct a square equivalent to a given parallelogram.

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Let ABCD be a parallelogram, b its base, and a its altitude.

To construct a square equivalent to the ABCD.

Construction. Upon the line MX take MN= a, and NO=b.
Upon MO as a diameter, describe a semicircle.

At N erect NP 1 to MO, to meet the circumference at P.

Then the square R, constructed upon a line equal to NP, is equivalent to the

Proof.

ABCD.

MN: NPNP: NO,

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(a 1 let fall from any point of a circumference to the diameter is a mean proportional between the segments of the diameter).

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390. COR. 1. A square may be constructed equivalent to a given triangle, by taking for its side a mean proportional between the base and one-half the altitude of the triangle.

391. COR. 2. A square may be constructed equivalent to a given polygon, by first reducing the polygon to an equivalent triangle, and then constructing a square equivalent to the triangle.

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