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BC2 AC2 AB2

m2

n2; therefore the square con

structed upon BC will be equivalent to the difference of the

squares constructed upon m and n.

EXERCISE.

Problem.-Upon a given straight line, to construct a rectangle equivalent to a given rectangle.

PROPOSITION XIV.-PROBLEM.

27. To construct a rectangle, having given its area and the sum of two adjacent sides.

Let MN be equal to the given sum of the adjacent sides of the required rectangle; and let the given area be that of the square whose side is AB.

Upon MN as a diameter semicircle. At M erect MP

describe a AB per

P

B

Q

M

R N

pendicular to MN, and draw PQ parallel to MN, intersecting the circumference in Q. From Q let fall QR perpendicular to MN; then MR and RN are the base and altitude of the required rectangle. For, by III., Proposition IX., Corollary, MR × RN = QR2 — PM2 — AB2.

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PROPOSITION XV.-PROBLEM.

28. To find two straight lines in the ratio of the areas of two given polygons.

Let squares be found equal in area to the given polygons respectively (23 and 24). Upon the sides of the right angle ACB, take CA and CB equal to the sides

A

C

B

of these squares, join AB, and let fall CD perpendicular to

AB. Then, by (III., 26), we have AC AD X AB, CB2

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therefore AD and DB are in the ratio of the areas of the

given polygons.

EXERCISE.

Problem. To find a square which shall be to a given square in the ratio of two given straight lines. (v. 28.)

PROPOSITION XVI.-PROBLEM.

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

Find M and N, the sides of squares respectively equal in area to P and Q (23 and 24).

Let a be any side of P, and find a fourth proportional a' to M, N, and a; upon a', as a homologous side to a, construct the polygon P'

Р

Q

a

M

P'

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N

similar to P; this will be the required polygon. For, by construction,

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therefore, taking the letters P, Q, and P', to denote the areas

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But, the polygons P and P' being similar, we have, by (Proposition IX.),

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Therefore the polygon P' is similar to the polygon P and equivalent to the polygon Q, as required.

EXERCISE.

Problem.-To construct a polygon similar to a given polygon, and whose area shall be in a given ratio to that of the given polygon. (v. 28, Exercise, and III., 43.)

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EXERCISES ON BOOK IV.

THEOREMS.

1. Two triangles are equivalent if they have two sides of the one respectively equal to two sides of the other, and the included angle of the one equal to the supplement of the included angle of the other.

2. The two opposite triangles formed by joining any point in the interior of a parallelogram to its four vertices are together equivalent to one-half the parallelogram.

3. The triangle formed by joining the middle point of one of the non-parallel sides of a trapezoid to the extremities of the opposite side is equivalent to one-half the trapezoid. (v. I., Exercise 24.)

4. The figure formed by joining consecutively the four middle points of the sides of any quadrilateral is equivalent to one-half the quadrilateral. (v. I., Exercise 32.)

5. If in a rectangle ABCD we draw the diagonal AC, inscribe a circle in the triangle ABC, and from its centre draw OE and OF perpendicular to AD and DC respectively, the rectangle OD will be equivalent to one-half the rectangle ABCD.

6. The area of a triangle is equal to one-half the product of its perimeter by the radius of the inscribed circle.

E

A

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G

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D

FO

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7. The area of a rhombus is one-half the product of the diag onals.

8. The straight line joining the middle points of the parallel sides of a trapezoid divides it into two equivalent figures.

9. Any line drawn through the point of intersection of the diagonals of a parallelogram divides it into two equal quadrilaterals. 10. In an isosceles right triangle either leg is a mean proportional between the hypotenuse and the perpendicular upon it from the vertex of the right angle.

11. If two triangles have an angle in common, and have equal areas, the sides about the equal angles are reciprocally proportional.

12. The perimeter of a triangle is to a side as the perpendicular from the opposite vertex is to the radius of the inscribed circle. (v. Exercise 6.)

13. Two quadrilaterals are equivalent when the diagonals of one are respectively equal and parallel to the diagonals of the other.

14. The sum of the perpendiculars from any point within an equilateral convex polygon upon the sides is constant.

Suggestion. Join the point with the vertices of the polygon. 15. The lines joining two opposite vertices of a parallelogram with the middle points of the sides form a parallelogram whose area is one-third the area of the given parallelogram.

M

E

G

P

F

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16. The sum of the squares on the segments of two perpendicular chords in a circle is equivalent to the square on the diameter. 17. Let ABC be any triangle, and upon the sides AB, AC, construct parallelograms AD, AF, of any magnitude or form. Let their exterior sides DE, FG, meet in M; join MA, and upon BC construct a parallelogram BK, whose side BH is equal and parallel to MA. Then the parallelogram BK is equivalent to the sum of the parallelograms AD and AF. (v. Proposition I.)

D

B

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H

K

From this deduce the Pythagorean Theorem.

18. Prove, geometrically, that the square described upon the sum of two straight lines is equivalent to the sum of the squares described on the two lines plus twice their rectangle.

Note. By the "rectangle of two lines" is here meant the rectangle of which the two lines are the adjacent sides.

19. Prove, geometrically, that the square described upon the difference of two straight lines is equivalent to the sum of the squares described on the two lines minus twice their rectangle.

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