Page images
PDF
EPUB

PROPOSITION III. THEOREM

443. The maximum of isoperimetric triangles on the same base is the one whose other two sides are equal.

[blocks in formation]

Hyp. ABC and ABD have equal perimeters, and AC=CB.

[blocks in formation]

Proof. Draw median CE and DF AB, meeting CE in F. Join FA and FB.

Then CE is the perpendicular bisector of AB.

[blocks in formation]

(Why?)

(442)

.. FE <CE.

(311)

or

.. area AFB < area ACB,

(unequal altitudes),

area ADB < area ACB.

Q.E.D.

444. COR. Of all isoperimetric triangles, the equilateral

has the maximum area.

PROPOSITION IV. THEOREM

445. Of all polygons having all sides given but one, the maximum can be inscribed in a semicircle having the undetermined side as diameter.

E

F

Q

B

Hyp. Polygon ABCDEF is the maximum of all polygons having given the sides AF, FE, ED, DC, and CB.

To prove ABCDEF can be inscribed in a semicircle whose diameter is AB.

Proof.

Join any vertex, as D, with A and B.

Then ▲ ADB must be the maximum of all triangles that can be formed with sides AD and DB. For otherwise by making ZADB a right one without changing the sides AD and DB, we could increase A ADB without altering the remaining parts AFED and DCB of the polygon. Or polygon ABCDEF would be increased, which is contrary to the hypothesis, since ABCDEF is a maximum.

.. A ADB is the maximum of all triangles having AD and DB given.

Then

ZADB is a right angle,

(Why ?)

and D is on a semicircumference that can be constructed on AB. For the same reason every vertex of the polygon must lie on the semicircumference.

Q.E.D.

Ex. 1017. To inscribe an angle in a given semicircle so that the sum of its arms is a maximum.

PROPOSITION V. THEOREM

446. Of all polygons constructed with the same given sides, that which can be inscribed in a circle is the maximum.

[merged small][merged small][merged small][ocr errors]

Hyp. Polygon ABCDE, which is inscribed in a circle, is mutually equilateral with polygon A'B'C'D'E', which cannot be inscribed in a circle.

To prove

area ABCDE> area A'B'C'D'E'.

Proof. From A draw diameter AF, and join F to the two nearest vertices C and D.

On C'D', construct AC'D'F' equal to ▲ CDF, and join A'F'.

[blocks in formation]

Then

A CFDA C'F'D'.

area ABCDE > area A'B'C'D'E'. (Ax. 5.) q.e.d.

PROPOSITION VI. THEOREM

447. Of all isoperimetric polygons of the same number of sides, the equilateral is the maximum.

[blocks in formation]

Hyp. ABCDE is the maximum of all isoperimetric polygons of the same number of sides.

To prove

AB BC= CD = DE = EA.

=

Proof. The polygon must be equilateral, for suppose any two sides AB and BC were not equal.

Then construct on the diagonal AC the isosceles triangle AB'C, isoperimetric with ABC.

Area AB'C would be greater than area ABC.

(443)

.. polygon AB'CDE would be greater than the isoperimetric polygon ABCDE, which would contradict the hypothesis.

Hence

AB =

BC.

Q.E.D.

448. COR. Of all isoperimetric polygons of the same number of sides, the maximum is regular.

Ex. 1018. In a given segment to inscribe an angle so that the sum of its arms is a maximum.

Ex. 1019. In a given semicircle to inscribe a trapezoid whose area is a maximum.

PROPOSITION VII. THEOREM

449. Of two isoperimetric regular polygons, that which has the greater number of sides has the greater

[blocks in formation]

Proof. Let F be any point in CD.

ABCFDE may be considered a hexagon, having one of its angles equal to a straight angle.

.. area H> area ABCDE.

(Why?)

Q.E.D.

450. COR. The area of a circle is greater than the area of any polygon whose perimeter equals the circumference of the circle.

Ex. 1020. In a given circle to inscribe a triangle, having a maximum perimeter.

Ex. 1021. The circumscribed regular polygon has a smaller area than any other polygon circumscribed about the same circle.

Ex. 1022. Of all equivalent parallelograms on the same base, which has the minimum perimeter ?

« PreviousContinue »