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5. Divide an angle of an equilateral triangle into five equal parts. Describe a o about the A, then use (453).

6. The square inscribed in a semicircle is equal to twofifths the square inscribed in the whole circle.

7. The area of a given circle is 314.16; if this circle be circumscribed by a square, find the area of the part between the circumference and the perimeter of the square.

8. The area of a circle is 40 feet; find the side of the inscribed square.

9. Find the angle subtended at the centre of a circle by an arc 6 feet long, if the radius is 8 feet long.

10. Find the length of the arc subtended by one side of a regular octagon inscribed in a circle whose radius is 20 feet.

11. Find the area of a circular sector whose arc contains 18°, the radius of the circle being 4 feet.

12. Find the area of a circular sector, the chord of half the arc being 20 inches, and the radius 45 inches.

13. The radius of a circle is 5 feet : find the area of a circle ry times as large.

14. The radius of a circle is y feet : find the radius of a circle 16 times as large.

15. Find the height of an arc, the chord. of half the arc being 10 feet, and the radius 16 feet.

16. Find the area of a segment whose height is 4 inches, and chord 30 inches.

17. Find the area of a segment whose height is 16 inches, the radius of the circle being 20 inches.

18. Find the area of a segment whose arc is 100°, the radius being 12 feet.

19. Find the area of a sector, the radius of the circle being 24 feet, and the angle at the centre 30°.

20. A circular park 500 feet in diameter has a carriageway around it 25 feet wide : find the area of the carriageway.

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MAXIMA AND MINIMA. 463. DEF. A maximum quantity is the greatest quantity of the same kind; and a minimum quantity, is the least quantity of the same kind.

Thus, the diameter of a circle is a maximum among all inscribed straight lines; and a perpendicular is a minimum

; among all the straight lines drawn from a given point to a given straight line.

464. Isoperimetric figures are those which have equal perimeters.

We give here a few simple but important propositions bearing on this part of Geometry.

Proposition 18. Theorem.

465. Of all triangles formed with two given sides, that in which these sides are perpendicular to each other is the maximum.

Hyp. Let ABC, A'BC be two As A having the sides AB, BC, respectively equal to A'B, BC; and let Z ABC be a rt. L.

Δ ABC > A A'BC. Proof. Draw A'D I to BC. Since the 1 is the shortest distance from a pt. to a line,(58)

.. A'B> A'D.

To prove

B

But
AB = A'B.

(Hyp.) .. AB > A'D. But AB and A'D are the altitudes respectively of the As ABC, A'BC. .:. A ABC > A A'BC.

(369) Q.E.D.

Proposition 19. Theorem.

466. Of all triangles having equal perimeters and the same base, the isosceles triangle is the maximum. Hyp. Let the As ABC, ABD

H have equal perimeters and the same base AB, and let the A ABC be

G isosceles.

A ABC > A ABD. A Proof. Produce AC to H, so that

M CH = CB = CA, and join HB. Then

LABH is a rt. L,

To prove

E

=

D

a

being inscribed in the fo whose cent. is C and radius is CB (240). Produce HB, take DL = DB, join AL, and draw CG,

= DM 11, and CE, DF 1, to AB.

Then AD + DL= AD + DB = AC + CB = AH.

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But 1 BH = BG = CE, and {BL = BM = DF.

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(61)

.. CE > DF,

and

..A ABC > A ABD.

(369) Q.E.D.

467. COR. Of all the triangles of the same perimeter, that which is equilateral is the maximum.

For, the maximum triangle having a given perimeter must be isosceles whichever side is taken as the base.

Proposition 20. Theorem.

468. Of all isoperimetric polygons having the same number of sides, the maximum polygon is equilateral. Hyp. Let ABCDE be the maximum

B B'
polygon of given perimeter and given
number of sides.

A
To prove it is equilateral.
Proof. Join AC.

If possible, let the sides AB, BC of
the A ABC be unequal, and let AB'C
be an isosceles having the same base AC, and

E

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If the diagonals of a quadrilateral are given in magnitude, the area of the quadrilateral is a maximum when the diagonals are at right angles to each other.

Proposition 21. Theorem.

469. Of all polygons formed of sides all given but one, the maximum can be inscribed in a semicircle with the undetermined side for its diameter.*

D

BA

Hyp. Let ABCDEF be the maximum polygon formed of the given sides AB, BC, CD, DE, EF.

To prove ABCDEF can be inscribed in a semicircle.

A

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Proof. Join any vertex, as D, with the extremities A and F of the side not given.

Then the AADF must be the max. of all As formed with the given sides AD and FD; for, if it is not, by increasing or decreasing the ADF, keeping the sides AD and FD unchanged in length, the A ADF may be increased, while the rest of the polygon, ABCD, DEF, remains unchanged; so that the whole polygon ABCDEF is increasec.

But this is contrary to the hypothesis that the given polygon is a maximum.

... the A ADF must be the max. of As formed with the given sides AD and FD.

,, the LADF is a rt. L.

(465)

.', D is on the semioce whose diameter is AF.

(240)

.. any vertex is on the semi oce whose diameter is AF.

Q. E.D.

* Peirce's Geometry, p. 92.

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