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Note.—The perpendicular height of the parallelogram is equal to the area divided by the basc.
1. Required the area of the square ABCD whose side is 5 feet 9 inches.
Here 5 ft. 9 in.=5.75: and 5.7512=5.75 X 5.75=33.0625 feet=33 fe. Oin. 9pa.=area required.
2. Required the area of the rectangle ABCD, whose length AB is 13.75 chains, and breadth BC 9.5 chains. D
squares each equal to the superficial measuring unit, and that the number of these squares, or the area of the figure, is equal to the number of linear measuring units in the length, as often repeated as there are linear measuring units in the breadth or height, that is, equal to the length multiplied by the height, which is the rule.
And since a rectangle is equal to an oblique parallelogram standing upon the same base, and between the same parallels, (Euc. I. 35,) the rule is true for any parallelogram in general. Q. E. D.
RULE II. If any two sides of a parallelogram be multiplied together, and the product again by the natural sine of the included angle, the last product will give the area of the parallelogram. That is ABXBC =nat. sine of the angle B=area.
Note.—Because the angles of a square and rectangle are each 90°, whose natural sine is unity, or 1, the rule in this case is the same as that given in the text.
130.625 Here 13.75 X 9.5=130.625; and
10 ac.=13 ac. O ro. 10 po.=area required.
3. Required the area of the rhombus ABCD, whose length AB is 12 feet 6 inches, and its height DE 9 feet 3 inches.
Here 12 fe. 6 in.=12.5, and 9 fe. 3 in.=9.25.
Whence 12.5 x 9.25=115.625 fe.=115 fe. 7 in. 6 pa. =area required.
4. What is the area of the rhomboides ABCD whose length AB is 10.52 chains, and height DE 7.63 chains ?
80.2676 Here 10.52 x 7.63= 80.2676 ; and
10 =8 ac. O ro. 4 po.=area required. 5. What is the area of a square whose side is 35.25 chains ?
ac. ro. po.
Ans. 124 1 1 6. What is the area of a square whose side is 8 feet 4 inches?
fe. in. pa
Ans. 69 5 4
7. What is the area of a rectangle whose length is 14 leet 6 inches, and breadth 4 feet 9 inches? fe. in. pa.
Ans. 68 10 6 8. Required the area of a rhombus, the length of whose side is 12.24 feet, and height 9.16 feet. fe. in. pa.
Ans. 112 1 5 9. Required the area of a rhomboides whose length is 10.51 chains, and breadth 4.28 chains.
ac. ro. pa.
Ans. 4 1 39.7 10. What is the area of a rhomboides whose length is 7 feet 9 inches, and height 3 feet 6 inches?
fe. in. pa.
Ans. 27 1 6 11. To find the area of a rectangular board, whose length is 121 feet, and breadth 9 inches.
Ans. 9 feet.
RULE.* Multiply the base by the perpendicular height, and half the product will be the area.
Note.—The perpendicular height of the triangle is equa. to twice the area divided by the base.
1. Required the area of the triangle ABC, whose base AB is 10 feet 9 inches, and height DC 7 feet 3 inches.
* A triangle is half a parallelogram of the same base and altitude (Cuc. I. 41,) and therefore the truth of this rule is evident.
Here 10 fe. 9 in.=10.75, and 7 fe. 3 in.=7.25,
77.9375 Whence 10.75 x 7.25=77.9375,and
2 feet=38 fe. 11 in. 7} pa.=area required.
2. What is the area of a triangle whose base is 18 feet 4 inches, and height 11 feet 10 inches? fe. in. pa.
Ans. 108 5 8 3. What is the area of a triangle whose base is 16.75 feet, and height 6.24 feet?
fe. in. pa.
Ans. 52 3 1 4. Required the area of a triangle whose base is 12.25 chains, and height 8.5 chains.
ac. ro. po.
Ans. 5 0 33 5. What is the area of a triangle whose base is 20 feet, and height 10.25 ?
Ans. 102.5 fe. PROBLEM III. To find the area of a triangle whose three sides only are
RULE.* 1. From half the sum of the three sides subtract each side severally.
* Demon. Let ac=A, AB= =b, bc=c, and AD=x ; (See preceding fig.) Then, since BD=b-X, we shall have c-b-x=cno=a-x", or c-b2+2bx-=-,
a’ +62-02 from which x is found
by trans, and reduction.
26 But cdo ACP AD2 = AC + AD X AC AD = (a a + 62 ca
a? + b .
2ab + a + b2 — C 26
26 2ab a’ — 13 + c
(a + b)
26 1 Whence cd = Va+ob — c) * (c
a— 6%), and the 2b trea į abx cd = £ v (a + b2 — c) * (c_ā — b?)
2. Multiply the half sum and the three remainders con. tinually together, and the square root of the product will be the area required.
1. Required the area of the triangle ABC, whose three sides BC, CA, and AB are 24, 36, and 48 chains respectively.
=54=} sum of the sides ;
= IV (a + b + cxa+b-cxcta obxe -a+b)
a + b + c a + bc cta b C-a + b - VG
2 which, by making s = 1 x (a + b + c) becomes =
V (88-cxs-bx s—a)=algebraical expression for the rule, as was to be demonstrated.
Cor. 1. If s be put equal to a+b, and d=b un a, the rule is v (92-c2)(c2d2).
Cor. 2. If all the sides be equal, the rule will become #a? /3, or $ ax 1.732 for the equilateral triangle whose side is a. Cor. 3. If the triangle be right angled, a being the
-a+b+c hypothenuse, the rule will be
2 1 px} p-a, putting p for the perimeter.
RULE II. Any two sides of a triangle being multiplied together, and the product again by half the natural sine of their included angle, will give the area of the triangle.
That is, ac X CB X nat. sine of the angle c=twice area.