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1. To find the area of a triangle, whose base is 625 and altitude 520 feet. Ans. 162500 sq. ft. 2. To find the number of square yards in a triangle, whose base is 40 and altitude 30 feet. Ans. 66.

3. To find the number of square yards in a triangle, whose is 49 and altitude 25 feet. Ans. 68.736].

bas

CASE II.

When two sides and their included angle are given.

RULE. Add together the logarithms of the two sides and the logarithmic sine of their included angle; from this sum subtract the logarithm of the radius, which is 10, and the remainder will be the logarithm of double the area of the triangle. Find, from the table, the number answering to this logarithm, and divide it by 2; the quotient will be the required area.

Let BAC be a triangle, in which there are given BA, BC, and the included angle B.

From the vertex A draw AD, perpendicular to the base BC, and represent the area of the triangle by Q. Then,

B

D

R: sin B:: BA: AD (Trig. Th. I.) :
BA
AD=
× sin B

hence,

R

(Book IV. Prop. VI.) ;

A

BCX AD

But, Q=

2

hence, by substituting for AD its value, we have
BCX BA x sin B
BCX BA x sin B
Q:
or 2Q=2
9

2R R Taking the logarithms of both numbers, we have log. 2Q=log. BC+log. BA+log. sin B—log. R; which proves the rule as enunciated.

+log. BC
+ log. BA
+log. sin B 57° 25′
—log. R

1. What is the area of a triangle whose sides are, BC= 125.81, BA=57.65, and the included angle B=57° 25'?

....

125.81 . . . . 2.099715 57.65

Then, log. 2Q=

1.760799 9.925626

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-10.

log. 2Q

and 2Q=6111.4, or Q=3055.7, the required area.

3.786140

2. What is the area of a triangle whose sides are 22 and 40 and their included angle 28° 57' ? Ans. 290.427.

3. What is the number of square yards in a triangle of which the sides are 25 feet and 21.25 feet, and their included angle Avs. 20.8694.

45° ?

CASE III.

When the three sides are known.

RULE.-1. Add the three sides together, and take half their sum 2. From this half-sum subtract each side separately.

3. Multiply together the half-sum and each of the three remainders, and the product will be the square of the area of the triangle. Then, extract the square root of this product, for the required area. ··

Or, After having obtained the three remainders, add together the logarithm of the half-sum and the logarithms of the respective remainders, and divide their sum by 2: the quotient will be the logarithm of the area.

Let ABC be the given triangle. Take CD equal to the side CB, and draw DB; draw AE parallel to DB, meeting CB produced, in E: then CE will be equal to CA. Draw CFG perpendicular to AE and DB, and it will bisect them at the points G and F. Draw FHI parallel to AB, meeting CA in H, and EA produced, in I. Lastly, with the centre H and radius HF, describe the circumference of a circle, meeting CA produced in K: this circumference will pass through I, because AI-FB-FD, therefore, HF=HI; and it will also pass through the point G, because FGI is a right angle.

K

H

[ocr errors]

A

=

D

G

Now, since HA=HD, CH is equal to half the sum of the sides CA, CB; that is, CH=CA+CB; and since HK is equal to IFAB, it follows that

CK=AC+CB+÷AB={S,

by representing the sum of the sides by S.
Again, HK HI=IF=AB, or KL=AB.
Hence, CL=CK—KL=4S—AB,
and
AK=CK-CA-IS-CA,
AL=DK=CK-CD-4S--CB.

and

Now, AGCG= the area of the triangle ACE, and AGFG the area of the triangle ABE; therefore, AGX CF the area of the triangle ACB

B

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Also, by similar triangles,

AG: CG:: DF: CF, or AI : CF ; therefore, AG × CF= triangle ACB=CG×DF=CG× AI; consequently, AG × CF × CG × AI= square of the area ACB. CGX CF=CKX CL=S(S-AB),

X

But and AG× AI =AK × AL=(¿S-CA) × (S—CB) ; therefore, AG x CFx CG x AIS(SAB) x (S-CA) × (S-CB), which is equal to the square of the area of the triangle ACB.

2)90

1. To find the area of a triangle whose three sides are 20, 30, and 40.

20

30

40

45 half-sum.

45

20

25 1st rem.

45

30

[ocr errors]

15 2d rem.

PROBLEM III.

To find the area of a trapezoid.

45 half-sum.

40

[ocr errors]

5 3d rem

Then, 45 x 25 x 15 x 5=84375.

The square root of which is 290.4737, the required area. 2. How many square yards of plastering are there in a triangle whose sides are 30, 40, and 50 feet? Ans. 663.

RULE. Add together the two parallel sides: then multiply their sum by the altitude of the trapezoid, and half the product will be the required area (Book IV. Prop. VII.).

1. In a trapezoid the parallel sides are 750 and 1225, and the perpendicular distance between them is 1540; what is the area? Ans. 152075.

2. How many square feet are contained in a plank, whose length is 12 feet 6 inches, the breadth at the greater end 15 inches, and at the less end 11 inches? Ans. 131 sq. ft.

3. How many square yards are there in a trapezoid, whose parallel sides are 240 feet, 320 feet, and altitude 66 feet?

Ans. 2053.

PROBLEM IV.

To find the area of a quadrilateral.

RULE.-Join two of the angles by a diagonal, dividing the quadrilateral into two triangles. Then, from each of the other angles let fall a perpendicular on the diagonal: then multiply

the diagonal by half the sum of the two perpendiculars, and the product will be the area.

1. What is the area of the quadrilateral ABCD, the diagonal AC being 42, and the perpendiculars Dg, Bb, equal to 18 and 16 feet?

Ans. 714.

PROBLEM V.

To find the area of an irregular polygon.

2. How many square yards of paving are there in the quadrilateral whose diagonal is 65 feet, and the two perpendiculars let fall on it 28 and 33 feet? Ans. 222.

RULE.-Draw diagonals dividing the proposed polygon into trapezoids and triangles. Then find the areas of these figures separately, and add them together for the content of the whole polygon.

b

1. Let it be required to determine the content of the polygon ABCDE, having five sides.

A.

Let us suppose that we have measured the diagonals and perpendiculars, and found AC=36.21, EC= 39.11, Bb=4, Dd=7.26, Aa=4.18, required the area. Ans. 296.1292.

E

B

a

Let AEea be an irregular figure, having for its base the right line AE. At the points A, B, C, D, and E, equally distant from each other, erect the perpendiculars Aa, Bb, Cc, Dd, Ee, to the

PROBLEM VI.

To find the area of a long and irregular figure, bounded on one side by a right line.

RULE.-1. At the extremities of the right line measure the perpendicular breadths of the figure, and do the same at several intermediate points, at equal distances from each other. 2. Add together the intermediate breadths and half the sum of the extreme ones: then multiply this sum by one of the equal parts of the base line: the product will be the required area. very nearly.

a b

base line AE, and designate them respectively by the letters a, b, c, d, and e.

Then, the area of the trapezoid ABba=!

the area of the trapezoid BCcb=

the area of the trapezoid CDdc=

and the area of the trapezoid DEed=

2

nence, their sum, or the area of the whole figure, is equal to b+c c+d d+e + + + since AB, BC, &c. are equal to each other. But this sum is also equal to

(a+b

+d+e) x

2

2

2

2(16.8

8.4 mean of the extremes.

7.4

9.2

10.2

35.2 sum.

=

35.2

a + b

10

2

b+c

2

c+d

2

d+e

× AB,

(2 +b+c+d+) x AB,

which corresponds with the enunciation of the rule.

1. The breadths of an irregular figure at five equidistant places being 8.2, 7.4, 9.2, 10.2, and 8.6, and the length of the hase 40, required the area.

8.2

8.6

× AB,

sum.

× BC,

4)40

10 one of the equal parts.

352=area.

× CD,

× DE;

2. The length of an irregular figure being 84, and the breadths at six equidistant places 17.4, 20.6, 14.2, 16.5, 20.1. and 24.4; what is the area? Ans. 1550.64.

PROBLEM VII.

To find the area of a regular polygon.

RULE 1.-Multiply half the perimeter of the polygon by th apothem, or perpendicular let fall from the centre on one of the sides, and the product will be the area required (Book V Prop. IX.)

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