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As 3 : 2 :: 18: 12=AE ? the parts required.
A farmer having three sons, whose ages where 20, 25, and 30 years, respectively, proposes dividing the triangular farm ABC among them in such a manner, that each shall have a part proportionable to his age; Now supposing AB 71, BC 61, and AC 63 chs. What will be each son's share.
A. R. P.
The area will be found equal to 180 1 20.3
A. R. P. ; ; 184 1 20
A. R P.
To divide a triangle into any number of parts, equal or uneqnal, by right lines drawn from a given point in one of its sides.
From the given point let fall perpendiculars on the other two sides; and by trigonometry find their lengths. Divide twice the area to be cut off adjacent to any of these sides, by the length of the perpendicular falling thereon; the quotient will give the base.
In the triangle ABC, are given AB 426 ; AC 365;
and BC 230; to divide it into 3 equal parts by lines drawn from the middle of the base AB.
By trigonometry the angle A=32° 29' and B=
Let AB be 45 chains; BC 40, and AC 20, to divide the whole into four equal parts by lines drawn from a point in AB. 10 chains from A.
The area will be found equal to 399.95 chs.
B = 26 23
C 90 53
To divide a three-sided field into any assigned parts, by lines drawn parallel to one of its sides.
Call that side parallel to which the lines are to be drawn, the base, then say, as the whole area is to the square of either side, so is the area to be cut off towards the verticle angle, to the square of its corresponding side.
To divide the triangular field ABC into two equal parts by a line drawn parallel to AB; supposing AB 45, BC 36, and AC 27 chains.
As the whole area 486 : 27||
Let it be required to divide the field ABC (whose sides are AB 40; BC 50; and AC 60 perches) into three parts, which will be to each other as the num.
Answer. BD=18.8. BE=29.8. and the area equal 992 perches.
Through a given point in a triangle to draw a line to cut off a given area, if possible.
Let ABC be the triangle, and P the point through which the line must pass; let PD, perpendicular, and PE parallel to AC be found, either by actual measurement, or by trigonometry Divide the area to be cut off, by the length of PD, and set off the quotient from C to f; subtract PE from Cf, square the remainder from which take the square of PE; the square root of the remainder, set off from f to n, will give the position of the line nPm, required.
Let DP=10, and PE 11 perches; and the area te