2. Given the side AB, Fig. 15, of a parallelogram, equal 20 ch. and the angle A 63° 30'; required the side AC, that the content may be 21; acres. SAB 20 ch.

Ar. Co. 8.69897 As ABxsin. A sin. A 63° 30'

0.04821 : the given area 215 sq. ch.

2.33244 10.00000

:: rad.

: AC 12.01 ch.

1.07962

3. Given one side of a triangle, equal 30 perches, an angle adjacent to this side 71° 15', and the area 2 acres; required the other side adjacent to the given angle. Ans. 22.53 perches.

4. Given one side of a parallelogram, equal to 32.26 ch., an angle adjacent to this side 83° 30', and the area 74 acres; required the other side adjacent to the given angle. Ans. 23.09 Ch.

PROBLEM VI.

The area and base of a triangle being given, to cut off a

given part of the area by a line running from the angle opposite the base.

RULE.

As the given area of the triangle,
Is to the area of the part to be cut off;
So is the given base,
To the base corresponding to that area*.

1

EXAMPLES.

1, Given the area of the triangle ABC, Fig. 88, equal 650 square perches and the length of the base AB 40 perches; it is required to cut off 290 perches towards the angle A, by a line running from the angle C to the base.

ABC. ADC. AB. AD.

As 650 : 290 :: 40 : 17.85 per. 2. In a triangle ABC, there are given the area 27 A. 1 R. 16 P. and the base AB 35.20 ch., to cut off 10 acres towards the angle B, by a line CD running from the angle C to the base : the part BD of the base is required. Ans. 12.87 Ch.

PROBLEM VII.

The area and two sides of a triangle being given, to cut

off a triangle containing a given area, by a line running from a given point in one of the given sides and falling on the other.

RULE.

As the given area of the triangle,
Is to the area of the part to be cut off ;
So is the rectangle of the given sides,
To a fourth term.

Divide this fourth term by the distance of the given point from the angular point of the two given sides; the quotient will be the distance of the required point from the same angle*.

* DEMONSTRATION. From the demonstration to prob. 3, sect. 1, Measuring Land, we have, Fig. 89, rad. : sin. A:: ABXAC : 2ABC and rad. : sin. A :: APXAG: 2APG; therefore(11 & 16.5) 2 ABC: 2 APG :: ABXAC: APXAG, or (15.5) ABC : APG :: AB X AC: APXAG; hence the truth of the rule is manifest.

EXAMPLES.

a

1. Given the area of the triangle ABC, Fig. 89, 5 acres, the side AB 50 perches, the side AC 40 perches, and the distance of a point P from the angle A, 36 perches, it is required to find a point G to which, if a line be drawn from the point P, it shall cut off a triangle APG containing 3 A. O R. 20 P. As the triangle ABC 800 sq. p.

Ar. Co. 7.09691 :. the triangle APG 500

2.69897

1.69897 :: ABXAC,

ŞAB 50
LAC 40

1.60206

: APXAG

AP 36

3.09691 1.55630

log.

AG 34.72 per.

1.54061

2. Given the area of a triangle ABC. 12 A. 1 R. 23 P. the side AB 20 ch., the side AC 16.25 ch., and the distance of a point P in the side AB, from the angle A 8.50 ch.; it is required to find the distance AG of a point G

. in the line AC, so that a line drawn from P to G may cut off a triangle APG containing 3 acres. Ans. 9.25 ch.

PROBLEM VIII.

The area and base of a triangle being given, to cut off a triangle containing a given area by a line running pa

a rallel to one of the sides.

RULE.

As the given area of the triangle,
Is to the area of the triangle to be cut off;
So is the square of the given base,

The square root of which will be the base of the required triangle*.

EXAMPLES.

1. Given the area of the triangle ABC, Fig. 90, 500 square perches, and the base AB .40 perches; it is required to cut off 120 sq. per. towards the angle A, by a line DG running parallel to the side BC.

2. Given the area of a triangle ABC, 10 acres, and the base AB 25 ch., to find BD a part of the base, so that a line DG running from the point D, parallel to the side AC, may cut off a triangle BDG containing 4į acres. Ans. BD=16.77 cht.

a

:

The truth of this rule is manifest from 19.6. This problem may be neatly constructed as follows : Let ABC, Fig. 90, be the given triangle and AB the given base; on A B describe the semicircle AEB, and take AF to AB in the ratio of the part to be cut off, to the whole triangle ; draw FE perpendicular to AB, meeting the semicircle in E, join AE, and make AD equal to AE ; from D draw DG parallel to BC and the thing is done. For, join EB and we have, by similar triangles, AB: AE :: AE: AF; therefore (20.6. cor. 2.) AB : AF :: AB*:AE? (AD) :: [19.6.] ABC: ADG.

If it be required to produce two sides of a given triangle so far that the triangle formed by these sides produced, and a line drawn between them pa, rallel to the third side, may contain a given area, it may be done by the above rule. Thus, Fig. 90, ADG : ABC :: ADS : ABP.

PROBLEM IX.

The bearings of two adjacent sides, AD, AE, Fig. 91, of

a tract of land being given, to cut off a triangle ABC containing a given area by a line BC running a given

course.

RULE.

From the given bearings of the lines, find the angles A, B, and C; then,

As the rectangle of the sines of the angles A and B,
Is to the rectangle of radius and sine of the angle C;
So is twice the given area,
To the square of the side AB*. .

In like manner the other sides may be found ; or having found one side, the others may be found by trig.case 1.

EXAMPLES.

1. Let the bearing of AD, Fig. 91, be N. 87° 30' E., and of AE, N. 27° 30' E.; it is required to cut off 10 acres by a line BC running N 380 W.

* The truth of this rule is evident from the demonstration to prob. 4, sect. 1, Measuring Land.

Construction. Draw AD, AE, Fig. 92, according to the given bearings, and in AD take AF equal the square root of the given area, and on it describe the square

AFGH; make IE=AI, and draw ED of the same bearing as the division line BC, meeting AD in D; on AD describe a semicircle, and produce GF to meet it in K, join AK and make AB equal to it; draw BC parallel to DE, and ABC will be the triangle required. For join IF, EF and KD, then (31.3. and cor. 8.6) AD : AK (AB) :: AK (AB): AF; or (cor. 19.6.) AD: AF :: ADE : ABC, but (1.6.) AD : AF :: ADE : AFE ; therefore (11.5.) ADE : ABC :: ADE : AFE, and consequently (9.5.) ABC=AFE; but because Al=IE, AFE=2.AFI=(41.1.) AFGH ; therefore ABC=AF