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Problem 5.

To reduce a Parallelogram to a Square equivalent in Area to it.

A Man having a Pannel of his Cupboard 12 Inches fquare broken out, and having by him a Board 9 Inches broad, and 16 Inches long, which is equal in Area to it, wants to know how this Board must be cut through into 2 Pieces, fo as exactly to fit the faid Hole of 12 Inches fquare.

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Operation. Firft, divide the Breadth A C into 3 equal Parts, and the Length into 4 equal Parts, and draw Lines across as in the Figure. Then, if the Board be cut through the Lines a, b, c, d, &c. and the Point B brought under the Place marked 12, the other Parts of it will fall into fuch Places as fhall form one complete Square of 12 Inches exactly.

Problem 6.

To increase the Surface of a Geometrical Parallelogram. Let ABCD be the Parallelogram to be increased; and fuppofe its Breadth AC is 3 Inches, and its Length A B 10 Inches.

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Operation. Firft, divide the fhorter Side AC into 3 equal Parts, and the longer Side A B into ten equal Parts, and draw the Lines across as is done in the Figure, so will the whole Surface be divided into 30 equal Parts, or 30 fquare Inches. Then from A to D draw the Diagonal Line A D, which will cut the Figure into two equal Triangles. Again, cut thofe Triangles into 2 Parts, through the Lines EF and GH, i. e. through the 4th Line from the bigger End, by which Means there will be produced 2 Triangles and 2 Trapeziums, which joined together as in the following Figures, will make 32 Squares, instead of the 30 drawn in the first Figure, so that here is apparently an Increase of 2 Square Inches more.

Note. If you draw in each Square of the firft Figure the reverse Side of a Guinea or Shilling, there will be only 30 Pieces, but immediately joining them, as in the ad and 3d Figures, you will have 32; that is, 2 Pieces more than before.-Good Gain.

Problem 7.

To find the Area of an oblique plain Triangle without falling a Perpendicular.

Rule.

From Half the Sum of the three Sides, fubtract each particular Side; then multiply the half Sum and the three Differences continually; the Square Root of the last Product will be the Area of the Triangle.

Example.

Suppofe A B C be a Triangle, whofe three Sides are as follow, viz. AB 20 Inches, AC 18 Inches, and BC 16 Inches, what is its Area?

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Operation. The Sum of the three Sides is 54; the

Half is 27. Then 27

1611; and 27 x 7 whofe Square Root is

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207; 27 18 = 9; 27 189 × 9 = 1701 × 11 = 18711, 136.78 Inches, the Area required.

By this Problem the Content of a Piece of Ground may be found without the Surveyor's going into it.

Problem 8.

If 100 Hurdles will fold 100 Sheep, how many will 10 Hurdles fold? *

Suppofe the Parallelogram ABCD to reprefent the Situation of the Hurdles when they fold 100 Sheep, where the Sides A B and CD are 49 Hurdles each, and the Ends AC and BD only Hurdle each.

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Then it is very evident, that if to each of the Ends A C and BD, there be added 1 Hurdle more, the Parallelogramic Space will be just as large again. Whence it follows, that if the former Space ABCD will fold 100 Sheep, the additional Parallelogram CDEF will fold 100 more, confequently the Whole, with the Addition of two Hurdles, will fold 200 Sheep.

This is generally called the Shepherd's Problem.

Problem 9.

To divide the Area of a Circle into any Number of equal Parts by concentric Circles.

Suppofe ABDE be à Circle whofe Area is required to e divided into 5 equal Parts by the concentric Circles GHI

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Operation. Firft, divide the Semi-diameter A e into 5 equal Parts, as A 1, 2, 3, 4, 5, and on the middle Point at e, as a Center, with the Radius or Opening e A, describe the Semi-circle A, a, b, c, d, e. Then, from the Points of equal Divifion at 1, 2, 3, 4, &c. raife Perpendiculars they meet the Semi-circle in the Points a, b, c, d, through which Points draw the concentric Circles E, F, G, H, and it is done.

Note. Suppofe 5 Smiths fhould agree to purchase a Grinding Stone among them, each paying an equal Share of the Price, and that each Man fhould have the Use of the Stone to wear off a fifth Part till it come to the laft Man, who was to wear it out. The first Man fhould wear the Stone from E to F; the fecond from F to G; the third from G to H; the fourth from H to I; and the fifth from I to the Center or Axis at e.

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