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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,
9 and 16 Inches long, which is equal in Area to it, wants to know how this Board must be cut through into 2 Pieces, so as exactly to fit the said Hole of 12 Inches square.
Operation. First, 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 such Places as shall 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 fuppose its Breadth AC is 3 Inches, and its Length' AB 10 Inches.
'B Operation. First, divide the shorter 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 square Inches. Then from A to D draw the Diagonal Line A D, which will cut the Figure into two equal Triangles. Again, cut those Triangles into 2 Parts, through the Lines E F 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 first 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.
From Half the Sum of the three sides, subtract 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.
Suppose A B C be a Triangle, whose three Sides are as follow, viz. AB 20 Inches, AC 18 Inches, and BC 16 Inches, what is its Area?
Operation. The Sum of the three Sides is 54 ; the Half is 27. Then 27 - 20 = 7; 27 - 18 = 9; 27 16 = 11; and 27 x 7 = 189 X 9 = 1701 X 11 = 18711, whose Square Root is = 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 108 Hurdles fold ? *
Suppose the Parallelogram ABCD to represent 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.
Then it is very evident, that if to each of the Ends AC and B D, there be added i 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, consequently 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.
Suppose ABDE be a Circle whose Area is required to he divided into 5 equal Parts by the concentric Cireles G HI.
Operation. First, 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 Division at 1, 2, 3, 4, &c. raise Perpendiculars
they meet the Semi-circle in the Points a, b, , d, through which Points draw the concentric Circles E, F, G, H, I, and it is done.
Note. Suppose 5 Smiths should agree to purchase a Grinding Stone among them, each paying an equal Share of the Price, and that each' Man should have the Use of the Stone to wear off a fifth Part till it come to the last Man, who was to wear it out. The first Man should wear the Stone from E to F; the second 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.