Mount Auburn Cemetery

210

Jany. 23 [186]5

Capt. Winsor,

Dear Sir,

There is an error in casting the contents of lot no. 3499 on Dianthus Path, that of Norris Fines, which I think you will, percieve on a moment's reflection, [Image: diagram with measurments] You have given the contents as 123 feet, which figures can be obtained only be adding the opposite sides, and multiplying the halves together, ie, 14 x 9 & one-half = 133. But in a figure of [duel?] unequal sides, this makes a considerable error, as you see if you divide the lot into a rectangle and triangle, which gives you but 123 & one-half feet, or 9 & one-half less than you returned.-

This figure is a trapezoid, i.e, one with four sides only two of which are parallel. The rule for finding the area of a trapezoid is:-

"Multiply half the sum of its parallel sides by the distance between them:" - for it is equal to two triangles whose bases are the two parallel sides of the trapezoid, and whose altitude is the distance between them. To prove it in this case, - the lot [Image: Trapezoid diagram with figures.] A.B.C.D. is divided into the two triangles A.B.C and A.C.D - The base of A.B.C is A.B., it's altitude is B.C, The base of A.C.D. is A.D its altitude E.D.- A.B.C = 13 x 3 = 39. A.C.D = 13 x 6 & one-half = 84 & one-half total 123 & one-half Or take A.B.C.E as a rectangle and the area is 13 x 6 = 78. Add triangle C.E.D = 13 x 3 & one-half = 45 & one-half - total 123 and one-half ft. But the bases of the triangles must always be the parallel sides of the traphezoid.

What will you do? report say 124f and let the Treas^r refund $9.?

Yours Respectfully

A. J. Coolidge

The above rule is an excellent one. In this case where the parallel sides are A.D. and B.C "half the distance betwen them" ( meaning of course the shortest distance) is either A.B or the dotted line E.C-