Fig. 7/7 shows two similar constructions for enlarging 8 pentagon so that its new area is twice that of the original.
Select a point P. (This may be on a corner, or within the outline of the pentagon, or outside the outline although this is not shown because the construction is very large)
Let A be a corner of the given pentagon
Join PA and produce it.
Orawa semi-circle, centre P. radius PA.
From P. drop a perpendicular to PA to meet the semicircle in S
Mark off PR: PQ in the required ratio, in this case 2:1.
Bisect AR in O, and erect a semi-circle, radius OR to cut PS produced inT.
Join SA and draw TA' parallel to SA
Although Fig 7/7 shows a pentagon, the construction applies to any plane figure and can be used to increase and decrease a plane figure in a known ratio of areas. Fig 7/8 shows a figure reduced to 4/7 its original size. The construction is identical to that used for Fig. 7/6 except that the ratio PR: PQ is 4; 7. Note that if there is a circle or part circle in the outline, the position of its centre is plotted.
1. From B, tho apex of the triangle, drop a perpendicular to meet the base in F.
3. From A and C erect perpendiculars to meet the bisected line in D and E.
It ahould be obvious from the shading that the part of the triangle that is outside the rectangle is equal in area to that part of the rectangle that overleps the triangle.
To construct a square equal in area to a given rectangle. ABCD (Fig. 7/10)
1. With centre 0. radius OC. draw an arc to meet AO produced in E.
2. Bisect AE and erect a semi-circle, radius AF. centre F.
3. Produce DC to meet the semi-circle in G. DG is one side of the square. (For the construction of a square, given one of the sides, see Ch. 2.)
This construction can be adapted to find thesquere root of a number. Fig. 7/11 shows how to find ^/fT
Since the area of the rectangle equals that of the square, then tfa always - 1. then b-c' otjb-c
Thus, always draw the original rectangle with one side equal to one unit and convert the rectangle into a square of equal area.
To construct a square aqual In area to a given triangle (Fig. 7/12)
This construction is a combination of those described in Figs. 7/9 and 7/10. First change the uiangle into a rectangle of equivalent area and then change the rectangle Into a square of equivalent area.
To construct a triangle equal in aree to a given polygon ABCDE (Fig. 7/13)
1. Join CE and from 0 draw a line parallel to CE to meet AE produced in F.
Since DF is parallel to CE, triangles COE and CFE have the same base and vertical height and therefore the same area. The polygon ABCDE now has the same area as the quadrilateral ABCF and the original five-sided figure has been reduced to a four-sided figure of the same area.
3. Join CA and from B draw a line parallel to CA to meet EA produced in G.
Since BG is parallel to CA. triangles CBA and CGA have the same base and vertical height and therefore the same area. The quadrilateral ABCF now has the same area as the triangle GCF and the original five-eided figure has been reduced to a three-sided figure of the same area. GCF is the required triangle. q
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