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OBAPTER VII-XEASUREMENT OF ARBAS.
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Bubtracted. Then, on carrying out the process of sankramina with the square root (of this resulting differenco) in relation to half the measure of the perimeter, the values of the required) base and the perpendioular-side are indeed obtained.
An example in illustration thereof.
In a derived longish quadrilateral figure, the measure of the perimeter is 170; the measure of the given area is 1,500. Tell me the values of the perpendicular-side and the base (thereof).
The rule for arriving at the respective pairs of (required) longish quadrilateral figures, (1) when tho numerioal measures of the perimeter are equal, and the area of the first figure is double that of the socond ; or, (2) when the areas of both the figures aro equal, and the numerical measure of the perimeter of the second figure is twice the numerical measuro of that of the first figure ; or, (3) (again) when, in relation to the two required figures, the numerical measure of the perimeter of the second figure is twice the numerical measure of the perimeter of the first figuro, and tho area of the first figure is twice the area of the second tiguro :
1317--133. (The larger numbers in the givoa ratios of) the perimeters as also (of) the areas (relating to the two required longish quadrilateral figures,) are divided by the smaller (numbero) corresponding to them. The resulting quotients) are multiplied (between themselves) and (then) squared. This same quantity.)
1814 to 1.33. If and y roprenent the two adjaount sides of the Arst rectangle, and a and & the two adjacent video of the second rootangla, the conditions mentioned in the three kinds of problems proposed to be solved by this role may be represented thus :
(1) +y=a+:
xy = 2ab. (2) 2(2+ y) = a +b:
by = ab. (8) 2 (8+ y) =a +b:
zy = 2ab. The solation given in the role seems to be correot only for the particular 01909 given in the problems in stan 186 to 186.
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