पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/४३७

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CHAPTER VII--MEASUREMENT OF AREAS.

An example in illustration thereof.

174${\tfrac {1}{2}}$. The base (of the given quadrilateral figure) is 14 ; each of the (two equal sides is 18; the top-side is 4; the Perpendicular is 12; and the optionally given number is 10. What is that other quadrilateral figure with two equal sides, the accurate measure (of the area) of which is the same as (the accurate measure of the area of (this given quadrilateral) ?

When an area with a given practically approximate measure is divided into any required number of parts, the rule for arriving at the numerical measure of the bases of those various parts of the quadrilateral figure with two equal sides, as also at The numerical measure of the sides as measured from the various division-points thereof, the numerical measure of the practically approximate value of the area of the quadrilateral figure with two equal sides being given:-

175${\tfrac {1}{2}}$. The difference between the squares of the (numerical) values of the base and the top-side (of the given quadrilateral figure with two equal sides) is divided by the total value of the (required) proportionate parts. By the quotient (so obtained),

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175${\tfrac {1}{2}}$. If ABCD be a quadrilateral with two equal sides, and if EF, GH and KL divide the quadrilateral so that the divided portion are in the proportion of m,n,p and q in respect of area, then according to the rule,

$EF={\sqrt {{\tfrac {d^{2}-b^{2}}{m+n+p+q}}\times m+b^{2}}}$;

$GH={\sqrt {{\tfrac {d^{2}-b^{2}}{m+n+p+q}}\times (m+n)+b^{2}}}$;

$KL={\sqrt {{\tfrac {d^{2}-b^{2}}{m+n+p+q}}\times (m+n+p)+b^{2}}}$;

and so on.

Similarly $AE={\tfrac {\left(a\times {\tfrac {d+b}{2}}\right)\times {\tfrac {m}{m+n+p+q}}}{\tfrac {EF+AD}{2}}}$; $EG={\tfrac {\left(a\times {\tfrac {d+b}{2}}\right)\times {\tfrac {n}{m+n+p+q}}}{\tfrac {GH+EF}{2}}}$;

$GK={\tfrac {\left(a\times {\tfrac {d+b}{2}}\right)\times {\tfrac {p}{m+n+p+q}}}{\tfrac {GH+EF}{2}}}$; and so on.

It can be easily shown that ${\tfrac {AB}{AE}}={\tfrac {BC-AD}{EF-AD}}$