Question 1025182
Problem: A triangular lot has two sides of length 100m and 48m as indicated in the figure below.
 The length of the perpendicular from a corner of the lot to the 48m side is 96m.
 A fence is to be erected perpendicular to the 48m side so that the area of the lot is equally divided.
 How far from A along segment AB should this perpendicular fence be constructed?
 Give your answer in simplest radical form AND rounded to the nearest tenth of a meter.
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Continuing what you have done already. On the diagram label the point on AD where the perpendicular line is as E, and the top point on the perpendicular line as F. Forming a triangle AEF which we know has an area of half the total area which is 1152 sq/m.
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Find the area of ADC: {{{1/2}}}*28*96 = 1344 sq/meters
We are dealing two similar triangles; ADC and AEF, 
The relationship of the areas of these two similar triangle is the ratio of the square of the two sides
let x = the distance from A to E, we know distance for A to D is 28
{{{x^2/28^2}}} = {{{1152/1344}}}
Cross multiply
1344x^2 = 28^2 * 1152
1344x^2 = 784 * 1152
x^2 = 903168/1344
x^2 = 672
x = {{{sqrt(672)}}} ~ 25.9 m is AE
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let me know if this made sense, ankor@att.net