SOLUTION: First off, sorry if I put this under the wrong section, wasn't 100% sure where it belongs. Problem: A triangular lot has two sides of length 100m and 48m as indicated in the fi

Question 1025182: First off, sorry if I put this under the wrong section, wasn't 100% sure where it belongs.
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.
Image (figure that is given): https://gyazo.com/95d78114652ee0c58a6a2214bcbb4dc3
Help on this would be VERY much appreciated. I'm quite confused on the wording itself, all I've found out so far is that the area of the triangle is 2304m and that segment AD is 28m using pythag. thm. From this I also know that segment DB is 20m and that segment CB must be square root of 9616. This is where I get stuck and don't know how to advance ;l
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
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.
:
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.
:
Find the area of ADC: *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
=
Cross multiply
1344x^2 = 28^2 * 1152
1344x^2 = 784 * 1152
x^2 = 903168/1344
x^2 = 672
x = ~ 25.9 m is AE
:
:
let me know if this made sense, ankor@att.net

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