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 fig

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Question 1025152: 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 Theo(13342) (Show Source):
You can put this solution on YOUR website!
you found AD = 28 and you found DB = 20.

find the area of triangle ADC and the area of triangle DBC.

the area of triangle ADC is equal to 1/2 * 28 * 96 = 1344.

the area of triangle DBC is equal to 1/2 * 20 * 96 = 960.

you already found the area of triangle ABC = 1/2 * 48 * 96 = 2304.

half of that area is equal to 1152.

since the area of triangle ADC is greater than that, you want to construct your perpendicular line to the left of CD.

that perpendicular line will form another triangle.

that triangle will be similar to triangle ADC.

since that triangle is similar to triangle ADC, then the corresponding sides will have to be proportional.

let the horizontal segment of that triangle = x and let the vertical segment of that triangle = y.

your ratio of corresponding sides being proportional will be:

x/y = 28/96.

the area of this new triangle will be equal to 1/2 * x * y = 1152.

solve for x in the equation of x/y = 28/96 and you get:

x = 28 * y / 96.

replace x in the equation of 1/2 * x * y = 1152 and you get:

1/2 * 28 * y / 96 * y = 1152.

combine like terms and simplify to get:

28 * y^2 / 192 = 1152.

solve for y to get y = sqrt(1152 * 192 / 28) = 88.87872958.

replace y with 88.87872958 in the equation of 1/2 * x * y = 1152 to get:

1/2 * x * 88.87872958 = 1152.

solve for x to get x = 1152 * 2 / 88.87872958 = 25.92296279.

the horizontal segment of yuur new triangle is equal to 25.92296279.

the vertical segment of your new triangle is equal to 88.87872958.

the area of your new triangle is 1/2 * 25.92296279 * 88.87872958 = 1152.

your vertical line splits the area of triangle ABC in half.

1152 is in the left half and 2304 - 1152 = 1152 is the area in the right half.

the following diagram shows the approximate location of your new vertical line.

the area of the triangle formed by the legs of x and y has an area equal to 1152 which is half the area of triangle ABC.