The new perpendicualar must be parallel to CD
Let MN be that perpendicular with M on AC and N on AC
I agree that AD = 28, so we can set up a ratio:
MN/AM = 96/28 = 24/7
MN = 24AM/7
Area of triangle AMN = (1/2)(2304) = 1152
= (1/2)(AM)(MN)
(1/2)(AM)(24AM/7) = 1152
AM^2 = 672
am=sqrt(672)
= appr 25.9 m
check:
AM=25.9
MN = 24(25.9)/7 = 88.8
Area of triangle AMN
=(1/2)(25.9)(88.8)
= 1149.96
which is close enough to 1152, allowing for round-off
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): gyazo(dot)com(forward slash)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
1 answer
2 answers