+0

# In triangle ABC, AB = 10 and AC = 17. Let D be the foot of the perpendicular from A to BC.

+1
53
3

In triangle ABC, AB = 10 and AC = 17. Let D be the foot of the perpendicular from A to BC. If BD:CD = 3:5, then find AD.

Nov 5, 2020

#1
0

a2 + b2 = c2

a2 + b2 = 102

a2 + b2 = 172

Using the info that it is a 3:5 ratio, you can determine that the maximium numbers it can be is 6:10 without being 10.

All other 3:5 ratios under 10 not using decimals are 3:5, since that would give AD = 16.7 and 8.6 which don't match up, so it isn't a whole number.

Next I used a simultaneus equation: 289 - y = 100 - x    and    3y = 5x.

What I got was x = 567/2 or 283.5 and y = 945/2 or 472.5

Squaring them gave me the answer: x = 16.8 and y = 21.7 which still didn't make sense.

So then I thought about it and it's impossible.

Nov 6, 2020
#2
+2

The question as written leads to a complex value for AD!!!!  Hoever, if BD/CD = 2:5 then AD has a more sensible value (=8).

Nov 6, 2020
#3
+1

If the ratio of BD:CD = 2 : 5    then                  102 - (2x)2 = 172 - (5x)2

x = 3          side  BC = 21

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

In case of  3 : 5 ratio, we have:                    102 - (3x)2 = 172 - (5x)2

x = 3.43693

BC = 27.49544    Impossible!

jugoslav  Nov 6, 2020
edited by jugoslav  Nov 6, 2020
edited by jugoslav  Nov 6, 2020