Home > Shapes
Side Length Settlement 2 (Posted on 2015-04-11 )

Points S and T are inside an equilatral triangle PQR such that:
ST = 1, SP = TP =√7, and:
SQ = TR = 2.
Find the length of PQ.

Solution
Comment 1 of 1

Let
PN be an altitude of triangle PQR, crossing ST at M.
Using PN as an axis of symmetry, / PSM = 90 deg. and
Pythagoras gives:
PM = sqrt(7 – ¼) = 3*sqrt(3)/2
and PN = a*sqrt(3)/2 where a = |PQ|
In triangle TRA (where TMNA is a rectangle):
TA^{2} + AR^{2}
= TR^{2}
(a*Sqrt(3)/2 – 3*sqrt(3)/2)^{2} + (a/2 -1/2)^{2} = 2^{2}
3(a – 3)^{2} + (a – 1)^{2}
= 16
a^{2} – 5a + 3 = 0
|PQ| = a = (1/2)(5 + sqrt(13)) = 4.302…
Posted by Harry
on 2015-04-13 18:36:05

Please log in:
Forums (0)
Newest Problems
Random Problem
FAQ |
About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox: