1)
In
, we know that
and
Find 
![[asy] pair A,B,C;B = (0,0.5);C = (sqrt(3)/2,0);A = -C;draw(B--C--A--B);label(](/api/ssl-img-proxy?src=https%3A%2F%2Flatex.artofproblemsolving.com%2Fb%2F6%2Ff%2Fb6fd493d4ea8d4b229da69ff752e6e80b22cff59.png)
2)
In quadrilateral
and
We know that
and
What is 
![[asy] size(5cm);pair A,B,C,D;A=(0,0);B=(sqrt(13),0);D=(0,sqrt(5));C=(2.8,2.9);draw(A--B--C--D--A);draw(rightanglemark(B,A,D)^^rightanglemark(D,C,B));label(](/api/ssl-img-proxy?src=https%3A%2F%2Flatex.artofproblemsolving.com%2F5%2F8%2Fe%2F58ecefdf14e263ad277554ed38326ef7f09f1abf.png)
3)
Two diagonals of quadrilateral
are perpendicular to each other at
We know
and
What is 
![[asy] size(5cm);pair A,B,C,D,O;A=(-1,0);B=(0,1);C=(3,0);D=(0,-3);draw(A--B--C--D--A--C^^B--D);draw(rightanglemark(C,O,B));label(](/api/ssl-img-proxy?src=https%3A%2F%2Flatex.artofproblemsolving.com%2Fb%2Ff%2F7%2Fbf72367d371fab3a64f6b933905f8a304edefc1c.png)
4)
In
, we have
,
, and
. Point
is on
such that
bisects
. If
, then what is 
![[asy] pair P,Q,R,X;R = (0,0);Q=(0,0.5);P = (sqrt(3)/2,0);X = (0.5/sqrt(3),0);draw(X--Q--P--R--Q);label(](/api/ssl-img-proxy?src=https%3A%2F%2Flatex.artofproblemsolving.com%2Fc%2F0%2F4%2Fc044c956236d7dc6a44ab85727c6f4e2734df37b.png)
5)
Points
and
are on side
of rectangle
such that
and
trisect
. If
and
, then what is the area of
?
![[asy] pair A,B,C,D,SS,T;C = (0,0);T = (0.3,0);SS = (1.5,0);D = (2.1,0);B = (0,0.6*sqrt(3));A = B+D;draw(T--A--SS--C--B--A--D--SS);label(](/api/ssl-img-proxy?src=https%3A%2F%2Flatex.artofproblemsolving.com%2Fc%2Fc%2Fe%2Fcce3c90dcfa09daf5610e4efbe8d333bcd0dafb5.png)
1) Draw altitude BD.....this will bisect angle ABC
Then angle BCA = 30°
And since triangle BDC is a 30 - 60- 90 triangle...then CD = [√3 / 2] [ 6 √3] = 9
And CD = (1/2) AC
So AC = 2CD = 18
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3) AO = BO and CO = DO
Then the angles opposite these sides = 45°
And since triangles ABO and DCO are 45 - 45 - 90 triangles
Then DO = DC/√2 and BO = AB / √2
So DO + BO = DC/√2 + AB/√2 = [ DC + AB ] / √2
And DO + BO = BD = [ DC + AB ] / √2 = [ 5 √2 ] / √2 = 5
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4) Since QX bisects PQR, then angles PQX = RQX = QPX = 30°
Then angle QXP = 120°...so angle QXR = 60°
And triangles QRP and QRX are 30 - 60 - 90 triangles
So.........since QR is opposite a 30° angle, it is (1/2)PQ = (1/2) 4√3 = 2√3
So.......QX = [ 2 / √3 ] * [ 2√3 ] = 4
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5) Since BAD is trisected, angle SAD = 30°
And triangle SAD is a 30 - 60 - 90 triangle
So.......AD = √3SD = √3 * 1 = √3
And angle TAD = 60° so triangle ATD is also a 30 - 60 - 90 triangle
And TD = √3 AD = √3 √3 = 3
So CD = CT + TD = 2√3 - 3 + 3 = 2√3
So......the area of ABCD = CD * AD = 2√3 * √3 = 6 units^2
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