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1)     

In $\triangle ABC$, we know that $AB = BC = 6\sqrt 3$ and $\angle ABC = 120^\circ.$ Find $AC.$
[asy] pair A,B,C;B = (0,0.5);C = (sqrt(3)/2,0);A = -C;draw(B--C--A--B);label(

 

 

2)

 

In quadrilateral $ABCD,$ $\angle A = \angle C =90^\circ,$ and $BC=CD.$ We know that $AB=\sqrt{13}$ and $AD=\sqrt{5}.$ What is $BC?$
[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(

 

 

 

3)

 

Two diagonals of quadrilateral $ABCD$ are perpendicular to each other at $O.$ We know $AO=BO,CO=DO,$ and $AB+CD=5\sqrt{2}.$ What is $BD?$
[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(

 

 

 

4)

 

In $\triangle PQR$, we have $\angle P = 30^\circ$$\angle RQP = 60^\circ$, and $\angle R=90^\circ$. Point $X$ is on $\overline{PR}$ such that $\overline{QX}$ bisects $\angle PQR$. If $PQ = 4\sqrt 3$, then what is $QX?$

[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(

 

 

5)

Points $S$ and $T$ are on side $\overline{CD}$ of rectangle $ABCD$ such that $\overline{AS}$ and $\overline{AT}$ trisect $\angle DAB$. If $CT = 2\sqrt{3}-3$ and $DS = 1$, then what is the area of $ABCD$?

[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(

Guest Oct 28, 2017
 #1
avatar+87635 
+2

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

 

 

cool cool cool

CPhill  Oct 28, 2017
 #2
avatar+87635 
+2

2)  Draw DB

Then DB   is a hypotenuse of triangles DBA, DBC

And its length  is  √  [  (√13)^2  + (√5)^2 ]  =  √18  = 3√2

 

Then because BC  = CD  then angle CDB  =   angle CBD  = 45°

 

So because  triangle  DBC  is a 45 - 45 - 90 triangle, BC  =   [ 3√2  ] / √2  =  3

 

 

cool cool cool

CPhill  Oct 28, 2017
edited by CPhill  Oct 28, 2017
 #3
avatar+87635 
+1

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

 

 

 

cool cool cool

CPhill  Oct 28, 2017
 #4
avatar+87635 
+1

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

 

 

cool cool cool

CPhill  Oct 28, 2017
 #5
avatar+87635 
+1

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

 

 

cool cool cool

CPhill  Oct 28, 2017

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