+0  
 
0
138
1
avatar

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
0

Guys dont do this proble but do the other one one question beneath this one!!!

From the Creator of this Problem

Guest Oct 28, 2017

8 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.