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1)     in the diagram below, we have $\overline{QR}\parallel\overline{ST}$$PQ = 8$$QS = 6$, and $PT = 10.5$. Find $PR$.

[asy] size(5cm);pair P,Q,R,SS,T;P = (0,0);Q = (0.5,0.6);R = (0.6,0);SS = 1.75*Q;T = 1.75*R;draw (Q--R--T--SS--P--R);label(

 

2)    In the diagram below, we have $\overline{MN}\parallel\overline{PQ}$. We also know that $OM = 9$$OQ = 6$, and $PN = 25$. Find $OP$

[asy] size(5cm);pair O,M,NN,P,Q;M = (-0.3,0.5);NN = (0.6,0.5);P = -0.7*NN;Q = -0.7*M;label(

 

 

 

 

3)       In the diagram, angles $ABC$ and $ADE$ are right angles. If $AC = 35$$AE = 11$, and $BE = 10$, then what is $AD$?

[asy] pair A,B,C,D,EE;B = (0,0);A = (1,0);C = (0,0.7);EE = (0.4,0);D = foot(EE,A,C);draw(D--EE--A--C--B--E);draw(rightanglemark(A,B,C,1.5));draw(rightanglemark(EE,D,A,1.5));label(

 

 

4)        $E$ is a point on side $\overline{CD}$ of square $ABCD$ such that $DE=2$ and $EC=5.$ The extensions of $\overline {AD}$ and $\overline {BE}$ intersect at $F.$ Find $DF.$

[asy] size(5cm);pair A,B,C,D,EE,F;A=(0,0);B=(6,0);D=(0,6);C=B+D;EE=(C+2D)/3;F=3D/2;draw(B--C--D--F--B--A--D);label(

 

 

 

5)          In the diagram below, we have $\triangle MNO\sim\triangle NRO\sim\triangle QPO.$ What is $\angle MNR$ in degrees?

[asy] pair O,M,NN,P,Q,R;M = (-0.4,0.5);R=0.75M;NN = (0.3,0.5);P = -0.9*NN;Q = -0.9*M;label(

 

 

 

6)       Segments $\overline{AC}$ and $\overline{BD}$ intersect at $O$. We know that $\dfrac{OA}{OB}=\dfrac{OD}{OC}$. Which angles below are equal to $\angle CDB$?

(w) $\angle ADB$
(x) $\angle CAB$
(y) $\angle BCA$
(z) $\angle ABD+\angle ACD$
[asy] size(5cm);pair A,B,C,D,O;A = 2*dir(15);B = 4*dir(140);C = -3A;D = -0.75B;draw(A--B--C--D--A--C^^B--D);label(

Guest Oct 28, 2017
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2+0 Answers

 #1
avatar+5229 
+2

1)     Triangle PQR  is similar to triangle PST , so

 

PQ / PS  =  PR / PT          And  PQ  =  8 ,  PS  =  8 + 6 ,  and  PT  =  10.5 .

 

8 / [8 + 6]  =  PR / 10.5     Multiply both sides of this equation by  10.5 .

 

10.5 * 8 / [8 + 6]  =  PR     Simplify.

 

6  =  PR

 

 

2)     Notice that we have two parallel lines cut by a transversal, and their alternate angles are congruent. So we can be sure that  triangle OMN  is similar to triangle OQP  , which means

 

OQ / OM  =  OP / ON      And  OQ  =  6 ,  OM  =  9 , and  ON  =  25 - OP

 

6 / 9  =  OP / [25 - OP]    Solve for  OP .

 

10  =  OP

 

 

3)     Angle CAB  is a part of both triangles, and both triangles have a right angle.

 

So  triangle ABC  is similar to  triangle ADE  by angle-angle similarity.

 

AB / AC  =  AD / AE      And  AB  =  11 + 10  , AC  =  35 ,  and AE  =  11

 

[11 + 10] / 35  =  AD / 11

 

11 * 21/35  =  AD

 

 

4)     Notice that angle DEF and angle CEB are vertical angles, angle BFD and angle FBC are alternate, and the other two are also alternate. So  triangle DEF  is similar to  triangle CEB .

 

DF / DE  =  CB / CE     Since  CD and CB are sides of the same square, CD = CB = 2 + 5

 

DF / 2  =  [2 + 5] / 5      I'll let you finish solving this for  DF .

 

 

5)     Triangle MNO  is similar to  triangle QPO , so  ∠MNO  =  ∠QPO  =  60°

 

There are  180°  in every triangle. ∠RNO  =  ∠PQO  =  180° - 60° - 66°  =  54°

 

∠MNR  =  ∠MNO - ∠RNO  =  60° - 54°  =  6°

hectictar  Oct 28, 2017
 #2
avatar+78590 
+2

Here's the last one

 

Since    OA / OB  = OD / OC

 

We must have that triangle  OAB   is similar to triangle ODC

 

And  angle  CDB  = angle CDO  = angle BAO = angle BAC = angle CAB

 

 

cool cool cool

CPhill  Oct 29, 2017

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