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1.Circle O is tangent to AB at A, and angle ABD = 90. If AB= 12 and CD= 18, find the radius of the circle

2. In cyclic quadrilaterla ABCD, AB=2, BC=3, CD=10, and DA=6. Let P be the intersection of lines AB and CD. Find the length BP.

 Mar 18, 2019
 #1
avatar+105347 
0

First one

 

From O.....draw a perpendicular bisector to chord DC...to meet DC at E

 

OE =  BA = 12

And DE  =  9

 

And....using the Pythagorean Theorem.....radius  OD =

 

sqrt ( OE^2 +  DE^2 )  =   sqrt (12^2 + 9^2 )  =  sqrt ( 144 + 81)  =  sqrt (225) =  15

 

 

cool cool cool 

 Mar 18, 2019
 #2
avatar+105347 
0

Unsure about this...but....here's my best attempt

 

Angle ABC  =  (1/2)(16/21)(360)  =  960/7°

Angle PBC = [ 180 - 960/7]° = [300/7]°

Angle DCB =  (1/2)(8/21)(360)  = 480/7°

Angle PCB = [ 180 - 480/7] = [780/7]°

Angle BPC =  180 - [ 300/7] - [ 780/7]  = [180/7]°

 

 

3/sin (BPC)  = PB / sin (PCB)

 

PB  =   3*sin(780/7) / sin(180/7)  ≈  6.44

 

 

cool  cool  cool

 Mar 18, 2019
 #3
avatar+23526 
+2

2.

In cyclic quadrilaterla ABCD, AB=2, BC=3, CD=10, and DA=6.

Let P be the intersection of lines AB and CD.

Find the length BP.

\(\text{Let $BP =x$} \\ \text{Let $PA =2+x$} \\ \text{Let $PC =y$} \\ \text{Let $PD =10+y$} \)


\(\text{Let $\angle BPC = P $} \\ \text{Let $\angle CBP = B $} \\ \text{Let $\angle ABC = 180^\circ -B $} \\ \text{Let $\angle ADC = 180^\circ - \angle ABC =180^\circ-(180^\circ -B)=B $} \)

 

1.

Intersecting secants theorem:

see: https://en.wikipedia.org/wiki/Intersecting_secants_theorem

\(\begin{array}{|rcll|} \hline \mathbf{BP\cdot PA} &\mathbf{=}& \mathbf{PC\cdot PD} \quad | \quad \text{Intersecting secants theorem} \\\\ x\cdot (2+x) &=& y\cdot (10+y) \\ \hline \end{array} \)

 

2.
sin-theorem:

\(\begin{array}{|lrcll|} \hline 1. & \dfrac{\sin(P)}{3} &=& \dfrac{\sin(B)}{y} \\ & \dfrac{\sin(P)}{\sin(B)} &=& \dfrac{3}{y} \\\\ 2. & \dfrac{\sin(P)}{6} &=& \dfrac{\sin(B)}{2+x} \\ & \dfrac{\sin(P)}{\sin(B)} &=& \dfrac{6}{2+x} \\ \hline & \dfrac{\sin(P)}{\sin(B)} = \dfrac{3}{y} &=& \dfrac{6}{2+x} \\\\ & \dfrac{3}{y} &=& \dfrac{6}{2+x} \\\\ & \dfrac{y}{3} &=& \dfrac{2+x}{6} \\\\ & y &=& \dfrac{3}{6}\cdot(2+x) \\\\ & \mathbf{y} & \mathbf{=}& \mathbf{\dfrac{1}{2}\cdot(2+x)} \\\\ & x\cdot (2+x) &=& y\cdot (10+y) \quad | \quad y = \dfrac{1}{2}\cdot(2+x) \\\\ & x\cdot (2+x) &=& \dfrac{1}{2}\cdot(2+x)\cdot \left(10+\dfrac{1}{2}\cdot(2+x)\right) \\\\ & x &=& \dfrac{1}{2} \cdot \left(10+ \dfrac{1}{2}\cdot(2+x) \right) \quad | \quad \cdot 2 \\\\ & 2x &=& 10+ \dfrac{1}{2}\cdot(2+x) \quad | \quad \cdot 2 \\\\ & 4x &=& 20+ 2+x \\\\ & 3x &=& 22 \\\\ & x &=& \dfrac{22}{3} \\\\ & \mathbf{ x } & \mathbf{=} & \mathbf{7.\bar{3}} \\ \hline \end{array}\)

 

laugh

 Mar 19, 2019

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