+0

# plz helpand explain these challenge problems thank you!!! you guys are awesome

0
216
6

1. Given that AF=4sqrt(3) and FC=5sqrt(3), what is BC?

2.  The diagram is not drawn to scale, but the measurements of the line segments and the right angles are correctly labeled. As shown in the diagram, BE=EC. Are the red and blue triangles similar? If so, enter the side of the blue triangle that corresponds to AB. If not, enter "no."

3.  The lengths of four sides of a trapezoid have ratios 1:1:1:2. The area of the trapezoid is 48sqrt(3). What is its perimeter?

4. We know Angle A=45 degrees, Angle B=15 degrees and BC=sqrt(6). What is AB?

5. In Triangle ABC, the circumcenter and orthocenter are collinear with vertex A. Which of the following statements must be true?

(1)  Triangle ABC must be an isosceles triangle.
(2)  Triangle ABC must be an equilateral triangle.
(3)  Triangle ABC must be a right triangle.
(4)  Triangle ABC must be an isosceles right triangle.

Enter your answer as a comma-separated list. If there is no correct option, write "none".

6.

Let H be the orthocenter of the equilateral triangle ABC. We know the distance between the orthocenters of Triangle AHC and Triangle BHC is 12. What is the distance between the circumcenters of Triangle AHC and Triangle BHC?

7.  As shown in the diagram, points B and D are on different sides of line AC. We know that Angle B=2*Angle D=60 degrees and that AC=4sqrt(3). What is the distance between the circumcenters of Triangle ABC and Triangle ADC?

Dec 11, 2018
edited by Guest  Dec 11, 2018

#1
+101778
+1

4. We know Angle A=45 degrees, Angle B=15 degrees and BC=sqrt(6). What is AB?

Angle C =  120°

Using the Law of Sines

BC / sin A   =  AB/ sin C

sqrt (6) /  (1/sqrt(2) ) =  AB / ( sqrt (3) / 2)

sqrt (12) = 2 AB / sqrt (3)

sqrt (3) * sqrt (12) = 2 AB

sqrt (36)  = 2 AB

6  =  2AB

AB =  3

Dec 11, 2018
#2
+101778
0

The first one is answered, here :

https://web2.0calc.com/questions/plz-help-and-explain-thoroughly-helpp

Dec 11, 2018
#3
+101778
+1

2.  The diagram is not drawn to scale, but the measurements of the line segments and the right angles are correctly labeled. As shown in the diagram, BE=EC. Are the red and blue triangles similar? If so, enter the side of the blue triangle that corresponds to AB. If not, enter "no."

It's obvious that, if the triangles are similar, then ΔABE is similar to ΔECD

Note that....if similar, the scale factor is  120/35 = 24/7

Since...if similar, then DC =  (7/24)BE = (7/24)CE

Let CE = x   ....so DC = (7/24)x

Then, by the Pythagorean Theorem,

sqrt ( x^2 + (7x/24)^2 ] = 35

x *sqrt [ 576 + 49] / 24  =  35

x * sqrt (625) = 24 * 35

x * 25 =  840

x = 840 /25 =  33.6  =  CE  = BE

So....by the Pythagorean Theorem......

sqrt (BE^2 + AB^2 ) =

sqrt (BE^2  + (24 CE/ 7)^2 ) =

sqrt ( 33.6^2 + [24 (33.6)/ 7]^2 )  =

sqrt ( 1128.96 + 13.271.04) =

sqrt ( 14400) =  120 = AE

So.....they are similar

AB  is similar to EC

Dec 11, 2018
#4
+22500
+10

7.

As shown in the diagram, points B and D are on different sides of line AC.

We know that Angle B=2*Angle D=60 degrees and that AC=4sqrt(3).

What is the distance between the circumcenters of Triangle ABC and Triangle ADC?

line segments:

$$\text{Let AC =4\sqrt{3} } \\ \text{Let AG=GC =2\sqrt{3} } \\ \text{Let AE=EC =r } \\ \text{Let AF=FC =R } \\ \text{Let FE = EG+GF  }$$

The distance between the circumcenters of Triangle ABC and Triangle ADC $$= FE$$

angle:

$$\text{Let \angle ABC = 60^{\circ}  } \\ \text{Let \angle ADC = \frac{\angle ABC}{2} = 30^{\circ}  } \\ \text{Let \angle AEC = 2*\angle ABC = 120^{\circ}  } \\ \text{Let \angle AFC = 2*\angle ADC = 60^{\circ}  }$$

$$\mathbf{EG = \ ?}$$

$$\begin{array}{|rcll|} \hline AC^2 &=& 2r^2\Big(1-\cos(120^{\circ})\Big) \quad \text{\cos-rule} \quad | \quad \cos(120)^{\circ} = -0.5 \\ (4\sqrt{3})^2 &=& 2r^2(1+0.5) \\ 48 &=& 2r^2(1.5) \\ 24 &=& r^2(1.5) \\ r^2 &=& 16 \\\\ AG^2 + EG^2 &=& r^2 \\ (2\sqrt{3})^2 + EG^2 &=& 16 \\ 12 + EG^2 &=& 16 \\ EG^2 &=& 4 \\ \mathbf{EG} & \mathbf{=}& \mathbf{2} \\ \hline \end{array}$$

$$\mathbf{GF = \ ?}$$

$$\begin{array}{|rcll|} \hline AC^2 &=& 2R^2\Big(1-\cos(60^{\circ})\Big) \quad \text{\cos-rule} \quad | \quad \cos(60)^{\circ} = 0.5 \\ (4\sqrt{3})^2 &=& 2R^2(1-0.5) \\ 48 &=& 2R^2(0.5) \\ 24 &=& R^2(0.5) \\ R^2 &=& 48 \\\\ AG^2 + GF^2 &=& R^2 \\ (2\sqrt{3})^2 + GF^2 &=& 48 \\ 12 + GF^2 &=& 48 \\ GF^2 &=& 36 \\ \mathbf{GF} & \mathbf{=}& \mathbf{6} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline FE &=& EG+GF \\ &=& 2+6 \\ \mathbf{FE} & \mathbf{=} & \mathbf{8} \\ \hline \end{array}$$

The distance between the circumcenters of Triangle ABC and Triangle ADC is 8

Dec 11, 2018
#5
+101778
+1

Well done, heureka!!!!

I always learn a lot from your geometric proofs   !!!

CPhill  Dec 11, 2018
#6
+22500
+9

Thank you, CPhill

heureka  Dec 12, 2018