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https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_19

Let ABCD be a trapezoid with bases BC and AD. In this trapezoid, the angle DBA is 105 degrees, with angle BAD is equal to 43 degrees. Finally, angle BCD is 141 degrees so what is angle CBD?

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In trapezoid ABCD, BC || AD, angle ABD =105°, angle A=43°, and angle C=141°. Find angle CBD, in degrees.

ok, the first thing I did is tried to work out what the trapezium looked like. 

THEN the answer became obvious.

BUT

it also became ovvious that I did not need to know what the trapezium looked like in ordet to answer this question.

Draw ANY trapezium where BC||AD

Label 

Label

37 is to 37 sqrt3    as   37 sqrt3 is to x

37 / 37 sqrt3     =  37 sqrt3 / x     solve for 'x'

Use cosine rule

a^2 + b^2 - 2ab cos = c^2

3² + 7²  - 2 (3)(7) cos A = 5^2    solve for cos A 

The arc cos     

Let line AC be perpendicular to line CE. Connect A to the mid-point of CE which is D, and connect E to the mid-point of AC which is B.  If AD and EB intersect at point F, BC=CD=15cm, then find the area of △DFE.

Hello Guest!

\(\overline{BE}=\sqrt{(15^2+30^2)cm^2}=\sqrt{1125cm^2}=33.541cm\\ \overline{EF}=\frac{1}{2}\cdot \overline{BE}\)

\(\large \frac{h_{△DFE}}{15cm}=\frac{\frac{1}{2}\cdot \overline{BE} }{\overline{BE}}=\frac{1}{2}\)

\(h_{△DFE}=7.5cm\)

\(A_{△DFE}=\frac{1}{2}\cdot \overline{ED}\cdot h_{△DFE}=\frac{1}{2}\cdot 15cm\cdot 7.5cm\)

\(A_{△DFE}=56.25cm^2\)

  !

So the diameter is 6 and the radius is 3.

3^2pi=9pi

Hope this helped!

yeah...? do you have an answer by any chance?

Im guessing AOPS?

If anyone could help I would really appreciate it..

Let's finish labelling the vertices of the triangle:  let C be the vertex at angle(x) and let D be the upper vertex.

Arc(BC) = 160^o because it is double the angle.

Arc(CD) = arc(BC) because their chords are congruent. Therefore, arc(CD) = 160^o.

This means arc(BD) = 360^o - 160^o - 160^o = 40^o.

Angle(x) is one-half the arc it intercepts, so it is 20^o.

2x + 6y = 7    divide this equation by 2

x + 3y = 3.5     now you can see    A = 1    B = 3.5

No problem!

If you don't understand how I found the ranges, just ask me :)

Yay! That is correct! Thank you so much!

Oh! Okay I see. So, the integers MUST be negative for this to work out, or else they will not make any sense. So, we should find the range of both equations and use the overlap.

SO...

The range of |x| + 1 > 7 is (-inf, -6) U (6, inf)

The range of |x+1| <= 7 is [-8, 6]. 

Therefore, the only values that work are -8 and -7. So, the answer would be -15.

That is ok Wolf, you did not sound rude at all.

I was just making an observation in my own blunt way.

Thank you for explaining.    

Hmm... It says its incorrect. But i'm not sure why..

The answer should be 0: here is why.

Here is a question for you to think about: Is the absolute value just for positive integers, or for both sides of 0?

In other words, if one number works, then it's negation will to, since the absolute value will be the same.

Therefore, if you add all those 0s up, you will get 0.

(If I'm wrong about this, please let me know!)

Points A,B,C and D are midpoints of the sides of the larger square. If the smaller square has area 60, what is the area of the bigger square?   

First, for reference, let's name one of the corners of the large square. 

Any of the corners will do, so let's just choose the top left one.  Call it V. 

Since the area of the small square is 60, then each side is length sqrt(60). 

DVA is an isoceles right triangle, with hypotenuse length sqrt(60).  

Therefore, VA is length sqrt(60) divided by sqrt(2), and the entire side is twice that. 

                                                      2 • sqrt(60)               

The side of the large square is    ——————    

                                                           sqrt(2)   

                                                                                                    4 • 60

Square that side, and you have the area of the large square    ————   thus  A_largesquare = 120

                                                                                                        2   

_.

Thank you Melody for clarifying my slightly hurried and slightly incorrect answer. :) I was about to go to online class. :) I really needed some more reliance since I can't always rely on myself. And yes you are correct that Guest should be able to finish it him/herself. \(:)\)

Use the same methodology as in "Find z" two threads down. 

Your 4-digit number = 2,149

Since  WX || YZ   --->   angle(WXY) + angle(XYZ)  =  180^o  

                            --->             (3x - 5) + (90) + (2x)  =  180^o

                            --->                                 5x + 85  =  180^o

                            --->                                         5x  =  95^o

                            --->                                           x  =  19^o 

A student is chosen at random from a class with 17 girls and 18 boys.  

The probability of choosing a boy is?  

The probability of choosing either a girl or a boy is?  

The probability of choosing a boy ~  

This is the number of boys divided by the total number of students:   51.43%  this is 18/35   

The probability of choosing either a girl or a boy ~             either one:   100%  this is 35/35   

_.