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These are mathcounts warmup questions, all of them require background knowledge, in which I don't have. I couldn't think of any. I was doing warmup practices and I was stumped on these problems

 

1. Right triangle ABC with AC = 3 units, BC = 4 units and AB = 5 units is rotated 90° counterclockwise about M, the midpoint of side AB, to create a new right triangle A'B'C'. What is the area of the shaded region where triangles ABC and A'B'C' overlap? Express your answer as a common fraction.

All I found was that MA is perpendicular to B'A'. I could "cheat" by using a protracter and graph paper, but that is not how they want you to solve the problem.

 

 

2. The mean of the first five terms of an arithmetic sequence is 27, and the mean of the first eight terms of the same sequence is 39. What is the absolute difference between the first and second terms of this sequence?

 

Sorry I have trouble with this type I have never learned how to do them. I can't think of anything but explicit formula.

 Sep 21, 2019
edited by CalculatorUser  Sep 21, 2019
edited by CalculatorUser  Sep 21, 2019
 #1
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+1

2nd question:

"2. The mean of the first five terms of an arithmetic sequence is 27, and the mean of the first eight terms of the same sequence is 39. What is the absolute difference between the first and second terms of this sequence?" By the way, what do you mean by the absolute difference between the first & second terms of this sequence? so we subtract the first mean from the second mean? That's what I did.. or we subtract the first term from the second term?

 

I think:

Let the first 5 terms of the arithmetic sequence be:   X , Y , Z , K, L

We know:

(X+Y+Z+K+L) /5 =27

"The mean of the first eight terms of the same sequence is 39" So I understood that it is the extention of the first mean 

Let the new 3 terms be F,S,Q

(X+Y+Z+K+L)/5 + F+S+Q=39 

we know that (X+Y+Z+K+L)/5=27

27+F+S+Q=39

F+S+Q=12

 

Ok so "Difference between the first and second terms of this sequence" 

first is 27-12=15

Not so sure.. maybe I understood it wrong.

 Sep 21, 2019
 #2
avatar+2417 
+3

When they say mean of 27 with first 5 terms, remember how to calculate for mean, that means the SUM of the first 5 terms is 27 * 5= 135

 

You gave me the SOLVING idea though!
 

You forgot how mean is calculated but solved it in the right way.

 

If solved correctly, the answer should be 8, which is what the answer sheet says.

 

I think it utilizes explicit formula. If someone else could solve it for other people who view this problem it would be great!

CalculatorUser  Sep 21, 2019
edited by CalculatorUser  Sep 21, 2019
 #3
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2)   5/2 * [2F + D(5 - 1)] =135.............(1)

       8/2 * [2F + D(8 - 1) ] =312...........(2), solve for F, D

F = 11 - which is the first term and D = 8 - which is the common difference.

The sequence is: 11, 19, 27, 35, 43, 51, 59, 67 . The mean of the first 5 terms is 27 and of all 8 terms is 39. The absolute difference between the first and the second is: abs[11 - 19] = 8, Or: 19 - 11 =8.

 Sep 21, 2019
edited by Guest  Sep 21, 2019
 #4
avatar+105683 
+2

Hi CalculatorUser,

 

Question 1

I have not answered this question. I have not even tried yet

I did draw it with geogebra, because that looked fun.     wink

 

This is the graphical approximation.

 

 Sep 21, 2019
 #5
avatar+105683 
+1

Question 1

I do not know how to do question one but I'd like to see someone else's solution.

It has probably got something to do with similarity....  maybe ...    indecision

 Sep 21, 2019
 #6
avatar+104969 
+6

2. The mean of the first five terms of an arithmetic sequence is 27, and the mean of the first eight terms of the same sequence is 39. What is the absolute difference between the first and second terms of this sequence?

 

Let the first five terms be  

a , (a + d), (a + 2d), (a + 3d), (a + 4d)..where "a" is the first term and "d"  is the common difference between terms....the sum of these  =  5a + 10d

 

So....

 

(5a + 10d)

________  =   27     →     a + 2d  =  27        (1)

      5

 

And the next three terms  are      (a + 5d), (a + 6d), (a + 7d)

So......the sum of the first 8 terms  =   8a + 28d

 

So.....

 

(8a + 28d)

_______      =  39   →     a + (7/2)d  =  39        (2)

      8

 

 

Subtract  (1)  from (2)    and we get that    (3/2)d   =  12    →    d = 8

 

So...the absolute difference between the first and second terms is just "d"    =  l  8  l   =   8

 

 

 

cool cool cool

 Sep 21, 2019
 #7
avatar+104969 
+5

1. 

 

Here's a pic : 

 

 Call the intersection of BC and  B'A' , N

Note that triangle BMN   is similar to triangle BCA

So  

MN/ BM  = AC / BC

MN / (5/2)  = 3/4

MN  = 15/8

 

So....the area of triangle BMN  =  (1/2) (product of the legs) = (1/2)(BM)(MN) = (1/2)(5/2)(15/8)  = 75/32

 

So BN  = sqrt [ BM^2 + MN^2]  = sqrt [ (5/2)^2 +(15/8)^2 ] =  25/8 

 

So B'N  =   B'M - MN =   (5/2) - (15/8)  =  20/8 - 15/8 =  5/8

So  A'N  = A'M + MN  =   5/2  + 15/8  =   20/8 + 15/8 = 35/8

 

And call the intersection of BC  and B'C', P

Because BC  is parallel to A'C'.....then

A'N / A'B'   = C'P / C'B'

(35/8) /  5  =  C'P /4

35/40  = C'P /4

C'P  = 3.5

Then B'P  = B'C' - C'P  =  4 -3.5  = .5

 

So....using the Pythagorean Theorem....NP  =  sqrt [B'N^2 - B'P^2] =  sqrt [ (5/8)^2 - (4/8)^2 ]  = 3/8

Then  BP  =  BN + NP  =  25/8 + 3/8  =  28/8 =  7/2

 

And call the intersection of BA and B'C', Q

 

And because PQ  is parallel to AC

 

Then     BP / BC  = PQ/CA

(7/2) / 4 =  PQ/ 3

(7/8) = PQ/3

(21/8)  = PQ

 

So  the area of triangle BPQ  =  (1/2) (BP)(PQ)  = (1/2) (7/2) (21/8)  =  147/32

 

So... the shaded area  = [area of triangle BPQ ]  - [area of triangle BMN ] =

 

[ 147 - 75 ] / 32  =  72/32 =  9/4  =  2.25 units^2

 

 

cool cool cool

 Sep 21, 2019
 #9
avatar+105683 
+2

Thanks Chris,

Another one for me to study   laugh

 

Ok I have studied it. 

I need to work on similar figure geometry questions too,  (Along with CalculatorUser.)

 

In summary you have said

Area desired = area of BPQ - area of BNM

These are similar triangles.

 

To find these areas you need length MN and length PQ and PB

 

You have then used a series of similar figures ratios and pythagoras theorum to do this.

 

All sorted.    laugh

Melody  Sep 21, 2019
edited by Melody  Sep 21, 2019
 #8
avatar+2417 
+3

Thanks CPhill, Melody, and Guest. I think Triangle similarity is something I need to work on more!

 Sep 21, 2019

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