## Poll

4 votes (57.14%) | |||

No votes (0%) | |||

1 vote (14.28%) | |||

No votes (0%) | |||

2 votes (28.57%) | |||

1 vote (14.28%) | |||

2 votes (28.57%) | |||

2 votes (28.57%) | |||

1 vote (14.28%) | |||

4 votes (57.14%) |

**7 members have voted**

January 26th, 2020 at 7:18:43 AM
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Given the larger figure is a square, solve for x.

Usual rules apply:

1. No searching

2. No posting links to solutions elsewhere.

3. Please put answers and solutions in spoiler tags until I announce a satisfactory solution.

4. Beer to the first satisfactory answer and solution.

5. Those who have won a beer in the last year may not play for the 24 hours following this posting.

Usual rules apply:

1. No searching

2. No posting links to solutions elsewhere.

3. Please put answers and solutions in spoiler tags until I announce a satisfactory solution.

4. Beer to the first satisfactory answer and solution.

5. Those who have won a beer in the last year may not play for the 24 hours following this posting.

Last edited by: Wizard on Jan 26, 2020

It's not whether you win or lose; it's whether or not you had a good bet.

January 26th, 2020 at 7:33:37 AM
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Is the yellow triangle an equilateral triangle? or does that have to be determined?

January 26th, 2020 at 7:56:17 AM
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No. It can't be.Quote:rsactuaryIs the yellow triangle an equilateral triangle? or does that have to be determined?

Take care out there.
Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..

January 26th, 2020 at 8:44:30 AM
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GGgggaaahhhhh.

I'm doing it, but somebody is sure to beat me to it.

I'm doing it, but somebody is sure to beat me to it.

Take care out there.
Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..

January 26th, 2020 at 9:21:48 AM
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Quote:WizardThis post was removed in the interests of giving others who haven't won a beer a chance at it. I copied and pasted the text to Don. Those who haven't won a beer yet have until 24 hours since the OP to solve it.

Hopefully nobody will cheat who may remember Don's post.

Last edited by: unnamed administrator on Jan 26, 2020

January 26th, 2020 at 9:52:08 AM
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Quote:rsactuaryIs the yellow triangle an equilateral triangle? or does that have to be determined?

It has to be determined.

It's not whether you win or lose; it's whether or not you had a good bet.

January 26th, 2020 at 1:23:52 PM
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Lol. Are the numbers the area or a side of the triangles?

The Terror of Casinos.

January 26th, 2020 at 1:43:10 PM
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Quote:BigJerLol. Are the numbers the area or a side of the triangles?

I think it has to be area.

January 26th, 2020 at 2:19:25 PM
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15.49

Let's say a side of the square has length L. The value of x is equal to the square area minus that of the triangles, or:

x = L^2 - 20

So, the key is solving for L^2, which we can get by solving a set of equations for the areas of the three triangles. The blue and green triangles both have a height of L. We can call their respective bases b1 and b2.

1/2 L x b1 = 8

1/2 L x b2 = 5

Solving each for the base in terms of L, we get:

b1 = 16/L

b2 = 10/L

The base and height of the red triangle would be L - b1 and L - b2. So, its area can be represented by:

1/2 (L - b1) x (L - b2) = 7 or

L^2 - Lb1 - Lb2 + 10/L x 16/L = 14

Substituting from above, we get:

L^2 - L(16/L) - L(10/L) + (16/L)(10/L) = 14

Which, after reducing, rearranging, and finally multiplying all terms by L^2, we get:

L^4 - 40L^2 + 160 = 0

Using the quadratic equation**, we can solve for L^2, which would be an area for the square of 35.49. Subtracting out the area of the known triangles, we get x = 15.49.

** FULL DISCLOSURE: I mis-remembered the formula for the quadratic equation (it's been quite a while...), and when my first answer didn't make sense, I had to look up the formula. Not sure if this would disqualify me for the beer.

x = L^2 - 20

So, the key is solving for L^2, which we can get by solving a set of equations for the areas of the three triangles. The blue and green triangles both have a height of L. We can call their respective bases b1 and b2.

1/2 L x b1 = 8

1/2 L x b2 = 5

Solving each for the base in terms of L, we get:

b1 = 16/L

b2 = 10/L

The base and height of the red triangle would be L - b1 and L - b2. So, its area can be represented by:

1/2 (L - b1) x (L - b2) = 7 or

L^2 - Lb1 - Lb2 + 10/L x 16/L = 14

Substituting from above, we get:

L^2 - L(16/L) - L(10/L) + (16/L)(10/L) = 14

Which, after reducing, rearranging, and finally multiplying all terms by L^2, we get:

L^4 - 40L^2 + 160 = 0

Using the quadratic equation**, we can solve for L^2, which would be an area for the square of 35.49. Subtracting out the area of the known triangles, we get x = 15.49.

** FULL DISCLOSURE: I mis-remembered the formula for the quadratic equation (it's been quite a while...), and when my first answer didn't make sense, I had to look up the formula. Not sure if this would disqualify me for the beer.

"Dealer has 'rock'... Pay 'paper!'"

January 26th, 2020 at 2:30:28 PM
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I think I have it

edit, I got my labelling wrong when I scribbled it out. correcting...

Let base side of square be y

Let short redgreen side be g

Let short blue side be b

Let short green side be g1 red side be r1

Let long green side be g2 red side be r2 (Not hypoteneuse)

Using area of triangle = 1/2 base x height

5=g.y/2

y=10/g

8=b.y/2

y=16/b

Looking at red triangle of area =7=r1.r2/2

14=r1.r2

but r1=y-b and r2=y-g

14=(y-b)(y-g)

14=(y-16/y).(y-10/y)

14=y^2-16-10+160/y^2

40=y^2+160/y^2

Now y^2 is the area of the square. Call y^2 S

0=S+160/S-40

multiply both sides by S

0=S^2-40S+160

a simple quadratic with solutions

S=35.49193338483

or

S=4.5080666151703

Now we reject the second answer because the total area is at least the sum of 8,5,7

So total square area is 35.49

therefore area x=35.49-8-7-5=15.49

I'm off to check my workings now. Just wanted to get it posted

edit, I got my labelling wrong when I scribbled it out. correcting...

Let base side of square be y

Let short redgreen side be g

Let short blue side be b

Let short green side be g1 red side be r1

Let long green side be g2 red side be r2 (Not hypoteneuse)

Using area of triangle = 1/2 base x height

5=g.y/2

y=10/g

8=b.y/2

y=16/b

Looking at red triangle of area =7=r1.r2/2

14=r1.r2

but r1=y-b and r2=y-g

14=(y-b)(y-g)

14=(y-16/y).(y-10/y)

14=y^2-16-10+160/y^2

40=y^2+160/y^2

Now y^2 is the area of the square. Call y^2 S

0=S+160/S-40

multiply both sides by S

0=S^2-40S+160

a simple quadratic with solutions

S=35.49193338483

or

S=4.5080666151703

Now we reject the second answer because the total area is at least the sum of 8,5,7

So total square area is 35.49

therefore area x=35.49-8-7-5=15.49

I'm off to check my workings now. Just wanted to get it posted

Last edited by: OnceDear on Jan 26, 2020

Take care out there.
Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..