four triangles in a square
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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
"My life is spent in one long effort to escape from the commonplace of existence. These little problems help me to do so." -- Sherlock Holmes
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?
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
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.
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
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.
"My life is spent in one long effort to escape from the commonplace of existence. These little problems help me to do so." -- Sherlock Holmes
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
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
January 26th, 2020 at 2:31:21 PM
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Joeman. You absolute ( insert expletive) :o) You beat me by 11 minutes, but we seem to agree.
I struggled for a long while because I tried to solve for y instead of y^2 and so I ended up with 4th order polynomial.
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
January 26th, 2020 at 2:35:22 PM
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I honestly thought someone else would post the answer while I was typing my response! Hey, you can't drink the beer anyway, right? ;-)Quote: OnceDearJoeman. You absolute ( insert expletive) :o)
"Dealer has 'rock'... Pay 'paper!'"
January 26th, 2020 at 2:39:09 PM
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So did I, but I was right!Quote: JoemanI honestly thought someone else would post the answer while I was typing my response! Hey, you can't drink the beer anyway, right? ;-)
Pah! I never received the last pint he sent me. I keep checking the mail for US parcel. I guess it never cleared customs.
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
January 26th, 2020 at 5:59:26 PM
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Quote: BigJerLol. Are the numbers the area or a side of the triangles?
Area.
"My life is spent in one long effort to escape from the commonplace of existence. These little problems help me to do so." -- Sherlock Holmes
January 26th, 2020 at 6:02:55 PM
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Quote: OnceDearJoeman. You absolute ( insert expletive) :o) You beat me by 11 minutes, but we seem to agree.
Yes, I agree that Joe got it first. Cheers, Joe, I owe you a beer. Commiserations to OD. I'll still buy you one, preferably in the wee village of Shackleford, but not for this reason.
"My life is spent in one long effort to escape from the commonplace of existence. These little problems help me to do so." -- Sherlock Holmes
