Probably use the partition of the square shown in the image...I mean, lol...the area will be such that x^2 = Area of right triangle with hypotenuse 3 + right triangle with hypotenuse of 4 + right triangle with hypotenuse of 5, etc (note that 3, 4, 5 is a Pythagorean triple, so it's a right triangle too, area is 4x3/2=6, etc)...
Right sides of right triangle with hypotenuse 3 will be a, x-b, where right sides of right triangle with hypotenuse 4 are be b, x, etc...making sure that those triangles are right triangles will already permit us to express x in terms of a and b, etc, using Pythagoras...actually, express a and b in terms of x, etc, and then solve for x when we add all the triangles areas, and set their sum equal to x^2, etc...
Before doing anything, we know that x needs to be less than 4 but more than 3. So this narrows it down quite a bit. We also know that x = y + z, so y and z each need to be less than x. We know that x² + y² = 5², so x² needs to be more than 12½ while y² needs to be less than 12½. Let us begin by labelling the diagram: - the bottom side is x - the lower portion of the lefthand side is w - the upper portion of the lefthand side is v - the righthand portion of the top side is y - the lefthand portion of the top side is z We then get our equations: x² + y² = 5² x² + w² = 4² z² + v² = 3² w + v = x y + z = x There's probably some way to solve this, but I'm not quite sure how. Nevertheless, we can get the answer by way of guess and check, beginning with x = 3.6 (i.e. a little above √12½). If x = 3.6 then y is around 3.47, and z is around 0.13, which means that v is around 2.997, which means that w is around 0.603. And 0.603² + 3.6² is around 13.324, which is too low (needs to be 16). Try again with x = 3.7. If x = 3.7 then y is around 3.363, so z is around 0.337, so v is around 2.981, so w is around 0.719. And 0.719² + 3.7² is around 14.207, which is still too small. Ok, it looks like maybe we need to bring up x by a lot. Try again with x = 3.95. If x = 3.95 then y is around 3.066, so z is around 0.884, so v is around 2.867, so w is around 1.083. And 1.083² + 3.95² is around 16.775, which is a bit too high. Try again with x = 3.9. If x = 3.9 then y is around 3.129, so z is around 0.771, so v is around 2.899, so w is around 1.001. And 1.001² + 3.9² is around 16.212... getting close. Try again with x = 3.85. If x = 3.85 then y is around 3.19 so z is around 0.66, so v is around 2.927, so w is around 0.923. And 0.923² + 3.85² is around 15.674... needs to be higher. Try again with x = 3.88. Actually, since I just noticed that 3.88² is very close to 15, and since these sorts of problems sometimes use nice even numbers, maybe we should test x = √15. If x = √15 then y is √10, so z is around 0.711, so v is around 2.915, so w is around 0.958. But 0.958² is not 1, so x cannot be √15. Still, it's very close. Let's try x = 3.88. If x = 3.88 then y is around 3.154, so z is around 0.726, so v is around 2.911, so w is around 0.969. And 0.969² + 3.88² is around 15.994, so we are burning up. Now here's a thought. Let us suppose that y were pi. If this were the case, then x would be just under 3.89. If x is just under 3.89, and y is pi, then z is around 0.748, so v is around 2.905, so w is around 0.985. And 0.985² + 3.89² is around 16.102. There is still one nice number we haven't yet tried. Maybe x is three and eight ninths. If x = 3.8̅ then y is around 3.143, so z is around 0.746, so v is around 2.906, so w is around 0.983. And 0.983² + 3.8̅² is still a bit too high. Therefore, let us conclude that x is somewhere between 3.88 and 3.8̅. Maybe I could just round it off to 3.885 and be done with it. (If I needed a better approximation then I could just keep narrowing it down, and I'd probably write a program to do that so I wouldn't have to keep typing it into the calculator, but this is probably "close enough".) *Final guess: 3.885* - Checking it against the actual answer, I see my answer is indeed "pretty close". Maybe I should have tried to pinpoint that third decimal but oh well. (I didn't even think to check if those two triangles were similar...)
We have x=4.cos(a) (bottom side) and x = 3.cos(a) + 4.sin(a) (left side) So cos(a)=4.sin(a) So tan(a)=1/4. Besides , we have (cos(a))^2 + (sin(a))^2=1 or (cos(a))^2 + ((1/4).cos(a))^2=1 or (1+1/16).(cos(a))^2=1 or cos(a)=sqrt(16/17)=4/sqrt(17) So x=4.cos(a)=16/sqrt(17).
this was a bit tricky due to negative root, which was avoided by lx= (see line 30) 10 print "mind math enigmas-what is the size of the square" 20 dim x(2,2),y(2,2):l1=3:l2=4:l3=5:sw=.01:lx=l2-l1:goto 60 30 l4=sqr(l2^2-lx^2):l5=sqr(l1^2-(lx-l4)^2):rem if disl5*l1 then stop 80 lx2=lx:gosub 30:if dg1*dg>0 then 70 90 lx=(lx2+lx1)/2:gosub 30:if dg1*dg>0 then lx1=lx else lx2=lx 100 if abs(dg)>1E-10 then 80 110 print lx:print "die flaeche=";lx^2 120 mass=800/lx:goto 140 130 xbu=x*mass:ybu=y*mass:return 140 x(0,0)=0:y(0,0)=0:x(0,1)=lx:y(0,1)=0:x(0,2)=0:y(0,2)=l4:x(1,0)=0:y(1,0)=l4 150 x(1,1)=l5:y(1,1)=lx:x(1,2)=0:y(1,2)=lx:x(2,0)=lx:y(2,0)=0:x(2,1)=lx:y(2,1)=lx 160 x(2,2)=l5:y(2,2)=lx:for a=0 to 2:x=x(a,0):y=y(a,0):gosub 130:xba=xbu:yba=ybu:gcola 170 for b=1 to 3:ib=b:if ib=3then ib=0 180 x=x(a,ib):y=y(a,ib):gosub 130:xbn=xbu:ybn=ybu:goto 200 190 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 200 gosub 190:next b:next a 210 l2r=sqr(lx^2+l4^2):fe=l2/l2r:fe=(1-fe)*100:print "der fehler=";fe;"%" mind math enigmas-what is the size of the square 3.88057 die flaeche=15.0588235 der fehler=0% > run in bbc basic sdl and hit ctrl tab to copy from the results window
@@joso5554 He wrote a BASIC program (and apparently he's German?) Without taking the time trying to decipher his program, I'm not entirely sure what it's doing or how it works. But the answer looks correct.
That is not true. x = squareroot(256/17) = 3.88057000058. The other side, call it y, is SquareRoot(169/17) = 3.15296312547. x^2 + y^2 = 5^2. I don't know where you are getting 4 from. If an hypotenuse is 5, it doesn't mean that either of the two sides has to be greater than 4.
Just looking at the image without doing any calculations demonstrates that is incorrect. The triangle formed at the bottom of the square has a hypotenuse of length 4, which means, by definition, both other sides of the triangle must be less than 4. Since one of those other sides is also the side of the square, the square's sides must be shorter than 4, thus the area cannot be 16.
Define c = cos(alpha) and s = sin(alpha)
Then x = 4c = 4s + 3c -> c = 4s -> c^2 = 16(1-c^2)
c = sqrt(16/17) --> x = 16/sqrt(17)
its on the right silly
No, That's wrong you just did black magic.
What level math is this?
The 3-4-5 triangle is right-angled. Therefore, the upper left triangle with hypotenuse 3 and the lower triangle with hypotenuse 4 are similar.
That’s the trick. But not the answer…
You skipped a step or two to derive your answer.
Probably use the partition of the square shown in the image...I mean, lol...the area will be such that x^2 = Area of right triangle with hypotenuse 3 + right triangle with hypotenuse of 4 + right triangle with hypotenuse of 5, etc (note that 3, 4, 5 is a Pythagorean triple, so it's a right triangle too, area is 4x3/2=6, etc)...
Right sides of right triangle with hypotenuse 3 will be a, x-b, where right sides of right triangle with hypotenuse 4 are be b, x, etc...making sure that those triangles are right triangles will already permit us to express x in terms of a and b, etc, using Pythagoras...actually, express a and b in terms of x, etc, and then solve for x when we add all the triangles areas, and set their sum equal to x^2, etc...
Before doing anything, we know that x needs to be less than 4 but more than 3. So this narrows it down quite a bit.
We also know that x = y + z, so y and z each need to be less than x. We know that x² + y² = 5², so x² needs to be more than 12½ while y² needs to be less than 12½.
Let us begin by labelling the diagram:
- the bottom side is x
- the lower portion of the lefthand side is w
- the upper portion of the lefthand side is v
- the righthand portion of the top side is y
- the lefthand portion of the top side is z
We then get our equations:
x² + y² = 5²
x² + w² = 4²
z² + v² = 3²
w + v = x
y + z = x
There's probably some way to solve this, but I'm not quite sure how. Nevertheless, we can get the answer by way of guess and check, beginning with x = 3.6 (i.e. a little above √12½).
If x = 3.6 then y is around 3.47, and z is around 0.13, which means that v is around 2.997, which means that w is around 0.603. And 0.603² + 3.6² is around 13.324, which is too low (needs to be 16). Try again with x = 3.7.
If x = 3.7 then y is around 3.363, so z is around 0.337, so v is around 2.981, so w is around 0.719. And 0.719² + 3.7² is around 14.207, which is still too small. Ok, it looks like maybe we need to bring up x by a lot. Try again with x = 3.95.
If x = 3.95 then y is around 3.066, so z is around 0.884, so v is around 2.867, so w is around 1.083. And 1.083² + 3.95² is around 16.775, which is a bit too high. Try again with x = 3.9.
If x = 3.9 then y is around 3.129, so z is around 0.771, so v is around 2.899, so w is around 1.001. And 1.001² + 3.9² is around 16.212... getting close. Try again with x = 3.85.
If x = 3.85 then y is around 3.19 so z is around 0.66, so v is around 2.927, so w is around 0.923. And 0.923² + 3.85² is around 15.674... needs to be higher. Try again with x = 3.88.
Actually, since I just noticed that 3.88² is very close to 15, and since these sorts of problems sometimes use nice even numbers, maybe we should test x = √15.
If x = √15 then y is √10, so z is around 0.711, so v is around 2.915, so w is around 0.958. But 0.958² is not 1, so x cannot be √15. Still, it's very close. Let's try x = 3.88.
If x = 3.88 then y is around 3.154, so z is around 0.726, so v is around 2.911, so w is around 0.969. And 0.969² + 3.88² is around 15.994, so we are burning up.
Now here's a thought. Let us suppose that y were pi. If this were the case, then x would be just under 3.89.
If x is just under 3.89, and y is pi, then z is around 0.748, so v is around 2.905, so w is around 0.985. And 0.985² + 3.89² is around 16.102.
There is still one nice number we haven't yet tried. Maybe x is three and eight ninths.
If x = 3.8̅ then y is around 3.143, so z is around 0.746, so v is around 2.906, so w is around 0.983. And 0.983² + 3.8̅² is still a bit too high.
Therefore, let us conclude that x is somewhere between 3.88 and 3.8̅. Maybe I could just round it off to 3.885 and be done with it. (If I needed a better approximation then I could just keep narrowing it down, and I'd probably write a program to do that so I wouldn't have to keep typing it into the calculator, but this is probably "close enough".)
*Final guess: 3.885*
-
Checking it against the actual answer, I see my answer is indeed "pretty close". Maybe I should have tried to pinpoint that third decimal but oh well.
(I didn't even think to check if those two triangles were similar...)
He did not really answer the question
We have x=4.cos(a) (bottom side)
and x = 3.cos(a) + 4.sin(a)
(left side)
So cos(a)=4.sin(a)
So tan(a)=1/4.
Besides , we have
(cos(a))^2 + (sin(a))^2=1
or
(cos(a))^2 + ((1/4).cos(a))^2=1
or
(1+1/16).(cos(a))^2=1
or
cos(a)=sqrt(16/17)=4/sqrt(17)
So x=4.cos(a)=16/sqrt(17).
this was a bit tricky due to negative root, which was avoided by lx= (see line 30)
10 print "mind math enigmas-what is the size of the square"
20 dim x(2,2),y(2,2):l1=3:l2=4:l3=5:sw=.01:lx=l2-l1:goto 60
30 l4=sqr(l2^2-lx^2):l5=sqr(l1^2-(lx-l4)^2):rem if disl5*l1 then stop
80 lx2=lx:gosub 30:if dg1*dg>0 then 70
90 lx=(lx2+lx1)/2:gosub 30:if dg1*dg>0 then lx1=lx else lx2=lx
100 if abs(dg)>1E-10 then 80
110 print lx:print "die flaeche=";lx^2
120 mass=800/lx:goto 140
130 xbu=x*mass:ybu=y*mass:return
140 x(0,0)=0:y(0,0)=0:x(0,1)=lx:y(0,1)=0:x(0,2)=0:y(0,2)=l4:x(1,0)=0:y(1,0)=l4
150 x(1,1)=l5:y(1,1)=lx:x(1,2)=0:y(1,2)=lx:x(2,0)=lx:y(2,0)=0:x(2,1)=lx:y(2,1)=lx
160 x(2,2)=l5:y(2,2)=lx:for a=0 to 2:x=x(a,0):y=y(a,0):gosub 130:xba=xbu:yba=ybu:gcola
170 for b=1 to 3:ib=b:if ib=3then ib=0
180 x=x(a,ib):y=y(a,ib):gosub 130:xbn=xbu:ybn=ybu:goto 200
190 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return
200 gosub 190:next b:next a
210 l2r=sqr(lx^2+l4^2):fe=l2/l2r:fe=(1-fe)*100:print "der fehler=";fe;"%"
mind math enigmas-what is the size of the square
3.88057
die flaeche=15.0588235
der fehler=0%
>
run in bbc basic sdl and hit ctrl tab to copy from the results window
What is this BS ???
@@joso5554 He wrote a BASIC program (and apparently he's German?) Without taking the time trying to decipher his program, I'm not entirely sure what it's doing or how it works. But the answer looks correct.
having x 4.123 just makes the problem wrong since it is greater than 4 which is the hypotenuse of right angle triangle
How did you get 4.123? x is the square root of 256/17, which is less than 4.
@@blessedchild4941 x=16/√17 and
256/17 will give you x^2 and not the value of x
*The triangle at x side having one side of 5 it also include a right angle so the x or second bigger side is 4.*
The area of the square is 4×4=16
In fact there's another secret option: other sides could be 63/13 and 16/13 but math teacher would never tell about it.
That is not true. x = squareroot(256/17) = 3.88057000058. The other side, call it y, is SquareRoot(169/17) = 3.15296312547. x^2 + y^2 = 5^2. I don't know where you are getting 4 from. If an hypotenuse is 5, it doesn't mean that either of the two sides has to be greater than 4.
No it’s not.
Just looking at the image without doing any calculations demonstrates that is incorrect. The triangle formed at the bottom of the square has a hypotenuse of length 4, which means, by definition, both other sides of the triangle must be less than 4. Since one of those other sides is also the side of the square, the square's sides must be shorter than 4, thus the area cannot be 16.
= Every place every living for a week post address proof of time to be x draw the location of the text ramayan and mahashivratri
don't know!
You should…
The square isn't a rectangle square.
x=4
Nope. The square is a square. The answer is not an integer !!!