Відео

Well, by quick inspection we have the following equations: r*sin(A1) = 7.5 r*sin(A2) = 3.5 4*A1 + 2*A2 = 180 So right away we have three equations in three unknowns. Manipulate the third equation: 4*A1 + 2*A2 = 180 2*A1 + A2 = 90 A2 = 90 - 2*A1 Substitute into the first to equations: r*sin(A1) = 7.5 r*sin(90-2*A1) = 3.5 --> r*cos(2*A1) = 3.5 --> r*(1-2*sin^2(A1)) = 3.5 Now let u = sin(A1). Then r*u = 7.5 --> u = 7.5/r r*(1-2*u^2) = 3.5 --> r*(1-2*(7.5/r)^2) = 3.5 r - 2*7.5^2/r = 3.5 r^2 - 3.5*r - 2*7.5^2 = 0 Roots are 12.5 and -9. A negative radius won't do, so r=12.5, d = 25. Q.E.D.

a=12, b=15, c=20. Did it in my head. If you want to see how to work it out, just write the Pythagorean theorem down in all the ways you can: b^2 = a^2 + 81 c^2 = a^2 + 256 b^2 + c^2 = 625 Reorganize: a^2 - b^2 = -81 a^2 - c^2 = -256 b^2 + c^2 = 625 There you have three equations in three unknowns, a^2, b^2, and c^2. Just solve them and take the square roots. I didn't do all that in my head, obviously, but this is how you'd solve the problem "in general." The particular one offered here just has almost the most obvious possible solution. "First guess," so to speak.

Well, that is a right triangle because 3, 4, 5 is a Pythagorean triple. Because of this, the angle between the 4-long side and the horizontal axis is the same as the angle between the 3-long side and the vertical axis. This tells us that the 3-long side hits the vertical axis at x/4. So, that bottom triangle has is a right triangle with side lengths x/4, x, and 4 (the hypotenuse). Using the Pythagorean theorem, (x/4)^2 + x^2 = 4^2 (17/16)*x^2 = 16 17*x^2 = 256 x^2 = 256/17 x = sqrt(256/17) = 3.8806 Q.E.D. Note, I figure out how to solve these in my head before watching the videos. Then I do the arithmetic, and then check the end of the video to make sure I got it right.

The problem does not specify where the left side of the rectangle lies along the horizontal axis. Presuming this to be a well-posed problem, this must mean that the answer is independent of that parameter. So we can choose it to be wherever we like. I choose it to end at the centerline of the semicircle. Therefore the rectangle is a square of side length S, and we see that S^2 + S^2 = 10^2 2*S^2 = 100 S^2 = Area = 50 Q.E.D. One should always be on the lookout for the opportunity to exploit an unspecified parameter when in a "formal testing" situation. If the horizontal location of the left edge of the rectangle had been specified, then we'd have had to do the work presented here in the video, but since it wasn't we were able to dodge that completely.

I like to try to solve these before watching the video, and to think about them in my head until I see an "elegant path." In this case I hate that that "Think outside" comment is up there - I'd prefer to not have hints like that. We can use the Pythagorean theorem to solve this. First application - complete the right triangle formed by the 24-long and 7-long sides. That hypotenuse is length 25. It's mirror image in the right quarter circle will extend the 7-long line to the other side of the semicircle, so we see that that full side (including the 7-long line) is length 32. Now apply the the Pythagorean theorem again to that large triangle (with sides 24 and 32) to find that the semicircle diameter is 40. So the semicircle radius is 20. Now we have a final right triangle with hypotenuse 25, side length 20, and side length the quantity we seek. Apply the Pythagorean theorem one last time to find that H = sqrt(25^2 - 20^2) = sqrt(225) = 15. Q.E.D.

I get .2499 for y length. The two tangent lines, AD and truncated line AE, will create an angle of 126.86 at the midpoint of line DC. Which sets up an angle of 53.13 at Midpoint DC--breakline,EC. Spllitting the angle again gives 26.56 degrees at midpoint of a new triangle, Midpoint, EC. Going pythagorean at this point useing midpoint circle diameter DC , and length of .5,(right half of line DC), as cosine, gives .2499 as sine. Or Y. So, length of red line is .2499 + 1 or 1.25 as rounded. Sorry I couldn't explain this better. But that's what I got.

Dont forget Euklid

the result can also be written like this : 3 + 2√2

What the actual fuck this is beyond my capabilities

Simple, cler and clever. Nice.

Method using Thales theorem and trigonometric ratio: 1. Construct right-angled triangle with side of length 10 and diameter of semicircle by Thales theorem. Let the right bottom angle of this triangle be theta. 2. Let radius of semicircle be R and width of rectangle be W. Height of rectangle = R as top of rectangle is tangent to circle. 3. In original right-angled triangle W = 10 cos(theta) In constructed right- angled triangle 2R = 10/cos(theta) 4. Area of rectangle = WR = 10 cos(theta) x 5/cos(theta) = 50.

awesome! thanks for the knowledge! I appreciate all you do for the math community Mind Math #Enigmas! Whoo hoo

Thanks for sharing the fun power of math! Here is a video solving a similar problem using a free, browser-based modeling tool. ua-cam.com/video/BLDpTQ5ediM/v-deo.html

Love a challenge Three "right" triangles are involved. Every triangle's angles sum to 180°. Pythagoreans theorem applies: a^2+b^2=c^2 (but not these a, b, c's. Three unknowns a, b, c Three equations: eq.1 9^2+a^2=b^2 eq.2 16^2+a^2=c^2 eq.3. (9+16)^2=b^2+c^2

Good

Multiply both elements of the base together, get the square root of the product. That's the height. Formula works if the opposite angle of the base is 90 degrees. Which it is in this case. Euclid's height theorem Short and sweet.

This is a very poorly presented problem.

I used the intersecting chords theorem, then the chord bisector theorem to derive a right triangle with legs ⁷/₂ and 2, and hypotenuse 𝒓. It's then just a case of Pythagoras to calculate 𝒓.

This explains how Joe Biden graduated from MIT. Hahahahahahahahaha

30 degree

I solve it like your way

Not a particularly hard problem since AD is the tangent at D and new point T is the tangent between AE, thus AT is one. Thus OA°E is the tan of (1/2)/1 = 0.5. But DA°E is twice the angle So what we need is to reduce this to the unit circle 1/2^+ 1 = 1.25 = OA^2 = OA = SQRT(5)/2 Sin angle = 1/2 / SQRT(5)/2 = SQRT(5)/5 Cosine angle = 2 * SQRT(5)/5 If we take DAO and move it so it butts AT along the side that is 1 the radius becomes a chord in an esoteric circle. The chord is 2 * SQRT(5)/5 . This is not what we want What we want is the half chord of double the angle and its bisector. The formula for doubling a chord on unit circle is is 2 chord angle * bisector Which in this case it’s (2* SQRT(5)/5)^2 this gives 2 4*5/25 = 40/25 or 8/5 If we seek the half chord it’s 4/5. I am now going to focus on the triangle AEB which is a right triangle this is the reciprocal angle ADE, since AD and BC are parallel the value on the unit circle represent by 4/5 is AB and AB°E is a right angle meaning the triangle has proportions of 3-4-5 (3/5-4/5-5/5). 4/5 * scale = 1 therefore scale = 1/(4/5) = 5/4. That is the “Length = ?”

the result, only if the base of the rectangular is 3/4 of the diameter, or?

Love your sense of humor.

I did it by similar triangles ratio of sides…Y/0.5= 0.5/1, Y=0.5x0.5, Y=0.25

My solution for this problem was way too long ... your solution is so much better

A very similar problem to this has already been presented elsewhere on you tube this is just a repeat. Let’s label so points A and B are the right and left ends, respectively, of the diameter of the semicircle with origin 0. AC is a chord of length 10 u with C a point between A and B on the semicircle. Let’s label two more points D, E and F. D is point (0,r), E is point (r,r) and F is point (r- horizontal traverse of AC,r). Length EF then equals (r,r) - (r- horizontal traverse of AC) = horizontal traverse of AC. How do we define a chord in terms of radius. r * chord θ (on unit circle) = length In this case r must be greater than 5 so that r * chord θ = 10 How do we define r there are two ways. Chord θ calculation of Chord θ = 2 sin (θ/2) Therefore r = 10/ chord θ or 10/(2 sin (θ/2)) Another useful value are halfchords and bisectors r = 10/( 2 * halfchord) The horizontal traverse of a chord with one point on the horizon is r * 2 * halfchord/ r)^2 = 2*(chord/(2r))^2 = 2*chord^2/4r^2 2*chord^2 / 4r^2 = 0.5 chord^2/ r = 0.5 * 10^2 / r = 0.5 * 100/r = 50/r So the height of the rectangle of base at y = 0 is D - O = (0,r) - (0,0) = r length Height times width = r u * 50u / r = 50 Domain of this answer is for all r >=5 and for all 0° < θ <=180°. The reason it cannot be zero is when that occurs r approaches infinity. To test we make chord = 5 * chord 180° = 5 * 2 = 10 since r is 5 and horizontal traverse is 10 the area is 50. Let’s test r * chord 60° = 10 = r*1 = 10 then r = 10 The horizontal traverse of 60° on the unit circle is 2 * halfchord^2 = 1/2 Rescaling to radius gives us 5 5*10 is 50 We test at 90° chord 90° = SQRT(2) r = 10/SQRT(2) the rectangle is a square of sides r r^2 = (10/SQRT(2)) = 100/2 = 50 We test at 120°, which has a chord 120° = SQRT(3) 10/SQRT(3) = r. The horizontal traverse = 2r * SQRT(3)^2 * 1/4 = 1.5 r 1.5 (10/SQRT(3))^2 = 1.5 * 100/3 = 50 How about a tiny angle say 0.00001 degree Chord 0.00001° We don’t know what that is but we do know the bisector is nearly 1. If that is the case 0.0000001 degrees the chord 0.0000001° = pi/180 * 1/10E7 Thus r = 100/ (pi/(180*10E7) = 100*180*10E7/pi = 1.8E11/pi The traversed area r * 2 * (chord/2r)^2 = 200r/4r^2 = 50/r (1.8E11/pi) * 50/ (1.8E11/pi) = 50 Let’s see Angle = 0. How many chords are in a circled subdivided by zero degree angles. The chord of 0° is zero. Thus there are infinite numbers of chords. Moreover 10/0 is undefined. The traverse 360/0.00001 = 36,000,000 of its fraction of 2pi 2pi/36,000,000 is the chord 0.00001° r = 100/(2 pi/ 36000000) = 50*36000000/pi So the bisector The horizontal traverse is 50*36000000*2* ((2pi/36,000,000)/(2*50*3600000))^2= (50*36000000/pi)*2(2pi^2/36000000^2)/(100*36000000^2) = (100*36000000/pi) * (2pi^2/100) = 360000

He did not really answer the question

😮

True difference of two square number is always odd number

😢

Wouldnt it be easier to do a 6 8 10 triangle?

What level math is this?

Sorry man, the other guy shows and explains it better.

Area of green region = 270-81π square units.

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...)

whats wrong with pythagorus? . Much simpler

Very good teacher

You 're so wrong.🙄

tks a lot!

Draw a regular nonagon. It is so easy to see the result.

3{4}2 > 3^^4

The explanation was way too fast

The solution and the ratios are completely wrong. The area varies depending on the angle alpha. By symmetry APB and BPC are always 135 degrees. The angle alpha can be 30 degrees and this time AC has length two times of AP which is 20. Then the area becomes 200

How 😂

3.6 hrs

HOW TO FIND THE EXACT EXACT EXACT CIRCUMFERENCE OF CIRCLE . Procedure Step 1. Draw a square, 2 diagonals and inscribe a circle in the square. Thereby side and diameter will be the same Step 2 : Subtract 2 diagonals from the perimeter of square Step3. Divide step2 with 8 Step4. Add 3 times of the side. to Step 3 At the end we get the EXACT circumference of the inscribed circle

HOW TO FIND THE EXACT EXACT EXACT CIRCUMFERENCE OF CIRCLE . Procedure Step 1. Draw a square, 2 diagonals and inscribe a circle in the square. Thereby side and diameter will be the same Step 2 : Subtract 2 diagonals from the perimeter of square Step3. Divide step2 with 8 Step4. Add 3 times of the side. to Step 3 At the end we get the EXACT circumference of the inscribed circle

HOW TO FIND THE EXACT EXACT EXACT CIRCUMFERENCE OF CIRCLE . Procedure Step 1. Draw a square, 2 diagonals and inscribe a circle in the square. Thereby side and diameter will be the same Step 2 : Subtract 2 diagonals from the perimeter of square Step3. Divide step2 with 8 Step4. Add 3 times of the side. to Step 3 At the end we get the EXACT circumference of the inscribed circle

= Every place every living for a week post address proof of time to be x draw the location of the text ramayan and mahashivratri