Thank you, sir, for posting this question and I had been following for almost a year!

@PreMath

8 ай бұрын

Glad it was helpful! You are very welcome! You are the best, Luckshmana dear. Keep it up 👍

Just another solution: Let the legs of the triangle be a and b; then {a^2+b^=s^2=400 and a*b/2=96}; so a=12, b=16 or vice versa. semiperimeter of the triangle p = (20+12+16)/2=24; radius of a circle inscribed in a triangle r = Atr/p= 96/24=4 Acirc = pi*r^2=16*pi

@noreldenzenky1527

8 ай бұрын

This is my solution it is faster and easir

@PreMath

8 ай бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

I'm afraid your solution has some problems. If we find the area of the outer square; 400 + 4 * 96 = 784. This is 28^2. Presumably triangle EDH has sides of ED=14, DH=14 and EH=20. The only problem is that such an isosceles triangle has an area of 97.98 and not 96. For your problem statement to work the inner square would have to be slightly rotated (1.17... degrees) and then the triangle EDH would not be isosceles and the rectangle POND would not be square. The problem can still be solved, but not as you showed.

@batchrocketproject4720

8 ай бұрын

[I'm wrong here, sorry] The diagram is wrong (see my comment above) the area of the blue square must be 1/2 the whole area if its corners touch the outer square as shown, making the corner triangles 100cm^2.

@rey-dq3nx

8 ай бұрын

You are assuming ED and DH are equal, they are not. in fact, given the area is 96 cm² and hypotenuse is 20 cm you can solve for the legs. This triangle is a Pythagorean triple, 12,16,20

@marioalb9726

8 ай бұрын

Side of blue square: b = √400 = 20 cm Height of white right triangle: A = ½ b.h = 96 h = 2A/b = 2. 96 / 20 h = 9,6 cm Radius of Circle: h = r + r/cos45° = 9,6 h = r ( 1 + √2) r = h / (1+√2) r = 3,976 cm Area of circle: A = π r² A = 49,67 cm² ( Solved ✓ ) Note that radius gives me 3,976 and not 4,00 There is a mistake in the statement of this exercise, areas 400 cm² and 96 cm² are not compatible

@harryragland7840

8 ай бұрын

Yes, I see my error. His method and solution are correct but the diagram could be better.

@rey-dq3nx

8 ай бұрын

@@marioalb9726, let me ask you 2 questions What base and height of a right triangle would give you an area of 96 cm² ? What base and height of the same right triangle would give you a hypotenuse of 20 cm? I’ll set up the 2 eq 1/2b*h= 96. eq1 b² +h²= 20^. eq2 solve for the base and height you can find r using two tangents on a circle theorem, b-r+h-r= 20 2r= b+h-20 r=?

Great explanation.👍 Thanks for sharing😊

By analyzing what we have, Area of EFGH = 400 and not considering the given area of each of the four triangles as 96. Then HF of the inner square = 20 * sqrt(2). Then the outer square area is the square of HF which equals 800. Then the outer square area - inner square area equals the area of the four triangles. That is, 800 - 400 = 400, Since there are four congruent triangles, each one is 400 / 4 = 100. This will give r = 4.142 and an area for the circle = 17.157PI = 53.901 square cm.

Thank you for this concept..i make sure to silve these evwryday..

Congratulations on solving the problem without first finding the lengths of the sides of the corner triangles! The straightforward approach might be to label the sides x and y and solve the set of equations xy/2 = 96 and x² + y² = 20². We take an educated guess that the triangle might be the Pythagorean triple 3-4-5 scaled up by a factor of 4, so 12-16-20. So, x = 12 and y =16 or x = 16 and y = 12. Then, xy/2 = (12)(16)/2 = 96 and our guess is correct. Then, note that the hypotenuse of ΔEDH = x + y - 2r = 20 by tangents from the same point to a circle have equal length. The tangents are x - r and y - r and must sum to the length of the hypotenuse, which also equals the side of the square, or 20. So 12 + 16 -2r = 20 and r = 4. From there, we solve for the area in the circle as shown in the video, A = πr² = π(4)² = 16π.

@MarieAnne.

8 ай бұрын

You can actually solve using your method without finding values for x and y. Since all corner triangles are congruent with legs x and y, then larger square has side length = x + y. We can calculate area of larger square as follows: (x + y)² = 400 + (4 * 96) = 784 → x + y = 28 So using hypotenuse of ΔEDH = x + y - 2r = 20 gives us 28 - 2r = 20 and therefore r = 4.

@Nothingx303

8 ай бұрын

Bro you can get the values of x and y by following this steps (X+ y)² -2xy =400 X+y = 28 Now I will use the algebraic identity That is (x+y)² -(x-y)² =4xy We get x-y =4 And after solving these equations we get x = 16 and y=12 Easy no quadratic only linear equation

@rehadeniz

4 ай бұрын

better :)

Corner triangles of larger square are congruent right triangles (as shown in video) with legs = x, y. Let AF = BG = CH = DE = x AE = DH = CG = BF = y So larger square has side length = x+y Area of larger square = (x+y)² = 400 + (4 * 96) = 784 → x+y = 28 Using two-tangent theorem, we get: DP = DN = r EP = EM = a HN = HM = b and this gives us: DE = DP + EP = r + a = x DH = DN + HN = r + b = y EH = EM + HM = a + b = 20 Now we add first two equations, and subtract the third one to get: (r + a) + (r + b) - (a + b) = x + y - 20 = 28 - 20 2r = 8 r = 4 For those interested, since xy/2 = 96 (xy = 192) and x+y = 28, then x, y = 12, 16 But as in the solution shown in video, we don't actually need to find this. We just need to know x+y.

@ieeaswaran

8 ай бұрын

My steps, somewhat mirroring yours @MarieAnne. were: a) We know from sqrt400 = 20 cms that the hypotenuses of the four corner right triangles are 20 cms each. b) Prove that the four corner right triangles are congruent using ASA and the fact that alpha + beta + 90 degs = 180 degs, where alpha and beta are the angles as identified in the diagram at one point in the video. The right angles of the inner and the outer squares can be used conveniently to establish ASA correspondences for all four corner triangles. c) Find out the side length of the outer square = sqrt784 = 28 cms. d) And since all four corner right triangles are congruent, we can conclude that the two non-hypotenuse sides of each corner right triangle add up to 28 cms. e) I think we now have all we need, since it's given that the area of each corner right triangle is 96 sq cms. f) Perimeter of each corner right triangle = 48 cm, so semiperimeter = 24 cms. g) Inradius = Area of triangle/semiperimeter, therefore it's 96 sq cms/24 cms = 4 cms. h) Area of incircle = pi x r squared = 16 pi. I found this is a simple and straightforward approach. There's no need for complicating the solution with any so-called "out-of-the-box" thinking (as the solution offered in the video does). I think I should NOT mention the completely MISLEADING diagram which has put so many off track!! 😜😜😜😜 My questions: a) Am I missing something? Is it at all necessary to find out the lengths of the non-hypotenuse sides? Do we miss something by not doing that? b) Is it at all necessary to use the two tangent theorem or construct triangles using the inradius, as has been done in the video?

Thanks for video.Good luck sir!!!!!!!!!!!!!!!!

Very good your explanation 👍

Thank you!

Thank you for good solution.

HD=14, HM=10=HN=HD-r=14-r ; r=14-10=4

@PreMath

8 ай бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

Let EF = FG = GH = HE = c, let AF = BG = CH = DE = b, let AE = BF = CG = DH = a So we have c = √400 = 20 and ½ ab = 96. Also, by Pythagoras, a² + b² = 20². Then b² = 20² − a² and b² = 2² 96² / a² ⇒ a⁴ − 20² a² + 192² = 0 ⇒ a = 12; b = 16. The radius (r) of the yellow circle is ½ (a + b − c) = 4 hence its area = π r² = 16π.

@PreMath

8 ай бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@Chriib

8 ай бұрын

This is the way I did it but after concluding that the sides of the right angle triangle is 12, 16 and 20 I took 12*16/(12+16+20)=4 following the formula for the radius r of a circle inscribed in a triangle with area A and perimeter p, A=pr/2 ---> r=2A/p

@ybodoN

8 ай бұрын

@@Chriib of course, r = ab / (a + b + c) is also correct to find the radius of a circle inscribed in a right triangle. r = ½ (a + b − c) follows from the other formula r² = (s - a)(s - b)(s - c) / s which follows from Heron's formula.

@Chriib

8 ай бұрын

@@ybodoN Actually my formula works on every triangle whilst yours only works on right aangle triangles. In this case it won't matter though since we do have a right angle triangle.

A quick glance at this and two squares are formed, each of area 2 * 96 = 192cm ^2. sqrt(192) =13.86 giving the outer square of equal sides 27.72 , this contradicts the comment from @ybodoN below who disputes AE and AF are equal with two unequal lengths of 12 and 16. I Think I will watch the video later.

@benv6875

8 ай бұрын

That was my first attempt also, but it doesn't work. Please look at my solution I just posted. Thanks

@tombufford136

8 ай бұрын

Yes I was not giving a solution but commenting on one of the other comments given that while they are likely expert their description was 'awkward'. Thank you for replying.@@benv6875

1/Green square side = 20 cm 2/Four right triangles at the 4 corners are congruent (angle-side-angle), 3/ Calculating the radius r of the inscribed circle in the triangle EDH Let DE, DH and EH be a,,b, and c respectively, we have: a.b= =2x96= 192 (1) sq(a+b) = sq a +sq b + 2a.b= sq c + 2.ab = 400 + 2x192=784= sq 28------> a+b =28 (2) -------->1/2 r(a+b+c)= 96------> 1/2 r(48) = 96 ----> r=192/48=4cm Area of the yellow circle= 16 pi sq cm

I have watched the video and it is an awkward problem. After some thought the solution makes sense and the video includes some useful skills. Two Tangent congruency and others. Summing all the different shapes, rectangle, square and triangles using equalities (equal triangles) before calculating the circle area requires some care. Nice Video !

a^2+b^2=400, (ab/2)((20-a)/b)^6=96 to solve for a, b ,and then the radius can be determined, so its area.😅

@PreMath

8 ай бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

Thanks for your solution - but in this solution, we have forgot to take into account twox1/8 squares portion to either side of corner square DO - Kindly correct if I am wrong.

Yay! I solved the problem.

Am I being very dumb, or shouldn't each of the 96 sq cm triangles actually be 100cm sq... ie a quarter of the area of the blue square.

@fadetoblah2883

8 ай бұрын

Not having watched the video yet, I think the unspoken implication to be derived from the given measurements is that the smaller square is slightly tilted, rather than perfectly "perpendicular" to the larger square. As a result, the two side lengths of the triangles (let's label them a and b) are not equal. Knowing that a^2 + b^2 = 400 and that ab = 192 (i.e. 96 x 2), you can solve for the values of a and b, which are a = 12, b = 16 or vice versa. From there, using the two tangent theorem, you can easily figure out the radius of the circle: (12 - r) + (16 - r) = 20 28 - 2r = 20 r = 4 So the area of the circle should (hopefully) be 16 pi. [Yup, it checks out.]

@batchrocketproject4720

8 ай бұрын

yes, the blue square has to be exactly half the outside area and each triangle 1/8th the total area. That's the only way the inner square can touch the outer one on all four of its conrners.

@rey-dq3nx

8 ай бұрын

@@batchrocketproject4720 have you tried to construct the figure on graphing paper? Draw a 28x28 square. Then draw triangles in each corner with legs measuring 12x16 units. The 4 hypotenuses will form a square that has a side length of 20 units. Incidentally, 12² + 16² = 20²

@marioalb9726

8 ай бұрын

This drawing is RUDELY OUT OF SCALE !!!! The rotation angle is not 45°. Half of the other comment assumed rotation angle is 45° and are finding differents values for radius of circle Let's calculate that rotation angle, for compatibilizing those areas 400cm² and 96cm². 96 cm² = ½ b.h b.h = 192 b²+h²=20² b.(20²-b²)^½ = 192 b² (20²-b²) = 192² 400b² - (b²)² = 192² (b²)² -400 b² + 192² = 0 Clearing: b = 16 cm h = 12 cm tan α = 12/16. α = 36,87° instead of 45° tan α/2 = r/(16-r) (16-r)/r = 1 / tan α/2 16/r - 1 = 1 / tan α/2 16/r = 1 + 1/tan α/2 r = 16 / (1 + 1/tan α/2) r = 4 cm. ( Verified √ ) Rotation angle is 36,87° instead of 45° !!! RUDELY OUT OF SCALE

@ieeaswaran

8 ай бұрын

Blame it on the diagram! It's a con, a yorker, a blooper, a dodge...call it what you may! 😀😀😀😀😀😀😀😀

Something wrong in calculations. If lenght is 14 then half lenght area will be 98 n not 96

@vr2209

8 ай бұрын

Moreover, if the side is 14 then the diagonal is 14√2, not 20

@rey-dq3nx

8 ай бұрын

The lengths are not 14 cm. Try drawing the figure graphically. Draw the outer 28x28 square on a graphing paper. Then at each corner draw a triangle whose legs are 12 and 16. After you draw the 4 hypotenuse, you have just formed a square with 20 units side lengths.

@marioalb9726

8 ай бұрын

There's a mistake in the statement of this exercise Areas 400 an 96 are not compatible The ratio of these areas has to be 4 400 / 4 = 100 (and not 96) or 96 x 4 = 384 (and not 400) or 392 / 98 = 4 (and not 400 neither 96)

Let the two sides of the triangles are a and b a^2+b^2=400 , ab=192 , a^2+b^2+2ab=784,(a+b)^2=784,a+b=28 we can observe that the the two sides EP and NH are equal to a-r and b-r and by using the two tangent theorem we can said that the side of the blue square = a+b-2r=20 ,2r=8, r=4 ,area of the circle =16r

@ahmedhakim3106

8 ай бұрын

*pi

@PreMath

8 ай бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@PreMath

8 ай бұрын

Got it! You meant 16pi

@ahmedhakim3106

8 ай бұрын

​@@PreMath❤

EH=√400=20 =a+b → Lados ∆EDH: (a+r)/(r+b)/(a+b) → Área ∆EDH =96=ar+br+r²→r²+20r-96=0 → r=4 → Área amarilla =πr² =16π Gracias y un saludo cordial.

تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .

@PreMath

8 ай бұрын

Thanks dear❤️

Because of the angles of the inset squares, all four triangles are the same, just rotated 90 degrees. The total area of the big square is 400 + 4 * 96 = 784. The side length is the square root of that: 28. The side length of the blue square is the square root of 400: 20. Break the triangle DEH based on the tangents with the circle. Because they are tangent to a circle and meet at points, we can pair the segment lengths and label them: DE = r + a DH = r + b EH = a + b because of the 90 degree angles making a square, we know r is the radius of the circle. DE + DH - EH = r + a + r + b - a - b DE + DH - EH = r + r DE + DH - EH = 2r Because the triangles are the same, DE = HC. HC + DH - EH = 2r HC and DH are two parts of the longer line DC. DC - EH = 2r 28 - 20 = 2r 8 = 2r 4 = r Area of circle: A = Pi r^2 A = Pi * 4^2 A = 16 Pi

@Copernicusfreud

8 ай бұрын

That is how I solved the problem. If the vertices of the inner square touch the lines of the outer square, then the vertices of the inside square must be at the midpoint of the edges of the outside square. I agree with the other posts in that the area of the outside triangles should probably be 100 cm^2, instead of the 96 cm^2 given in the problem. The 96 cm^2 x4 +400 cm^2 makes the side a perfect 28 cm.

@TurquoizeGoldscraper

8 ай бұрын

@@Copernicusfreud The diagram looks like the inner square touches the outer square at the midpoints, but it's actually canted a bit. As the video says, the diagram may not be true to scale; so unless you are told it's at the midpoints, don't assume it.

@Copernicusfreud

8 ай бұрын

My thinking is that if the interior polygon is slightly off, then it cannot be a square. The interior polygon is a quadrilateral, but not quite square. All the small triangles are the same area means that it must be square. If it is square, it must not be canted.

Triangles outside blue square should be 100 each so part of the computation should 100 not 96. Area of big square is 800 minus 400 hundred blue square remaining 400 so each corner triangle is 100 not 96.

Though it appears to be the case, it is unclear whether the vertices of the blue square/diamond EFGH are coincident with the midpoints of the sides of square ABCD. Luckily we can use the helpfully supplied area of triangle ∆GCH to verify that. The area of EFGH is 400, so the sides are each √400 = 20. As ∆GCH _should_ be an isosceles right triangle if the EFGH vertices are at the midpoints of the sides of ABCD, GC and CH should be equal to (1/√2)20 = 10√2 and give a value of 96 when the area is calculated. A = bh/2 = (10√2)²/2 = 200/2 = 100 96 ≠ 100 As it turns out, the vertices of EFGH are not at the midpoints. We can determine the total area of the large square and find its side length from that. A = S² 400 + 4(96) = S² S² = 400 + 384 S = √784 = 28 Since it doesn't particularly matter, we'll assume that GC > CH, and as all four triangles share straight lines and are separated by right triangles, they are all congruent, the the two legs of each triangle add to 28. Let GC = x and CH = y. CH² + GC² = HG² (28-x)² + x² = 20² 784 - 56x + x² + x² = 400 2x² - 56x + 384 = 0 x² - 28x + 192 = 0 (x-12)(x-16) = 0 x = 12 ❌ | x = 16 CH y = 28-(12) | y = 28-(16) y = 16 ❌ | y = 12 CH So GC = HD = EA = FB = 16 and CH = DE = AF = BG = 12. As HD = 16 = 4(4), DE = 12 = 3(4), and EH = 20 = 5(4), ∆HDE (and the other three triangles) are 4:1 ratio Pythagorean triple triangles. Let S, T, and U be the three points of tangency for circle O, counterclockwise from the top. As SD and DT are converging tangents from the same circle, SD = DT = r, since ∠DTO = ∠OSD = 90°, due to intersecting radii and tangents. Thus HS = 16-r and TE = 12-r. Also due to converging tangents (this time converging on H and E), EH equals the sum of those two segments. RH = HS + TE 20 = (16-r) + (12-r) 20 = 28 - 2r 2r = 28 - 20 r = 8/2 = 4 A = πr² = π4² = 16π cm²

Professor"s answer is right, radius is 4 cm indeed and area of the circle is 16*PI. I was determined to proof this using trigonometry. There is only one right triangle with long side 20 cm and area 96 cm^2. It's other sides are 12 cm and 16 cm. See step by step solution below. I will solve set of two equations. 1. 12/16 = tang x = 2 tang (x/2)/(1-tang^2(x/2)); 2. R/(16-R) = tang (x/2); 12/16 = 2 R/(16-R)*(1-[R/(16-R]^2); 12/16 = 2 R/(16-R)*(1-[R^2/(256-32R+R^2]); 3/4 = 2 R/(16-R)*[(256-32R+R^2-R^2)/(256-32R+R^2)]; 3/4 = 2 R*(256-32R+R^2)/[(16-R)*(256-32R)]; 3/4 = (512R-64R^2+2R^3)/(4096-512R-256R+32R^2); 3/1 = (2R^3-64R^2+512R)/(8R^2-192R+1024); 2R^3-64R^2+512R-24R^2+576R-3072 = 0; 2R^3-88R^2+1088R-3072 = 0; R^3-44R^2+544R-1536 = 0; This is cubic equation. I try R1 = 4. (R-4)(R^2-40R+384) = 0; R-4 = 0, R1=4 cm; This is first root. Or R^2-40R+384 = 0; This is quadratic equation. (R-16)(R-24) = 0; R-16 = 0 Or R-24 = 0; R2 = 16 cm Or R3 =24 cm; Both roots are rejected. they are to big. Final answer is: R = 4; Area = 16*PI; I myself suspected problem with professor's answer, but radius is 4 cm indeed. He is correct. Thanks for reading.

Here's a somewhat simpler treatment. All letter-point references are to your diagram at 8:00. The blue square is 400 square units, so its side EH is √400 = 20. The big square is 400 + (4)(96) = 400 + 384 = 784, so its side DC is √784 = √((4)(4)(49)) = √((4^2)(7^2)) = (4)(7) = 28 (had a hunch it was an integer). This means that MH = NH = 20/2 = 10; DH = 28/2 = 14; and r = DN = (14 - 10) = 4. From there we get the area to be 16π, as you did. But I first tried to do it by saying the distance MD = (1 + √2)r and also by Pythagoras, MD = √(14^2 - 10^2) = √(196 - 100) = √96; so r should equal (√96)/(1 + √2). On the calculator, this keeps coming out 4.058, approximately 1.5% high. One of life's little mysteries.... Cheers. 🤠

@ybodoN

8 ай бұрын

Mysteriously, on the diagram, DEH is an isosceles right triangle but its area isn't 100 as should be 😉

@marioalb9726

8 ай бұрын

THERE IS A MISTAKE in the statement of this exercise Sometimes r= 4 Sometimes r= 4,058 Sometimes r= 4,14 Sometimes r= 3,976 depending on the way to solving !!! Area of blue square 400 cm² is not compatible with area 96 cm² of isosceles right triangle

@williamwingo4740

8 ай бұрын

@@marioalb9726 Looking at it again, by symmetry (reflecting each white triangle about its base) the sum of the four white triangles should be 400, the same as the blue square; so they should be 100 each, not 96. And the diameter of the blue square should be 20√2 rather than 28. That's what you get for assuming a problem is stated correctly. But that's all right: I made a mistake myself once. Years later, I thought I'd made another; but it turned out I was wrong about that. 🤠

@ybodoN

8 ай бұрын

Remember that "this diagram may not be 100% true to the scale!" ⇒ we cannot assume that CG = CH or DE = DH. A right triangle with an area of 96 cm² and a hypotenuse of 20 cm can only be a (12, 16, 20) Pythagorean triangle.

Triangle DEH has two 45 degree angels. Hence ED=DH=10 sqrt(2). The area of the corner triangle is therefore (10*sqrt(2)*(10*sqrt(2) divided by 2= 100 cm2. This will give a radius of about 4.14 cm.

@marioalb9726

8 ай бұрын

Side of blue square: b = √400 = 20 cm Height of white right triangle A = ½ b.h h = 2A/b = 2. 96 / 20 h = 9,6 cm Radius of Circle: h = r + r/cos45° h = r ( 1 + √2) r = h / (1+√2) r = 3,976 cm Area of circle: A = π r² A = 49,67 cm² ( Solved ✓ ) THERE IS A MISTAKE in the statement of this exercise Sometimes r= 4 Sometimes r= 4,14 Sometimes r= 3,976 depending the way to solving !!!

@marioalb9726

8 ай бұрын

It is not possible areas 400 cm² and 96 cm², at the same time. Or one or the other !!

@MarieAnne.

8 ай бұрын

@@marioalb9726 It is possible. You're assuming that blue square has been rotated 45 degrees from the horizontal/vertical, and that the vertices of the blue square are halfway along the sides of the white square. This is not so. The corner triangles have legs = 12 and 16 and hypotenuse = 20. You cannot just ignore one of the facts given, and add a fact of your own that is not given (isosceles right triangles).

@MarieAnne.

8 ай бұрын

*@leowald1* You're assuming that the right triangles in the corners of the white square are isosceles. They are not.

@Abby-hi4sf

8 ай бұрын

@@MarieAnne. do you agree the diagonal of the blue triangle is the length or the width of the outer square ABCD? If yes the blue square vertex are on the mid point of the outer ABCD sides lines. So the only way the blue side vertex touch all sides of ABCD is tilted 45. Then all triangles are congruent and isoceless. And their area can not be 96 cm

Can we not say DE = DH , as DP=DN and PE =PM, and applying Pythagoras in DEH gives DE and hence radius.

Let's solve the system of equations: DE×DH= 192; DE^2+DH^2=400. DE=12; DH=16. Inscribed circle radius r= 2×96/(12+16+20)= 4 Area of a circle A= π×4^2= π×256. Thank you sir

Setting the area of the right triangle. 96 is a false assumption, It should be 100.

radius x (a+b+c)/2 = area of triangle = 96, so radius = 4. Why are we not solving this simply with the inscribed circle formula? If a circle is inscribed (exactly touching the three sides) in any triangle, the formula for the triangle's area uses the circle's radius, and that radius is all we need to solve the total problem. This video shows why the four corner triangles are the same, so the area of the outer square is (4x96) + 400 = 784, which is 28x28, so the side length of the outer square is 28, so the corner triangles have sides summing to (28 + 20) = 48. Radius x 48 / 2 = 96, so radius = 4.

I have a doubt professor. Square area plus rectangle area = EDH area ? There are 2 outer triangles ( exterior triangles) . After extending the lines from the square ( 400 sq units ).

Simpler way r =(a+b-c)/2 Where a ,b,c is the sides of triangle a^2+b^2=c^2=400 ,ab=192 so a=12 ,b=16,c=20, So r=4 ,area=16pi

@PreMath

8 ай бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

Poor diagram. The blue square is tilted to the right forming a 12 by 16 right triangle. but the diagram does not look any way near that 12 * 16 = 192 192/2= 96 and that is why the corner is 96 cm^2 12^2 + 16^2 = 20^2 . The 20^2 = 400 cm and that is why the area of the blue is 400 144 + 256 = 400 Let the base of the triangle = 12 and the height = 16 Draw a perpendicular line from the BASE of the triangle to the center of the yellow circle to the hypotenuse of the triangle. This length is 12- r ( tangent circle theorem). Hence part of the hypotenuse is 12 - r To calculate the remaining part, draw a perpendicular line from the HEIGHT of the triangle to the center of the yellow to the hypotenuse of the triangle. This length is 16- r ( tangent circle theorem). Hence the remaining part of the hypotenuse is 16-r The combined length of the hypotenuse in terms of r is 28 - 2r (16-r + 12-r) So 12, 16, and 28- r will be used to find r using the Pythagorean Theorem in which 28- r is the hypotenuse 12^2 + 16^2 = (28-r)^2 400 = 4r^2 + 784- 56 r - 56 r 400 = 4r^2 + 784 -112 r 0 = 4r^2 + 384 - 112 r 0 = r^2 + 96 - 28r 0 = (r-4)(r-24) r=4 and r=24 can't use 28 too big r=4 Area of circle = r^2 pi = 4^2 pi = 16 pi answer

AS others may have pointed out, the diagram is MISLEADING (someone wrote RUDELY misleading!) -- till you realise that 96x4 = 384 sq cm does not add up to 400 sq cm, hence the four corner triangles aren't right isoceles triangles. From the given figure, one is likely to jump to the conclusion that each side of the outer square is 20xsqrt2, and the (presumed) equal sides of the corner right triangles are 20sqrt2/2. An accurate hand drawn diagram would have helped, but the 96x4 not equal to 400 check would do the trick. Alas, many will miss it. The caveat at the beginning says 'The diagram MAY not be 100% to scale,' but the person explaining the problem does not at any point say that the side length of the outer square cannot be assumed to be 20sqrt2 because the diagonals of the blue square are NOT perpendicular to the sides of the outer square. In fact, in this case at least, he should have stated at the outset that the diagram is OUTRIGHT INACCURATE and that the outer square is therefore NOT double the area of the inner square, as many would hastily conclude. An interesting problem, but with a yorker (some would say, con) thrown in! 😆😆😆😆

I think the area of triangle should be 100: diagonal of inside square is the side of outer square which is equal to sqrt of 20sq + 20sq =sqrt of800: therefore the area of outer square =sqrt800 xsqrt800=800. Therefore area of triangle =[800-400]/4=100

The side of square is 400+4x96 under root=28. Perimeter of triangle would be 28+20=48. So r=96x2/48=4. So simple.

Sir by an educated guess we get the side of the each triangle 🔺️ as 12 cm , 16 cm and 20 cm So to find the radius i will use this Ar(🔺️DEH)= semi-perimeter × radius We get r = 4 cm Then we can easily find out the area of the given circle please try make your solutions as simpler as possible.

Sir by the way from which country do u belong from ?

@PreMath

8 ай бұрын

The United States! Take care dear.

@arnavkange1487

8 ай бұрын

Love from India sir

@PreMath

8 ай бұрын

@@arnavkange1487 Thanks dear ❤

I got upto 4.98, but I knew the solution was close by, just didn't give it enough time - I am having dinner at the same time! I would say if you could put at the start of your video the caption "More than one way to solve". This is so viewers can know they reached the same destination, through different paths.🌴

Method N°2: In this calculation I did't use as data, the area 96 cm² Isosceles right triangle: Hypotenuse = √400 = 20 Side = 20 cos45° = 20/√2 = 10√2 Taking the appropriate right triangle, with vertex in the centre of the circle: tan (45°/2) = r / ( 10√2 - r) ( √2-1 ) . (10√2 - r) = r 5,858 - 0,4142 r = r r ( 1 + 0,4142) = 5,858 r = 5,858 / √2 r = 4,142 cm Three differents results for the same problem: Video : r = 4 cm Method 1 : r = 3,976 cm Method 2 : r = 4,142 cm There is some mistake in the statement of this exercise, areas 400 cm² and 96 cm² are not compatible

@MarieAnne.

8 ай бұрын

The mistake is yours. You cannot assume that the corner triangles are isosceles, since nowhere is this stated in the problem. The corner right triangles have legs = 12 and 16, and hypotenuse = 20.

@marioalb9726

8 ай бұрын

​@@MarieAnne. This drawing is RUDELY OUT OF SCALE !!!! The rotation angle is not 45°. It's a mistake of the author, or It seems a yoke towards us !!!! Half of the other comment assumed rotation angle is 45° and are finding differents values for radius of circle Let's calculate that rotation angle, for compatiblizing those areas 400cm² and 96cm². 96 cm² = ½ b.h b.h = 192 b²+h²=20² b.(20²-b²)^½ = 192 b² (20²-b²) = 192² 400b² - (b²)² = 192² (b²)² -400 b² + 192² = 0 Clearing: b = 16 cm h = 12 cm tan α = 12/16. α = 36,87° instead of 45° tan α/2 = r/(16-r) (16-r)/r = 1 / tan α/2 16/r - 1 = 1 / tan α/2 16/r = 1 + 1/tan α/2 r = 16 / (1 + 1/tan α/2) r = 4 cm. ( Verified √ ) Rotation angle is 36,87° instead of 45° !!! RUDELY OUT OF SCALE

@KenFullman

4 ай бұрын

IF the inner square was rotated by exactly 45 degrees then the four corner triangles would add up to exactly 400 cm² (so each triangle would have an area of 100) This was your clue that the inner square is not rotated by exactly 45 degrees. Ther route I took was: Obviously all four small triangles are congruent. We can deduce the hypotenuse of △EHD from the square root of the blue square (20 cm). We also know the area of △EHD is 96 Put these together and we derive a quadratic equation that gives two answers 12,16 So that gives us the shorter sides of the triangle. Then using the side lengths of 12,16 and 20 we find the inscribed circle has a radius of 4 From that we can then get the area of 16π

400+96＊4=784 784=28^2 28/2=14 400=20^2 20/2=10 r^2+(14-r)^2=r^2+10^2 r^2-28r+96=0 (r-4)(r-24)=0 0

@PreMath

8 ай бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@marpaub

8 ай бұрын

Can you please explain this equation: r^2+(14-r)^2=r^2+10^2 ?

ab=192, a^2+b^2=400, a^2+(192/a)^2=400, a^4-400a^2+192^2=0, a^2=256 or 144, r=16 or 12, (s-16)/t=3/4, (s-16)^+t^2=400, 4u=3t, u^2+t^2=400, 9t^2+16t^2=400, 25t^2=6400, t=16, (s-16)=(3/4)(16)=12, s=16+12=28, that is the side of the larger square, and all triangles are 12x16, therefore r=4, and the answer is 16 pi=50.265482 approximately. 😅

Thnx. But is there any easier way to solve this problem?

@MarieAnne.

8 ай бұрын

Here is my solution: Corner triangles of larger square are congruent right triangles (as shown in video) with legs = x, y. Let AF = BG = CH = DE = x AE = DH = CG = BF = y So larger square has side length = x+y Area of larger square = (x+y)² = 400 + (4 * 96) = 784 → x+y = 28 Using two-tangent theorem, we get: DP = DN = r EP = EM = a HN = HM = b and this gives us: DE = DP + EP = r + a = x DH = DN + HN = r + b = y EH = EM + HM = a + b = 20 Now we add first two equations, and subtract the third one to get: (r + a) + (r + b) - (a + b) = x + y - 20 = 28 - 20 2r = 8 r = 4 For those interested, since xy/2 = 96 (xy = 192) and x+y = 28, then x, y = 12, 16 But as in the solution shown in video, we don't actually need to find this. We just need to know x+y.

@ashrafsafwat3688

8 ай бұрын

Your solution is direct much easier way and I consider it the model answer. Thanks a lot and I am indebted for your explanation. By the way , I am a consultant physician but maths is my favourite subject because it is a pure intellectual ability.

The given figure is highly misleading, the inner square should be shown somewhat tilted. Considering the symmetry of the figure, one finds out no matter what side lengths (as their sum of squares should be less than 400) of triangle are taken (as to conjure the area of 96), the answer would remain same. Thus making the most obvious choice of side lengths 12 and 16 to accrue the area of 96. If radius be r, then by symmetry, one gets (12-r)+(16-r) = 20 => r = 4, giving the area of 16pi.

96(1+r^2+r^4+r^6)=96(r^8-1)/(r^2-1)=400, 96r^8-400r^2+304, 6r^8-25r^2+19=0, (r^2-1)(6r^6+6r^4+6r^2-19)=0, .....😅

@PreMath

8 ай бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

I think 10+r=8sqrt(3) and r=8sqrt(3)-10 r^2=64(3)+100-160sqrt(3) r^2=292-160sqrt(3) A=pi(292-160sqrt(3))

Side of blue square: b = √400 = 20 cm Height of white right triangle A = ½ b.h h = 2A/b = 2. 96 / 20 h = 9,6 cm Radius of Circle: h = r + r/cos45° h = r ( 1 + √2) r = h / (1+√2) r = 3,976 cm Area of circle: A = π r² A = 49,67 cm² ( Solved ✓ )

@marioalb9726

8 ай бұрын

Method 3: Isosceles right triangle: Area = 96 cm² = ½ s² s = 13,8564 cm This side is different respect to method 2. Taking the appropriate right triangle, with vertex in the centre of the circle: tan (45°/2) = r / ( 13,8564 - r) ( √2-1 ) . (13,8564- r) = r 5,7395 - 0,4142 r = r r ( 1 + 0,4142) = 5,7395 r = 5,7395 / √2 r = 4,058 cm In this calculation I did't use as data, the area 400 cm² Four differents results for the same problem: Video : r = 4,000 cm Method 1 : r = 3,976 cm Method 2 : r = 4,142 cm Method 3 : r = 4,058 cm There is some mistake in the statement of this exercise, areas 400 cm² and 96 cm² are not compatible

@Abby-hi4sf

8 ай бұрын

@@marioalb9726 The picture is confusing, but the triangles are 30, 60 , 90 let the blue side = S, X² + Y² = S² = 400, and XY/2 = 96 the sides of the triangle X, Y, S, = 12, 16 20 (x+y) ² = x ² + y ² + 2xy = 400 + 2(192)= 784 x+ y= 28 xy/2 = 96 the sides are X, Y, S, = 12, 16 20 BY TANGENT CIRCLE THEOREM. side S = 20 = (12- r) + ( 16 - r) = r = 4

@marioalb9726

8 ай бұрын

@@Abby-hi4sf Good observation !!! But in that case, the picture is rudely out of scale !!!, any bad drawing is a mistake of the author. Students have found four differents values of 'r' , since 3,97cm until 4,14cm, depending on the method of resolution. Most assumed that the rotation angle was 45°.

@phungpham1725

8 ай бұрын

@@marioalb9726 the four white right triangles at the 4 corners are congruent, but not isosceles. That is why the problem is tricky! If they are isosceles, the area of each must be 100 sq cm.

@marioalb9726

8 ай бұрын

@@phungpham1725 Please, read my last comment. The figure is grossly out of scale, This is an exercise very misleading, half of subscribers fell into the trap. Those triangles are not isosceles, but are drawn as such. Also in other comment, I have calculated that rotation angle and it was 36,7° instead of 45°. Grossly out of scale

Very difficult for the smaller square is not sited in the middle.😮

@PreMath

8 ай бұрын

Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

What can’t you assume EM to be 20/2=10, angle OEM to be 45/2, Tan(OEM)=r/10? This gives me an r of 4.14.

@aghoover

8 ай бұрын

I guess the blue square could be slightly tilted within the large square.

Is there a mistake in the specification? If the outer shape is a also square, then the area of the rectangle can not be 96 cm². It has to be 100 cm².

@alainpeugny1146

8 ай бұрын

The geometrical figure is not as symmetrical as you may think. DC=28cms DH=12cms and ED=HC=16cms. In other words, the triangle EDH is not isosceles.

Wow! Why so complicated? This is easily done without Phyagorus, all the complicated triangles, and quadratic equations by simply using tangents to the circle. You get to the point of EM = EP =10. The greater outside square area is 784 by simply adding up the individual areas.Therefore each side is 28. ED is then 14, minus 10 (EP) which leaves 4 , which is r. What do you think? Please let me know. Thanks

How is that right triangle supposed to be 96cm^ and not 100cm^? The blue square is not simmetric inside the big square?

why not to find sides of the triangle with the area 96 and then use the formula for the radius of an inscribed in a triangle circle? It will be more elegant solution. sides of the triangle are 12, 16, 20. The formula is: r=2x Area/a+b+c. The area of a circle is 3.14xr^2.

@ZbigniewRocki-se2ep 1 second ago I found that radius is not 4. It is more than 4. Triangle sides are 20,16,12. 16*12/2 = 96. 12/16 = tang a = 2 tang (a/2) R/(16-R) = tang (a/2) 12/16 = 2R/(16-R) 32R =192-12R 44R = 192 R = 4.(36), A = PI * R *R = ~59.82 [cm2]

The picture is confusing, but the triangles are 30, 60 , 90 let the blue side = S, X² + Y² = S ² = 400, and XY/2 = 96 the sides are X, Y, S, = 12, 16 20 BY TANGENT CIRCLE THEOREM. side S = 20 = (12- r) + ( 16 - r) = r = 4

According to the picture, 4 x 96 cm^2 = 400 cm^2 ??? 😮😮😮😮😮😮😮😮

@marioalb9726

8 ай бұрын

There is a mistake !! in the statement of this exercise.

@bullerheden

8 ай бұрын

​@@marioalb9726Premath has never stated that the blue square is rotated 45 degrees. He has stated that the diagram is not true to scale.

@marioalb9726

8 ай бұрын

@@bullerheden This is the WORST exercise of PreMath. He should remove this video. If there is no mistake, then the drawing is RUDELY out of scale !!! and it would be a joke or mockery towards us. Just read the others comments, half of them finding differents values for 'r'.

Just use the information given in the diagram. Don't assume that it is to scale and you will find the right answer r is 4cm😊.

asnwer=180 isi t

THERE IS A MISTAKE in the statement of this exercise Sometimes r= 4,000 Sometimes r= 4,142 Sometimes r= 3,976 Sometimes r= 4,058 depending on the way to solving !!! Area of blue square 400 cm² is not compatible with area 96 cm² of isosceles right triangle

@marioalb9726

8 ай бұрын

The ratio of these areas has to be 4 400 / 4 = 100 (and not 96) or 96 x 4 = 384 (and not 400) or 392 / 98 = 4 (and not 400 neither 96)

@marioalb9726

8 ай бұрын

​​This drawing is RUDELY OUT OF SCALE !!!! The rotation angle is not 45°. It's a mistake of the author, or it seems a yoke towards us !!!! Half of the others comments assumed rotation angle is 45° and are finding differents values for radius of circle Let's calculate that rotation angle, for compatibilizing those areas 400cm² and 96cm². 96 cm² = ½ b.h b.h = 192 b²+h²=20² b.(20²-b²)^½ = 192 b² (20²-b²) = 192² 400b² - (b²)² = 192² (b²)² -400 b² + 192² = 0 Clearing: b = 16 cm h = 12 cm tan α = 12/16. α = 36,87° instead of 45° tan α/2 = r/(16-r) (16-r)/r = 1 / tan α/2 16/r - 1 = 1 / tan α/2 16/r = 1 + 1/tan α/2 r = 16 / (1 + 1/tan α/2) r = 4 cm. ( Verified √ ) Rotation angle is 36,87° instead of 45° !!! RUDELY OUT OF SCALE

S=16π≈50,24

????...(20/sqrt2)-r=10...96???

This drawing is GROSSLY OUT OF SCALE !!!! The rotation angle is not 45°. It seems a yoke towards us !!!! Half of the other comments assumed rotation angle is 45° and are finding differents values for radius of circle If rotation angle is 45°, areas 400cm² and 96cm² are not compatible. Just for exercising, let's calculate that rotation angle, for compatiblizing those areas. 96 cm² = ½ b.h b.h = 192 b²+h²=20² b.(20²-b²)^½ = 192 b² (20²-b²) = 192² 400b² - (b²)² = 192² (b²)² -400 b² + 192² = 0 Clearing: b = 16 cm h = 12 cm tan α = 12/16. α = 36,87° instead of 45° tan α/2 = r/(16-r) (16-r)/r = 1 / tan α/2 16/r - 1 = 1 / tan α/2 16/r = 1 + 1/tan α/2 r = 16 / (1 + 1/tan α/2) r = 4 cm. ( Verified √ ) Rotation angle is 36,87° instead of 45° !!!

@ybodoN

8 ай бұрын

00:44 _"Please keep in mind that this diagram may not be 100% true to the scale"_ 😉

@marioalb9726

8 ай бұрын

@@ybodoN Yes, exactly, that means lightly out of scale, slightly out of scale, not intentioned out of scale, not 100% but almost 100%. That doesn't mean RUDELY out of scale, intenctionaly out of scale !!! 70% different scales, one figure one scale, the other figure other scale. Doesn't mean this. Besides of that, Most subscribers use to solve the exercise without watching the video !!! This is the worst video of PreMath

@ieeaswaran

8 ай бұрын

That raises the question: Is it INTENTIONALLY out of scale? :-) And in this case, "out of scale" is the wrong choice of words; the diagram's actually MISLEADING.

@marioalb9726

8 ай бұрын

@@ieeaswaran Yes, it is. If you read the others comments, half of them assumed rotation angle 45°. Students have to be prepared, for misleading diagrams.

I know I'm not living in a fantasy world ...and not only am I sure that the Yellow circle has a negative radius, but that all circles have a negative radii. It's not a negative radius at all, but an imaginary radius that if squared gives a Real radius with a Real center. Welcome too planet Earth! 🙂

@PreMath

8 ай бұрын

Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[...next day. leaving comment as a reminder not to jump to conclusions, but it is wrong! apologies] The blue square cannot touch the outer square on all four of its corners unless the area of the outer square is exactly double the area of the blue square. In other words, the diagram is wrong as the corner triangles have to be exactly 1/4 of the blue square area i.e. 100 cm^2. If you have a square and fold the corners to meet at the centre, you end up with a square of exactly half the original area and four triangles exactly 1/8th of the original area each.

@batchrocketproject4720

8 ай бұрын

Apologies, I'm wrong about this. I'd assumed the corner triangles were isosceles. Sorry to anyone I confused, and to the video maker (great problem by the way, thank you)

This problem is wrong, Sqr(96)=4 Sqr(6) >> diagonal =8 Sqr(2) is not equal 20

Your solution makes no sense. The angles that you called alpha and beta are equal, and 45 degrees. You said ABCD is a square. So everything is symmetrical inside the big square and therefore you have 45 degree angles in the corner triangles. And the height of the corner triangle becomes r(1+sqrt(2)), and r= 9.6 / (1 + sqrt(2)).

@rey-dq3nx

8 ай бұрын

The professor didn’t reply because he wants you to stay ignorant

@bullerheden

8 ай бұрын

The blue square is rotated a little from the 45 degree.

@dariosilva85

8 ай бұрын

​@@bullerheden That is not possible, then some corner wouldnt touch the sides.

@bullerheden

8 ай бұрын

Note that the corner triangles are smaller than 100

@dariosilva85

8 ай бұрын

​@@bullerheden I dont know what you meant by smaller than 100. The corner triangles are 90-45-45 degrees, right?

Radius is not 4 it is 8root3 minus 10 your answer is wrong

@ZbigniewRocki-se2ep

8 ай бұрын

You are right radius is not 4. It is more than 4. Triangle sides are 20,16,12. 12/16 = tang a = 2 tang (a/2) R/(16-R) = tang (a/2) 12/16 = 2R/(16-R) 32R =192-12R 44R = 192 R = 4.(36), A = PI * R *R = ~59.82 [cm2]

@talaljaston8891

8 ай бұрын

All you have made was inputted the area as 96 instead of 98 the area is 98 and not 96 this will give you a radius of 4 and your calculation becomes correct

Method 3: Isosceles right triangle: Area = 96 cm² = ½ s² s = 13,8564 cm Taking the appropriate right triangle, with vertex in the centre of the circle: tan (45°/2) = r / ( 13,8564 - r) ( √2-1 ) . (13,8564- r) = r 5,7395 - 0,4142 r = r r ( 1 + 0,4142) = 5,7395 r = 5,7395 / √2 r = 4,058 cm In this calculation I did't use as data, the area 400 cm² Four differents results for the same problem: Video : r = 4,000 cm Method 1 : r = 3,976 cm Method 2 : r = 4,142 cm Method 3 : r = 4,058 cm There is some mistake in the statement of this exercise, areas 400 cm² and 96 cm² are not compatible

Is it possible?😢 no offense.

Triangle hypotenuse is √400 = 20 Triangle side is 20/√2 Triangle area is ½(20/√2)² = 100

@Premath Quite a brouhaha your erroneous diagram has created!

EH not equal to GH not equal to 20 GH = 8×sqroot6 EH = (25/3 )×sqroot6 ***If EH=EF=FG=HG=20 triangleHCG' area will be 100 instead.

@marioalb9726

8 ай бұрын

There's a mistake in the statement of this exercise Areas 400 an 96 are not compatible The ratio of these areas has to be 4 400 / 4 = 100 (and not 96) or 96 x 4 = 384 (and not 400) or 392 / 98 = 4 (and not 400 neither 96)