@ PreMath, Sir. I am 83 years old, I did my Matriculation back in in mid 1950's . In math. I did not attempt any question in Algebra at that time in exame, but did not miss any in Geometry. that is 64 years ago. Now seeing your problem I got interested. The way you solved is also simple. But I will solve ' r ' other way OE/ EA = tan 22.5 that is r/(2- r) = tan 22.5 = .4142 . solving r =.5858 units and area= 1.078 Sq. units
I did it a similar way. I kept splitting right triangles and using trig functions until I had an isosceles triangle of 90/22.5/22.5 where the 2 equal sides were equal to r (.585786), which got me to an area of 1.078024.
This is the same method I used since I noticed the circle is inscribed in the triangle, which means the center of the circle is an incenter. The incenter is made up of the intersection of the three angle bisectors. We know it's a 45-45-90 triangle so we can use 22.5 degrees and tangent to solve for the radius.
Here's a different approach: the entire triangle is 45-45-90; so by Pythagoras, the hypotenuse AC = 2√2; Now CD = CF = 2 - r; but CF = 1/2 the hypotenuse = √2; so 2 - r = √2; and rearranging, r = 2 - √2. The area of the circle is π(2 - √2)^2 = π(4 - 4√2 + 2) = π(6 - 4√2). At first I did it trigonometrically by drawing the line CO and saying that r/(2 - r) = tan(45/2); and then using the half-angle formula {had to look that up}: tan(45/2) = (1 - cos(45))/sin(45) = (1 - 1/√2)/(1/√2) = √2*(1 - √2); so r/(2 - r) = √2*(1 - √2) = √2 - 2; and pretty much the same from there. That gives the same answer, but it's unnecessarily complicated. Cheers. 🤠
Perhaps slightly simpler: CFB is a right triangle with 45 degree corners. Therefore BF=sqrt(2). Also, BF= (sqrt(2)+1) * r. Thus r= sqrt(2)/(sqrt(2)+1) = sqrt(2)*(sqrt(2)-1)/((sqrt(2)+1)*(sqrt(2)-1)) = (2-sqrt(2)), and then now that you have r the rest is the same.
I'm trying to figure out how you got your answer. I got stomped after getting the BF = square root of 2. How did you manage to get the square root of 2 +1??? I mean I got your idea that yours is simpler, I am just trying to get head to get it. Please advise
@@mayac.1345 Look at the diagram at 6:29. BF = BO + OF. BO = r√2, because triangle BEO is 90°-45°-45°. OF = r. BF = r + r√2 = r (1 + √2) = r (√2 + 1) = (√2 + 1) * r.
I used the law of sines: Knowing that half the hypotenuse (AF) is sqrt of 2 in length and half of angle A is 22.5 degrees which must hit the center of the circle, forming another right triangle (AOF). Therefore, using the law of sines for the right triangle AOF, the radius (OF) divided by the sine of angle A (angle = 22.5) equals sqrt of 2 (length AF) divided by the sine of angle AOF (angle = 112.5). Solve for r.
You can also draw a small triangle joining O, C, and F, CF=sqrt(2), and angle OCF=22,5 deg. So r=sqrt(2)*tan(22.5), and the area equals 1,078 sq units.
Alternatively, join each vertex to the centre of the circle. The total area of the resulting three triangles will equal the area of the big triangle (each small one has an altitude of r). This equation leads quickly to r and hence the area.
These are always so much fun to watch. Simple, logical steps moving you through the problem to its solution. I wish that they taught me this when I was in skool way back when.
Good explanation. The only thing I wonder is why round off pi to two decimals and square root of 2 is rounded off to four decimals after the point, instead of rounding off both to the same amount of significant numbers.
That is quite bothersome. I think it’s because 3.14 is the *standard* given in a “setting,” like in school, a workplace, and science lab. There is the saying you won’t need more than either the first 3 digits or 9 digits of pi. I think the first 2 decimal places (first 3 digits) should’ve been fine, 1.41.
@@stealthgamer4620 I would just leave pi and root 2 in the final answer. Whoever needs the decimal value can calculate it to whatever precision they need for their use case
@@PattyManatty I would agree to leaving pi and sqrt 2 in the answer, it’s just this video is inconsistent with the decimal places for the decimated/calculated answer.
@@barclaymatheson8240 This is the silliest comment I've seen in a long time. You could say the same thing about the square. You can keep measuring the lengths with more and more accuracy, and get more and more decimal points. An exactly 3 inch piece of wood doesn't exist. Then there's the problem that at a molecular level these lines aren't actually straight. Plus their length, that is their exact length, is likely always in flux because atoms are moving and their exact length is not a well-defined idea. So yeah, if you get ridiculous about it. But as the calm mango said, we can do these calculations to essentially as arbitrary precision as our measuring tools will allow.
@@VolvoxSocks As this is an isosceles triangle if we bisect the triangle we have another right angle isosceles triangle where the short sides are root 2. If the radius in R then ......... Root of (R²+R²) +R = Root 2 = R+Root(2R²) Root 2 = R + R(Root 2) = R (1 + Root 2) So R = Root 2 over (1 + Root 2) Area of the circle then equals Pi * 2 over (3+2 Root2)
I solved this problem in a different and easy way Just make three triangles by joining OD , OF, and OE And find the areas of each triangles and add and equate to the area of big triangle ie. ABC which is 2 sq units which will give the radius of the circle And then you can easily calculate the area of circle
@@sanketmishraa659 Thank you for your reply. One of the things I like most in mathematics is finding more than one way to solve a problem. I can see you are thinking of a different way than solutions I already know, and I am trying to understand your solution. It looks like I would need more details and steps explained to understand your solution. But I still don't understand. That's okay. Again, thank you for sharing and trying to help me see your alternate solution.
Good job. An alternative it can be solved using triangle-inner circle. Triangle area= radius of inner circle times half of triangle perimeter. So you can get the radius and then calculate the circle area.
I used the incircle method (circle inside the triangle tangent to all three sides). The green circle was an incircle, and its radius is equal to the area of the triangle divided by the semiperimeter. The area of the triangle is bh/2, or 2x2 / 2 = 2. The hypotenuse is 2√2, so the semiperimeter is (2 + 2 + 2√2)/2 = 2 + √2. Therefore, radius is 2 / (2 + √2), which equates to 2 - √2. From there you can find the area of the circle.
Please, using a calculator, I see that it is true, but algebraically, how do you know 2/(2 + √2) = 2 - √2 ? I don't see how to change one expression into the other using algebra. Thank you in advance!
@@LiveHappy76 The trick is knowing (root(2)+2)*(root(2)-2) = 2+2root(2)-2root(2)-4 = 2-4 = -2, which means you can move the squareroot to the top of the fraction by multiplying by (root(2)-2)/(root(2)-2), which equals 1. So 2/(root(2)+2) = (2 * (root(2)-2))/((root(2)+2) * (root(2)-2)) = (2root(2) - 4)/(-2) = 2 - root(2)
Line BF is tangent to line AC, segmenting triangle ABC into two mirror triangles ABF and BCF. Focusing on triangle ABF, line BF is the same length of line AF. This provides a route to the solution. Line AF=line BF. Simple pythagorean application here: square root of AB = length of line AF and line BF. This also indicates the length of CF, as the hypotenuse of triangle ABC was cut precisely in half, therefore line CF = line AF. Line CF = line CD. Line OE = DB. DB =CB-CD. 2- sqrt(2). Area of the circle is Pi x (2-sqrt(2))^2 or approximately 1.078.
Triangle ABC has two sides of equal length (AB and BC) and a 90 degree angle (angle B) so we know the other two angles are 45 degrees each and so this is a so-called 45-45-90 triangle with sides a=1, b=1 and c=√2. In this case AB and BC are 2 and so the diagonal AC is 2√2. Now draw a line from point B through point O perpendicular to line AC and call the intersection with AC point F. We now have a triangle ABF, which also has a 90 degree angle and two 45 degree angles. So this too is a triangle with sides 1-1-√2. AF and BF have an equal length and AB as its diagonal. With AB=2 that means BF=AF=√2. If we now draw the two lines OD and OE, just like in step 3, we can see that triangle BOE is another 45-45-90 triangle. With BE=BD=r that means the diagonal BO=r√2. OF=r so BF=r√2+r=r(1+√2). We also know that BF=√2 so r(1+√2)=√2 which gives r=√2/(1+√2). The area of the circle then is πr²=π(√2/(1+√2))²=2π/(3+2√2) which is approximately 1.08.
Once done step 1, you may consider the area of the triangle ABC (2x2/2), which is also the sum of the areas of the triangles OAB, OBC and OAC expressed in term of base (AB, BC and AC) and heigth (r in each case).
You can solve without the two-tangent theorem by recognising that the circle intersects the hypothenusis at exactly half its lenth which is sqrt(2). Then recognise that the the distance BO is r×sqrt(2) which you did. Because it is a unilateral triangle we know that d(B,F)=d(A,F) thus r=sqrt(2)/(1+sqrt(2))
Interesting solution but it's faster to notice the geometrical properties of O. The 3 triangles OAB, OAC, OBC, have the same height, which is the radius of the circle r. But the sum of those 3 triangles areas, is the complete area of the ABC triangle. Therefore, Area(ABC) = 2*2/2 = 2 is also the sum of the 3 smaller triangles areas AB*r/2 + AC*r/2 + BC*r/2 = r*(AB+BC+CB)/2. So r*(2+2+2*sqrt(2))/2 = 2, which gives r = 2/(2+sqrt(2)) = 2*(2-sqrt(2))/(4-2) = 2-sqrt(2). Done. I guess everyone knows how to move from radius to surface of the circle ;-) !. Thanks for your channel.
I was going the right direction, similar to your answer..Even though the answer makes sense..it seems that there should be an easier way. Very logical. Tried to do it all in my head... needed to write things down. Reminds me of trigonometry homework...the extra credit kind. Nice video.
My solution FA = √2 So BF = √2 Let's call the distance between B and the nearest point of the circle x Now we have BF = x + 2r So that's √2 = x + 2r On the other hand, BO = x + r By using the same small triangle as in the video: BO = 2√r Thus, x + r = 2√r We now have a system of 2 equations x + r = 2√r √2 = x + 2r By solving for r using substitution we find that r = 2 - √2
Hello all, there is a much simpler way to demonstrate. - Lets call d the distance from point B to the small cercle. - lets write sr() the square root function A/ Then the heigh of the triangle is 2.r + d = sr(2) B/ The distance BO is r + d As diagonal of square BDOE it is r.sr(2) Then r + d = r.sr(2) C/ lets replace B/ into A/ 2r + d = sr(2) r + (r + d) = sr(2) r + (r.sr(2)) = sr(2) r.(1 + sr(2)) = sr(2) r.(1 + sr(2)) . (1 - sr(2)) = sr(2).(1 - sr(2)) r.(1-2) = sr(2) - 2 r = 2 - sr(2) Then the area of the cercle A = pi . (2-sr(2))²
An alternative approach, since this is a right-angled isosceles triangle, hence BF is not only the angle bisector of BF^2 = 4 - 2 = 2 => BF = √2 Also we know that: BO:OF::(BC+BA):AC => BO:OF::4:2√2 => BO = OF√2. And we know that: BF = BO + OF = √2 => OF(√2 + 1) = √2 => OF = √2(√2 - 1) And OF is nothing but the radius of the incircle. Therefore, the area of the circle is: A = π × OF^2 = π × 2 (√2 - 1)^2 = 3.14 × 2 × (3 - (2 × 1.41)) = 1.13 sq. units. I hope i am right.🙏
@@murdock5537 thank you for your reply 😊🙏 my God, it's been a month!😱 I have forgotten the problem too, my bad! 🙈😂 I will check and get back.. thanks again for your insights.. 😊🙏
@@murdock5537 Hello Mr. Murdock, terribly sorry for my delayed reply. As rightly pointed out by, i have indeed made a calculation error. Your calculation is absolutely correct, it is approximately 1.078 sq. units. Thank you. 😇🙏
PreMath, you win the prize for most complicated solution! Using the labels you added (D, E &F), Area = pi*r^2 r = AB -AE (radius-tangent theorem) AC = 2*sqrt(2) (pythag) AF = AC/2 = sqrt(2) (symmetry) AE = AF (two tangent theorem) AB - AE = 2 - sqrt(2) = r Area = pi*(2-sqrt(2))^2 = 1.078...
Symmetry could offer another solution: Let r be the circle's radius. Consider a mirror placed along AC. If D is the image of B, ABCD is a square; AC is tangent to the image circle at F. BD=2BF=2(BO+r) Sqrt(2^2+2^2)=2[Sqrt(r^2+r^2)+r] r=Sqrt(2)/[Sqrt(2)+1]=2-Sqrt(2) Area=[2-Sqrt(2)]^2 pi=[6-4Sqrt(2)] pi
The circle is the incentre which comes from the angle bisectors. Write the equations for the angle bisectors: for angle B it is y = x (assume it is on a set of axes). Prepare and equation for angle c. y=-2.414+2 (angle c = 45 and half of this is 22.5 so take tan (90+22.5) Solve these equations simultaneously which will give the co-ordinates of the incentre. The x value will be the radius and calculate the area form there. Fairly quick.
Area of ABC is 2*2/2 = r*(2+2+2sqrt2)/2. The last is the sum of areas of BOA, BOC and AOC. In this way I solved without pen and paper for a minute or two.
Thanks Plamen for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
Here's my attempt just solving it from the thumbnail. I'll edit this comment if I'm wrong: 1. Draw 3 segments. Each connecting the circle's center, **O**, to the 3 tangent points between the circle and the triangle. 2. Call the tangent point on the left leg **P**. The top vertex of the triangle **Q**. The tangent point on the hypotenuse **S**. The bottom right vertex **T**. The tangent point on the bottom **U**. 3. Because **OP** , **OS**, and **OU** are radii of the same circle, they are **congruent** to each other. Because each radii intersects each leg at the point of tangency, each radii is **perpendicular** to each leg. This makes **O** the **incenter** of the triangle. 3_1. Because **O** is the incenter, it bisects both 45° angles into 4 congruent angles. This makes angles POQ, QOS, SOT, TOU congruent (each 67.5°) as well. Triangles POQ, QOS, SOT, TOU have their **shortest sides congruent** since its the radius of the circle. In addition, they have **congruent corresponding angles**, making the triangles congruent to each other. 4. This makes S the midpoint of the hypotenuse. This makes QS = QP also. QS = 2 * SQRT(2) ÷ 2 = QP 5. The segment parallel to OU; on the left leg of the triangle; connecting the right angle to P = 2 - SQRT(2). 6. A square is formed at the bottom left corner because all angles are right angles and 2 segments are congruent, being radii of the same circle. Therefore the radius of the circle is 2 - SQRT(2) 7. Area is pi * r^2 which is **pi times (2 - SQRT(2))** Edit: I was correct but I forgot that the 2 tangent theorem was a valid proof. Thanks for the video. I took the long route.
Another approach : Let us consider the triangle OBE. Basic trigonometrics defines : cos(OBE) = OE/OB We know that OE = r OB = BF-r. As you have shown, BF = Sqrt(2). Hence, OB = Sqrt(2)-r It follows that cos(OBE) = r/(Sqrt(2)-r) We know that the angle OBE = 45° (that is, pi/4 radians). The value of cos(pi/4) is well-known, it is equal to Sqrt(2)/2 It follows that r/(Sqrt(2)-r)=Sqrt(2)/2. A bit of algebra and we show that r = 2/(2+Sqrt(2)) = 2 - Sqrt(2)
Nice and awesome, many thanks! Draw a perpendicular to AB from F = FG = 1 (since AF = √2). 1 = r + r√2/2 → r = 1/(1 + (√2/2)) = 2 - √2 → r² = 6 - 4√2 → Area circle = π(6 - 4√2) 🙂 or take the fast lane: r + r√2 = √2 → r = √2/(1 + √2) = 2 - √2 🙂
@@dibjr Not that familiar with programming languages, I tend to use ^ for the exponentiation. I would say A = pi * (s - 0.5 (2 * s^2)^0.5 )^2 not sure if we would get the same outcome for all positive s.
We can use the formula for radius of in circle of a triangle which is (Area of triangle/ semiperimeter of triangle) = radius of in circle. Area of triangle = 2*2/2 = 2 Semi perimeter = (2+2+2*sqrt(2))/2 = 2+sqrt(2) Radius of in circle = 2/(2+sqrt(2)) =2-sqrt(2) Area of circle = pi*square (2-sqrt(2))
Thanks dear for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
This question is taught in 9th standard in India Radius of inscribed circle is called inradius and in case of right angle triangle inradius = (base + perpendicular - hypotenuse )/2
Yes, it is taught in the USA too as the inradius is twice the area of triangle divided by the perimeter of triangle. That work for any triangle. For the case of a right triangle, the inradius is half the sum of the legs minus the hypotenuse. You are correct.
My solution: as the middle of the inner circle is cross point of the angle-halfing lines and as the hypothenusis of the triangle is, by pythagoras, squareroot of 8, (2^2+2^2=8) we find a triangle with the radius, (drawn directly to the hypothenusis) half of the hypothenusis and the halfing line of the corner's angle of the big triangle, this angle has 22,5 degrees, because 90 plus two times 45 makes 180 in the big triangle. So half of the corner's angle is 22.5. In that small triangle tan22,5 equals radius divided through half of hypothenusis, so radius equals tan22,5 multiplied by one half of squareroot 8. r equals 0,586. Put into area formula for circles pi x r^2 equals 1,078. This solution came to my mind firstly. I'm not sure it's easier or more complicated to someone else...
It's easy. For a circle with a radius r, inscribed in the triangle ABC, the area T of the triangle is equal to the perimeter of the triangle times the radius of the circle divided by 2 T=(AB * r + BC * r + CA * r) / 2 = (AB + BC + CA) * r / 2 = L * r / 2 L = AB + BC + CA is the perimeter of a triangle. (By the way: The formula T = r * (L / 2) applies to every polygon inside which a circle is inscribed, and also to the disc itself, if we consider the circle to be inscribed "inside" the disc.) So we have: Area of the triangle Tt = r * (L / 2) => r = 2 * Tt / L; Area of a disc Td = (r * 2 * Pi * r) / 2 = pi * r ^ 2 = pi * (2 * Tt / L) ^ 2 In the problem, the triangle is rectangular with the legs equal to a, so: Triangel area - Tt = α ^ 2/2; 2Tt = a ^ 2 Triangel perimeter - L = a + a + a * sgrt (2) = a * (2 + sgrt (2)) r = 2 * Tt / L = a / ((2 + sgrt (2)) = a * ((2-sgrt (2)) / 2) for a = 2 r = 2-sgrt (2)
by the formula of incentre directly r = area / semi perimeter ===> area = 2 semi perimeter = 2+sqrt2 and r = 2- root2 . Then finding area by the formula of circle 1.08 .
AB=2sqrt2 (pythagorem) if you connect the middle of the circle to the points where it hits the triangle, and connect it to point B, youll get 2 triangles with legs r,r and a 90* angle so the other side is rSqrt2, if you also connect the middle pf the circle to side AB, the distance of AC to point B is r+rSqrt2. the area of a triangle =(bh)/2, we already know one b and h, AB and BC, 2 and 2. so the area of the triangle is (2•2)/2=2. we also know the base could be AC and the height d(AC->B)=r(1+sqrt2) so the area is also ((r(1+sqrt2))•2sqrt2)/2 and we knwo the area is 2 so ((r(1+sqrt2))•2sqrt2)/2=2. that is a linear equation in r so we can solve: (r(1+sqrt2))•2sqrt2=4 , r(1+sqrt2)=4/2sqrt2 , r= (4/2sqrt2)/1+2sqrt2=4/(2sqrt2+8)=(4(2sqrt2-8)/(8-64)=(8(sqrt2-4))/(8(-7))=(sqrt2-4)/(-7). the area of a circle is πr^2=π((18-8sqrt2)/49)
@@jankoodziej877 If you pause the video at 4:35 you can see visually that it's r plus the diagonal of the square with side length r. It is also the height of the triangle if you rotate the figure such that AC is on the bottom. Then you equate the 2
Draw the 3 lines from the circle center to each of the triangle tangent points with the circle. It then becomes obvious that the bottom edge of the triangle = r + sqrt2, so r = 2 - sqrt2 and so the area = pi r ^2. Done.
I offer another solution: ABC is isosceles, symmetrical about the y=x line. The circle's defined by its center(x,y) and its radius. The center is on that line y=x and its radius reaches to the edges, so the entire circle can be described as an r. We know the distance of the center to a wall is equal to the distance of the center to (1,1), since that's all just the radius. These distances can be written as the equations d=r and d=(1-r)*√2, that second distance is a little weirder but it's just turning dist-from-0 into dist-from-1 and then converting side length of a right isosceles triangle to hypotenuse length. We can now transitive-property combine those expressions into r=(1-r)*√2, and solve for r. r=(1-r)*√2 r=√2-√2*r r+√2*r=√2 r(1+√2)=√2 r=√2/(1+√2) And then you use the radius to get the area with πr² π*(√2/(1+√2))² π*(2/(1+2√2+2)) 2π/(3+2√2) , Which does indeed evaluate out to 1.078
ABC right isoceles triangle such that AB=BC=2. O center of the circle inscribed in ABC radius of inscribed circle=r. height from B to [AC] cuts the latter at a point F. Since ABC is an isoceles triangle, [BF] is also the median and the bissector of the angle B. Since the center of the inscribed circle is the intersection of the angle bissectors, O is on [BF]. Since ABC is also a right angle triangle, the median equals half the hypothenuse. Thus BF=AC/2. In a right angle/isoceles triangle of side a, the hypothenuse equals a×sqrt(2) Thus, AC=2sqrt(2) BF=sqrt(2). OF=r (for obvious reasons) BO=sqrt(2)-r But [AB] and [BC] are the tangents from B to the circle, thus there is a square with diagonal [OB] and side length r BO=r×sqrt(2) sqrt(2)-r=r×sqrt(2) r(1+sqrt(2))=sqrt(2) r=sqrt(2)/(sqrt(2)+1)=sqrt(2)×(sqrt(2)-1)/(2-1)=2-sqrt(2) Area of inscribed circle=pi×(6-4sqrt(2))
Very well done series. One thing, however…when you say, for instance, “2A+2B=R” you say “2A + 2B equals TO R”. That ‘TO’ is not needed and in fact is confusing as it sounds as if you are saying that 2A + 2B equals 2R. Simply put, two thing equal another. No “to” needed here. I hope this helps…thanks for the math education…Steve Z
I used a different method. I found the area and the hypotenuse. I connected the centre of the circle to the vertices of the triangle and subsequently found the areas of 3 triangles. I added them up with the variable as the radius..and equated to the area of the traingle. Hence you get the radius and you can directly find the area
I went about that somewhat differently, but got the same answer: Since it's right-iso, a radius from circle-center to hypotenuse-center is tilted by 45 deg, which means its height above circle-center is R/sqrt(2), or R*sqrt(0.5) add that to the height of the circle-center itself and it's R+R*sqrt(0.5), or R*(1+sqrt(0.5)) but the middle of the hypotenuse is also halfway up the length-2 side of the triangle, which means that R*(1+sqrt(0.5))=1, so R=1/(1+sqrt(0.5) which is 0.5857864376 (the same thing as 2/sqrt(2)) then plugged that into area = pie r squared from other comments, it looks like there's a variety of ways to do that...
I remember the old equation A + B = C + D from high school geometry class! A and B are the sides of the right triangle, C is the hypotenuse, and D is the diameter of the inscribed circle. So you can easily calculate the diameter as 2 + 2 - 2 sqrt(2). The radius is half of that, and then A = pi R^2.
I found it easier to bisect one of the 45 degree angles, then bisect the 90 degree angle and they will intersect at the center of the circle by construction. Then use the new right triangle that has opposite side length r and adjacent side length root two. Then simple trig to find r. This uses a few less theorems.
Simple. Center of circle is (r,r). Tangent point on long side of triangle is (1,1). Distance from (r,r) to (1,1) is r. Solve for r gives r = 2-√2. Area follows by applying A=πr².
Draw line OB. This line is R + y. Length y is from where line OB cuts the circle to point B. Solve for y in terms of R: (y + R) x cos(45) = R so y = 0.414213 x R Draw line FB. This line is 2 x R + y = 2 x R + 0.414213 x R = 2.414213 x R. Line FB is also 2 x sqrt(2) / 2 = sqrt(2) which gives R = sqrt(2) / 2.414213 = 0.585786 Area of green disc is PI x R x R = 1.07802416 Did anybody read this? ;-)
Solved it with formula of an area of triangle inscribed near a circle. We have that area of a triangle is 2*2*0.5 = 2 and we have that area of a triangle also is p*r, where p is a semiperimetr of a triangle. So, it comes to 2 = (2 + 2 + 2*sqrt(2)) * 0.5 * r.
drop perpendicular from circle's center to the hypotenuse and base of length of r , hypotenuse will be divided into two parts,both's lengths will be √2. and AE(look in the video) will be (2-r) now join Center of circle to Angle A, use Pythagoras theorem for both triangles to find their hypotenuses and equate them
Another way: the point o is the intersection of the two bisectors of the angles in A and in C If we put a system of Cartesian axes with origin in B, we know the angular coefficients of the two bisectors and and the points they pass through Calculate equations of straight lines passing through the corresponding points and then calculate the intersection of them The coordinates of the intersection are the radius of the circumference--> area I think it’s way easier
Another simple way to go about is the formula Δ=rs, where Δ is the area of the triangle, s is the semi-perimeter and r is the inradius. Area=0.5*2*2=2, s=0.5(2+2+2√2) =2+√2 r=Δ/s= 2/(2+√2) =2-√2 Then use πr² for area
This could be solved WAY easier than you did. You just needed to realise that the distance from tangency points to the bisector are equal, so the "R" must be "2-(2^1/2)". Then you just get the Area and that's all. 🤷
Using geometry only, solution boils down to - Tangent theorem: r = BC - CD = 2 - root2 = 0.586 Area = 1.078. If trig is allowed Lumkile's solution is elegant (r=root2 x tan 22.5)
Line AC's equation is Y = 2 - X and Line BF's is Y = X these 2 lines intersect where X = 2 - X or X = 1 and Y = 1 Line AC is tangent to the circle at point = (1 , 1) // Circle has its center at (r , r) // Circle's equation is r^2=(x - r)^2 + (y - r)^2 // Solving the circle equation for point of tangency = (1 , 1) will give us the radius value r = 2 - SQRT(2)
Because BC is tangent to OD, so there is a right angle, the same with BA and OE. There is another right angle(ABC), so angle EOD is right too. All the sides are same, all are radius of circle. Hope it helped!
Thanks Simon for such an elegant explanation. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
@@simons2006 I intentionally asked for his followers to benefit. I have been following him for a long time because I like his contents. Though this is not the answer I was expecting. IT'S A SQUARE ONLY BECAUSE THE TWO TANGENTS MEET AT A RIGHT ANGLE.
point B is (0,0), point F is (1,1) equation of circle : (x-r)^2 + (y-r)^2 = r^2 plug in x=1, y=1 then solve for r via quadratic formula. r= 2 +/- √2 r = 2 - √2 is the only one of the two that makes sense based on the problem. then use area = pi * r^2
It's easier to set up the Pythagorean for the half of the BDOE square by drawing the line BF. BF is √2 and BO is √2-r Then let's take the BOE triangle: r²+r²=(√2-r)² 2r²=2+r²-2√2r r²+2√2r-2=0 Using the quadratic formula, you get the same result but in a simpler way.
Area of △ABC is 2 (∵2x2/2=2) Also area of △ABC is sum of area □BDOE, △AOE, △AOF, △COD, △COF. It is r^2 + (2-r)r/2 x 4. Therefore, a quadratic equation is deducted. r^2 - 4r+2 = 0
We have: AC = ABsqrt(2) = 2sqrt(2) p = (The perimeter of triangle)/2 = (AB + BC + AC)/2 = 2 + sqrt(2) S = AB.BC.(1/2) = 2 Apply the formula for the relationship between the area of a triangle and the radius of the circle inscribed in the triangle, we have: S = pr => r = S/p = 2 - sqrt(2) => The area of the circle is: S = r^2.pi = (6 - 4sqrt(2))pi
I love this kind of problems in maths ……till 10th I was brilliant in solving all this but unfortunately I took PCB in 11th to pursue medicine I still miss old days of solving maths problems
I solved it in an other way and found A=2*pi/(3+2*sqrt(2)) that seems different from video result... but I found that is the same result written in other form, I found that: 3-2*sqrt(2) = 1/(3+2*sqrt(2)) and this is quite interesting. In the same way I found that: sqrt(2)+1 = 1/(sqrt(2)-1) This is a simple application of (a+b)*(a-b) = a^2 - b^2
there is no tangent to a circle theorem stating the angle between a tangent and a radius is perpendicular. The correct statement of the theorem is 'A line that is tangent to a circle is perpendicular to the radius drawn at the point of tangency. The other theorem, we make use of, states: Center of the inner tangent circle of a triangle is at the intersection of bisectors. A simple solution is: 1-) Calculating AC=Sqrt(8) from Pythagorean theorem on Triangle CBA. 2-We draw a line through B-O-F. Now CBF and FBA are isosceles triangles due to 45 degree (PI/4 radians) angles at A, B and C therefore CF is Sqrt(8)/2 = Sqrt(2). 2-) Since angle COF is 1/2(45) =22.5 degrees (PI/8 radians).We can look up for tan 22.5 degrees (PI/8 rad.) Or we can calculate from tan (PI/4) = 1 = 2 tan(PI/8)/(1-tan^2(PI/8)) as SQRT(2)-1. Here, if we use half angle formula we must discard the negative solution for tan(PI/8) as this is not in the 1'st quadrant 3-)Considering COF triangle we have tan(PI/8) = Sqrt(2)-1 = OF/CF = r/Sqrt(2) giving r = 2-Sqrt(2). .4-) From this we calculate area S = PI * r^2 = 1.078 units.
OEB is a 1 … 1 … sqrt(2) triangle, with sides r, r, r * sqrt(2). OF is also r. AF = CF = 2 sqrt(2) / 2 = sqrt(2) AF = BF, OF + OB = BF r * sqrt(2) + r = sqrt(2) r (sqrt(2) + 1) = sqrt(2) using conjugate r(sqrt(2) + 1)(sqrt(2) - 1) = sqrt(2)(sqrt(2) - 1) r(2 - sqrt(2) + sqrt(2) -1) = 2 - sqrt(2) r = 2 - sqrt(2)
hello, at 1:38 you say c²=8, which is true, but you can't say that sqrt(c²) = sqrt(8) and then C=sqrt(8). You should have said that by taking square root of each term we've got C=sqrt(8). Thanks
@ PreMath, Sir. I am 83 years old, I did my Matriculation back in in mid 1950's . In math. I did not attempt any question in Algebra at that time in exame, but did not miss any in Geometry. that is 64 years ago. Now seeing your problem I got interested. The way you solved is also simple. But I will solve ' r ' other way
OE/ EA = tan 22.5 that is r/(2- r) = tan 22.5 = .4142 . solving r =.5858 units
and area= 1.078 Sq. units
Hello old person
I did it a similar way. I kept splitting right triangles and using trig functions until I had an isosceles triangle of 90/22.5/22.5 where the 2 equal sides were equal to r (.585786), which got me to an area of 1.078024.
This is the same method I used since I noticed the circle is inscribed in the triangle, which means the center of the circle is an incenter. The incenter is made up of the intersection of the three angle bisectors. We know it's a 45-45-90 triangle so we can use 22.5 degrees and tangent to solve for the radius.
Same here, using true values at each stage rather than algebra.
I actually really enjoyed that it was solved by only geometry and algebra.
An exact solution, probably first used many years ago.
Here's a different approach:
the entire triangle is 45-45-90; so by Pythagoras, the hypotenuse AC = 2√2;
Now CD = CF = 2 - r; but CF = 1/2 the hypotenuse = √2; so
2 - r = √2; and rearranging,
r = 2 - √2.
The area of the circle is π(2 - √2)^2
= π(4 - 4√2 + 2)
= π(6 - 4√2).
At first I did it trigonometrically by drawing the line CO and saying that r/(2 - r) = tan(45/2); and then using the half-angle formula {had to look that up}:
tan(45/2) = (1 - cos(45))/sin(45) = (1 - 1/√2)/(1/√2) = √2*(1 - √2); so
r/(2 - r) = √2*(1 - √2) = √2 - 2; and pretty much the same from there. That gives the same answer, but it's unnecessarily complicated.
Cheers. 🤠
That was my initial approach when I saw the problem
..nice...but its incorrect...
CD not equal to CF
r = (1/4) * ( √(1 + 8 * √(2) ) - 1 )
Solution:
CA = 2 * √(2)
FC = FA = BF = √(2)
BF = 2r^2 + r
2r^2+r - √(2) = 0
r = 1/4 ( √(1 + 8√(2) ) - 1) [only the positive root of the quadratic formula]
r ≈ 0.63
Area of cirle = π * r^2 = ( ( 1/4 ) * ( √(1 + 8*√(2) ) - 1) )^2 * π
= (1/16) * ( √(1 + 8 * √(2) ) - 1 )^2 * π
= (1/16) * ( √(1 + √(128) ) - 1 )^2 * π
A ≈ 1.236
Note: added parenthesises for ppl relying on PEMDAS without considering jinxed position
Perhaps slightly simpler: CFB is a right triangle with 45 degree corners. Therefore BF=sqrt(2). Also, BF= (sqrt(2)+1) * r. Thus r= sqrt(2)/(sqrt(2)+1) = sqrt(2)*(sqrt(2)-1)/((sqrt(2)+1)*(sqrt(2)-1)) = (2-sqrt(2)), and then now that you have r the rest is the same.
Great tip!
Thanks Lance for the feedback. You are awesome 👍 Take care dear and stay blessed😃
I'm trying to figure out how you got your answer. I got stomped after getting the BF = square root of 2. How did you manage to get the square root of 2 +1??? I mean I got your idea that yours is simpler, I am just trying to get head to get it. Please advise
@@mayac.1345 Look at the diagram at 6:29. BF = BO + OF. BO = r√2, because triangle BEO is 90°-45°-45°. OF = r. BF = r + r√2 = r (1 + √2) = r (√2 + 1) = (√2 + 1) * r.
@@LiveHappy76 yes so simple
What you just did is doing math in a proper way. Excellent explanation.
I used the law of sines: Knowing that half the hypotenuse (AF) is sqrt of 2 in length and half of angle A is 22.5 degrees which must hit the center of the circle, forming another right triangle (AOF). Therefore, using the law of sines for the right triangle AOF, the radius (OF) divided by the sine of angle A (angle = 22.5) equals sqrt of 2 (length AF) divided by the sine of angle AOF (angle = 112.5). Solve for r.
You can also draw a small triangle joining O, C, and F, CF=sqrt(2), and angle OCF=22,5 deg. So r=sqrt(2)*tan(22.5), and the area equals 1,078 sq units.
How is ocf 22.5⁰?
@@MKD1101 angle ACB =DCF is 45° so 1/2 ACB = oCF = 22.5°
I should have scrolled down! I got the same answer but used A.
How would you know O without using his method
Yep, I did it that way. So much quicker and easier than that in the video.
Alternatively, join each vertex to the centre of the circle. The total area of the resulting three triangles will equal the area of the big triangle (each small one has an altitude of r). This equation leads quickly to r and hence the area.
It is the best answer.
These are always so much fun to watch. Simple, logical steps moving you through the problem to its solution. I wish that they taught me this when I was in skool way back when.
Please explain your ideaway to solution
Good explanation. The only thing I wonder is why round off pi to two decimals and square root of 2 is rounded off to four decimals after the point, instead of rounding off both to the same amount of significant numbers.
That is quite bothersome. I think it’s because 3.14 is the *standard* given in a “setting,” like in school, a workplace, and science lab. There is the saying you won’t need more than either the first 3 digits or 9 digits of pi. I think the first 2 decimal places (first 3 digits) should’ve been fine, 1.41.
@@stealthgamer4620 I would just leave pi and root 2 in the final answer. Whoever needs the decimal value can calculate it to whatever precision they need for their use case
@@PattyManatty I would agree to leaving pi and sqrt 2 in the answer, it’s just this video is inconsistent with the decimal places for the decimated/calculated answer.
Because it's bullsh*t math. Humans haven't figured out how to calculate the area of ANY circle. Infinite points=infinite decibels.
@@barclaymatheson8240 This is the silliest comment I've seen in a long time.
You could say the same thing about the square. You can keep measuring the lengths with more and more accuracy, and get more and more decimal points. An exactly 3 inch piece of wood doesn't exist.
Then there's the problem that at a molecular level these lines aren't actually straight. Plus their length, that is their exact length, is likely always in flux because atoms are moving and their exact length is not a well-defined idea.
So yeah, if you get ridiculous about it. But as the calm mango said, we can do these calculations to essentially as arbitrary precision as our measuring tools will allow.
Why would anyone vote thumbs down for this well-done video with very clear explanations?
Look how he labeled the triangle.
I think there is a more elegant solution as the line BF is also root 2. So Root 2 = r+Root(r²+r²)
How do you solve for r from there?
@@VolvoxSocks As this is an isosceles triangle if we bisect the triangle we have another right angle isosceles triangle where the short sides are root 2.
If the radius in R then ......... Root of (R²+R²) +R = Root 2 = R+Root(2R²)
Root 2 = R + R(Root 2) = R (1 + Root 2)
So R = Root 2 over (1 + Root 2)
Area of the circle then equals Pi * 2 over (3+2 Root2)
@@alanwiltshire2139 Ah thanks. Saw how you got the rt(2)=r+rt(2r^2). Didn't think of converting it to r(1 + rt(2)). Cheers!
Agreed
Side minus half of diagonal is much easier and faster way: r=2-(2√2)/2=2-√2. In other way: r=BE=AB-AE=AB-AF=2-√2
I solved this problem in a different and easy way
Just make three triangles by joining OD , OF, and OE And find the areas of each triangles and add and equate to the area of big triangle ie. ABC which is 2 sq units which will give the radius of the circle
And then you can easily calculate the area of circle
You write OF twice. I don't follow your solution.
*Hey i just written what i think i can easily do and i m not forcing you to follow my solution it is up to you.*
@@sanketmishraa659 Thank you for your reply. One of the things I like most in mathematics is finding more than one way to solve a problem. I can see you are thinking of a different way than solutions I already know, and I am trying to understand your solution. It looks like I would need more details and steps explained to understand your solution. But I still don't understand. That's okay. Again, thank you for sharing and trying to help me see your alternate solution.
Excellent explanation with simple and detailed calculation 👍👍👍
Good job. An alternative it can be solved using triangle-inner circle. Triangle area= radius of inner circle times half of triangle perimeter. So you can get the radius and then calculate the circle area.
Triangle perimeter = 2 + 2 + 2√2 = 4 + 2√2. Half of triangle perimeter = (4 + 2√2)/2 = 2 + √2. Triangle area = bh/2 = 2*2/2 = 2. Now you say 2 = r(2 + √2), so r = 2/(2 + √2). Circle area = πr² = π(2/(2 + √2))(2/(2 + √2)) = 1.374. No, circle area = 1.078.
bro u have to rationalise that value
2/(2+_/2) take root 2 कॉमन ull get root 2 upon root 2 + 1
now multiply with root 2 -1 / root 2 -1
u will get थे radius 2-root2
Excellent. Thx a lot.
You did it the hard way. It is much simpler.
I used the incircle method (circle inside the triangle tangent to all three sides). The green circle was an incircle, and its radius is equal to the area of the triangle divided by the semiperimeter. The area of the triangle is bh/2, or 2x2 / 2 = 2. The hypotenuse is 2√2, so the semiperimeter is (2 + 2 + 2√2)/2 = 2 + √2. Therefore, radius is 2 / (2 + √2), which equates to 2 - √2. From there you can find the area of the circle.
Please, using a calculator, I see that it is true, but algebraically, how do you know 2/(2 + √2) = 2 - √2 ? I don't see how to change one expression into the other using algebra. Thank you in advance!
@@LiveHappy76 The trick is knowing (root(2)+2)*(root(2)-2) = 2+2root(2)-2root(2)-4 = 2-4 = -2, which means you can move the squareroot to the top of the fraction by multiplying by (root(2)-2)/(root(2)-2), which equals 1.
So 2/(root(2)+2) = (2 * (root(2)-2))/((root(2)+2) * (root(2)-2)) = (2root(2) - 4)/(-2) = 2 - root(2)
@@LiveHappy76 rationalising the denominator
Wonderful solution! I passed through the triangles’ areas
Very well channel to learn math in simply way.
Line BF is tangent to line AC, segmenting triangle ABC into two mirror triangles ABF and BCF. Focusing on triangle ABF, line BF is the same length of line AF. This provides a route to the solution. Line AF=line BF. Simple pythagorean application here: square root of AB = length of line AF and line BF. This also indicates the length of CF, as the hypotenuse of triangle ABC was cut precisely in half, therefore line CF = line AF. Line CF = line CD. Line OE = DB. DB =CB-CD. 2- sqrt(2). Area of the circle is Pi x (2-sqrt(2))^2 or approximately 1.078.
perfect explanation.I appreciate it
Triangle ABC has two sides of equal length (AB and BC) and a 90 degree angle (angle B) so we know the other two angles are 45 degrees each and so this is a so-called 45-45-90 triangle with sides a=1, b=1 and c=√2. In this case AB and BC are 2 and so the diagonal AC is 2√2.
Now draw a line from point B through point O perpendicular to line AC and call the intersection with AC point F. We now have a triangle ABF, which also has a 90 degree angle and two 45 degree angles. So this too is a triangle with sides 1-1-√2. AF and BF have an equal length and AB as its diagonal. With AB=2 that means BF=AF=√2.
If we now draw the two lines OD and OE, just like in step 3, we can see that triangle BOE is another 45-45-90 triangle. With BE=BD=r that means the diagonal BO=r√2. OF=r so BF=r√2+r=r(1+√2). We also know that BF=√2 so r(1+√2)=√2 which gives r=√2/(1+√2).
The area of the circle then is πr²=π(√2/(1+√2))²=2π/(3+2√2) which is approximately 1.08.
Cool
Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
Once done step 1, you may consider the area of the triangle ABC (2x2/2), which is also the sum of the areas of the triangles OAB, OBC and OAC expressed in term of base (AB, BC and AC) and heigth (r in each case).
You can solve without the two-tangent theorem by recognising that the circle intersects the hypothenusis at exactly half its lenth which is sqrt(2). Then recognise that the the distance BO is r×sqrt(2) which you did. Because it is a unilateral triangle we know that d(B,F)=d(A,F) thus r=sqrt(2)/(1+sqrt(2))
Interesting solution but it's faster to notice the geometrical properties of O. The 3 triangles OAB, OAC, OBC, have the same height, which is the radius of the circle r. But the sum of those 3 triangles areas, is the complete area of the ABC triangle.
Therefore, Area(ABC) = 2*2/2 = 2 is also the sum of the 3 smaller triangles areas AB*r/2 + AC*r/2 + BC*r/2 = r*(AB+BC+CB)/2.
So r*(2+2+2*sqrt(2))/2 = 2, which gives r = 2/(2+sqrt(2)) = 2*(2-sqrt(2))/(4-2) = 2-sqrt(2). Done. I guess everyone knows how to move from radius to surface of the circle ;-) !. Thanks for your channel.
I was going the right direction, similar to your answer..Even though the answer makes sense..it seems that there should be an easier way. Very logical. Tried to do it all in my head... needed to write things down. Reminds me of trigonometry homework...the extra credit kind. Nice video.
Excellent
Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
There is one formula which you can use to find r in rectangular triangle: a+b-c /2 and in this case r=2+2-2root2 /2 = 2-2root2
My solution
FA = √2
So BF = √2
Let's call the distance between B and the nearest point of the circle x
Now we have BF = x + 2r
So that's √2 = x + 2r
On the other hand,
BO = x + r
By using the same small triangle as in the video:
BO = 2√r
Thus, x + r = 2√r
We now have a system of 2 equations
x + r = 2√r
√2 = x + 2r
By solving for r using substitution we find that r = 2 - √2
You are the best sir.
Thank you.
It is isosceles right angle triangle. So when we draw angle bisector of B, BF it divide AC in √2:√2. Thus AD=AF=√2 & DB = 2-√2.
Hello all,
there is a much simpler way to demonstrate.
- Lets call d the distance from point B to the small cercle.
- lets write sr() the square root function
A/ Then the heigh of the triangle is
2.r + d = sr(2)
B/ The distance BO is r + d
As diagonal of square BDOE it is r.sr(2)
Then r + d = r.sr(2)
C/ lets replace B/ into A/
2r + d = sr(2)
r + (r + d) = sr(2)
r + (r.sr(2)) = sr(2)
r.(1 + sr(2)) = sr(2)
r.(1 + sr(2)) . (1 - sr(2)) = sr(2).(1 - sr(2))
r.(1-2) = sr(2) - 2
r = 2 - sr(2)
Then the area of the cercle
A = pi . (2-sr(2))²
An alternative approach, since this is a right-angled isosceles triangle, hence BF is not only the angle bisector of BF^2 = 4 - 2 = 2
=> BF = √2
Also we know that:
BO:OF::(BC+BA):AC
=> BO:OF::4:2√2
=> BO = OF√2.
And we know that:
BF = BO + OF = √2
=> OF(√2 + 1) = √2
=> OF = √2(√2 - 1)
And OF is nothing but the radius of the incircle. Therefore, the area of the circle is:
A = π × OF^2 = π × 2 (√2 - 1)^2 = 3.14 × 2 × (3 - (2 × 1.41)) = 1.13 sq. units.
I hope i am right.🙏
You made it (almost)🙂 - A = 1,078... (simply do the math in the last line...)
@@murdock5537 thank you for your reply 😊🙏 my God, it's been a month!😱 I have forgotten the problem too, my bad! 🙈😂 I will check and get back.. thanks again for your insights.. 😊🙏
@@murdock5537 Hello Mr. Murdock, terribly sorry for my delayed reply. As rightly pointed out by, i have indeed made a calculation error. Your calculation is absolutely correct, it is approximately 1.078 sq. units. Thank you. 😇🙏
@@imonkalyanbarua Great, many thanks for your reply.
@@murdock5537 😊👍
PreMath, you win the prize for most complicated solution! Using the labels you added (D, E &F),
Area = pi*r^2
r = AB -AE (radius-tangent theorem)
AC = 2*sqrt(2) (pythag)
AF = AC/2 = sqrt(2) (symmetry)
AE = AF (two tangent theorem)
AB - AE = 2 - sqrt(2) = r
Area = pi*(2-sqrt(2))^2 = 1.078...
Symmetry could offer another solution:
Let r be the circle's radius. Consider a mirror placed along AC. If D is the image of B, ABCD is a square; AC is tangent to the image circle at F.
BD=2BF=2(BO+r)
Sqrt(2^2+2^2)=2[Sqrt(r^2+r^2)+r]
r=Sqrt(2)/[Sqrt(2)+1]=2-Sqrt(2)
Area=[2-Sqrt(2)]^2 pi=[6-4Sqrt(2)] pi
The circle is the incentre which comes from the angle bisectors. Write the equations for the angle bisectors: for angle B it is y = x (assume it is on a set of axes). Prepare and equation for angle c. y=-2.414+2 (angle c = 45 and half of this is 22.5 so take tan (90+22.5) Solve these equations simultaneously which will give the co-ordinates of the incentre. The x value will be the radius and calculate the area form there. Fairly quick.
Area of ABC is 2*2/2 = r*(2+2+2sqrt2)/2. The last is the sum of areas of BOA, BOC and AOC. In this way I solved without pen and paper for a minute or two.
Thanks Plamen for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
Here's my attempt just solving it from the thumbnail. I'll edit this comment if I'm wrong:
1. Draw 3 segments. Each connecting the circle's center, **O**, to the 3 tangent points between the circle and the triangle.
2. Call the tangent point on the left leg **P**. The top vertex of the triangle **Q**. The tangent point on the hypotenuse **S**. The bottom right vertex **T**. The tangent point on the bottom **U**.
3. Because **OP** , **OS**, and **OU** are radii of the same circle, they are **congruent** to each other.
Because each radii intersects each leg at the point of tangency, each radii is **perpendicular** to each leg. This makes **O** the **incenter** of the triangle.
3_1. Because **O** is the incenter, it bisects both 45° angles into 4 congruent angles. This makes angles POQ, QOS, SOT, TOU congruent (each 67.5°) as well. Triangles POQ, QOS, SOT, TOU have their **shortest sides congruent** since its the radius of the circle. In addition, they have **congruent corresponding angles**, making the triangles congruent to each other.
4. This makes S the midpoint of the hypotenuse. This makes QS = QP also. QS = 2 * SQRT(2) ÷ 2 = QP
5. The segment parallel to OU; on the left leg of the triangle; connecting the right angle to P = 2 - SQRT(2).
6. A square is formed at the bottom left corner because all angles are right angles and 2 segments are congruent, being radii of the same circle. Therefore the radius of the circle is 2 - SQRT(2)
7. Area is pi * r^2 which is **pi times (2 - SQRT(2))**
Edit: I was correct but I forgot that the 2 tangent theorem was a valid proof. Thanks for the video. I took the long route.
Wow!Lately I found the problem like this.But with Pythagorus at triangle.THX for this vid.I can beat the problem.😆
Another approach :
Let us consider the triangle OBE.
Basic trigonometrics defines : cos(OBE) = OE/OB
We know that OE = r
OB = BF-r. As you have shown, BF = Sqrt(2). Hence, OB = Sqrt(2)-r
It follows that cos(OBE) = r/(Sqrt(2)-r)
We know that the angle OBE = 45° (that is, pi/4 radians). The value of cos(pi/4) is well-known, it is equal to Sqrt(2)/2
It follows that r/(Sqrt(2)-r)=Sqrt(2)/2.
A bit of algebra and we show that r = 2/(2+Sqrt(2)) = 2 - Sqrt(2)
Nice and awesome, many thanks! Draw a perpendicular to AB from F = FG = 1 (since AF = √2).
1 = r + r√2/2 → r = 1/(1 + (√2/2)) = 2 - √2 → r² = 6 - 4√2 → Area circle = π(6 - 4√2) 🙂
or take the fast lane: r + r√2 = √2 → r = √2/(1 + √2) = 2 - √2 🙂
It is interesting to solve for r in terms of the side of the triangle as s, rather than the specific length 2. You end up with A = 𝝅s**2(3 - 2√2)/2
Thanks David for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards
"**" ? The double multiplying symbol was a typo?
@@BlacksmithTWD No, ** is exponentiation in programming languages.
@@dibjr
Not that familiar with programming languages, I tend to use ^ for the exponentiation.
I would say A = pi * (s - 0.5 (2 * s^2)^0.5 )^2 not sure if we would get the same outcome for all positive s.
We can use the formula for radius of in circle of a triangle which is (Area of triangle/ semiperimeter of triangle) = radius of in circle.
Area of triangle = 2*2/2 = 2
Semi perimeter = (2+2+2*sqrt(2))/2 = 2+sqrt(2)
Radius of in circle = 2/(2+sqrt(2))
=2-sqrt(2)
Area of circle = pi*square (2-sqrt(2))
Thanks dear for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
Bullseye! You read my mind
Professor is a Master in Analytical Geometry.
Thanks Ramani dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
Love and prayers from the USA!
Just use the area formula A=pr/2, where p is a perimeter, and r is the radius. Solving get r=2-sqrt(2). Same result, just much shorter.
Very Nice explanation!!! Congrats!!!👍👍👍
This question is taught in 9th standard in India
Radius of inscribed circle is called inradius and in case of right angle triangle inradius = (base + perpendicular - hypotenuse )/2
Yes, it is taught in the USA too as the inradius is twice the area of triangle divided by the perimeter of triangle. That work for any triangle. For the case of a right triangle, the inradius is half the sum of the legs minus the hypotenuse. You are correct.
My solution:
as the middle of the inner circle is cross point of the angle-halfing lines
and
as the hypothenusis of the triangle is, by pythagoras, squareroot of 8, (2^2+2^2=8)
we find a triangle with the radius, (drawn directly to the hypothenusis) half of the hypothenusis and the halfing line of the corner's angle of the big triangle,
this angle has 22,5 degrees, because 90 plus two times 45 makes 180 in the big triangle. So half of the corner's angle is 22.5.
In that small triangle tan22,5 equals radius divided through half of hypothenusis,
so radius equals tan22,5 multiplied by one half of squareroot 8. r equals 0,586.
Put into area formula for circles pi x r^2 equals 1,078.
This solution came to my mind firstly. I'm not sure it's easier or more complicated to someone else...
It's easy.
For a circle with a radius r, inscribed in the triangle ABC, the area T of the triangle is equal to the perimeter of the triangle times the radius of the circle divided by 2
T=(AB * r + BC * r + CA * r) / 2 = (AB + BC + CA) * r / 2 = L * r / 2
L = AB + BC + CA is the perimeter of a triangle.
(By the way: The formula T = r * (L / 2) applies to every polygon inside which a circle is inscribed, and also to the disc itself, if we consider the circle to be inscribed "inside" the disc.)
So we have:
Area of the triangle Tt = r * (L / 2) => r = 2 * Tt / L;
Area of a disc Td = (r * 2 * Pi * r) / 2 = pi * r ^ 2 = pi * (2 * Tt / L) ^ 2
In the problem, the triangle is rectangular with the legs equal to a,
so:
Triangel area - Tt = α ^ 2/2;
2Tt = a ^ 2
Triangel perimeter - L = a + a + a * sgrt (2) = a * (2 + sgrt (2))
r = 2 * Tt / L = a / ((2 + sgrt (2)) = a * ((2-sgrt (2)) / 2)
for a = 2
r = 2-sgrt (2)
by the formula of incentre directly r = area / semi perimeter ===> area = 2 semi perimeter = 2+sqrt2 and r = 2- root2 . Then finding area by the formula of circle 1.08 .
AB=2sqrt2 (pythagorem) if you connect the middle of the circle to the points where it hits the triangle, and connect it to point B, youll get 2 triangles with legs r,r and a 90* angle so the other side is rSqrt2, if you also connect the middle pf the circle to side AB, the distance of AC to point B is r+rSqrt2. the area of a triangle =(bh)/2, we already know one b and h, AB and BC, 2 and 2. so the area of the triangle is (2•2)/2=2. we also know the base could be AC and the height d(AC->B)=r(1+sqrt2) so the area is also ((r(1+sqrt2))•2sqrt2)/2 and we knwo the area is 2 so ((r(1+sqrt2))•2sqrt2)/2=2. that is a linear equation in r so we can solve: (r(1+sqrt2))•2sqrt2=4 , r(1+sqrt2)=4/2sqrt2 , r= (4/2sqrt2)/1+2sqrt2=4/(2sqrt2+8)=(4(2sqrt2-8)/(8-64)=(8(sqrt2-4))/(8(-7))=(sqrt2-4)/(-7). the area of a circle is πr^2=π((18-8sqrt2)/49)
I found it easiest to calculate the length of BF, then get it in terms r and solve. Thanks for the vid!
Thanks Jack for the visit!
You are awesome 👍 Take care dear and stay blessed😃 Kind regards
Love and prayers from Arizona, USA!
I agree, much simpler approach.
How do you get r from BF exactly? BF is larger by what seems to me to be unknown value.
@@jankoodziej877 If you pause the video at 4:35 you can see visually that it's r plus the diagonal of the square with side length r.
It is also the height of the triangle if you rotate the figure such that AC is on the bottom. Then you equate the 2
@@JackRule16 ahh, thanks!
Very convoluted explanation.
Draw the 3 lines from the circle center to each of the triangle tangent points with the circle. It then becomes obvious that the bottom edge of the triangle = r + sqrt2, so r = 2 - sqrt2 and so the area = pi r ^2. Done.
Great work sir ,We can also use tangent theorem!❤
So nice of you Aravind dear!
Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
Super Explaination PREMATH SIR 🙏👍👌😃
So nice of you dear!
Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
I offer another solution:
ABC is isosceles, symmetrical about the y=x line. The circle's defined by its center(x,y) and its radius. The center is on that line y=x and its radius reaches to the edges, so the entire circle can be described as an r.
We know the distance of the center to a wall is equal to the distance of the center to (1,1), since that's all just the radius.
These distances can be written as the equations d=r and d=(1-r)*√2, that second distance is a little weirder but it's just turning dist-from-0 into dist-from-1 and then converting side length of a right isosceles triangle to hypotenuse length.
We can now transitive-property combine those expressions into r=(1-r)*√2, and solve for r.
r=(1-r)*√2
r=√2-√2*r
r+√2*r=√2
r(1+√2)=√2
r=√2/(1+√2)
And then you use the radius to get the area with πr²
π*(√2/(1+√2))²
π*(2/(1+2√2+2))
2π/(3+2√2)
, Which does indeed evaluate out to 1.078
ABC right isoceles triangle such that AB=BC=2.
O center of the circle inscribed in ABC
radius of inscribed circle=r.
height from B to [AC] cuts the latter at a point F.
Since ABC is an isoceles triangle, [BF] is also the median and the bissector of the angle B.
Since the center of the inscribed circle is the intersection of the angle bissectors, O is on [BF].
Since ABC is also a right angle triangle, the median equals half the hypothenuse.
Thus BF=AC/2.
In a right angle/isoceles triangle of side a, the hypothenuse equals a×sqrt(2)
Thus, AC=2sqrt(2)
BF=sqrt(2).
OF=r (for obvious reasons)
BO=sqrt(2)-r
But [AB] and [BC] are the tangents from B to the circle, thus there is a square with diagonal [OB] and side length r
BO=r×sqrt(2)
sqrt(2)-r=r×sqrt(2)
r(1+sqrt(2))=sqrt(2)
r=sqrt(2)/(sqrt(2)+1)=sqrt(2)×(sqrt(2)-1)/(2-1)=2-sqrt(2)
Area of inscribed circle=pi×(6-4sqrt(2))
Very well done series. One thing, however…when you say, for instance, “2A+2B=R” you say “2A + 2B equals TO R”. That ‘TO’ is not needed and in fact is confusing as it sounds as if you are saying that 2A + 2B equals 2R. Simply put, two thing equal another. No “to” needed here.
I hope this helps…thanks for the math education…Steve Z
I used a different method. I found the area and the hypotenuse. I connected the centre of the circle to the vertices of the triangle and subsequently found the areas of 3 triangles.
I added them up with the variable as the radius..and equated to the area of the traingle.
Hence you get the radius and you can directly find the area
thank you i am slowly grasping it......
Brilliant video!
Thanks a lot!
Very well explained.
I went about that somewhat differently, but got the same answer:
Since it's right-iso, a radius from circle-center to hypotenuse-center is tilted by 45 deg,
which means its height above circle-center is R/sqrt(2), or R*sqrt(0.5)
add that to the height of the circle-center itself and it's R+R*sqrt(0.5), or R*(1+sqrt(0.5))
but the middle of the hypotenuse is also halfway up the length-2 side of the triangle,
which means that R*(1+sqrt(0.5))=1, so R=1/(1+sqrt(0.5) which is 0.5857864376
(the same thing as 2/sqrt(2))
then plugged that into area = pie r squared
from other comments, it looks like there's a variety of ways to do that...
Connect B-F, calculate length using pythagorus, divide by 2 to get radius, area = pi * r^2. Takes about 30 seconds. Same answer.
I remember the old equation A + B = C + D from high school geometry class! A and B are the sides of the right triangle, C is the hypotenuse, and D is the diameter of the inscribed circle. So you can easily calculate the diameter as 2 + 2 - 2 sqrt(2). The radius is half of that, and then A = pi R^2.
The best solution.
No, it is not. Point "o" is not in the middle of F-B. If it would be ,the circle would have "bridged" the BC and BA sides.
I found it easier to bisect one of the 45 degree angles, then bisect the 90 degree angle and they will intersect at the center of the circle by construction. Then use the new right triangle that has opposite side length r and adjacent side length root two. Then simple trig to find r. This uses a few less theorems.
Much easier
Yes. In this type of triangle, Pythagorean theorem and law of sines are enough.
Simple. Center of circle is (r,r). Tangent point on long side of triangle is (1,1). Distance from (r,r) to (1,1) is r. Solve for r gives r = 2-√2. Area follows by applying A=πr².
Draw line OB. This line is R + y. Length y is from where line OB cuts the circle to point B. Solve for y in terms of R: (y + R) x cos(45) = R so y = 0.414213 x R
Draw line FB. This line is 2 x R + y = 2 x R + 0.414213 x R = 2.414213 x R. Line FB is also 2 x sqrt(2) / 2 = sqrt(2) which gives R = sqrt(2) / 2.414213 = 0.585786
Area of green disc is PI x R x R = 1.07802416
Did anybody read this? ;-)
Solved it with formula of an area of triangle inscribed near a circle. We have that area of a triangle is 2*2*0.5 = 2 and we have that area of a triangle also is p*r, where p is a semiperimetr of a triangle. So, it comes to 2 = (2 + 2 + 2*sqrt(2)) * 0.5 * r.
drop perpendicular from circle's center to the hypotenuse and base of length of r , hypotenuse will be divided into two parts,both's lengths will be √2.
and AE(look in the video) will be (2-r)
now join Center of circle to Angle A,
use Pythagoras theorem for both triangles to find their hypotenuses and equate them
Thank you so much... I love ur vdos
Radius is half of hypotenuse if it's a right angled triangle. So area = pi r square
Very nice explanation sir. Do you have any video about all different kinds of theorem?
Another way: the point o is the intersection of the two bisectors of the angles in A and in C
If we put a system of Cartesian axes with origin in B, we know the angular coefficients of the two bisectors and and the points they pass through
Calculate equations of straight lines passing through the corresponding points and then calculate the intersection of them
The coordinates of the intersection are the radius of the circumference--> area
I think it’s way easier
Another simple way to go about is the formula Δ=rs, where Δ is the area of the triangle, s is the semi-perimeter and r is the inradius. Area=0.5*2*2=2, s=0.5(2+2+2√2) =2+√2
r=Δ/s= 2/(2+√2) =2-√2
Then use πr² for area
This could be solved WAY easier than you did.
You just needed to realise that the distance from tangency points to the bisector are equal, so the "R" must be "2-(2^1/2)".
Then you just get the Area and that's all. 🤷
Elegant solution!
Using geometry only, solution boils down to - Tangent theorem: r = BC - CD = 2 - root2 = 0.586 Area = 1.078.
If trig is allowed Lumkile's solution is elegant (r=root2 x tan 22.5)
Same solution I used, just the other side, r = AB - AE = 2 - √2. Most other people solved using AF = BF = OF + OB.
Nicely done, and an excellent review of geometry with algebra! Thank you.
Line AC's equation is Y = 2 - X and Line BF's is Y = X these 2 lines intersect where X = 2 - X or X = 1 and Y = 1 Line AC is tangent to the circle at point = (1 , 1) // Circle has its center at (r , r) // Circle's equation is r^2=(x - r)^2 + (y - r)^2 // Solving the circle equation for point of tangency = (1 , 1) will give us the radius value r = 2 - SQRT(2)
Really enjoyed the video
I didnt solve but, like always, your explanation is awesome.
So nice of you dear! You are awesome 👍 I'm glad you liked it! Take care dear and stay blessed😃
I used the nradio theorem,
a+b=c+2r
2+2=2sqrt2 +2r
2-2sqrt2=r
But it's a good demonstration
one could also prove CFO and CDO to be congruent, and say:
tangent of angle OCF =r/(2-r).
nonetheless, very well explained!
Here it is a very useful result in case of right angled triangle, radius of incircle = P+B-H/2
Where P=perpendicular
B= base
H=hypotenuse of triangle
The question doesn't specify that quadrilateral OEBD is a square, how did you know that it's a square ?
Because BC is tangent to OD, so there is a right angle, the same with BA and OE. There is another right angle(ABC), so angle EOD is right too. All the sides are same, all are radius of circle. Hope it helped!
Thanks Simon for such an elegant explanation. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
@@simons2006 I intentionally asked for his followers to benefit. I have been following him for a long time because I like his contents.
Though this is not the answer I was expecting. IT'S A SQUARE ONLY BECAUSE THE TWO TANGENTS MEET AT A RIGHT ANGLE.
@@tambuwalmathsclass thats what i basically said😁. Chill...
@@simons2006 Yes 😁 but that is too much explanation for the layman 👍
point B is (0,0), point F is (1,1)
equation of circle : (x-r)^2 + (y-r)^2 = r^2
plug in x=1, y=1 then solve for r via quadratic formula. r= 2 +/- √2
r = 2 - √2 is the only one of the two that makes sense based on the problem. then use area = pi * r^2
we can use formula r = L/s, where L is the area of triangle and s = 0.5 x (perimeter of triangle)
Radius is half of hypotenuse if it's right angled triangle
a very simple problem made very complicated!!!
It's easier to set up the Pythagorean for the half of the BDOE square by drawing the line BF.
BF is √2 and BO is √2-r
Then let's take the BOE triangle:
r²+r²=(√2-r)²
2r²=2+r²-2√2r
r²+2√2r-2=0
Using the quadratic formula, you get the same result but in a simpler way.
Came here to say this!
Area of △ABC is 2 (∵2x2/2=2)
Also area of △ABC is sum of area □BDOE, △AOE, △AOF, △COD, △COF.
It is r^2 + (2-r)r/2 x 4.
Therefore, a quadratic equation is deducted. r^2 - 4r+2 = 0
We have:
AC = ABsqrt(2) = 2sqrt(2)
p = (The perimeter of triangle)/2 = (AB + BC + AC)/2 = 2 + sqrt(2)
S = AB.BC.(1/2) = 2
Apply the formula for the relationship between the area of a triangle and the radius of the circle inscribed in the triangle, we have:
S = pr => r = S/p = 2 - sqrt(2)
=> The area of the circle is: S = r^2.pi = (6 - 4sqrt(2))pi
inradius = ∆/s
∆= 1/2bh = 1/2*2*2 = 2 (area of triangle)
Semiprimeter = 2+2+√(4+4) = 2+√2
Finally inradius = 2/(2+√2) = (2-√2)
Now area of circle =πr^2 = 3.14(6-4√2)
I love this kind of problems in maths ……till 10th I was brilliant in solving all this but unfortunately I took PCB in 11th to pursue medicine I still miss old days of solving maths problems
I solved it in an other way and found A=2*pi/(3+2*sqrt(2)) that seems different from video result...
but I found that is the same result written in other form, I found that:
3-2*sqrt(2) = 1/(3+2*sqrt(2))
and this is quite interesting.
In the same way I found that:
sqrt(2)+1 = 1/(sqrt(2)-1)
This is a simple application of (a+b)*(a-b) = a^2 - b^2
there is no tangent to a circle theorem stating the angle between a tangent and a radius is perpendicular. The correct statement of the theorem is 'A line that is tangent to a circle is perpendicular to the radius drawn at the point of tangency. The other theorem, we make use of, states: Center of the inner tangent circle of a triangle is at the intersection of bisectors. A simple solution is: 1-) Calculating AC=Sqrt(8) from Pythagorean theorem on Triangle CBA. 2-We draw a line through B-O-F. Now CBF and FBA are isosceles triangles due to 45 degree (PI/4 radians) angles at A, B and C therefore CF is Sqrt(8)/2 = Sqrt(2). 2-) Since angle COF is 1/2(45) =22.5 degrees (PI/8 radians).We can look up for tan 22.5 degrees (PI/8 rad.) Or we can calculate from tan (PI/4) = 1 = 2 tan(PI/8)/(1-tan^2(PI/8)) as SQRT(2)-1. Here, if we use half angle formula we must discard the negative solution for tan(PI/8) as this is not in the 1'st quadrant 3-)Considering COF triangle we have tan(PI/8) = Sqrt(2)-1 = OF/CF = r/Sqrt(2) giving r = 2-Sqrt(2). .4-) From this we calculate area S = PI * r^2 = 1.078 units.
AC=AB*SQR(2) (diagonale of a square) --> AF=AE=0,5*AB*SQR(2) --> AE= SQR (2) --> AB=r+AE, --> 2 =r+SQR(2) --> r=2-SQR(2)
OEB is a 1 … 1 … sqrt(2) triangle, with sides r, r, r * sqrt(2). OF is also r.
AF = CF = 2 sqrt(2) / 2 = sqrt(2)
AF = BF, OF + OB = BF
r * sqrt(2) + r = sqrt(2)
r (sqrt(2) + 1) = sqrt(2)
using conjugate
r(sqrt(2) + 1)(sqrt(2) - 1) = sqrt(2)(sqrt(2) - 1)
r(2 - sqrt(2) + sqrt(2) -1) = 2 - sqrt(2)
r = 2 - sqrt(2)
hello, at 1:38 you say c²=8, which is true, but you can't say that sqrt(c²) = sqrt(8) and then C=sqrt(8). You should have said that by taking square root of each term we've got C=sqrt(8). Thanks
Radius is 2-√2
Hence area of circle is π×(2-√2)²
3.14×(2-1.414)²
3.14×(0.586)²
3.14×0.343396
1.0782