Easy. 54= the area of Triangle. So 54=15×r/2+3r^2+r(12-3r)/2+3r(9-r)/2===>r=2
@alfal4239
10 күн бұрын
So 54=15*r/2+12*r/2 + 9*3r/2
@nandisaand5287
11 күн бұрын
I used trig: Angle ABC= arctan(9/12) =36.87 Draw a line PB. Since Triangle QPB=EPB, angle PBQ=½ ABC =½(36.87)=18.435 Tan(18.435)=r/(12-3r) ⅓=r/(12-3r) X-Multiply: 3r=12-3r 6r=12 r=2 Area=2•[Pi(2)²] =8•Pi
@jimlocke9320
11 күн бұрын
You beat me to it! Alternatively, you can use the tangent double angle formula to determine that tan(
@ThePhantomoftheMath
11 күн бұрын
Great job!!!
@gnanadesikansenthilnathan6750
3 күн бұрын
Yes but there can he another method.find the hypotaneous . Area = 1/2 bh for larger triangle. Draw a tangent that divides the triangle into two halves . This tangent is equal to the diameter of two circles. Find the oenth of tangent . With we fan know the radius. Combined area = πr^r + πr^2 = 2πr^2.
@rangaweerakkody165
9 күн бұрын
drop a perpendicular to BC from P, PD. PDB and POB triangles are congrunet. If b^ = 2x tan 2x = 3/4. PBD = tan x. OB is 12 - 3r Using tan formula, 3/4 = 2 tan x / 1 - tan^2 x -> 3tan^2 x + 8 tan x - 3 = 0, tan x = 1/3 or -3 since x > 0, tan x = 1/3 tan x = Po / OB = r/(12- 3r) = 1/3, r = 2 Area = 8pi
@luiscostacarlos
10 күн бұрын
genius solution, without using a calculator Nice. Solução genial, sem uso de calculadora.👏👏👏
@ThePhantomoftheMath
10 күн бұрын
@@luiscostacarlos Thank you! ❤️
@rogerphelps9939
9 күн бұрын
The long side of the equation has equation y = 9-3x/4. the circumference of the right hand circle has (y-r)^2 + (x-3r)^2 = r^2. It is straightforward to determine r from these equaqtions and the area of the circles follows easily.
@bjorntorlarsson
9 күн бұрын
I bisected the angle at B to get the two congruent right triangles BPQ and BPE that have the radius as one side. And from there on. I won't type it out here because I then went about it more roundabout than what is probably necessary. The crux is to find a relationship between the triangle and a circle center.
@ThePhantomoftheMath
9 күн бұрын
That's a really nice approach!
@quigonkenny
9 күн бұрын
Assume triangle vertices are A, B, C, labeled counterclockwise from right angle. Left circle center is O, right circle center is P, all as in video. (Paused video here) Let the point of tangency between circle O and AC be S, and the points of tangency between circles O and P and AB be M and N respectively. Let the point of tangency between circle P and BC be V, and the point of tangency between the two circles will be T. Let the radius of the two congruent circles be r. As AC = 9 = 3(3) and AB = 12 = 4(3), it's clear that ∆ABC is a 3:4:5 Pythagorean triple right triangle and BC = 5(3) = 15. As NB and BV are tangent to circle P and intersect at B, NB = BV = x. As SA and AM are tangent to circle O and intersect at A, SA = AM. As ∠OSA = ∠AMO = 90° as S and M are points of tangency and OS and OM are radii, and as ∠SAM = 90°, ∠MOS = 90° as well and SAMO is a square. As the point of tangency between two circles and their centers is always collinear, OP contains point T, so OP = OT + PT = 2r. As PN and OM are both perpendicular to AB and thus parallel to each other, OMNP is a rectangle with a width of 2r (OP, MN) and height of r (OM, PN). As AB = 12 and AN = r+2r = 3r, then as x = NB, x = 12-3r. VC thus equals 15-(12-3r) = 3+3r. Draw PV, and extend NP to L on BC. As ∠BVP = ∠LNB = 90° (radius at tangent), and ∠VLP = ∠BLN = ∠BCA (corresponding angles, as NL is parallel to AC), then ∆PVL and ∆LNB are similar to ∆ABC. On ∆PVL, as PV = r corresponds to the long leg of the 3:4:5 triangle, then VL = 3r/4 and LP = 5r/4. BL = BV + VL BL = x + 3r/4 = 12 - 3r +3r/4 BL = 12 - 9r/4 LN = 5r/4 + r = 9r/4 Triangle ∆LNB: BL/LN = BC/AC (12-9r/4)/(9r/4) = 15/9 = 5/3 3(12-9r/4) = 5(9r/4) 36 - 27r/4 = 45r/4 72r/4 = 36 18r = 36 r = 36/18 = 2 Combined circle area: A = πr² + πr² = 2πr² A = 2π2² = 8π sq units
@ThePhantomoftheMath
9 күн бұрын
Very very nice!
@Ray3-d4v
9 күн бұрын
Really appreciate your reminders of pertinent theorems as you solved the problem. I saw a 3-4-5 triangle and used trig to get r but I really like your purely geometric approach.
@ProfessorDBehrman
9 күн бұрын
Actually, this problem is most easily solved using the formula for the radius of the in-circle of a right triangle: R = (1/2)( a + b - c ) In this case we have b = 12 - 2R, a = (3/4)b , c = (5/4)b, and we get a linear equation for R. No need to use Pythagorean Theorem.
@devondevon4366
10 күн бұрын
The area of each circle is 4pi, and the area of both circles = is 8pi or 25. 133 Answer Draw a perpendicular line through the two red circles to form another right triangle 3-4-5 and a trapezoid. This triangle will be similar to the 9, 12, and 15 triangles above or the 3-4-5 scaled up by 3 since both are right triangles and share another angle The new triangle's horizontal base will = 12 - 2r since it is a diameter away ( 2 radii = 1 diameter) from the new triangle Hence, its vertical base will = 3/4 ( 12- 2r ) = 9 - 1.5 r (Recall, the original triangle is a scaled-up 3-4-5; hence the vertical side will equal 3/4 of the horizontal side) Hence its hypotenuse of the vertical side= 5/3 (9 - 1.5 r ) = 15 - 2.5 r HYPOTENUSE But, according to the tangent circle theorem, the hypotenuse of a triangle ( focus on the new triangle ) when a circle is inscribed can be determined by the sum of the distance from the point of tangency from both the vertical and horizontal base Hence, the hypotenuse = (12- 2r - r ) + ( 9-1.5 r - r) = 21-5.5 r So, the hypotenuse in terms of r is not only 15- 2.5 r but 21-5.5 r Hence, to find r set both to equal each other 15 - 2.5 r = 21-5.5 r - 2.5 r + 5.5 r = 21 -15 3 r = 6 r = 6/3 r = 2 The radius of the red circle = 2 Hence, area = pi r^2 = 4 pi Hence, the area of both circles = 8 pi
@ThePhantomoftheMath
10 күн бұрын
@@devondevon4366 Very nice one!
@devondevon4366
10 күн бұрын
@@ThePhantomoftheMath Thanks. I tried something new, hoping it would work, and it did by forming another 3-4-5 scaled up by 2, hence 6-8-10, and a trapezoid, height 4, bases ( 9 and 6). Then, I checked if both = the area of the original triangle's area of 54, and it did as The area of the original triangle (54) is equal to the area of the trapezoid (30) and the area of the 6-8-10 triangle (24).
@ThePhantomoftheMath
10 күн бұрын
@@devondevon4366 That's really clever!
@firstnamelastname307
10 күн бұрын
The radius of inscribed circle in famous 3-4-5 triangle is 1 (not hard to prove). The result follows by reflecting that circle via side length with 3 (in fact we need a picture 2 times proportional to arrive at given 9-12).
@KipIngram
10 күн бұрын
The angle at x=12 is arctan(0.75) = 36.8699 degrees. If the red circles have radius r, then the center of the rightmost circle is at x = 3*r, y = r. To get to the tangent point of the rightmost circle with the triangle hypotenuse, we need to increase x by r*sin(36.8699) and y by r*cos(36.8699). So for the tangent point x = (3+sin(36.8699))*r y = (1+cos(36.8699))*r This point must also be on the line y = 9 - 0.75*x. So 9 - 0.75*(3+sin(36.8699))*r) = (1+cos(36.8699))*r [ (1+cos(36.8699)) + 0.75*(3+sin(36.8699)) ]*r = 9 This solves out to r = 2. Therefore, each circle has area 4*pi, so the total is Area = 8*pi. Q.E.D.
@marioalb9726
11 күн бұрын
tan α = 9/12 --> α = 36,97° tan ½α = r / (12-3r) = 1/3 12/r - 3 = 3 r = 2 cm A = 2 (πr²) = 8π cm² ( Solved √ )
@UAPch
8 күн бұрын
In an algebra class depending on the instructor, using trig may get you marked partial credit, even with the right answer a sound methodology LOL 💯
@marioalb9726
8 күн бұрын
@@UAPch Thanks four your comment. I didn't understand that of "partial credit". I didn't know that there were credits. Using trigonometry, all is easier !!! This videos are not just for students, are for everyone, that says the goal statement of this channel !! I just solve these exercises, for fun
@constantinfedorov2307
8 күн бұрын
Как то это сложно все. 12 - 4r = 2r, r = 2. Все решение. Для тех, кто не понял (хотя трудно таких представить). Внутренняя общая касательная (она же радикальная ось) отсечет от треугольника подобный ему треугольник, для которого одна из окружностей, - правая, - вписанная. Так как и тот и другой треугольники подобны треугольнику со сторонами 3,4,5, у которого радиус вписанной окружности 1, то стороны отсеченного радикальной осью треугольника можно записать, как 3r, 4r, 5r, а разница между большими катетами равна 2r. И никаких длинных уравнений. :)
@bkp_s
11 күн бұрын
Nice vdo sir
@ThePhantomoftheMath
11 күн бұрын
Thank you sir!
@lab483
11 күн бұрын
Crafty I like it
@ThePhantomoftheMath
11 күн бұрын
Thanks. Glad you liked it!
@JSSTyger
10 күн бұрын
A = 8π in total
@krwada
8 күн бұрын
I cheated. I used trig. One only needs to notice that only the right triangle needs to be solved is that which is formed by right triangle PQB. The acute angle of PQB is simply the 1/2 angle of the CAB acute angle. so ... r/(12-3r) = tan(1/2 angle of CAB) In the example, r = 2. The rest is just using the pi-r-squared formula. I wonder if there is a way to do this puzzle from pure geometry without Pythagoras? Maybe, using similar to the trig solution I show above. After all, angles and half-angles and their relationships are well defined with respect to the corresponding ratios of the sides of the right triangle no?
@ThePhantomoftheMath
8 күн бұрын
@@krwada No, you didn't "cheat". Trig is completely valid way of solving this problem.
@IllllllllllIIlIIIIlIlllI
10 күн бұрын
This is really great, and thank you for providing the Arabic translation 🤍
@ThePhantomoftheMath
10 күн бұрын
Thanks! I will try to keep providing those translations in the future as well.
Пікірлер: 39