By 345 Pythagoras gets the Day Off Unknown U = R/4 you subtract the 4 side from the 5 side to get unknown. 5 = (5/3 + 5/4) R or 1 = (1/3 + 1/4) 4U 1= (4/3 + 3/3) U 1= 7/3. U U =3/7
In right triangle ABC, AC = 3 and BC = 4, so AB = 5 (3-4-5 triangle) Proceed as shown in video up to 3:58 Using similar triangles AOE and ABC we get r/(3-r) = 4/3 3r = 12 - 4r 7r = 12 r = 12/7 Again, using similar triangles AOE and ABC we get (r+x)/r = 5/4 4r + 4x = 5r 4x = r x = (1/4)r = (1/4)(12/7) = 3/7
Wow! Alternatively, OBD is similar to ABC which is a 3-4-5 triangle OD = r, BD = 4-r (4 - r) /r = 4/3 12 - 3r = 4r 7r = 12 r = 12/7 Also OB = (5/3)r = 20/7 x = 5 - OB - r x = 35/7 - 20/7 - 12/7 x = 3/7
Of course, @MrPwmiles has the most elegant solution, obvious within seconds. Once again, a YT 'teacher', Math Booster, lacks mathematical insight and any sense of symmetry.
At 4:15 you start area calculations. THIS IS EXCESSIVELY COMPLICATED! Solve for r knowing triangle BDO is similar to BCA. r/(4-r) = 3/4. From this 4r = 12-3r & r = 12/7 = OP. OA = (5/4)r = 15/7 X = PA = (15/7) - (12/7) = 3/7. No quadratic equation needed!
Невероятная сложность решения в ролике меня просто поразила. :)))))))))) Центр окружности - это основание биссектрисы прямого угла (потому что равноудален от сторон угла :)) ), то есть делит гипотенузу на отрезки 5*3/(3+4)=15/7 и 20/7; Центр, точки касания и вершина прямого угла - это вершины квадрата. За пределами квадрата остаются два треугольника, подобных исходному, с гипотенузами 15/7 и 20/7 (в сумме 5, разумеется), откуда радиус (меньший катет одного треугольника и больший у другого) 12/7 (тут обычно трудности с восприятием этой совершенно элементарной мысли, поэтому поясню - у треугольника с гипотенузой 15/7 и подобного "египетскому" (3,4,5) на самом деле известны и катеты 9/7 и 12/7, а не только гипотенуза :) больший катет как раз и будет радиус этой окружности) Осталось 15/7 - 12/7 = 3/7. Очень сложная задача, ну совсем олимпиадная. :)))
Que questão bonita! É possível determinar o raio utilizando a divisão do triângulo original em dois triângulos cujas alturas são iguais ao raio. A área do triângulo original é a soma das áreas dos triângulos cujas alturas são iguais ao raio. Parabéns pela escolha. Brasil - Outubro de 2024. What a beautiful question! You can determine the radius by dividing the original triangle into two triangles whose heights are equal to the radius. The area of the original triangle is the sum of the areas of the triangles whose heights are equal to the radius. Congratulations on your choice. Brazil - October 2024.
The problem can be solved much simpler. Two equations: 1) (r+x)/r = 5/4 2) (r+x)/(3-r) = 5/3 1) solves to r = 4x Filling in 4x for r in 2) solves to x = 3/7
draw a horizontal line and a perpendicular line from the center of the circle r / (r + x) = 4/5 5r = 4r + 4x x = r/4 r / (3 - r) = 4/3 3r = 12 - 4r r = 12/7 x = r/4 = 3/7
Alternate method with simple calculations in coordinate system: center of circle is the intersection of (A,B) and bisectiing line of angle BCA. Choose coordinte center in C. So we have x/4 + y/3 = 1 and y = -x. => O(-12/7;12/7) => r=12/7. Then x = dist(O,A) - r = 3/7. No rocket science, few geometry rules, simple doing even for pupils.
You did not mention in the question that A B passes by the center of the circle. Put the origine at C, the center of the circle is (-r , r). The line AB has the equation 4y - 3x = 12. So 7r = 12. Now x = SQRT (sqr(r) + sqr(3-r)) - r. Replace r by 12/7 and you got the answer.
By Pythagoras or secant tangent theorem, (x+r)² = r² + (3-r)² or x(x+2r) =(3-r)² eqn 1. By similar triangles OE/AO = BC/AB. By Pythagoras, AB = Sqrt(3² + 4²)= 5. Therefore r/(r+x) = 4/5 or r = 4x. Substituting r in eqn 1. yields x(x+8x) = 9 - 24x + 16x² or 7x² - 24x + 9 = 0. Factorising yields (7x - 3)(x-3) = 0 which gives x = 3/7 or x = 3. X = 3 yields r = 4 x 3 =12 (r > 5) not possible. Therefore x = 3/7.
can be easier using 3-4-5 triangle (and previously determining r = 12/7) AB = x + r + OB (with OB = sqrt(r²+(4-r)²) from Pythagorus) => 5 = x+ r + sqrt(r²+(4-r)²) => x = 5 - r - sqrt(r²+(4-r)²) => x = 5 - 12/7 - sqrt((12/7)²+(16/7)²) => x = 23/7 - sqrt(12²+16²)/7 (12²+16² = 20² still from homothetic 3-4-5 triangle) => x = 23/7 - sqrt(20²)/7 => x = 23/7 - 20/7 = 3/7 (no quadratic equation to solve)
El centro del circulo al cual pertenece el arco de circunferencia inscrito descansa obligadamente en la hipotenusa? No me queda claro. Si no es obligado entonces x es indeterminada.
It's well triangle OBD is similar to his counterpart OEA also ABC OD is Radius center of semicircle AC=3, r=AE=CE, Square ⬛️ =r*r=r² BC=4, BD+DC=4 , BD+r=4, BD=4-r Theorem (4-r)²+(r)²=16-8r+r²+r² =(BO)² AC=3, AE+EC=3, AE+r=3, 3-r=AE r/(4-r)=(3-r)/r, 12=6r-2r²+4r²+8r-2r² 12=7r, r=12/7 OP=r, QO=r QP=2r=2*12/7=24/7=QP Theorem Phythagorean(degree 90) in triangle BOD (12/7)²(4-12/7)²=(B0)² =(BO)²=8.1632653061 BO=sqrt(8.1632653061) In traingle AOE (r)²+(3-r)²=(r+x)² r²+9-6r+r²=r²+2rx+x² (3-r)²=x(2r+x) x²+24/7x=9-72/7+144/49 7x²+24=(441-504+144)/7 49x²+168=441-504+144 49x²+168=81 (7x)²+(13-1)(13+1)=(9)² x=-b±sqrt(b²-4ac)/2a)=x x=3/7
The main triangle is a classic 3-4-5 right triangle. Its easier to use side ratio. Exc triangle OEA OE:4::EA:3, then r/4=(3-r)/3 there you get r=12/7 And then OA:5::OE:4, then (r+x)/5=r/4, then 4r+4x=5r, then 4x=r, so x=r/4=3/7... done in less than 1 minute without hassle
As BC = 4 and CA = 3, ∆BCA is a 3-4-5 Pythagorean triple triangle and AB = 5. Let O be the center point of the semicircle (at the midpoint of PQ). As AP = x and OP = r, AO = x+r. Let M and N be the points of tangency between semicircle O and BC and CA respectively. Draw OM and ON. As radii, OQ = OM = ON = OP = r. First method: As BC and CA are tangent to semicircle O, ∠OMC = ∠CNO = 90°. As ∠MCN = 90° as well, then ∠NOM = 90° and OMCN is a square with side length r. As BC = 4 and CA = 3, BM = 4-r and NA = 3-r. As ∠BMO = ∠BCA = 90° and ∠B is common, ∆BMO and ∆BCA are similar. OB/BM = AB/BC OB/4-r = 5/4 OB = (4-r)5/4 = 5 - 5r/4 AB = 5 AO + OB = 5 (x+r) + (5-5r/4) = 5 x + r = 5r/4 x = 5r/4 - r = r/4 Triangle ∆BMO: OM² + BM² = OB² r² + (4-r)² = (5-5r/4)² = ((20-5r)/4)² r² + 16 - 8r + r² = (400-200r+25r²)/16 32r² - 128r + 256 = 25r² - 200r + 400 7r² + 72r - 144 = 0 7r² + 84r - 12r - 144 = 0 7r(r+12) - 12(r+12) = 0 (r+12)(7r-12) = 0 r = -12 ❌ | r = 12/7 x = r/4 = (12/7)/4 = 3/7 Second method: Draw OC. This creates two triangles, ∆OBC and ∆OCA. As BC and CA are tangent to semicircle O at M and N respectively, OM is perpendicular to BC and ON is perpendicular to CA. [ABC] = [OBC] + [OCA] bh/2 = bh/2 + bh/2 4(3)/2 = 4r/2 + 3r/2 12 = 7r r = 12/7 As OM is perpendicular to BC, ON is perpendicular to CA, and BC is perpendicular to CA, CN = OM = r. As CA = 3, NA = 3-r = 3-(12/7) = 9/7. As ON = 12/7, ∆ONA is a 3/7:1 ratio 3-4-5 Pythagorean triple triangle and AO = 5(3/7) = 15/7. AO = x + r 15/7 = x + 12/7 x = 15/7 - 12/7 = 3/7
O needs to be on the 45 deg diagonal through C, and belong also to the line AB. So it's coordinates are (12/7) (-1,1), where C is the origin. The distance of O to A is therefore (15/7). The distance x is therefore 15/7 - 12/7 (the radius of the circle) = 3/7
Let's consider the triangles OAC and OBC then solve the equation Area (ABC)=Area(Oac)+ area(obc) to find the radius Then Pythagoras theorem to calculate x
5! Satz des Phytagoras: Die Summe der Quadratflächen über den kurzen Seiten eines techtwinkligen Dreiecks, entspricht der Quadratfläche der längsten Seite. (BEWUSST ohne die Verwendung der Fachbegriffe formuliert!) Da die Kantenlänge eines Quadrats die Wurzel aus der Fläche ist: 3×3=9 und 4×4=16 9 +16=25. 5×5=25, also 5. 😂😂😂
Your problem not explicit enough. You are assuming that the center of the circle is on the line of ab. The problem doesn’t say that. Therefore the center of the circle is not necessarily be on the hypotenuse also reach the answer thru sin and co-sin is much easier
You used very long claculations To find r, after proving AEO is similar to ODB AE / OE = OD / BD (3 - r)/r = r/(4 - r) (3 - r) (4 - r) = r^2 12 - 7r + r^2 = r^2 12 = 7r r = 12/7 Now apply Pythagoras in AEO (3-r)^2 + r^2 = (r+x)^2 x = Root((3-r)^2 + r^2) - r Now input r = 12/7 and calculate for x x = 3/7 Alternatively, r / (r+x) = 4/5 (because we have 3-4-5 right triangle) 5r = 4r + 4x r = 4x x = r/4 x = 12/(7 x 4) x = 3/7
The way you this solve problem is way too tedious than it needs be. Just use similar triangles on AOE and OBD to solve for r. Then plug in the value of r in the eqn (r+x)^2=r^2+(3-r)^2 to get x directly. No need to invoke the quadratic formula.
Draw a perpendicular line from the point of tangency from both bases to the hypotenuse, forming a new triangle. The sides of these triangles are 3 - radius, radius, and a hypotenuse of sides 5/4 r since the new triangle is similar to the original 3-4-5 right triangle Hence 5/4 r = (3-r)^2 + r^2 25/16 r^2 = 9 + r^2 - 6r + r^2 25/16 r^2 = 9 + 2r^2 - 6 r 25 r ^2 = 144 + 32 r^2 - 96 r 0 = 7 r^2 - 96 r + 144 r = 1.71429 Hence, the length from the center of the semi-circle to the end of the triangle is 5/4 * 1.71429 or 2.1428625 Since r = 1.71429, then x = 2.1428625 - 1.71429 = 0.4285725
The sum of the area of triangle OBD + the area of square OECD + the area of triangle OAE must be equal to the area of triangle ABC. Means a(4-a) + a(3-a) + a^2 =12 This means: a^2 - 7a + 12 = 0 This means: a = 3 or a = 4 While a < 3. This problem is incorrect.
Let's reason about the symmetry of the image in relation to the vertical. The equation of the circle is (x-r)^2+(y-r)^2=r^2. the center is the intersection of the lines: y=x and y=ax+b (noted ∆) with 3=a(0)+b and 0=4a+b i.e.: b=3 and a=-3/4 d 'where y=(-3/4)x+3. For y=x, we will have: (4/4)x=3-(3/4)x or (7/4)x=3. So r=12/7=24/14. If the center is noted C and A(0,3) on ∆, ||AC||=15/7 let D be the point of intersection of the circle and ∆ in the segment AC. ||AC||=||AD||+||DC||=x+r. 15/7=x+12/7, x=3/12. Why ||AC||=15/7 ? ||AC||^2 = r^2+(3-r)^2 = 9-6r+2r^2 = 2(144/49) + (9*49)/49 - (6*12*7)/49 = (288+441-504)/49=(729-504)/49=225/49=(15/7)^2 and then ||AC||=15/7. Other solution : r=12/7 and according to the diagram, ABC and AOE are two identical triangles. Consequently: OE/BC=AO/AB, i.e. r/4=(r+x)/5. So, 5r/20=4r/20+4x/20. So r/20=4x/20, x=r/4=(12/7)/4=3/7.
This shape does not exist!!! The center of the circle in a triangle like this with sides 4, 3, has hypotenuse =5. The circle tangent to the sides does NOT have a center on the hypotenuse!!! Don't post fake problems! mercy!!! It is impossible that tο be an Olympiad problem!!! Prove it, tell us the year! I'm waiting
Дуже погане та нудне навчання блогерами Канади , яке не стимулює учнів : проста задача , яка вирішується двума формулами з подібності трикутників та відношенням квадрата дотичної до сікущої , проведених с однієї точки . OD/BD=AC/DC , r/4-r=3/4 , 4r=12-3r , r=12/7 . AE*2=AP х AQ , (3 - r)*2 = X x (X+2r) (3 - 12/7)*2=Х х (Х + 24/7) , (9/7)*2 = X*2 + 24/7X , X*2 + 24/7X - 81/49 , Х= -12/7 + \/ 144/49+81/49 = -12/7 + \/225/49 = -12/7 + 15/7 = 3/7. Другий корінь негативний , тому його вирішення не приводиться .
Use Similar Triangles ABC and AOE , x=r/4 and r=12/7,then x=3/7
YES. ITS NOT NECESSARY WHAT SHE DID. SIMILARITY IS SO MUCH EASIER AND QUICKER
I came to write the same thing and timestamp this onwards 3:57
3-r, r, r+x is 3,4,5 triangle
By 345 Pythagoras gets the Day Off
Unknown U = R/4 you subtract the 4 side from the 5 side to get unknown.
5 = (5/3 + 5/4) R or 1 = (1/3 + 1/4) 4U
1= (4/3 + 3/3) U
1= 7/3. U
U =3/7
Mene bhi yahi Kiya hai😊
1min without a Pen. Lol
In right triangle ABC, AC = 3 and BC = 4, so AB = 5 (3-4-5 triangle)
Proceed as shown in video up to 3:58
Using similar triangles AOE and ABC we get
r/(3-r) = 4/3
3r = 12 - 4r
7r = 12
r = 12/7
Again, using similar triangles AOE and ABC we get
(r+x)/r = 5/4
4r + 4x = 5r
4x = r
x = (1/4)r = (1/4)(12/7) = 3/7
Wow! Alternatively, OBD is similar to ABC which is a 3-4-5 triangle
OD = r, BD = 4-r
(4 - r) /r = 4/3
12 - 3r = 4r
7r = 12
r = 12/7
Also
OB = (5/3)r = 20/7
x = 5 - OB - r
x = 35/7 - 20/7 - 12/7
x = 3/7
This is the shortest way to the solution!
That is correct
Of course, @MrPwmiles has the most elegant solution, obvious within seconds. Once again, a YT 'teacher', Math Booster, lacks mathematical insight and any sense of symmetry.
Yes, that's how I did it!
3,4,5 rule its very handy! 👍
CONGRATULATON!YOU FIND THE MOST CUMBERSOME WAY TO SOLVE THIS PROBLEM!
At 4:15 you start area calculations. THIS IS EXCESSIVELY COMPLICATED! Solve for r knowing triangle BDO is similar to BCA. r/(4-r) = 3/4. From this 4r = 12-3r & r = 12/7 = OP. OA = (5/4)r = 15/7 X = PA = (15/7) - (12/7) = 3/7. No quadratic equation needed!
Exactly what i did.
No need of quadratic equation.
Solve using similar triangles
We don't have to use similarity when finding r.
Since ABC=OBC+OAC, 6=2r+3r/2. This is much easier.
Clever !
Nice question, this can also be solved using similar triangles. Would be faster and easier - dont need to deal with quadratics...
What a convoluted way to get to that answer!
Невероятная сложность решения в ролике меня просто поразила. :))))))))))
Центр окружности - это основание биссектрисы прямого угла (потому что равноудален от сторон угла :)) ), то есть делит гипотенузу на отрезки 5*3/(3+4)=15/7 и 20/7; Центр, точки касания и вершина прямого угла - это вершины квадрата. За пределами квадрата остаются два треугольника, подобных исходному, с гипотенузами 15/7 и 20/7 (в сумме 5, разумеется), откуда радиус (меньший катет одного треугольника и больший у другого) 12/7 (тут обычно трудности с восприятием этой совершенно элементарной мысли, поэтому поясню - у треугольника с гипотенузой 15/7 и подобного "египетскому" (3,4,5) на самом деле известны и катеты 9/7 и 12/7, а не только гипотенуза :) больший катет как раз и будет радиус этой окружности) Осталось 15/7 - 12/7 = 3/7. Очень сложная задача, ну совсем олимпиадная. :)))
Устная задача,из подобия тр след (3-R)/3=R/4 , след R=12/7, тк ОА=15/7(гипотенуза а катеты 12/7 и 9/7) то x=3/7.
Que questão bonita! É possível determinar o raio utilizando a divisão do triângulo original em dois triângulos cujas alturas são iguais ao raio. A área do triângulo original é a soma das áreas dos triângulos cujas alturas são iguais ao raio. Parabéns pela escolha. Brasil - Outubro de 2024. What a beautiful question! You can determine the radius by dividing the original triangle into two triangles whose heights are equal to the radius. The area of the original triangle is the sum of the areas of the triangles whose heights are equal to the radius. Congratulations on your choice. Brazil - October 2024.
The problem can be solved much simpler.
Two equations:
1) (r+x)/r = 5/4
2) (r+x)/(3-r) = 5/3
1) solves to r = 4x
Filling in 4x for r in 2) solves to x = 3/7
There is no O in the center of semi circle at the start of this task. Where did you get that? 'O' can be elsewhere..
And OP = r on what basis?
He's wrong.
draw a horizontal line and a perpendicular line from the center of the circle
r / (r + x) = 4/5
5r = 4r + 4x
x = r/4
r / (3 - r) = 4/3
3r = 12 - 4r
r = 12/7
x = r/4 = 3/7
Alternate method with simple calculations in coordinate system: center of circle is the intersection of (A,B) and bisectiing line of angle BCA. Choose coordinte center in C. So we have x/4 + y/3 = 1 and y = -x. => O(-12/7;12/7) => r=12/7. Then x = dist(O,A) - r = 3/7.
No rocket science, few geometry rules, simple doing even for pupils.
You did not mention in the question that A B passes by the center of the circle.
Put the origine at C, the center of the circle is (-r , r). The line AB has the equation 4y - 3x = 12. So 7r = 12.
Now x = SQRT (sqr(r) + sqr(3-r)) - r. Replace r by 12/7 and you got the answer.
Hello, how can we be sure that the center of the circle is on [AB] ?
Because it is a semicircle.
it is a semicircle
@@gandelve How do you know it's a semicircle?
@@rick57hart but it's not written anywhere
@@gandelve It's not written anywhere.
By Pythagoras or secant tangent theorem, (x+r)² = r² + (3-r)² or x(x+2r) =(3-r)² eqn 1. By similar triangles OE/AO = BC/AB. By Pythagoras, AB = Sqrt(3² + 4²)= 5. Therefore r/(r+x) = 4/5 or r = 4x. Substituting r in eqn 1. yields x(x+8x) = 9 - 24x + 16x² or 7x² - 24x + 9 = 0. Factorising yields (7x - 3)(x-3) = 0 which gives x = 3/7 or x = 3. X = 3 yields r = 4 x 3 =12 (r > 5) not possible. Therefore x = 3/7.
can be easier using 3-4-5 triangle (and previously determining r = 12/7)
AB = x + r + OB (with OB = sqrt(r²+(4-r)²) from Pythagorus)
=> 5 = x+ r + sqrt(r²+(4-r)²)
=> x = 5 - r - sqrt(r²+(4-r)²)
=> x = 5 - 12/7 - sqrt((12/7)²+(16/7)²)
=> x = 23/7 - sqrt(12²+16²)/7
(12²+16² = 20² still from homothetic 3-4-5 triangle)
=> x = 23/7 - sqrt(20²)/7
=> x = 23/7 - 20/7 = 3/7
(no quadratic equation to solve)
El centro del circulo al cual pertenece el arco de circunferencia inscrito descansa obligadamente en la hipotenusa? No me queda claro. Si no es obligado entonces x es indeterminada.
It's well triangle OBD is similar to his counterpart OEA also ABC
OD is Radius center of semicircle
AC=3, r=AE=CE, Square ⬛️ =r*r=r²
BC=4, BD+DC=4 , BD+r=4, BD=4-r
Theorem (4-r)²+(r)²=16-8r+r²+r²
=(BO)²
AC=3, AE+EC=3, AE+r=3, 3-r=AE
r/(4-r)=(3-r)/r, 12=6r-2r²+4r²+8r-2r²
12=7r, r=12/7 OP=r, QO=r
QP=2r=2*12/7=24/7=QP
Theorem Phythagorean(degree 90) in triangle BOD
(12/7)²(4-12/7)²=(B0)²
=(BO)²=8.1632653061
BO=sqrt(8.1632653061)
In traingle AOE
(r)²+(3-r)²=(r+x)²
r²+9-6r+r²=r²+2rx+x²
(3-r)²=x(2r+x)
x²+24/7x=9-72/7+144/49
7x²+24=(441-504+144)/7
49x²+168=441-504+144
49x²+168=81
(7x)²+(13-1)(13+1)=(9)²
x=-b±sqrt(b²-4ac)/2a)=x
x=3/7
The main triangle is a classic 3-4-5 right triangle. Its easier to use side ratio.
Exc triangle OEA
OE:4::EA:3, then
r/4=(3-r)/3 there you get r=12/7
And then OA:5::OE:4, then
(r+x)/5=r/4, then 4r+4x=5r, then 4x=r, so x=r/4=3/7... done in less than 1 minute without hassle
As BC = 4 and CA = 3, ∆BCA is a 3-4-5 Pythagorean triple triangle and AB = 5. Let O be the center point of the semicircle (at the midpoint of PQ). As AP = x and OP = r, AO = x+r. Let M and N be the points of tangency between semicircle O and BC and CA respectively. Draw OM and ON. As radii, OQ = OM = ON = OP = r.
First method:
As BC and CA are tangent to semicircle O, ∠OMC = ∠CNO = 90°. As ∠MCN = 90° as well, then ∠NOM = 90° and OMCN is a square with side length r. As BC = 4 and CA = 3, BM = 4-r and NA = 3-r.
As ∠BMO = ∠BCA = 90° and ∠B is common, ∆BMO and ∆BCA are similar.
OB/BM = AB/BC
OB/4-r = 5/4
OB = (4-r)5/4 = 5 - 5r/4
AB = 5
AO + OB = 5
(x+r) + (5-5r/4) = 5
x + r = 5r/4
x = 5r/4 - r = r/4
Triangle ∆BMO:
OM² + BM² = OB²
r² + (4-r)² = (5-5r/4)² = ((20-5r)/4)²
r² + 16 - 8r + r² = (400-200r+25r²)/16
32r² - 128r + 256 = 25r² - 200r + 400
7r² + 72r - 144 = 0
7r² + 84r - 12r - 144 = 0
7r(r+12) - 12(r+12) = 0
(r+12)(7r-12) = 0
r = -12 ❌ | r = 12/7
x = r/4 = (12/7)/4 = 3/7
Second method:
Draw OC. This creates two triangles, ∆OBC and ∆OCA. As BC and CA are tangent to semicircle O at M and N respectively, OM is perpendicular to BC and ON is perpendicular to CA.
[ABC] = [OBC] + [OCA]
bh/2 = bh/2 + bh/2
4(3)/2 = 4r/2 + 3r/2
12 = 7r
r = 12/7
As OM is perpendicular to BC, ON is perpendicular to CA, and BC is perpendicular to CA, CN = OM = r. As CA = 3, NA = 3-r = 3-(12/7) = 9/7. As ON = 12/7, ∆ONA is a 3/7:1 ratio 3-4-5 Pythagorean triple triangle and AO = 5(3/7) = 15/7.
AO = x + r
15/7 = x + 12/7
x = 15/7 - 12/7 = 3/7
Just use the triangle 3,4,5
You can find it easy.
3.4.5 it's very important and handy learn it
Who says it's a semicircle?
r/(4-r)=(3-r)/r》r=12/7
Potencia de A respecto a la circunferencia =X[(24/7)+X] =[3-(12/7)]^2》X=3/7
Gracias y saludos.
R(Radius) = 4K and AT = 3K (T is tangent from A), then OT(R)= 4K, but segment TC = R = 3-3K, so K = 3/7,......AO = 5k but is also 4k + x so x=K=3/7
O needs to be on the 45 deg diagonal through C, and belong also to the line AB. So it's coordinates are (12/7) (-1,1), where C is the origin.
The distance of O to A is therefore (15/7). The distance x is therefore 15/7 - 12/7 (the radius of the circle) = 3/7
You mean dist O to A is 15/7, so x is 3/7. Very clever method.
@@RAG981 You are correct. I tidied things up. Thank you.
Then you want to complicate what is simple. Triangle ABC is similar to AEO. (3-R)/3 = R/4 = (X+R)/5
Let's consider the triangles OAC and OBC then solve the equation
Area (ABC)=Area(Oac)+ area(obc) to find the radius
Then Pythagoras theorem to calculate x
the problem include semicircle? any where?
Very Nice !
5! Satz des Phytagoras: Die Summe der Quadratflächen über den kurzen Seiten eines techtwinkligen Dreiecks, entspricht der Quadratfläche der längsten Seite. (BEWUSST ohne die Verwendung der Fachbegriffe formuliert!) Da die Kantenlänge eines Quadrats die Wurzel aus der Fläche ist: 3×3=9 und 4×4=16 9 +16=25. 5×5=25, also 5. 😂😂😂
Con semejanza de triángulos se puede resolver mucho más rápido.
You could just used trig find the angle and r is easily calculated. Great job anyway for that long calculation👍👍
r=(A*B)/(A+B) and X=A/(A+B) - simple proportions
Your problem not explicit enough. You are assuming that the center of the circle is on the line of ab. The problem doesn’t say that. Therefore the center of the circle is not necessarily be on the hypotenuse also reach the answer thru sin and co-sin is much easier
I said in the beginning of the video that it is semicircle. So, centre will be on the line AB.
You used very long claculations
To find r, after proving AEO is similar to ODB
AE / OE = OD / BD
(3 - r)/r = r/(4 - r)
(3 - r) (4 - r) = r^2
12 - 7r + r^2 = r^2
12 = 7r
r = 12/7
Now apply Pythagoras in AEO
(3-r)^2 + r^2 = (r+x)^2
x = Root((3-r)^2 + r^2) - r
Now input r = 12/7 and calculate for x
x = 3/7
Alternatively,
r / (r+x) = 4/5
(because we have 3-4-5 right triangle)
5r = 4r + 4x
r = 4x
x = r/4
x = 12/(7 x 4)
x = 3/7
The way you this solve problem is way too tedious than it needs be. Just use similar triangles on AOE and OBD to solve for r. Then plug in the value of r in the eqn (r+x)^2=r^2+(3-r)^2 to get x directly. No need to invoke the quadratic formula.
Beautiful solution.
Draw a perpendicular line from the point of tangency
from both bases to the hypotenuse, forming a new triangle.
The sides of these triangles are 3 - radius, radius, and a hypotenuse
of sides 5/4 r since the new triangle is similar to the original
3-4-5 right triangle
Hence 5/4 r = (3-r)^2 + r^2
25/16 r^2 = 9 + r^2 - 6r + r^2
25/16 r^2 = 9 + 2r^2 - 6 r
25 r ^2 = 144 + 32 r^2 - 96 r
0 = 7 r^2 - 96 r + 144
r = 1.71429
Hence, the length from the center of the semi-circle
to the end of the triangle is 5/4 * 1.71429 or 2.1428625
Since r = 1.71429, then x = 2.1428625 - 1.71429 = 0.4285725
With Thales theorem
BD/BC=OD/AC
(4-R)/4=R/3
we obtain R=12/7
Another Thales theorem
AO/AB=OE/BC
(x+R)/5=R/4
x=R/4
x=12/4 7=3/7
So easy
Откуда следует, что центр окружности на гипотенузе? Кстати она равна пяти.
It is much easy to solve... but you went to calculate everything to find your solution.
Only the lower triangle solve it.
pleasu shut up you are a kid behind the screan who has nothing in his life and talk shit you are not able to solve this so be quit
The sum of the area of triangle OBD + the area of square OECD + the area of triangle OAE must be equal to the area of triangle ABC. Means a(4-a) + a(3-a) + a^2 =12 This means: a^2 - 7a + 12 = 0 This means: a = 3 or a = 4 While a < 3. This problem is incorrect.
Let's reason about the symmetry of the image in relation to the vertical. The equation of the circle is (x-r)^2+(y-r)^2=r^2. the center is the intersection of the lines: y=x and y=ax+b (noted ∆) with 3=a(0)+b and 0=4a+b i.e.: b=3 and a=-3/4 d 'where y=(-3/4)x+3. For y=x, we will have: (4/4)x=3-(3/4)x or (7/4)x=3. So r=12/7=24/14. If the center is noted C and A(0,3) on ∆, ||AC||=15/7 let D be the point of intersection of the circle and ∆ in the segment AC. ||AC||=||AD||+||DC||=x+r. 15/7=x+12/7, x=3/12.
Why ||AC||=15/7 ?
||AC||^2 = r^2+(3-r)^2 = 9-6r+2r^2 = 2(144/49) + (9*49)/49 - (6*12*7)/49 = (288+441-504)/49=(729-504)/49=225/49=(15/7)^2 and then ||AC||=15/7.
Other solution : r=12/7 and according to the diagram, ABC and AOE are two identical triangles. Consequently: OE/BC=AO/AB, i.e. r/4=(r+x)/5. So, 5r/20=4r/20+4x/20. So r/20=4x/20, x=r/4=(12/7)/4=3/7.
Устная задача,из подобия тр след (3-R)/3=R/4 ,R=12/7, x=3/7
Еле дошло, (3-R)/R=3/4)))))
There is a much easier way to find x by using similar triangles.
3/(3-r)=4/r=5/r+x
It’s also easy by using trignomatry method
Complicou demais.
Check Video Time > 11:00
Presh talwaker from mind your decisions made a formula for these types of problems
Similar triangles it is!!
It's not a semi circle.
Через подобие треугольников будет быстрее и легче!
THERE IS A MISTAKE IN THE ANSWER:X=378/98 IS THE CORRECT ANSWER
very lengthy Process....Just use similar triangle method and solve it then it is very easy to simplify..🤬🤣
I remember Very in portuguese .
3/7 ans
Phương Pháp giải hay, nhưng nói tiếng nước ngoài khó hiểu
This shape does not exist!!! The center of the circle in a triangle like this with sides 4, 3, has hypotenuse =5. The circle tangent to the sides does NOT have a center on the hypotenuse!!! Don't post fake problems! mercy!!! It is impossible that tο be an Olympiad problem!!! Prove it, tell us the year! I'm waiting
Με ομοιότητα βγαίνει αμμεσα.
Чешем левой ногой правое ухо😂😅
Thales Theorem it'll be easier.
Guruji, this solution is not upto your own high standards.
❤👍👋
用这种解法会被人嘲笑一年。
5
Дуже погане та нудне навчання блогерами Канади , яке не стимулює учнів : проста задача , яка вирішується двума формулами з подібності трикутників та відношенням квадрата дотичної до сікущої , проведених с однієї точки . OD/BD=AC/DC , r/4-r=3/4 , 4r=12-3r , r=12/7 .
AE*2=AP х AQ , (3 - r)*2 = X x (X+2r) (3 - 12/7)*2=Х х (Х + 24/7) , (9/7)*2 = X*2 + 24/7X ,
X*2 + 24/7X - 81/49 , Х= -12/7 + \/ 144/49+81/49 = -12/7 + \/225/49 = -12/7 + 15/7 = 3/7.
Другий корінь негативний , тому його вирішення не приводиться .
aaaarrrgghhhhhh just use a calculatorrr :))))
BO+OA=AB???? Why??????😂😂😂😂😂
note 3/7 = 0.4285725
0.4285725
5