I'm thinking line AB is 7.07. Find the incircle radius for both triangles surrounding each circle. Use the formula. ( radius = area/semiperimeter) I get radius 3 for circle B. And radius 4 for circle A. Added together, that forms the cosine of a triangle with a hypotenuse consisting of line AB. Thus cosine lenght EB is 7. The sine of that triangle would be 1. So, going Pythagorean, I'd say line AB is 7.07
@@Quantcircle Well, Area of B triangle = 54. The total perimeter = 36 ( 9 + 12 + 15). Semiperimeter = 18 Using the incircle formula for radius makes the inradius of B = 3. Sq root of Line AB works out to about 7.07. I know that sometimes they want the answer as a simple square root. Say, sq root of 50.
^=read as to the power *=read as square root QR=*(20^2 + 15^2) =*(400+225)=*625=25 Let SR=x So QS=25- x In triangle PQS, angle S =90 So, PS^2 =PQ^2 - QS^2 =(20^2) - (25 - x)^2 =400-{625 -50x+x^2) =400-625+50x- x^2 =50x -x^2 -225.....eqn1 Again in triangle PSR, angle S=90 So, PS^2=PR^2 - SR^2 =15^2 - X^2=225 -X^2....eqn2 Eqn1 & eqn2 are equal due to PS^2 SO, 50X -X^ -225 =225-X^2 50X=225+225=450 X= 450/50 = 9 SO, SR=9 QS= 25-9 =16 Now in triangle PQS PS=*(PQ^2 -QS^2) =*(20^2-16^2) =*(400-256)=*144=12 QS=16 SO area of PQS=(12×16)/2=96 Let R= radious of circle which stands in the interior of PQS triangle R=(PQ×QR×QS)/(4×area) =(20×16×12)/(4×96) )=10 Let r= radious of the circle which stands in the interior region of the triangle PSR Area of PSR =(PS×SR)/2 =(12×9)/2=54 r=(PR×SR×PS)/(4×Area ) =(15×9×12)/(4×54)=7.5 AB=R+r =10+7.5=17.5 Hence AB=17.5 Wrongly i have aplied the formula for exterior circle Hope this question will be as below In triangle PRS the circle intersects PS, SR, PR at the point m, n, o respectively Let PO=y So, RO=15-y Ro=Rn=15-y So Sn=9-(15-y)=y-6 Again po=Pm=y Sm=12-y Sm=Sn 12-y=y-6 2y=18 Y=18/2=9 Sm=r=12-y=12-9=3 In the triangle PQS the circle intersects at a, b, c to PQ, QS, PS respectively Let Pa=t Pa=Pc=t So, Sc=12-t Qa=Qb=20-t So, Sb=16-(20-t)=t-4 Sb=Sc t-4=12-t 2t=16 t=18/2=8 Sc=R=12-t=12-8=4 AB=R+r=4+3=7 Hence AB=7 Note :Draw the parallel line segment to SC from A to PS, simillarly from B to Ps with parallel to Sn( where they touch the circle are also the radious of the circle
@@ManojkantSamal but AB is not parallel to line SC or QR . Perpendicular distance of point A from line QR is 4 cm and perpendicular distance of point B from line QR is 3 cm. So you can easily conclude that line BE is parallel to QR . And in right triangle AEB because Line EB is parallel to QR and Line AC is perpendicular to line QR. Angle AEB will be 90 degrees. Here EB is equal to sum of Radii of both the incircles. ( PS and QR are tangent to both these incircles)
Calculate carefully. The smaller incircle touching line PS on a point located at line BE ( 3 cm from point S) but larger circle is touching PS at a distance of 4 cm from point S.
AB is not 7 cm because the circles are not tangent to each other. They don't touch at all. I'll show using a different method than used in video: coordinate geometry. We have radius of circle with center A = 4 cm and radius of circle with center B = 3 cm. Let PS be along y-axis, and let QR be along x-axis. Since circle with radius 3 is in 1st quadrant and tangent to both axes, it has coordinates B = (3, 3) Since circle with radius 4 is in 2nd quadrant and tangent to both axes, it has coordinates A = (−4, 4) Now we can use distance formula to find distance between points A(−4,4) and B(3,3) AB = √[(−4−3)² + (4−3)²] = √[(−7)² + (1)²] = √[49 + 1] =√(50) = 5√2 ≈ 7.07
No BE is not equal to 4√3, since AB is not 7 because the circles are not tangent to each other. They don't touch at all. I'll show using a different method than used in video: coordinate geometry. We have radius of circle with center A = 4 cm and radius of circle with center B = 3 cm. Let PS be along y-axis, and let QR be along x-axis. Since circle with radius 3 is in 1st quadrant and tangent to both axes, it has coordinates B = (3, 3) Since circle with radius 4 is in 2nd quadrant and tangent to both axes, it has coordinates A = (−4, 4) Now we can use distance formula to find distance between points A(−4,4) and B(3,3) AB = √[(−4−3)² + (4−3)²] = √[(−7)² + (1)²] = √[49 + 1] =√(50) = 5√2 ≈ 7.07 And E is located 1 unit below A, so E = (−4,, 3) and the distance from B(3,3) and E(−4,3) is indeed 7
@@AdarshSuna-d2v Hello! I've actually answered the same question for Mr. Pankaj Chawda in a previous comment. Please take a look at that comment for the explanation. Let me know if you have any further questions!
*@AdarshSuna-d2v* AB is not 7 cm because the circles are not tangent to each other. They don't touch at all. I'll show using a different method than used in video: coordinate geometry. We have radius of circle with center A = 4 cm and radius of circle with center B = 3 cm. Let PS be along y-axis, and let QR be along x-axis. Since circle with radius 3 is in 1st quadrant and tangent to both axes, it has coordinates B = (3, 3) Since circle with radius 4 is in 2nd quadrant and tangent to both axes, it has coordinates A = (−4, 4) Now we can use distance formula to find distance between points A(−4,4) and B(3,3) AB = √[(−4−3)² + (4−3)²] = √[(−7)² + (1)²] = √[49 + 1] =√(50) = 5√2 ≈ 7.07
@@Quantcircle when two circles touch each other externally , distance between their centres is equal to the sum of their radii. That will be equal to 3+4=7
But in this question, these two circles are not touching each other externally. Look at the line segments AC and BD, which are perpendicular to the line QR. BE is the distance between these two lines. And AB is the distance between the centers of these two circles.
AB=5√2≈7,075≈7,08
I'm thinking line AB is 7.07. Find the incircle radius for both triangles surrounding each circle. Use the formula. ( radius = area/semiperimeter) I get radius 3 for circle B. And radius 4 for circle A. Added together, that forms the cosine of a triangle with a hypotenuse consisting of line AB. Thus cosine lenght EB is 7. The sine of that triangle would be 1. So, going Pythagorean, I'd say line AB is 7.07
@@lasalleman6792 Check the radius of Circle B again.
@@Quantcircle Well, Area of B triangle = 54. The total perimeter = 36 ( 9 + 12 + 15). Semiperimeter = 18 Using the incircle formula for radius makes the inradius of B = 3. Sq root of Line AB works out to about 7.07. I know that sometimes they want the answer as a simple square root. Say, sq root of 50.
BE is 8cm. Look at it carefully using all the distances from the external points.
It’s wrong sir. The correct answer is 8.06cm
@@lasalleman6792It’s wrong sir. The correct answer is 8.06cm
^=read as to the power
*=read as square root
QR=*(20^2 + 15^2)
=*(400+225)=*625=25
Let SR=x
So QS=25- x
In triangle PQS, angle S =90
So,
PS^2 =PQ^2 - QS^2
=(20^2) - (25 - x)^2
=400-{625 -50x+x^2)
=400-625+50x- x^2
=50x -x^2 -225.....eqn1
Again in triangle PSR, angle S=90
So,
PS^2=PR^2 - SR^2
=15^2 - X^2=225 -X^2....eqn2
Eqn1 & eqn2 are equal due to PS^2
SO,
50X -X^ -225 =225-X^2
50X=225+225=450
X= 450/50 = 9
SO,
SR=9
QS= 25-9 =16
Now in triangle PQS
PS=*(PQ^2 -QS^2)
=*(20^2-16^2)
=*(400-256)=*144=12
QS=16
SO area of PQS=(12×16)/2=96
Let R= radious of circle which stands in the interior of PQS triangle
R=(PQ×QR×QS)/(4×area)
=(20×16×12)/(4×96) )=10
Let r= radious of the circle which stands in the interior region of the triangle PSR
Area of PSR =(PS×SR)/2
=(12×9)/2=54
r=(PR×SR×PS)/(4×Area )
=(15×9×12)/(4×54)=7.5
AB=R+r
=10+7.5=17.5
Hence AB=17.5
Wrongly i have aplied the formula for exterior circle
Hope this question will be as below
In triangle PRS the circle intersects PS, SR, PR at the point m, n, o respectively
Let PO=y
So, RO=15-y
Ro=Rn=15-y
So
Sn=9-(15-y)=y-6
Again po=Pm=y
Sm=12-y
Sm=Sn
12-y=y-6
2y=18
Y=18/2=9
Sm=r=12-y=12-9=3
In the triangle PQS the circle intersects at a, b, c to PQ, QS, PS respectively
Let Pa=t
Pa=Pc=t
So, Sc=12-t
Qa=Qb=20-t
So, Sb=16-(20-t)=t-4
Sb=Sc
t-4=12-t
2t=16
t=18/2=8
Sc=R=12-t=12-8=4
AB=R+r=4+3=7
Hence AB=7
Note :Draw the parallel line segment to SC from A to PS, simillarly from B to Ps with parallel to Sn( where they touch the circle are also the radious of the circle
@@ManojkantSamal but AB is not parallel to line SC or QR . Perpendicular distance of point A from line QR is 4 cm and perpendicular distance of point B from line QR is 3 cm. So you can easily conclude that line BE is parallel to QR . And in right triangle AEB because Line EB is parallel to QR and Line AC is perpendicular to line QR. Angle AEB will be 90 degrees. Here EB is equal to sum of Radii of both the incircles. ( PS and QR are tangent to both these incircles)
Calculate carefully. The smaller incircle touching line PS on a point located at line BE ( 3 cm from point S) but larger circle is touching PS at a distance of 4 cm from point S.
AB is not 7 cm because the circles are not tangent to each other. They don't touch at all.
I'll show using a different method than used in video: coordinate geometry.
We have radius of circle with center A = 4 cm and radius of circle with center B = 3 cm.
Let PS be along y-axis, and let QR be along x-axis.
Since circle with radius 3 is in 1st quadrant and tangent to both axes, it has coordinates B = (3, 3)
Since circle with radius 4 is in 2nd quadrant and tangent to both axes, it has coordinates A = (−4, 4)
Now we can use distance formula to find distance between points A(−4,4) and B(3,3)
AB = √[(−4−3)² + (4−3)²] = √[(−7)² + (1)²] = √[49 + 1] =√(50) = 5√2 ≈ 7.07
Sir BE is equal to 4√3
@@bidyutbarandas5612 provide your solution so that I can check it and then clarify the situation.
No BE is not equal to 4√3, since AB is not 7 because the circles are not tangent to each other. They don't touch at all.
I'll show using a different method than used in video: coordinate geometry.
We have radius of circle with center A = 4 cm and radius of circle with center B = 3 cm.
Let PS be along y-axis, and let QR be along x-axis.
Since circle with radius 3 is in 1st quadrant and tangent to both axes, it has coordinates B = (3, 3)
Since circle with radius 4 is in 2nd quadrant and tangent to both axes, it has coordinates A = (−4, 4)
Now we can use distance formula to find distance between points A(−4,4) and B(3,3)
AB = √[(−4−3)² + (4−3)²] = √[(−7)² + (1)²] = √[49 + 1] =√(50) = 5√2 ≈ 7.07
And E is located 1 unit below A, so E = (−4,, 3) and the distance from B(3,3) and E(−4,3) is indeed 7
h²=400-x²
h²=225-(25-x)²
h²=225-625+50x-x²
h²=-400+50x-x²
400-x²=-400+50x-x²
800=50x
x=16
h²=400-256=144
h=12
20=16-r1+12-r1
20=28-2r1
r1=4
15=9-r2+12-r2
r2=3
AB²=(4-3)²+(4+3)²
AB²=1²+7²
AB=√50
AB=5√2
Lösung:
Pythagoras: QR = √(20²+15²) = 25
Kathetensatz von Euklid: QS = 20²/QR = 20²/25 = 16 ⟹ SR = QR-QS = 25-16 = 9
Höhensatz von Euklid: PS = √(QS*SR) = √(16*9) = 12
a = Radius des linken Kreises
b = Radius des rechten Kreises
Fläche ∆PQS = PS*QS/2 = PQ*a/2+QS*a/2+PS*a/2 ⟹
12*16/2 = 20*a/2+16*a/2+12*a/2 ⟹
96 = 10*a+8*a+6*a ⟹
96 = (10+8+6)*a ⟹
96 = 24*a |/24 ⟹ a = 4
Fläche ∆PSR = PS*SR/2 = PS*b/2+SR*b/2+PR*b/2 ⟹
12*9/2 = 12*b/2+9*b/2+15*b/2 ⟹
54 = 6b+4,5b+7,5b ⟹
54 = (6+4,5+7,5)*b ⟹
54 = 18b |/18 ⟹ b = 3
Pythagoras:
AB = √[(a+b)²+(a-b²)] = √[(4+3)²+(4-3)²] = √[7²+1²] = √50 ≈ 7,0711[cm]
Solution:
Pythagoras: QR = √(20²+15²) = 25
Euclid’s Pythagorean Theorem:
QS = 20²/QR = 20²/25 = 16 ⟹ SR = QR-QS = 25-16 = 9
Euclid's altitude theorem: PS = √(QS*SR) = √(16*9) = 12
a = radius of the left circle
b = radius of the right circle
Area ∆PQS = PS*QS/2 = PQ*a/2+QS*a/2+PS*a/2 ⟹
12*16/2 = 20*a/2+16*a/2+12*a/2 ⟹
96 = 10*a+8*a+6*a ⟹
96 = (10+8+6)*a ⟹ 96 = 24*a |/24 ⟹ a = 4 area ∆PSR = PS*SR/2 = PS*b/2+SR*b/2+PR*b/2 ⟹ 12*9/2 = 12*b/2+9*b/2+15*b/2 ⟹ 54 = 6b+4.5b+7.5b ⟹ 54 = (6+4.5+7.5)*b ⟹ 54 = 18b |/18 ⟹ b = 3 Pythagoras: AB = √[(a+b)²+(a-b²)] = √[(4+3)²+(4-3)²] = √[7²+1²] = √50 ≈ 7.0711[cm]
@@gelbkehlchen Sie haben eine sehr gute Methode verwendet.
Length BE is not 7cm but 8cm. So the correct answer is AB = 8.06cm.
@@SamuelDonald-pr2uu Please provide the calculations to support your answer.
@@SamuelDonald-pr2uu There is no length BE = 8 cm. My calculation is correct.
Sir answer is wrong.
Correct answer is 7cm.
Mistake is : BE is not equal to 7..
Rather AB is equal to 7cm..
@@AdarshSuna-d2v Hello! I've actually answered the same question for Mr. Pankaj Chawda in a previous comment. Please take a look at that comment for the explanation. Let me know if you have any further questions!
*@AdarshSuna-d2v* AB is not 7 cm because the circles are not tangent to each other. They don't touch at all.
I'll show using a different method than used in video: coordinate geometry.
We have radius of circle with center A = 4 cm and radius of circle with center B = 3 cm.
Let PS be along y-axis, and let QR be along x-axis.
Since circle with radius 3 is in 1st quadrant and tangent to both axes, it has coordinates B = (3, 3)
Since circle with radius 4 is in 2nd quadrant and tangent to both axes, it has coordinates A = (−4, 4)
Now we can use distance formula to find distance between points A(−4,4) and B(3,3)
AB = √[(−4−3)² + (4−3)²] = √[(−7)² + (1)²] = √[49 + 1] =√(50) = 5√2 ≈ 7.07
円Aと円Ｂは接してないんですね
@@koun3866 この問題では、この2つの円は互いに外接していません。線分ACとBDを見てください。これらはQR線に垂直です。BEはこれら2本の線の間の距離です。また、ABはこの2つの円の中心間の距離です。小さい円は、BE上のある点でPS線に接しています。そして、大きい円は、その少し上でPS線に接しています。
まず、大きな円について話しましょう。QRとPSの両方がその接線であり、その半径は4 cmです。接点を中心Aと結ぶと、4 x 4の正方形が形成されます。つまり、大きな円が接線PSに触れる接点は、点Sから4 cmの距離にあることを意味します。同様に、小さな円が接線PSに触れる接点も、点Sから3 cmの距離にあります。この2つの円が、同じ接線PSに触れながら互いに接することはどうすれば可能でしょうか？
This is wrong answer , correct answer is 7. BEvis not equal to 7
Explain the reason behind your answer
Explain the method by which you are getting 7 as your answer...
@@Quantcircle when two circles touch each other externally , distance between their centres is equal to the sum of their radii. That will be equal to 3+4=7
But in this question, these two circles are not touching each other externally. Look at the line segments AC and BD, which are perpendicular to the line QR. BE is the distance between these two lines. And AB is the distance between the centers of these two circles.
The smaller circle touches line PS at some point on line BE. And larger circle touches line PS slightly above it.