Question

Two construction circle circle circle of radius 5 centimetre and 3 cm find the length of chord of a larger circle which towards a triangle smaller circle

Answer 1

Answer:

Let the radius of the bigger circle be 'R' and the radius of the smaller circle be 'r'.

Let the radius of the bigger circle be 'R' and the radius of the smaller circle be 'r'.It is given that R=5cm=OB and r=3cm=OD and AB is the chord whose length we have to find.

Let the radius of the bigger circle be 'R' and the radius of the smaller circle be 'r'.It is given that R=5cm=OB and r=3cm=OD and AB is the chord whose length we have to find.AD=BD and OD⊥AB(radius is perpendicular to a chord and it divides the chord into two equal parts)

Let the radius of the bigger circle be 'R' and the radius of the smaller circle be 'r'.It is given that R=5cm=OB and r=3cm=OD and AB is the chord whose length we have to find.AD=BD and OD⊥AB(radius is perpendicular to a chord and it divides the chord into two equal parts)therefore, ΔODB is a right angled triangle

Let the radius of the bigger circle be 'R' and the radius of the smaller circle be 'r'.It is given that R=5cm=OB and r=3cm=OD and AB is the chord whose length we have to find.AD=BD and OD⊥AB(radius is perpendicular to a chord and it divides the chord into two equal parts)therefore, ΔODB is a right angled trianglewhere, OD²+BD²=OB²

Let the radius of the bigger circle be 'R' and the radius of the smaller circle be 'r'.It is given that R=5cm=OB and r=3cm=OD and AB is the chord whose length we have to find.AD=BD and OD⊥AB(radius is perpendicular to a chord and it divides the chord into two equal parts)therefore, ΔODB is a right angled trianglewhere, OD²+BD²=OB² (3)²+BD²=(5)²

Let the radius of the bigger circle be 'R' and the radius of the smaller circle be 'r'.It is given that R=5cm=OB and r=3cm=OD and AB is the chord whose length we have to find.AD=BD and OD⊥AB(radius is perpendicular to a chord and it divides the chord into two equal parts)therefore, ΔODB is a right angled trianglewhere, OD²+BD²=OB² (3)²+BD²=(5)² 9+BD²=25

Let the radius of the bigger circle be 'R' and the radius of the smaller circle be 'r'.It is given that R=5cm=OB and r=3cm=OD and AB is the chord whose length we have to find.AD=BD and OD⊥AB(radius is perpendicular to a chord and it divides the chord into two equal parts)therefore, ΔODB is a right angled trianglewhere, OD²+BD²=OB² (3)²+BD²=(5)² 9+BD²=25 BD²=25-9

Let the radius of the bigger circle be 'R' and the radius of the smaller circle be 'r'.It is given that R=5cm=OB and r=3cm=OD and AB is the chord whose length we have to find.AD=BD and OD⊥AB(radius is perpendicular to a chord and it divides the chord into two equal parts)therefore, ΔODB is a right angled trianglewhere, OD²+BD²=OB² (3)²+BD²=(5)² 9+BD²=25 BD²=25-9 BD²=16

Let the radius of the bigger circle be 'R' and the radius of the smaller circle be 'r'.It is given that R=5cm=OB and r=3cm=OD and AB is the chord whose length we have to find.AD=BD and OD⊥AB(radius is perpendicular to a chord and it divides the chord into two equal parts)therefore, ΔODB is a right angled trianglewhere, OD²+BD²=OB² (3)²+BD²=(5)² 9+BD²=25 BD²=25-9 BD²=16 BD=4cm

Let the radius of the bigger circle be 'R' and the radius of the smaller circle be 'r'.It is given that R=5cm=OB and r=3cm=OD and AB is the chord whose length we have to find.AD=BD and OD⊥AB(radius is perpendicular to a chord and it divides the chord into two equal parts)therefore, ΔODB is a right angled trianglewhere, OD²+BD²=OB² (3)²+BD²=(5)² 9+BD²=25 BD²=25-9 BD²=16 BD=4cm Since AD=BD=4cm

Let the radius of the bigger circle be 'R' and the radius of the smaller circle be 'r'.It is given that R=5cm=OB and r=3cm=OD and AB is the chord whose length we have to find.AD=BD and OD⊥AB(radius is perpendicular to a chord and it divides the chord into two equal parts)therefore, ΔODB is a right angled trianglewhere, OD²+BD²=OB² (3)²+BD²=(5)² 9+BD²=25 BD²=25-9 BD²=16 BD=4cm Since AD=BD=4cm therefore, AB=AD+BD

Let the radius of the bigger circle be 'R' and the radius of the smaller circle be 'r'.It is given that R=5cm=OB and r=3cm=OD and AB is the chord whose length we have to find.AD=BD and OD⊥AB(radius is perpendicular to a chord and it divides the chord into two equal parts)therefore, ΔODB is a right angled trianglewhere, OD²+BD²=OB² (3)²+BD²=(5)² 9+BD²=25 BD²=25-9 BD²=16 BD=4cm Since AD=BD=4cm therefore, AB=AD+BD AB=4+4

Let the radius of the bigger circle be 'R' and the radius of the smaller circle be 'r'.It is given that R=5cm=OB and r=3cm=OD and AB is the chord whose length we have to find.AD=BD and OD⊥AB(radius is perpendicular to a chord and it divides the chord into two equal parts)therefore, ΔODB is a right angled trianglewhere, OD²+BD²=OB² (3)²+BD²=(5)² 9+BD²=25 BD²=25-9 BD²=16 BD=4cm Since AD=BD=4cm therefore, AB=AD+BD AB=4+4 AB=8cm