2 circles of radius 3 cm each is constructed such that centre of each touches the circumference of other as shown below.
$A C_{1} C_{2} is a/an$ .

isosceles triangle

No worries! We‘ve got your back. Try BYJU‘S free classes today!

equilateral triangle

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

scalene triangle

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B equilateral triangle
Since the radius of the circle is the same, we have
$A C_{2} = A C_{1} = r$

$C_{1} C_{2} is also equal to radius of the circle.$

$Hence, A C_{1} = A C_{2} = C_{1} C_{2} = r$

$∴ The triangle formed is an$ $equilateral triangle.$

Suggest Corrections

Similar questions

Two circles of radius 3 cm each is constructed such that centre of each touches the circumference of other as shown in figure. The triangle $A C_{1} C_{2}$ is

If the vertices of a triangle are (-4,0), (4,0) and (0,3), the triangle is a/an _______.

The points $(a, a)$ , $(- a, - a)$ and $(- \sqrt{3} a, \sqrt{3} a)$ form the vertices of an

Two tangents are drawn onto a circle with centre O and radius r, from an external point P. These tangents touch the circle at points A and B respectively. The triangle ABP is definitely a ______________ .

Three circles each of radius 5 cm is constructed in such a way that the centre of each touches the circumference of other as shown in figure. The length BC is equal to

Join BYJU'S Learning Program

2 circles of radius 3 cm each is constructed such that centre of each touches the circumference of other as shown below. AC1C2 is a/an .

The correct option is B equilateral triangleSince the radius of the circle is the same, we have AC2=AC1=r C1 C2 is also equal to radius of the circle. Hence, AC1=AC2=C1 C2=r ∴The triangle formed is an equilateral triangle.

2 circles of radius 3 cm each is constructed such that centre of each touches the circumference of other as shown below.
$A C_{1} C_{2} is a/an$ .

The correct option is B equilateral triangle
Since the radius of the circle is the same, we have
$A C_{2} = A C_{1} = r$

$C_{1} C_{2} is also equal to radius of the circle.$

$Hence, A C_{1} = A C_{2} = C_{1} C_{2} = r$

$∴ The triangle formed is an$ $equilateral triangle.$