A piece of wire is bent in the shape of an equilateral triangle each of whose sides measures 8.8 cm. This wire is rebent to form a circular ring. What is the diameter of the ring?
please explain step by step
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Answers
Given:–
- Measure of each side of an equilateral triangle created from a piece of wire = 8.8cm
To find:–
- Diameter of the ring
Formulas used:–
- Perimeter of traingle = a+b+c
- Circumference of circle = 2πR
where,
- π = 22/7
- R = radius
Step by step explaination:–
☯According to the question circumference of triangle
= circumference of circle.
a + b + c = 2πR
☯ Now evaluating values..
8.8 + 8.8 + 8.8 = 2 × 22/7 × R
26.4 = 44/7 × R
R = 26.4 × 7 / 44
R = 4.2 cm
☯ Now calculating diameter..
2R
2×4.2
8.4 cm
Answer:–
Diameter of ring is 8.4cm
Answer:
Diameter = 8.4 cm
Step-by-step explanation:
The length of the wire will remain the same all the time.
So, perimeter of triangle = circumference of circle
=> 3*(8.8 cm) [Equilateral triangle] = 2πr
=> 26.4 cm = 2(22/7)(r)
∴ r = (26.4) ÷ (22/7) ÷ 2
=> (26.4)*(7/22)*(1/2) (Dividing by a number is the same as multiplying by its reciprocal)
=> (26.4)(1/22)(7)(1/2)
=> (1.2)(7/2) => 3.5*1.2 => 4.2 cm is the radius.
We know that d = 2r
Thus, diameter = 2r => 2(4.2) => 8.4 cm.