Question

Mr.Javier designed an arch made of bent iron for the top of a school's main entrance.The 12 segments between the two concentric semicircles are each 0.8meter long.Suppose the diameter of the inner semicircle is 4 meters.What is the total length of the bent iron used to make this arch?​

Answer 1

Given :- Mr.Javier designed an arch made of bent iron for the top of a school's main entrance.The 12 segments between the two concentric semicircles are each 0.8meter long.Suppose the diameter of the inner semicircle is 4 meters. What is the total length of the bent iron used to make this arch ?

Solution :-

→ Length of each segment = 0.8 m .

so,

→ Length of 12 such segment = 0.8 * 12 = 9.6 cm.

now,

→ Radius of inner arc = Diameter / 2 = 4/2 = 2cm.

→ Radius of outer arc = D/2 = 5.6/2 = 2.8 cm.

then,

→ Length of inner arc = circumference of semi - circle = π * radius = 3.14 * 2 = 6.28 cm.

and,

→ Length of outer arc = circumference of semi - circle = π * radius = 3.14 * 2.8 = 8.792 ≈ 8.8 cm.

therefore,

→ The total length of the bent iron used to make this arch = 9.6 + 6.28 + 8.8 = 24.68 cm (Ans.)

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Answer 2

Answer:

24.68 meters

Step-by-step explanation:

There are 12 straight pieces measuring 0.8meters each, for a total of

12(0.8meters)=9.6meters.

There are also 2 half-circle arcs.

The inside arc has a radius of 4meters/2=2meters .

The outside arc has a radius of 2meters+0.8meters=2.8meters .

Since the circumference of a circle of radius R can be calculated as 2π(R) ,

the length of a semicircle of radius R can be calculated as  π(R) .

So, the length needed for the inside semicircle is  π(2meters)=about6.28meters,

and the length needed for the outside semicircle is  π(2.8)=about8.80meters ,

The total length of the bent iron used to make the arch is about

9.6meters+6.28meters+8.80meters=24.68meters