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If O is the centre of the above circle, ∠A + ∠B + ∠C + ∠D + ∠E + ∠F + [#permalink]
Could anyone please explain how the answer is 360.

I am also getting the answer 720.

Thanks in advance
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Re: If O is the centre of the above circle, ∠A + ∠B + ∠C + ∠D + ∠E + ∠F + [#permalink]
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Bunuel wrote:

If O is the centre of the above circle, ∠A + ∠B + ∠C + ∠D + ∠E + ∠F + ∠G + ∠H = ?

(A) 90°
(B) 180°
(C) 360°
(D) 540°
(E) 720°

Attachment:
2021-03-08_14-33-42.png


Property: The angle subtended by an arc at the Center is equal to the measure of the arc, while the angle subtended by an arc anywhere at the circumference of circle is equal to half the measure of the arc

∠A = \(\frac{DE+EF}{2}\)
∠B = \(\frac{FG+EF}{2}\)
∠C = \(\frac{FG+GH}{2}\)
∠D = \(\frac{GH+HA}{2}\)
∠E = \(\frac{HA+AB}{2}\)
∠F = \(\frac{AB+BC}{2}\)
∠G = \(\frac{BC+CD}{2}\)
∠H = \(\frac{DE+CD}{2}\)

When you add all of them,

∠A + ∠B + ∠C + ∠D + ∠E + ∠F + ∠G + ∠H = \(\frac{DE+EF}{2}+\frac{FG+EF}{2}+\frac{FG+GH}{2}+\frac{GH+HA}{2}+\frac{HA+AB}{2}+\frac{AB+BC}{2}+\frac{BC+CD}{2}+\frac{DE+CD}{2}\)
\(=\frac{2(DE+EF+FG+GH+HA+AB+BC+CD)}{2}=DEFGHABCD\), that is the circumference of the circle.

Measure of circumference of any circle is 360.


C
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Re: If O is the centre of the above circle, ∠A + ∠B + ∠C + ∠D + ∠E + ∠F + [#permalink]
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