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# If O is the centre of the above circle, ∠A + ∠B + ∠C + ∠D + ∠E + ∠F +

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

I am also getting the answer 720.

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