In the figure below, if ^@ \angle A + \angle B + \angle C + \ang | Math Question and Answer

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In the figure below, if $\angle A + \angle B + \angle C + \angle D + \angle E + \angle F + \angle G = x$ degrees, then what is $x ?$

Answer:

$540$

Step by Step Explanation:

Let us divide the shape into triangles with respect to a center $O.$
We observe, from the diagram, that seven triangles have formed $e.g. OAC, OBD,$ etc. We also observe that the angles around $O$ have been added twice.
Now, $\angle A + \angle B + \angle C + \angle D + \angle E + \angle F + \angle G = x$
$\implies$ Sum of the angles of the $7$ triangles $-$ Angle formed around $O = x$
$\implies 7 \times 180 - 2 \times 360 = x$
$\implies x = 540$
Hence, the value of $x$ is $540.$
Illustration of step 3:
According to the angle sum property of a triangle, the sum of angles of a triangle is $180^\circ.$
For $7$ triangles $AOF, GOE, FOD, EOC, DOB, COA, BOG,$
$( \angle FAO + \angle OFA + \angle AOF) + (\angle GOE + \angle OEG + \angle EGO) + (\angle FOD + \angle OFD + \angle ODF) \\ + (\angle EOC + \angle OEC + \angle OCE) + (\angle DOB + \angle ODB + \angle OBD) + (\angle COA + \angle OCA + \angle OAC) \\ + (\angle BOG + \angle OGB + \angle OBG) = 7 \times 180$
Now,
$\begin{align} & [ ( \angle CAO + \angle OAF) + (\angle DBO + \angle OBG) + (\angle ECO + \angle OCA) + (\angle FDO + \angle ODB) \\ & + (\angle GEO + \angle OEC) + (\angle AFO + \angle OFD) + (\angle EGO + \angle OGB) ] \\ & + [ (\angle AOG + \angle GOF) + (\angle GOF + \angle FOE) + (\angle FOE + \angle EOD) + (\angle EOD + \angle DOC) \\ & + (\angle DOC + \angle COB) + (\angle COB + \angle BOA) + (\angle BOA + \angle AOG) ] = 1260^\circ \\ \implies & \angle A + \angle B + \angle C + \angle D + \angle E + \angle F + \angle G + [2\angle AOG + 2\angle GOF + 2\angle FOE \\ & + 2\angle EOD + 2\angle DOC + 2\angle COB + 2\angle BOA] = 1260^\circ \\ \implies & \angle A + \angle B + \angle C + \angle D + \angle E + \angle F + \angle G + 2 [\angle AOG + \angle GOF + \angle FOE \\ & + \angle EOD + \angle DOC + \angle COB + \angle BOA] = 1260^\circ \\ \implies & \angle A + \angle B + \angle C + \angle D + \angle E + \angle F + \angle G + (2 \times 360^\circ) = 1260^\circ \\ \implies & \angle A + \angle B + \angle C + \angle D + \angle E + \angle F + \angle G = 1260^\circ - 720^\circ \\ \implies & \angle A + \angle B + \angle C + \angle D + \angle E + \angle F + \angle G = 540^\circ \end{align}$

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