Question

I a polygon, ABCDEF is a six cornered figure. Find ∆ ∆ ABC + ∆ BCD + ∆ CDE + ∆ DEF + ∆ EFA + ∆ FAB.​

Answer 1

Answer:

720°

Step-by-step explanation:

We know that sum of the angles of a triangle is 180°

Therefore in ∆ABC, we have

∠CAB + ∠ABC + ∠BCA = 180° …….. (i)

In△ ACD,

∠DAC + ∠ACD + ∠CDA = 180° …….. (ii)

In △ADE,

∠EAD + ∠ADE + ∠DEA =180° ………. (iii)

In △AEF,

∠FAE + ∠AEF + ∠EFA = 180° ………. (iv)

Adding (i), (ii), (iii), (iv) we get

∠CAB + ∠ABC + ∠BCA + ∠DAC + ∠ACD + ∠CDA + ∠EAD + ∠ADE + ∠DEA + ∠FAE + ∠AEF +∠EFA = 720°

Therefore ∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA = 720°

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