Question

14. In the six cornered figure, (Fig. 20), AC, AD and AE are joined. Find
FAB+ 2 ABC+ ZBCD + 2 CDE + ZDEF + ZEFA.
D
A
B
Fig. 20​

Answer 1

Answer:

We know that sum of the angles of a triangle is 180° Therefore in ∆ABC, we have ∠CAB + ∠ABC + ∠BCA = 180° …….. (i) In△ ACD, we have ∠DAC + ∠ACD + ∠CDA = 180° …….. (ii) In △ADE, we have ∠EAD + ∠ADE + ∠DEA =180° ………. (iii) In △AEF, we have ∠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°

Step-by-step explanation: