Question

13. In figure 6.16, find angle ABC + angle BCD + angle CDE + angle DEF + angle EFA + angle FAB.






Fig. 6.16​

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°Read more on Sarthaks.com - https://www.sarthaks.com/687662/in-the-six-cornered-figure-fig-ac-ad-and-ae-are-joined-find-fab-abc-bcd-cde-def-efa