Q8 of 24 Page 256

In the figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

(i) ar (∆ACB) = ar (∆ACF)


(ii) ar (AEDF) = ar (ABCDE)


i) Given AC || BF


Distance between two parallel lines is constant therefore if we consider AC as common base of ΔABC and ΔFAC then perpendicular distance between lines AC and BF will be same i.e height of triangles ΔABC and ΔFAC will be same


As ΔABC and ΔFAC are triangles with same base and equal height


⇒ area(ΔABC) = area(ΔFAC)


ii) since area(ΔABC) = area(ΔFAC)


add area(ACDE) to both sides


⇒ area(ΔABC) + area(ACDE) = area(ΔFAC) + area(ACDE) …(i)


From figure


area(ΔABC) + area(ACDE) = area(ABCDE) …(ii)


area(ΔFAC) + area(ACDE) = area(AFDE) …(iii)


using (ii) and (iii) in (i)


⇒ area(ABCDE) = area(AFDE)


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