Q3 of 24 Page 256

In the figure, ∆ABC and ∆ABD are two triangles on the same base AB. If line segment CD is bisected by at O, show that ar(∆ABC) = ar(∆ABD)

The Figure given in question does not match what the question says here is the correct figure according to the question



Consider ΔACD


AO is the median which divides the area(ΔACD) into two equal parts


⇒ area(ΔAOD) = area(ΔAOC) …(i)


Consider ΔBCD


BO is the median which divides the area(ΔBCD) into two equal parts


⇒ area(ΔBOD) = area(ΔBOC) …(ii)


Add equation (i) and (ii)


⇒ area(ΔAOD) + area(ΔBOD) = area(ΔAOC) + area(ΔBOC) …(iii)


From figure


area(ΔAOD) + area(ΔBOD) = area(ΔABD) and


area(ΔAOC) + area(ΔBOC) = area(ΔABC)


therefore equation (iii) becomes


area(ΔABD) = area(ΔABC)


More from this chapter

All 24 →
1

In a triangle ABC (see figure), E is the midpoint of median AD, show that

(i) ar ∆ABE = ar∆ACE


(ii) ar ∆ABE = ar(∆ABC)


 2

Show that the diagonals of a parallelogram divide it into four triangles of equal area.

 4

In the figure, ∆ABC, D, E, F are the midpoints of sides BC, CA and AB respectively. Show that

(i) BDEF is a parallelogram


(ii) ar(∆DEF) = ar(∆ABC)


(iii) ar(BDEF) = ar(∆ABC)


 5

In the figure D, E are points on the sides AB and AC respectively of ∆ABC such that ar(∆DBC) = ar(∆EBC). Prove that DE || BC.