Q. 34.4( 37 Votes )
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
AO is the median which divides the area(ΔACD) into two equal parts
⇒ area(ΔAOD) = area(ΔAOC) …(i)
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)
area(ΔAOD) + area(ΔBOD) = area(ΔABD) and
area(ΔAOC) + area(ΔBOC) = area(ΔABC)
therefore equation (iii) becomes
area(ΔABD) = area(ΔABC)
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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.
In the figure, diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar(∆AOD) = ar(∆BOC).
ABCD is a parallelogram. The diagonals AC and BD intersect each other at ‘O’. Prove that ar(∆AOD) = ar(∆BOC). (Hint: Congruent figures have equal area)
Find the area of trapezium ABCD as given in the figure in which ADCE is a rectangle. (Hint: ABCD has two parts)
In the figure, XY is a line parallel to BC is drawn through A. If BE || CA and CF || BA are drawn to meet XY at E and F respectively. Show that ar(∆ABE) = ar (∆ACF).
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)
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)
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)
In the figure, if ar ∆RAS = ar ∆RBS and [ar (∆QRB) = ar(∆PAS) then show that both the quadrilaterals PQSR and RSBA are trapeziums.