Q18 of 25 Page 9

A point O inside a rectangle ABCD is joined to the vertices. Prove that the sum of the areas of the pair of opposite triangles formed is equal to the sum of the other pairs of triangles.

Construction:

Draw POQ||AB and SOR||AD.



Proof:


ar(Δ AOD) + ar(Δ BOC) = 1/2 (AD× BD)+ 1/2 (BC× OQ)


∵ AD = BC (opp. Sides of rectangle)


⇒ ar(Δ AOD) + ar(Δ BOC) = 1/2 (AD× OP)+ 1/2 (AD× OQ)


= 1/2 (AD × (OP+OQ))


= 1/2 (AD× PQ)


= 1/2 (AB× AB)


= 1/2 ar(rect. ABCD) ….. (1)


And,


ar(Δ AOB)+ar(Δ COD) = 1/2 (AB× OS)+ 1/2 (CD× OR)


∵ AB = CD (opp. Sides of rectangle)


⇒ ar(Δ AOB)+ar(Δ COD) = 1/2 (AB× OS)+ 1/2 (AB× OR)


= 1/2 AB× (OS+OR)


= 1/2 (AB× SR)


= 1/2 (AB× AD)


= 1/2 ar(rect. ABCD) …… (2)


From (1) and (2) we get,


ar(Δ AOD) + ar(Δ BOC) = ar(Δ AOB)+ar(Δ COD)


More from this chapter

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16

In figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ).

35

17

P and Q are respectively the mid- points of sides AB and BC of of a triangle ABC and R is the mid- point of AP, show that
(i) ar (PRQ) = 1/2 ar (ARC)
(ii) ar (RQC) = 3/8 ar (ABC)
(iii) ar (PBQ) = ar (ARC)

 19

In a quadrilateral ABCD diagonals AC and BD intersect at O such that OB=OD. If AB=CD, show that ar(Δ AOD)=ar(Δ BOC)

 20

In Fig., ABCD is a trapezium in which AB||DC and DC=40 cm and AB=60 cm. If X and Y are, respectively, the mid points of AD and BC, Prove that


(i) XY =50 cm (ii) DCYX is a trapezium


(iii) ar(trap. DCYX) = ar(trap. XYBA)