Q24 of 25 Page 9

P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).

It can be observed that ΔBQC and parallelogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC


If a triangle and parallelogram are on the same base and have the same altitude,
the area of the triangle will be half that of the parallelogram.


Area (ΔBQC) = 1/2 Area (ABCD) (i)




Similarly,


If a triangle and parallelogram are on the same base and have the same altitude,
the area of the triangle will be half that of the parallelogram.


ΔAPB and parallelogram ABCD lie on the same base AB
and between the same parallel lines AB and DC




Area (ΔAPB) = 1/2 Area (ABCD) (ii)


From equation (i) and (ii), we get


Area (ΔBQC) = Area (ΔAPB)


More from this chapter

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21

What figure is formed by joining the mid-points of the adjacent sides of a rhombus?

 22

In Fig., ABC is a right triangle right angled at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX⊥ DE meets BC at Y. Show that:


(i) Δ MBC ≅ ABD


(ii)ar(BYXD) =2ar(Δ MBC)


(iii) ar(BYXD) = ar(ABMN)


(iv) Δ FCB ≅ Δ ACE


(v) ar(CYXE) = 2ar(Δ FCB)


(vi) ar(CYXE) = ar(ACFG)


(vii) ar(BCED) = ar(ABMN)+ ar(ACFG)

 23

In Fig., ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

 25

If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that

ar (EFGH) = 1/2 ar (ABCD)