Q25 of 25 Page 9

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)

Let us join HF





In parallelogram ABCD,


AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)




AB = CD (Opposite sides of a parallelogram are equal)


1/2 AD = 1/2 BC


And AH || BF


AH = BF and AH || BF (H and F are the mid-points of AD and BC)


Therefore, ABFH is a parallelogram




Since,


ΔHEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF


Therefore,




Area of triangle HEF = 1/2 × Area (ABFH) ….. (i)


Similarly,


It can also be proved that




Area of triangle HGF = 1/2 × Area (HDCF) …… (ii)


On adding (i) and (ii), we get


Area of triangle HEF + Area of triangle HGF = 1/2 Area (ABFH) + 1/2 Area (HDCF)


Area (EFGH)= 1/2 [Area (ABFH) + Area (HDCF)]


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


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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.

 24

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).