In Fig. 9.15, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

In Fig. 9.15, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

**Answer**

Given,

AB = CD = 16 cm (Opposite sides of a parallelogram)

CF = 10 cm and AE = 8 cm

Now,

Area of parallelogram = Base × Altitude

= CD × AE = AD × CF

⇒ 16 × 8 = AD × 10

⇒ AD = 128/10 cm

⇒ AD = 12.8 cm

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).

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).

**Answer**

Given,

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

To Prove,

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

Construction,

H and F are joined.

Proof,

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

⇒ 1/2 AD = 1/2 BC

Also,

AH || BF and and DH || CF

⇒ AH = BF and DH = CF (H and F are mid points)

Thus, ABFH and HFCD are parallelograms.

Now,

ΔEFH and ||gm ABFH lie on the same base FH and between the same parallel lines AB and HF.

∴ area of EFH = 1/2 area of ABFH --- (i)

also, area of GHF = 1/2 area of HFCD --- (ii)

Adding (i) and (ii),

area of ΔEFH + area of ΔGHF = 1/2 area of ABFH + 1/2 area of HFCD

⇒ area of EFGH = area of ABFH

⇒ ar (EFGH) = 1/2 ar(ABCD)

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).

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).

**Answer**ΔAPB and ||gm ABCD are on the same base AB and between same parallel AB and DC.

Therefore,

ar(ΔAPB) = 1/2 ar(||gm ABCD) --- (i)

Similarly,

ar(ΔBQC) = 1/2 ar(||gm ABCD) --- (ii)

From (i) and (ii),

we have ar(ΔAPB) = ar(ΔBQC)

In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that

(i) ar(APB) + ar(PCD) = 1/2 ar(ABCD)

(ii) ar(APD) + ar(PBC) = ar(APB) + ar(PCD)

[Hint : Through P, draw a line parallel to AB.]

**Answer**

(i) A line GH is drawn parallel to AB passing through P.

In a parallelogram,

AB || GH (by construction) --- (i)

Thus,

AD || BC ⇒ AG || BH --- (ii)

From equations (i) and (ii),

ABHG is a parallelogram.

Now,

In ΔAPB and parallelogram ABHG are lying on the same base AB and between the same parallel lines AB and GH.

∴ ar(ΔAPB) = 1/2 ar(ABHG) --- (iii)

also,

In ΔPCD and parallelogram CDGH are lying on the same base CD and between the same parallel lines CD and GH.

∴ ar(ΔPCD) = 1/2 ar(CDGH) --- (iv)

Adding equations (iii) and (iv),

ar(ΔAPB) + ar(ΔPCD) = 1/2 {ar(ABHG) + ar(CDGH)}

⇒ ar(APB) + ar(PCD) = 1/2 ar(ABCD)

(ii) A line EF is drawn parallel to AD passing through P.

In a parallelogram,

AD || EF (by construction) --- (i)

Thus,

AB || CD ⇒ AE || DF --- (ii)

From equations (i) and (ii),

AEDF is a parallelogram.

Now,

In ΔAPD and parallelogram AEFD are lying on the same base AD and between the same parallel lines AD and EF.

∴ ar(ΔAPD) = 1/2 ar(AEFD) --- (iii)

also,

In ΔPBC and parallelogram BCFE are lying on the same base BC and between the same parallel lines BC and EF.

∴ ar(ΔPBC) = 1/2 ar(BCFE) --- (iv)

Adding equations (iii) and (iv),

ar(ΔAPD) + ar(ΔPBC) = 1/2 {ar(AEFD) + ar(BCFE)}

⇒ ar(APD) + ar(PBC) = ar(APB) + ar(PCD)