Class Nine NCERT Solutions Chapter 9

Chapter 9: Areas of Parallelograms and Triangles (NCERT Solutions)


  1. Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.

 Sol. (i) Base DC, parallels DC and AB

(iii) Base QR, parallels QR and PS

(v) Base AD, parallels AD and BQ.


1. In the figure, ABCD is a parallelogram, AE DC and CF AD.

If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

 Sol. Area of parallelogram ABCD

= AB × AE = 16 × 8 cm2 = 128 cm2

Also, area of parallelogram ABCD = AD × FC = (AD × 10) cm2

∴ AD × 10 = 128

⇒ AD = 128/ 10 = 12.8 cm

2. 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).

Sol. Given: A parallelogram ABCD. E, F, G, H are mid-points of sides AB, BC, CD, DA respectively.

To Prove : ar (EFGH) = 1/2 ar (ABCD)

Construction: Join AC and HF.

Proof: In ΔABC,

E is the mid-point of AB.

F is the mid-point of BC.

⇒ EF || AC and EF = 1/2 AC … (i)

Similarly, in ΔADC, we can show that

HG || AC and HG = 1/2 AC … (ii)

From (i) and (ii)

EF || HG and EF = HG

∴ EFGH is a parallelogram.

[One pair of opposite sides is equal and parallel]

In quadrilateral ABFH, we have

HA = FB and HA || FB

[AD = BC ⇒ 12 AD = 1/2 BC ⇒ HA = FB]

∴ ABFH is a parallelogram. [One pair of opposite sides is equal and parallel]

Now, triangle HEF and parallelogram HABF are on the same base HF and between the same parallels

HF and AB.

∴ Area of ΔHEF = 1/2 area of HABF … (iii)

Similarly, area of ΔHGF = 1/2 area of HFCD … (iv)

Adding (iii) and (iv),

Area of ΔHEF + area of ΔHGF

= 1/2 (area of HABF + area of HFCD)

⇒ ar (EFGH) = 1/2 ar (ABCD) Proved.

3. 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).

Sol. Given: A parallelogram ABCD. P and Q are any points on DC and AD respectively.

To prove : ar (APB) = ar (BQC)

Construction : Draw PS || AD and QR || AB.

Proof: In parallelogram ABRQ, BQ is the diagonal.

∴ area of ΔBQR = 1/2 area of ABRQ ………. (i)

In parallelogram CDQR, CQ is diagonal.

∴ area of ΔRQC = 1/2 area of CDQR ……….. (ii)

Adding (i) and (ii), we have

area of ΔBQR + area of ΔRQC

= 12 [area of ABRQ + area of CDQR]

⇒ area of ΔBQC = 1/2 area of ABCD.

Again, in parallelogram DPSA, AP is diagonal.

∴ area of ΔASP = 1/2 area of DPSA ……… (iv)

In parallelogram BCPS, PB is diagonal.

∴ area of ΔBPS = 1/2 area of BCPS ………. (v)

Adding (iv) and (v)

area of ΔASP + area of ΔBPS

= 12 (area of DPSA + area of BCPS)

⇒ area of ΔAPB = 1/2 (area of ABCD) ……….. (vi)

From (iii) and (vi), we have

area of ΔAPB = area of ΔBQC. Proved.