#### Class Nine NCERT Solutions Chapter 9

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

**TEXTBOOK’S
EXERCISE – 9.1 **

**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.

**TEXTBOOK’S EXERCISE – 9.2 **

**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 cm^{2} = 128 cm^{2 }

Also, area of parallelogram ABCD = AD × FC = (AD × 10) cm^{2}

∴ 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