Students should go through these JAC Class 9 Maths Notes Chapter 9 Areas of Parallelograms and Triangles will seemingly help to get a clear insight into all the important concepts.

### JAC Board Class 9 Maths Notes Chapter 9 Areas of Parallelograms and Triangles

**Polygonal Region**

Polygon region can be expressed as the union of a finite number of triangular regions in a plane such that if two of these intersect, their intersection is either a point or a line segment. It is the shaded portion including its sides as shown in the figure.

**Parallelogram**

(a) Base and Altitude of a Parallelogram:

→ Base: Any side of parallelogram can be called its base.

→ Altitude: The perpendicular to the base from the opposite side is called the altitude of the parallelogram corresponding to the given base.

In the given Figure

→ DL is the altitude of ||^{gm} ABCD corresponding to the base AB.

→ DM is the altitude of ||^{gm} ABCD, corresponding to the base BC.

Theorem 1.

A diagonal of parallelogram divides it into two triangles of equal area.

Proof:

Given: A parallelogram ABCD whose one of the diagonals is BD.

To prove: ar(ΔABD) = ar(ΔCDB).

Proof: In ΔABD and ΔCDB;

AB = DC [Opp sides of a ||^{gm}]

AD = BC [Opp. sides of a ||^{gm}]

BD = BD [Common side]

∴ ΔΑΒD ≅ ΔCDB [By SSS]

ar(ΔABD) = ar (ΔCDB) [Areas of two congruent triangles are equal]

Hence, proved.

Theorem 2.

Parallelograms on the same base or equal base and between the same parallels are equal in area.

Proof:

Given: Two Parallelograms ABCD and ABEF are on the same base AB and between the same parallels AB and FC.

To Prove: ar(||^{gm} ABCD) = ar(||^{gm} ABEF)

Proof: In ΔADF and ΔBCE, we have

AD = BC [Opposite sides of a ||^{gm}]

and AF = BE [Opposite sides of a ||^{gm}]

AD || BC (Opposite sides of a parallelogram)

⇒ ∠ADF = ∠BCE (Alternate interior angles)

∴ ΔADF ≅ ΔBCE [By AAS]

∴ ar(ΔADF) = ar(ΔBCE) …..(i)

[Congruent triangles have equal area]

∴ ar (||^{gm} ABCD) = ar(ABED) + ar(ΔBCE)

= ar (ABED) + ar(ΔADF) [Using (1)]

= ar(||^{gm} ABEF).

Hence, ar(||^{gm} ABCD) = ar(||^{gm} ABEF).

Hence, proved.

Theorem 3.

The area of parallelogram is the product of its base and the corresponding altitude.

Proof:

Given: A ||^{gm} ABCD in which AB is the base and AL is the corresponding height.

To prove: Area (||^{gm} ABCD) = AB × AL.

Construction: Draw BM ⊥ DC so that rectangle ABML is formed.

Proof: ||^{gm} ABCD and rectangle ABML are on the same base AB and between the same parallel lines AB and LC.

∴ ar(||^{gm} ABCD) = ar(rectangle ABML)

= AB × AL

∴ area of a ||^{gm} = base × height.

Hence, proved.

**Area Of A Triangle**

Theorem 4.

Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

Proof:

Given: Two triangles ABC and PCB on the same base BC and between the same parallel lines BC and AP.

To prove: ar(ΔABC) = ar(ΔPBC)

Construction: Through B, draw BD || CA intersecting PA produced in D and through C, draw CQ || BP, intersecting line AP produced in Q.

Proof: We have, BD || CA (By construction) And, BC || DA [Given]

∴ Quad. BCAD is a parallelogram.

Similarly, Quad. BCQP is a parallelogram.

Now, parallelogram BOQP and BCAD are on the same base BC, and between the same parallels.

∴ ar (||^{gm} BCQP) = ar (||^{gm} BCAD)….(i)

We know that the diagonals of a parallelogram divides it into two triangles of equal area.

∴ ar(ΔPBC) = \(\frac{1}{2}\)(||^{gm} BCQP) …..(ii)

And ar (ΔABC) = \(\frac{1}{2}\)(||^{gm} BCAD)….(iii)

Now, ar (||^{gm} BCQP) = ar(||^{gm} BCAD) [From (i)]

\(\frac{1}{2}\)ar(||^{gm} BCAD) = \(\frac{1}{2}\)ar(||^{gm} BCQP)

Hence, \(\frac{1}{2}\)ar(ABC) = ar(APBC) (Using (ii) and (iii) Hence, proved.

Theorem 5.

The area of a trapezium is half the product of its height and the sum of the parallel sides.

Proof:

Given: Trapezium ABCD in which AB || DC, AL ⊥ DC, CN ⊥ AB and AL = CN = h (say). AB = a, DC = b.

To prove: ar(trap. ABCD) = \(\frac{1}{2}\)h × (a + b).

Construction: Join AC.

Proof: AC is a diagonal of quad. ABCD.

∴ ar(trap. ABCD) = ar(ΔABC) + ar(ΔACD)

= \(\frac{1}{2}\)h × a+\(\frac{1}{2}\)h × b= \(\frac{1}{2}\)h(a + b).

Hence, proved.

Theorem 6.

Triangles having equal areas and having one side of the triangle equal to corresponding side of the other, have their corresponding altitudes equal.

Proof:

Given: Two triangles ABC and PQR such that

(i) ar(ΔABC) = ar(ΔPQR) and

(ii) AB = PQ.

CN and RT and the altitudes corresponding to AB and PQ respectively of the two triangles.

To prove: CN = RT.

Proof: In ΔABC, CN is the altitude corresponding to the side AB

ar(ΔABC) = \(\frac{1}{2}\)AB × CN ……(i)

Similarly, ar(ΔPQR) = \(\frac{1}{2}\)PQ × RT ……(ii)

Since, ar(ΔABC) = ar(ΔPQR) [Given]

∴ \(\frac{1}{2}\)AB × CN = PQ × RT

Also, AB = PQ [Given]

∴ CN = RT Hence, proved.