This GSEB Class 9 Maths Notes Chapter 9 Quadrilaterals covers all the important topics and concepts as mentioned in the chapter.

## Quadrilaterals Class 9 GSEB Notes

**Planar region:**

The part of the plane enclosed by a simple closed figure is called a planar region corresponding to that figure.

**Area of a closed figure:**

The magnitude or measure of the planar region cqrresponding to a closed figure is called its area. The magnitude or measure is always expressed with the help of a positive real number (in some units).

**Areas of congruent figures:**

- If two figures A and B are congruent, they must have equal areas. However, the -converse of this statement is not true. In other words, two figure having equal areas need not be congruent.
- Symbol for the area of a figure: Area of figure A is denoted as ar(A) symbolically.

**Properties of area of figures:**

- If A and B are two congruent figures, then ar (A) = ar (B).
- If a planar region formed by a figure T is made up of two non-overlapping planar regions formed by figures P and Q, then ar (T) = ar (P) + ar (Q).

**Figures on the same base and between , the same parallels:**

Two figures are said to be on the same base and between the same parallels, if they have a common base (side) and the vertices (or the vertex) opposite to the common base of each figure lie on a line parallel to the base.

Note: Out of two parallel lines, one line must contain the common base.

**Parallelograms on the same base and between the same parallels:**

- Theorem 9.1: Parallelograms on the same base (or equal bases) and between the same parallels are equal In area.
- The area of a parallelogram Is the product of Its base and the altitude corresponding to that base.
- Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels, one of which contains the same (or equal) base / s.
- If a parallelogram and a triangle are on the same base and between the same parallels, then the area of the triangle Is half the area of the parallelogram.

Example 1:

In the given figure, ABCD is a parallelogram and EFCD is a rectangle. Also, AL ⊥ DC.

Prove that:

(i) ar (ABCD) = ar (EFCD)

(ii) ar(ABCD) = DC × AL

Answer:

(i) As a rectangle Is also a parallelogram, ar (ABCD) = ar(EFCD) (Theorem 9.1)

(II) From above result, ar (ABCD) = ar (EFCD) = DC × FC

(Area of the rectangle = length × breadth) …… (1)

Since, AL ⊥ DC. AFCL is also a rectangle.

∴ AL = FC … …(2)

Hence, ar(ABCD) = DC × AL [From (1) and (2)]

Example 2:

If a triangle and a parallelogram are on the same base and between the same parallels, then prove that the area of the triangle is equal to half the area of the parallelogram.

Answer:

Let, ABP and parallelogram ABCD be on the same base AB and between the same parallels AB and PC (See the given figure).

Draw BQ ∥ AP to obtain another parallelogram ABQR Now, parallelograms ABQP and ABCD are on the same base AB and between the same parallels AB and PC.

∴ ar(ABQP) = ar(ABCD) (Theorem 9.1) ………… (1)

But, ∆PAB ≅ ∆BQP

(Diagonal PB divides parallelogram ABQP into two congruent triangles.)

∴ ar(PAB)=ar(BQP) …………(2)

∴ ar(PAB)= \(\frac{1}{2}\)ar(ABQP) [From (2)] ………….. (3)

Hence, ar (PAB) = \(\frac{1}{2}\)ar (ABCD) [From (1) and (3))

**Triangles on the same base and between the same parallels :**

Theorem 9.2:

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

**Area of a triangle:**

The area of a triangle is half the product of its base (or any side) and the corresponding altitude (or height).

Theorem 9.3:

Two triangles having the same base (or equal bases) and equal areas lie between the same parallels, one of which contains the base (or bases).

**Property of a median of a triangle:**

Any median of a triangle divides the triangle into two triangles of equal area.

Example 1:

Show that a median of a triangle divides it into two triangles of equal areas.

Answer:

Let, ABC be a triangle and let AD be one of its medians (see the given figure).

Since the formula for area involves altitude, let us draw AN ⊥ BC.

Now, ar (ABD) = \(\frac{1}{2}\) × base × altitude (of ∆ ABD)

= \(\frac{1}{2}\) × BD × AN

= \(\frac{1}{2}\) × CD × AN (∵ BD = CD)

= \(\frac{1}{2}\) × base × altitude (of ∆ACD) = ar (ACD)

Example 2:

In the given figure, ABCD is a quadrilateral and BE ∥ AC and also BE meets DC produced at E. Show that area of ∆ADE is equal to the area of the quadrilateral ABCD.

Answer:

∆BAC and ∆EAC are on the same base AC and between the same parallels AC and BE.

∴ ar (BAC) = ar (EAC) [Theorem 9.2]

∴ ar (BAC) + ar (ADC) = ar (EAC) + ar (ADC) (Adding same area on both sides)

∴ ar (ABCD) = ar (ADE)