Suppose on a road there are four light posts. The sides of the road are like two lines perfectly parallel to each other. The light posts are on these two sides. Thus, two of them are on one of the lines, and two are on the other line.
Make a parallelogram by joining these four light posts, using them as vertices. Now, let’s make a trapezium, but using the same base as for the parallelogram. Finally, construct a triangle, on the same base, placing the third vertex on any point of the other line.
All these figures have a common base. Their opposite vertices lie on the line parallel to the base. Thus, they’re said to be on the same base and between the same parallels.

It’s important to note that the common base must lie on one of the two parallels. 

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Take a square between two parallel lines. Without moving the base, slide the top of the square to the right, so its vertical sides are slanting. The two figures formed are both parallelograms. Will their areas be equal? 

Name the first parallelogram as ABCD, the second as EFCD. With the base as DC, there are three figures: Two triangles AED and BFC, plus trapezium EBCD – a region common to both the parallelograms.

By corresponding angles test on 2 parallel lines and transversal,

Angle DAE = Angle CBF

Angle DEA = Angle CFB

Since opposite sides of a parallelogram are congruent

DE = CF 

DA = CB

Thus, the triangles AED and BFC are congruent by Side-Angle-Side test.

ar(AED) = ar(BFC)

Ar(ABCD) = Ar(AED) + Ar(EBCD)

Ar(EFCD) = Ar(BFC) + Ar(EBCD)

Thus, ar(ABCD) = ar(EFCD)

Therefore, Parallelograms with the same base, between the same parallels, have equal areas.

Ar(parallelogram) = (height)x(base)

If two parallelograms are between the same parallel lines with a common base, their altitudes measure equal. The altitudes are also parallel to each other.

The distance between two parallel lines is always constant at any point. Because the altitudes are perpendicular to the parallels, they make a figure like a square.

The above theorem can then be rephrased as follows:

Parallelograms placed on the same/equal base(s) and between the same parallels have equal areas.

The converse of this theorem is true.

Parallelograms on the same/equal base(s) with equal areas, lie between the same parallel lines. 

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Back to the road with the light posts. Make two triangles but with the same base CB. One of the posts on top is the vertex of the first triangle as point A. The other post is the vertex of the second triangle at point E. 

Two triangles ABC and EBC are formed. Notice that the second triangle is an obtuse-angled triangle. Construct the altitude of both triangles. The first triangle will have its altitude inside itself, while the second will have it outside. The heights of both these triangles lie between two parallel lines and are perpendiculars. The distance between two parallel lines taken at any point is always constant. Hence, they are equal as seen with the parallelograms earlier. 

Again, Area of a Triangle = ½ (Base) \( \times \) (Height)

Both the triangles have a common base and equal altitudes. Therefore, they have equal areas.

Theorem:

Two triangles on the same/equal base(s), between the same parallels, have equal areas.

From the formula for area, it was proven earlier that two triangles with the same/equal base(s), and with equal areas, will have equal corresponding altitudes.

The converse of the theorem is true as well:

If two triangles have the same base and have equal areas, they are situated between the same parallel lines.

1. Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

A rectangle is also a parallelogram; all four vertices are right angles.

Since parallelograms ABCD and ABEF are on the same base and have equal areas, they are lying between two parallel lines.

Perimeter is the sum of all sides of a figure.

The shortest distance between two points is a straight line. The vertical sides of the rectangle ABEF, sides BE and FA, are perpendicular to the parallel lines. 

Whereas in parallelogram ABCD, sides BC and DA are not perpendicular to the parallel lines.  Thus, they are longer than sides BE and FA. 

This proves that the perimeter of parallelogram ABCD is greater than that of rectangle ABEF.

2. D, E and F are respectively the mid-points of the sides BC, CA and AB of a Triangle ABC. Show: 

(i) BDEF is a parallelogram. (ii) ar(BDEF) = ½ ar(ABC)

When joining the midpoints of the sides, we obtain seg EF and seg DE.

Observe that EF is parallel to side BC, while DE is parallel to side AB. 

By default, EF is parallel to BD and DE is parallel to FB. 

In the figure BDEF, opposite sides are parallel. 

Therefore,

BDEF is a parallelogram

Observe that seg FD is a diagonal to BDEF. A diagonal of a parallelogram divides it into equal halves. FD divides BDEF into triangles FBD and FED.

Similarly, diagonal ED divides FDCE into halves; diagonal FE divides FAED into halves. 

Therefore, triangle ABC is divided into four equal triangles: FBD, FED, EDC and EFA. 

Thus, 

ar(BDEF) = ar(FBD) + ar(FED) = ½ar(ABC)

…

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1. What are the topics and theorems covered in the NCERT Solutions for the Class 9 Maths Chapter Areas of Parallelograms and Triangles ?

In this chapter, we see properties related to areas of two geometric shapes: a parallelogram and a triangle. The topics covered are as follows:

• Figures on the Same Base and Between the Same Parallels
• Parallelograms on the same Base and Between the same Parallels
• Triangles on the same Base and between the same Parallels

2. What are the formulae for areas in class 9 Maths chapter Area of parallelogram and triangle?

Area of triangle:  ½(base)x(height)

Area of a parallelogram: (base)x(height)

Height refers to the perpendicular segment drawn from any vertex to its opposite side.

3. Is Class 9 NCERT Solutions For Maths Areas Of Parallelograms And Triangles available for free?

Download the MSVGo app to get a free trial of the NCERT solutions, CBSE learning videos, and the learning concepts in the chapter. Solve Maths problems quickly, get quick revisions and interactive learning. Regular practice helps deepen the understanding of the concept and teaches how to solve the problems.