A Brief Overview of Chapter 9 Areas of Parallelograms and Triangles

We know about geometry from the previous chapters where you have learned the properties of triangles and quadrilaterals. Now we will find out how to calculate surface areas of parallelograms and triangles by applying our knowledge of their properties. It will help you to understand how knowledge of geometry can be applied to solve real-life problems.

For instance, the formula for area of a rectangle can be used to find out the area of a large rectangular field. Additionally, a fundamental knowledge of class 9 areas of parallelogram and triangles are also used by engineers and architects while designing and constructing buildings.

You may know that a section of a plane bounded within a simple closed figure is called planar region and the measure of this region is known as its area. Apart from this, it would help if you kept in mind while studying areas of parallelograms and triangles that congruent figures or figures which have the same shape and size also have equal areas. However, two figures having the same area may not be congruent.

Thus, an area of a figure may be defined as a number in units that are associated with the planar region of the same. This definition has been discussed in detail in our NCERT solutions for class 9th maths chapter 9 areas of parallelograms and triangles.

You have learnt in previous classes the properties and formulae to calculate the area of various geometric figures like squares, rhombus, and rectangles. In this section, you will learn how to calculate areas of parallelograms and triangles lying on the same base and within the same parallels by applying that knowledge.

So, when are two figures said to be on the same base?

According to NCERT solutions class 9 maths chapter areas of parallelograms and triangles, two figures are on the same base and within the same parallels, if they have the following properties –

A Common base or side

Common vertices or vertex opposite to the common base and lying on a line which is parallel to the base.

Given below are some theorems from 9th CBSE maths areas of parallelograms and triangles. It is based on the relation between two parallelograms lying on the same base and between the same parallels.

A thorough understanding of these theorems will enable you to solve subsequent exercises easily. You can cross-check your answers with our areas of parallelograms and triangles class 9 questions with answers.

Theorem 1:

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

A rectangle which is also a parallelogram lying on the same base and between same parallels also have the same area.

Hence the area of a parallelogram = base x height.

You can revise your answers with our areas of parallelograms and triangles class 9 exercise 9.2 solutions after attempting the questions on your own.

Theorem 2:

Two triangles which have the same bases and are within the same parallels have equal area. therefore -

Area of a triangle is ½ x base x height. You can practise questions in this theorem from areas of parallelograms and triangles exercise 9.3.

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

According to areas of parallelograms and triangles, Area of trapezium = ½ x (sum of parallel side) x (distance between them)

Area of a rhombus = ½ x product of the diagonals

Practise questions based on the theorem on your own and then check your answers with our areas of parallelograms and triangles class 9 exercise 9.3 solutions.

Theorem 3:

Triangles which have the same areas and lies on the same base, have their corresponding altitudes equal.

You can go through NCERT solutions for class 9th maths chapter 9 areas of parallelograms and triangles to gain more clarity on this theorem.

Our study materials on topics like areas of parallelograms and triangles are quite engaging and it aids students to learn and memorise important theorems and concepts easily.

Students can also sign up for our online interactive classes for doubt clearing and to know more about the topics such as areas of parallelograms and triangles answers. Now you can also download our Vedantu app for enhanced access.

FAQ (Frequently Asked Questions)

1. What Are The Steps To Prove That Area Of A Triangle Is Equal To ½ X Base X Height?

Let us consider a triangle ABC where BC is the base, and AL is the height. We are required to prove that area (ABC) = ½ x BC x AL.

To prove that, first we will draw a line segment AD parallel to BC and CD, which is parallel to BA. Now we have a parallelogram BCDA where AC is its diagonal which bisects it into two triangles having equal areas.

Therefore, ar (ABC) = ½ x ar (llgm BCDA)

= ½ x BC x AL

[according to areas of parallelograms and triangles, area of a parallelogram is base x height]

Hence it is proved that area of a triangle is ½ x base x height. You can refer to NCERT solutions for class 9 maths areas of parallelograms and triangles to learn how to prove theorems in detail.

2. How Can It Be Proved That Area Of A Rhombus Is ½ X Product Of Diagonals?

Let us consider a rhombus PQRS which have two diagonals. We will have to prove that ar (Rhombus PQRS) = ½ x PR x QS. We know that one of the properties of a rhombus is that diagonals intersect each other at right angles. Let QO PR and SO ⊥PR.

So area (Rhombus PQRS) = area (△PQR ) + area(△PSR)

= {½ x PR x QO} + {½ x PR x SO}

= ½ x PR x (QO + SO)

= ½ x PR x QS

= ½ x product of diagonals

Hence it is proved that area of rhombus is equal to ½ x product of diagonals. You can also refer to NCERT solutions class 9 maths areas of parallelograms and triangles exercise 9.3 for more solved questions like this.

3. What Are The Properties Of A Parallelogram?

According to NCERT 9th maths areas of parallelograms and triangles, properties of a parallelogram state that both pairs of opposite sides should be equal. So, in a parallelogram AB = DC and BC = AD. The opposite sides should be parallel to each other. Hence, in a parallelogram, AB II DC and BC II AD. The diagonals should bisect each other and divide the parallelogram in two congruent triangles, which means that in a quadrilateral ABCD, ∆ ABC ≅ ∆ CDA. Each of the angles of a parallelogram should be right angles. Therefore, the sum of all interior angles of a parallelogram is 360 degree.

Additionally, you can go through our areas of parallelograms and triangles answers which explain the properties with diagrams.

4. What Is Meant By Median, Altitude, Incentre, Orthocentre And Circumcentre Of A Triangle?

A line segment that connects the midpoints of a side with the opposite vertex is called the median corresponding to that side. The length of the perpendicular drawn from base to vertex is known as the altitude of a triangle. To learn more about these definitions, refer to areas of parallelograms and triangles class 9 NCERT solutions.

Orthocentre is the point of intersection of all three altitudes in a triangle. In a triangle, the intersection point of all its internal bisectors of all the angles of a triangle is called its incentre. Similarly, the point of intersection of perpendicular bisectors of sides of a triangle is known as its circumcentre.