RS Aggarwal Class 9 Solutions Chapter-8 Triangles

• The Perimeter of a Triangle: ½ × Length × Breadth.

• Area of a Triangle: Total sum of the lengths of a triangle.

• Internal Angle: 60°.

• Total Sum Internal Angles: 180°.

Types of Triangles:

Triangles can be distinguished on the following basis:

• According to the length of the sides

• According to the measure of internal angles 

• According to the Length of Sides:

1. Equilateral Triangle: It is a Triangle with all sides of the same length. All the internal angles are 60°. Let us suppose a Triangle of sides a, b and c

Equilateral Triangle: a = b = c

2. Isosceles Triangle: It has two sides of equal length and another side is of different measure.

Isosceles Triangle: a = b but not equal to c

3. Scalene Triangle: It has sides of unequal length which means all the sides are of different lengths. All the angles are also of different lengths.

• According to the measure of internal angles:

1. Right Angle Triangle: Any triangle with one interior angle of measure 90° is known as Right Triangle.

            Right angles obey Pythagoras Theorem which is a2 + b2 = c2

Where a, b, c are the sides of any Triangle

2. Obtuse Triangle: Any triangle with anyone interior angle greater than 90° is called an obtuse triangle.

3. Acute Triangle: Any triangle with interior angles less than 90° is called Acute  Triangle.

Let us now work on a few examples related to Triangle problems:

Problem 1: Consider a triangle with sides 3 cm, 6 cm, and 10 cm then find the perimeter of the triangle.

Ans: Perimeter of a Triangle = a + b + c

        Perimeter = 3 + 6 + 10

   = 19cm

Many questions based on Heron's Formula are asked in various examinations. So, let us study this very important formula.

We apply Heron's formula when we are asked to find the area of the Triangle but we are not provided with the height of the Triangle. In that case, we can easily solve the problem by using Heron's Formula.

We can solve the problem using Heron's Formula provided that all the sides of the triangle are given in the question.

Area of Triangle using Heron's Formula=

√s(s−a)(s−b)(s−c)

√s(s−a)(s−b)(s−c)

Here, a, b, c are sides of Triangle

And   s = (a + b + c)/2

Let’s solve a problem based on the above concept.

Problem 2: Determine the area of a triangle with sides 4, 6, and 8 units in length.

Answer: As we are not given the height of the Triangle so, we have to use Heron's Formula to determine the area of the Triangle 

Semi perimeter(s)= a + b + c/2

                                = 4 + 6 + 8/2 = 9

Now Area of Triangle= √

s(s−a)(s−b)(s−c)

s(s−a)(s−b)(s−c)

=√9(9−5)(9−4)(9−8)

√9(9−5)(9−4)(9−8)

=√9×4×5×1

√9×4×5×1

=√3×3×2×2×5

√3×3×2×2×5

= 6√5 square units 

RS Aggarwal Class 9 solutions chapter 8 triangles consist of an exercise named Exercise 8 with a total of 29 questions. After that, there are Multiple Choice Questions which are 12 in all.

RS Aggarwal Class 9 Solutions Chapter 8: Preparation Tips

Class 9 math RS Aggarwal chapter 8 solutions can be solved with an easy method. Students are advised to go through all the basic concepts related to Chapter-8 of Class 9 which will further help them to understand the solutions given by RS Aggarwal more easily. We advise the child to first solve simple and easy questions. When they can solve such questions only then they should try their hands for solutions to tough questions in RS Aggarwal. They should revise the same concept after a particular period to strengthen their learning power.

No doubt Triangle seems to be an easy topic but sometimes questions are confusing and students find themselves solving such questions.RS Aggarwal solutions class 9 math chapter 8 are compiled in such a manner that they cover various concepts. Students should thoroughly follow every solution of RS Aggarwal solutions class 9 math ch 8 to be able to solve the toughest questions.

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