RS Aggarwal Class 9 Solutions Chapter-14 Areas of Triangles and Quadrilaterals

The major concepts learned in RS Aggarwal Class 9 Maths Chapter 14 are:

• Area of different triangles and quadrilaterals

• Area of a quadrilaterals using diagonals

These two topics are essential in building a strong base in geometry and score well in exams. There are two exercises in the chapter:

• The first exercise consists of 44 problems that give various cncepts of finding the area of triangles and different quadrilateral figures using suitable formulae.

• The second exercise consists of multiple-choice questions which test the student’s grasp on the concepts learned by solving the first exercise.

The questions present in Class 9 Maths RS Aggarwal Solutions Chapter 14 are ideal for intense practice to excel in the exams.

The following are some important formulae that RS Aggarwal Solutions Class 9 Maths Chapter 14 comprises of:

1. The basic area of any triangle: ½  × b × h

where b is the base and h is the height of the triangle

2. Area of an equilateral triangle: An equilateral triangle is a triangle whose all the sides are equal.

Area of an equilateral triangle = \[\frac{\sqrt{3}}{4}\times a^{2}\]

Height of an equilateral triangle = \[\frac{\sqrt{3}}{4}\times a\]

where a is the measure of the two equal sides and b is the measure of the base.

3. Area of an isosceles triangle: \[\frac{{b}}{4}\times\sqrt{4a^{2}-b^{2}}\]

4. Area of a triangle having three different sides (Heron’s formula):- 

Suppose a triangle has sides measuring a, b and c. 

According to Heron’s formula:-

The first step includes finding the semi-perimeter of the triangle:

Semi-perimeter (S) = \[\frac{a+b+c}{2}\]

Using the above finding and implementing it in Heron’s formula for the area of a triangle constitutes the second step.

Area = \[\sqrt{s\lgroup s-a\rgroup\lgroup s-b\rgroup\lgroup s-c\rgroup}\]

On the basis of the measure of sides and angles, there are seven different types of quadrilaterals, each having a specific formula for its area.

1. Parallelogram: Its opposite sides are equal and parallel to each other.

Area of a parallelogram = base × height

2. Rectangle: It is a parallelogram with all four angles measuring 90⁰

Area of a rectangle = length × breadth

3. Square: It is a rectangle whose all sides are equal

Area of a square = side2

4. Rhombus: It is a parallelogram whose all sides are equal.

Area of a rhombus = base × height

Area of rhombus using diagonals = \[\frac{d_{1}\times d_{2}}{2}\]

5. Trapezium: A trapezium has one pair of sides parallel. 

Area of trapezium = \[\frac{1}{2}\lgroup a+b\rgroup\times h\]

where a and b are the measure of two parallel-sided and h is the height

6. Kite: It has two pairs of equal adjacent sides but unequal opposite sides.

Area of a kite = \[\frac{d_{1}\times d_{2}}{2}\]

Note: Perimeter of any figure is equal to the sum of its sides.

Some examples of problems are as follows:

Find the side and the height of an equilateral triangle whose area is 

100\[\sqrt{3}\] cm²

Ans:  Let the side be a

According to the problem:-

\[\frac{\sqrt{3}}{4}\times a^{2}=100\sqrt{3}cm^{2}\]

 \[a^{2}=\frac{100\sqrt{3}\times 4}{\sqrt{3}}cm^{2}\]

 \[a=\sqrt{400}\]

a = 20 cm

∴ Height of the triangle =  \[\frac{\sqrt{3}}{4}\times a =5\sqrt{3} cm\]

1. Find the Area of a Trapezoidal Pool Whose Parallel Sides are 25 m and 10 m and the Non-parallel Sides are 14 m and 13 m.

Ans:  Let the pool ABCD be divided into two figures i.e. a parallelogram and a triangle.

According to the figure, Δ BCF has sides 14m, 13m and 15m.

As the sides of the triangle are different, Heron’s formula will come into action.

S= \[\frac{14+ 15+13}{2}\] = 21 m

A=\[\sqrt{S\lgroup s-a\rgroup\lgroup s-b\rgroup\lgroup s-c\rgroup} = \sqrt{21\lgroup 21-14\rgroup\lgroup 21-15\rgroup\lgroup 21-13\rgroup}\] = \[\sqrt{21\times7\times6\times8} = 84m^{2}\]

Height of Δ BCF = Height of parallelogram AFCD

Area of Δ BCF = ½ × b × h = 84 m2

∴ ½ × 15 × h = 84 

h = 11.2 m

Area of the parallelogram AFCD = base × height = 10 × 11.2 m2 = 112 m2

∴ Area of the pool = (84 + 112) m2 = 196 m2

Tips to Prepare with RS Aggarwal Maths Class 9 Solutions Chapter 14

Vedantu’s RS Aggarwal Class 9 Maths Ch 14 Solutions are perfect for last-minute revision. In order to grow a better understanding of the chapter, one must maintain a formula list aside while going through the problems. This will give a clear picture of how to approach each and every problem with greater confidence. The solutions are designed in such a way so as to allow students to understand all the important details in the RS Aggarwal Class 9 Maths Chapter 14, enabling them to score well in the exams.

The experienced educators and subject experts with years of teaching experience have curated the Class 9 Maths RS Aggarwal Solutions Chapter 14 with a special focus on analytical foundation building. The Maths RS Aggarwal Class 9 Chapter 14 Solutions plays a key role in providing a better learning experience.