NCERT Solutions for Class 9 Maths Chapter 9: Areas of parallelograms and triangles

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The NCERT Solutions for Class 9 Mathematics are provided in this article. Areas of Parallelograms And Triangles deal with the area parallelograms and triangles have, and figures present on the same base and between the same parallels.

Class 9 Maths Chapter 9 Areas of Parallelograms And Triangles has a weightage of 14 marks in the Class 9 Maths Examination. NCERT Solutions for Class 9 Maths for Chapter 9 covers the following important concepts:

Download: NCERT Solutions for Class 9 Mathematics Chapter 9 pdf

NCERT Solutions for Class 9 Mathematics Chapter 9

NCERT Solution

Important Topics in Class 9 Maths Chapter 9 Areas of Parallelograms And Triangles

Important Topics in Class 9 Maths Chapter 9 Areas of Parallelograms And Triangles are elaborated below:

Area of a Paralellogram

The area of a parallelogram can be defined by the region a parallelogram bounds in a respective two-dimensional space.

The Area of a Parallelogram, Area = b × h (sq. units)
Here,

“b” is the base of the parallelogram
“h” is the height of the parallelogram.

Perimeter of a Parallelogram

The Perimeter of parallelogram is the same as the sum of all four sides of the respective parallelogram.

The Perimeter of a Parallelogram:
First, consider a Parallelogram with two adjacent sides, a and b.
Hence, the Perimeter of parallelogram = a + b + c + d
⇒ 2a + 2b = 2(a+b)
⇒ P = 2(a+b).

Area of a Trapezoid Formula

The area of a trapezoid can be defined as its total space covered by the sides. If the length of the sides are known, the area of the trapezoid can be evaluated by splitting it into smaller polygons, including rectangles and triangles.

Example: Determine the area of a trapezoid that has parallel sides 32 cm and 12 cm respectively. The height of the trapezoid is mentioned as 5 cm. Evaluate the area of the trapezoid.

Solution: As per the equation, the following data is known,
a = 32 cm
b = 12 cm
h = 5 cm
Thus, the area of the trapezoid is = A = ½ (a + b) h
A = ½ (32 + 12) × (5)
= ½ (44) × (5)
= 110 cm².

NCERT Solutions for Class 9 Maths Chapter 9 Exercises:

The detailed solutions for all the NCERT Solutions for Areas of Parallelograms and Triangles under different exercises are:

Also Read:

Unit Related Topics
Area of Rectangle	Areas of Parallelograms and Triangles Important Theorems	Trapezoids
Trapezoid Formula	Perimeter of a Trapezoid	Parallelogram Area

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CBSE X Related Questions

1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them
Monthly consumption
(in units)
Number of consumers
65 - 85
4
85 - 105
5
105 - 125
13
125 - 145
20
145 - 165
14
165 - 185
8
185 - 205
4

Monthly consumption (in units)	Number of consumers
65 - 85	4
85 - 105	5
105 - 125	13
125 - 145	20
145 - 165	14
165 - 185	8
185 - 205	4

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2.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

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3.
Find the sums given below :
\(7 + 10\frac 12+ 14 + ....... + 84\)
\(34 + 32 + 30 + ....... + 10\)
\(–5 + (–8) + (–11) + ....... + (–230)\)

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4.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

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5.
Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, . . . .
(ii) \(2, \frac{5}{2},3,\frac{7}{2}\), . . . .
(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . .
(iv) – 10, – 6, – 2, 2, . . .
(v) 3, \(3 + \sqrt{2} , 3 + 3\sqrt{2} , 3 + 3 \sqrt{2}\) . . . .
(vi) 0.2, 0.22, 0.222, 0.2222, . . . .
(vii) 0, – 4, – 8, –12, . . . .
(viii) \(\frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}\), . . . .
(ix) 1, 3, 9, 27, . . . .
(x) a, 2a, 3a, 4a, . . . .
(xi) a, \(a^2, a^3, a^4,\) . . . .
(xii) \(\sqrt{2}, \sqrt{8} , \sqrt{18} , \sqrt {32}\) . . . .
(xiii) \(\sqrt {3}, \sqrt {6}, \sqrt {9} , \sqrt {12}\) . . . . .
(xiv) \(1^2 , 3^2 , 5^2 , 7^2\), . . . .
(xv) \(1^2 , 5^2, 7^2, 7^3\), . . . .

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6.
If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos² A – sin²A or not

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NCERT Solutions for Class 9 Maths Chapter 9: Areas of parallelograms and triangles

NCERT Solutions for Class 9 Mathematics Chapter 9

Important Topics in Class 9 Maths Chapter 9 Areas of Parallelograms And Triangles

Area of a Paralellogram

Perimeter of a Parallelogram

Area of a Trapezoid Formula

NCERT Solutions for Class 9 Maths Chapter 9 Exercises:

CBSE X Related Questions

2.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

3.
Find the sums given below :
\(7 + 10\frac 12+ 14 + ....... + 84\)
\(34 + 32 + 30 + ....... + 10\)
\(–5 + (–8) + (–11) + ....... + (–230)\)

4.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

6.
If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos² A – sin²A or not

Similar Mathematics Concepts

Comments

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NCERT Solutions for Class 9 Maths Chapter 9: Areas of parallelograms and triangles

NCERT Solutions for Class 9 Mathematics Chapter 9

Important Topics in Class 9 Maths Chapter 9 Areas of Parallelograms And Triangles

Area of a Paralellogram

Perimeter of a Parallelogram

Area of a Trapezoid Formula

NCERT Solutions for Class 9 Maths Chapter 9 Exercises:

CBSE X Related Questions

1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare themMonthly consumption (in units) Number of consumers65 - 85 485 - 1055105 - 12513125 - 14520145 - 16514165 - 1858185 - 2054

2. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

3. Find the sums given below :\(7 + 10\frac 12+ 14 + ....... + 84\)\(34 + 32 + 30 + ....... + 10\)\(–5 + (–8) + (–11) + ....... + (–230)\)

4. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

6. If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos2 A – sin2 A or not

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

3.
Find the sums given below :
\(7 + 10\frac 12+ 14 + ....... + 84\)
\(34 + 32 + 30 + ....... + 10\)
\(–5 + (–8) + (–11) + ....... + (–230)\)

4.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

6.
If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos² A – sin²A or not