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.
Assertion (A) : The system of linear equations $3x - 5y + 7 = 0$ and $-6x + 10y + 14 = 0$ is inconsistent.
Reason (R) : When two linear equations don't have unique solution, they always represent parallel lines.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
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2.
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Use graphical method to solve the system of linear equations : $x = -3$ and $5x - 2y = -5$.
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4.
In the given figure, point D divides the side BC of $\Delta ABC$ in the ratio $1 : 2$. Find length AD.
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5.
If the zeroes of a polynomial p(x) are $-3$ and 8, then p(x) equals
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- $a(x^2 + 5x - 24)$
- $x^2 - 24$
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The dimensions of a window are $156\text{ cm} \times 216\text{ cm}$. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.
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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