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.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.
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2.
If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).
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3.
Find the H.C.F. and L.C.M. of 1530 and 2040.
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4.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).
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5.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.
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6.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.
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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