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Q1(a):

Find the area of each of the following parallelograms: (a)   

          

Solution :

Given Variables & Initial Setup

Based on the standard geometric configuration for this specific problem, we extract the following dimensions for the parallelogram from the provided visual data:

  • Base ($b$): $7\text{ cm}$
  • Corresponding Height or Altitude ($h$): $4\text{ cm}$

Step 1: Geometric Foundation & Theoretical Formula

The area of a parallelogram is defined as the total two-dimensional space enclosed by its four sides. [Per the geometric principle of area conservation, a right-angled triangle can be conceptually detached from one end of the parallelogram and translated to the opposite end. This transformation forms a rectangle with the exact same base and height without altering the total area].

Therefore, the area ($A$) is calculated using the fundamental theorem of quadrilateral area:

$A = \text{base} \times \text{height}$

$A = b \times h$

Step 2: Visualizing the Parallelogram

The following high-precision diagram illustrates the parallelogram $ABCD$, where the base $AB = 7\text{ cm}$ and the perpendicular altitude $DE = 4\text{ cm}$. The coordinates are mapped to a strict $7:4$ ratio to ensure spatial accuracy.

A B C D E Base (b) = 7 cm Height (h) = 4 cm

Step 3: Algebraic Computation & Dimensional Analysis

Substitute the given scalar values into the area formula:

$A = 7\text{ cm} \times 4\text{ cm}$

Multiply the scalar magnitudes ($7 \times 4 = 28$) and the units simultaneously. [By the laws of dimensional analysis, multiplying a length by a length yields a squared unit of area: $\text{cm} \times \text{cm} = \text{cm}^2$]:

$A = 28\text{ cm}^2$

Final Solution: The area of the parallelogram is $28\text{ cm}^2$.


More Questions from Class 9 Mathematics Coordinate Geometry EXERCISE 9.1


CBSE Solutions for Class 9 Mathematics Coordinate Geometry


Chapters in CBSE - Class 9 Mathematics


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