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Q1(c):
Find the area of each of the following parallelograms:
(c) 
Find the area of each of the following parallelograms:
(c)
Solution :
Initial Setup & Given Variables
Based on the standard geometric configuration for this specific problem, we extract the primary dimensions of the given parallelogram from the provided visual data:
- Base ($b$): $2.5 \text{ cm}$
- Corresponding Altitude/Height ($h$): $3.5 \text{ cm}$
Step 1: Theoretical Foundation
The area of a parallelogram is determined by the product of its base and its corresponding altitude. [Per the geometric principle of area equivalence, a parallelogram can be orthogonally decomposed and rearranged into a rectangle of identical base and height without any loss of area].
The governing mathematical formula is:
$\text{Area} = b \times h$
Step 2: Substitution of Dimensional Values
We substitute the given scalar quantities into the area formula. It is critical to ensure that both measurements share the same unit ($\text{cm}$) before proceeding with the multiplication.
$\text{Area} = 2.5 \text{ cm} \times 3.5 \text{ cm}$
Step 3: Arithmetic Computation & Analytical Breakdown
To ensure absolute precision and avoid floating-point errors, we convert the decimal values into their fractional equivalents prior to multiplication:
$2.5 = \frac{25}{10}$
$3.5 = \frac{35}{10}$
Multiplying the fractions algebraically:
$\text{Area} = \left( \frac{25}{10} \right) \times \left( \frac{35}{10} \right)$
$\text{Area} = \frac{25 \times 35}{10 \times 10}$
Calculating the numerator ($25 \times 35$):
$25 \times 30 = 750$
$25 \times 5 = 125$
$750 + 125 = 875$
Substituting the numerator back into the fraction:
$\text{Area} = \frac{875}{100}$
Converting the fraction back to a standard decimal format by shifting the decimal point two places to the left:
$\text{Area} = 8.75 \text{ cm}^2$
Final Solution: The area of the parallelogram is $8.75 \text{ cm}^2$.
More Questions from Class 9 Mathematics Coordinate Geometry EXERCISE 9.1
- Q1(a): Find the area of each of the following parallelograms: (a)
- Q1(b): Find the area of each of the following parallelograms: (b)
- Q1(d): Find the area of each of the following parallelograms: (d)
- Q1(e): Find the area of each of the following parallelograms: (e)
- Q2(a): Find the area of each of the following triangles: (a)
- Q2(b): Find the area of each of the following triangles: (b)
- Q2(c): Find the area of each of the following triangles: (c)
- Q2(d): Find the area of each of the following triangles: (d)
- Q3(a): Find the missing values: Base = $20$ cm, Height = ______, Area of the Parallelogram = $246$ cm$^2$.
- Q3(b): Find the missing values: Base = ______, Height = $15$ cm, Area of the Parallelogram = $154.5$ cm$^2$.
- Q3(c): Find the missing values: Base = ______, Height = $8.4$ cm, Area of the Parallelogram = $48.72$ cm$^2$.
- Q3(d): Find the missing values: Base = $15.6$ cm, Height = ______, Area of the Parallelogram = $16.38$ cm$^2$.
- Q4(a): Find the missing values: Base = $15$ cm, Height = ______, Area of Triangle = $87$ cm$^2$.
- Q4(b): Find the missing values: Base = ______, Height = $31.4$ mm, Area of Triangle = $1256$ mm$^2$.
- Q4(c): Find the missing values: Base = $22$ cm, Height = ______, Area of Triangle = $170.5$ cm$^2$.
- Q5(a): $PQRS$ is a parallelogram (Fig 9.14). $QM$ is the height from $Q$ to $SR$ and $QN$ is the height from $Q$ to $PS$. If $SR = 12$ cm and $QM = 7.6$ cm. Find: (a) the area of the parallegram $PQRS$.
- Q5(b): $PQRS$ is a parallelogram (Fig 9.14). $QM$ is the height from $Q$ to $SR$ and $QN$ is the height from $Q$ to $PS$. If $SR = 12$ cm and $QM = 7.6$ cm. Find: (b) $QN$, if $PS = 8$ cm.
- Q6: $DL$ and $BM$ are the heights on sides $AB$ and $AD$ respectively of parallelogram $ABCD$ (Fig 9.15). If the area of the parallelogram is $1470$ cm$^2$, $AB = 35$ cm and $AD = 49$ cm, find the length of $BM$ and $DL$.
- Q7: $\triangle ABC$ is right angled at $A$ (Fig 9.16). $AD$ is perpendicular to $BC$. If $AB = 5$ cm, $BC = 13$ cm and $AC = 12$ cm, Find the area of $\triangle ABC$. Also find the length of $AD$.
- Q8: $\triangle ABC$ is isosceles with $AB = AC = 7.5$ cm and $BC = 9$ cm (Fig 9.17). The height $AD$ from $A$ to $BC$, is $6$ cm. Find the area of $\triangle ABC$. What will be the height from $C$ to $AB$ i.e., $CE$?
CBSE Solutions for Class 9 Mathematics Coordinate Geometry
Chapters in CBSE - Class 9 Mathematics
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