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Q3(d):
Find the missing values: Base = $15.6$ cm, Height = ______, Area of the Parallelogram = $16.38$ cm$^2$.
Solution :
Given Variables & Initial Setup
We are tasked with determining the missing dimension of a parallelogram based on its area and base. The known parameters are defined as follows:
- Base ($b$): $15.6 \text{ cm}$
- Area ($A$): $16.38 \text{ cm}^2$
- Height ($h$): Unknown
Step 1: Governing Geometric Principle
The area of a parallelogram is mathematically defined as the product of its base and the corresponding perpendicular height. [Per standard Euclidean geometry principles for planar quadrilaterals, a parallelogram can be transformed into a rectangle of equal base and height without altering its area].
The fundamental formula is expressed as:
$A = b \times h$
Step 2: Algebraic Substitution
Substitute the given numerical values into the area formula to establish the algebraic equation:
$16.38 \text{ cm}^2 = 15.6 \text{ cm} \times h$
Step 3: Isolation of the Unknown Variable
To solve for the height ($h$), we must isolate the variable by dividing both sides of the equation by the base ($15.6 \text{ cm}$). [By the Division Property of Equality]:
$h = \frac{16.38 \text{ cm}^2}{15.6 \text{ cm}}$
Step 4: Arithmetic Computation & Decimal Normalization
To perform the division with precision, normalize the decimals by multiplying both the numerator and the denominator by $100$ (shifting the decimal point two places to the right):
$h = \frac{16.38 \times 100}{15.60 \times 100} = \frac{1638}{1560}$
Now, execute the division:
- $1638 \div 1560 = 1$ with a remainder of $78$.
- Add a decimal point and append a zero to the remainder: $780$.
- $780 \div 1560 = 0.5$.
Combining these yields the exact quotient:
$h = 1.05 \text{ cm}$
Geometric Visualization
Below is a rigorously scaled vector representation of the parallelogram. The aspect ratio of the base to the height in the drawing is exactly $15.6 : 1.05$ (approximately $14.86 : 1$), demonstrating the highly elongated nature of this specific geometric figure.
Final Solution: The missing height of the parallelogram is exactly $1.05 \text{ cm}$.
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(c): Find the area of each of the following parallelograms: (c)
- 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$^2$.
- 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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