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Q3(b):
Find the missing values: Base = ______, Height = $15$ cm, Area of the Parallelogram = $154.5$ cm$^2$.
Solution :
Given Variables & Initial Setup
We are tasked with determining the missing base dimension of a parallelogram given its height and total area. The known parameters are defined as follows:
- Height ($h$): $15 \text{ cm}$
- Area ($A$): $154.5 \text{ cm}^2$
- Base ($b$): Unknown
Theoretical Foundation
The area of a parallelogram is defined as the total two-dimensional space enclosed within its four sides. Geometrically, a parallelogram can be rearranged into a rectangle of the same base and height. Therefore, the governing formula for the area of a parallelogram is:
$A = b \times h$
[Per the geometric postulate of area equivalence under translation, the area of a parallelogram is strictly the product of its base and the corresponding perpendicular height].
Step 1: Algebraic Substitution
Substitute the given numerical values into the standard area formula:
$154.5 = b \times 15$
Step 2: Isolation of the Variable
To solve for the unknown base ($b$), we must isolate it on one side of the equation. We achieve this by dividing both sides of the equation by the height ($15$).
$b = \frac{154.5}{15}$
[Applying the Division Property of Equality, which states that dividing both sides of an equation by the same non-zero number preserves the equality].
Step 3: Arithmetic Computation & Decimal Manipulation
To perform the division with precision, we can eliminate the decimal in the numerator by multiplying both the numerator and the denominator by $10$:
$b = \frac{154.5 \times 10}{15 \times 10}$
$b = \frac{1545}{150}$
Now, simplify the fraction by dividing both terms by their common factors. First, divide by $5$:
$\frac{1545 \div 5}{150 \div 5} = \frac{309}{30}$
Next, divide by $3$:
$\frac{309 \div 3}{30 \div 3} = \frac{103}{10}$
Converting the simplified fraction back into a decimal yields:
$b = 10.3 \text{ cm}$
Geometric Visualization
Below is a scaled, mathematically accurate representation of the parallelogram, demonstrating the relationship between the base, the perpendicular height, and the enclosed area.
Final Verification
To ensure absolute accuracy, we substitute the calculated base back into the original area formula:
$A = 10.3 \text{ cm} \times 15 \text{ cm}$
$A = 154.5 \text{ cm}^2$
The calculated area matches the given area perfectly, confirming the validity of the derived base.
Final Solution: The missing value for the Base is $10.3 \text{ cm}$.
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(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
Top Tutors who teach Coordinate Geometry
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