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Q3(a):
Find the missing values: Base = $20$ cm, Height = ______, Area of the Parallelogram = $246$ cm$^2$.
Solution :
Initial Setup & Given Variables
We are tasked with determining the missing dimension of a parallelogram given its area and base. Let us define the known scalar quantities:
- Base ($b$) = $20 \text{ cm}$
- Area ($A$) = $246 \text{ cm}^2$
- Height ($h$) = Unknown
Theoretical Foundation
The area of a parallelogram is defined as the region enclosed by its four sides in a two-dimensional plane. [Per the standard geometric theorem for the area of a parallelogram], the area is equal to the product of its base and the corresponding perpendicular height. The formula is expressed as:
$A = b \times h$
Where:
- $A$ represents the total area.
- $b$ represents the length of the base.
- $h$ represents the perpendicular height (altitude) drawn to that base.
Geometric Visualization
Below is a precise, scaled vector representation of the parallelogram, illustrating the relationship between the base, the perpendicular height, and the total enclosed area.
Step 1: Setting up the Equation
By substituting the given values into the area formula, we establish a linear equation with one variable ($h$).
$246 = 20 \times h$
Step 2: Algebraic Manipulation
To isolate the variable $h$, we apply the Division Property of Equality. We divide both sides of the equation by the coefficient of $h$, which is $20$.
$h = \frac{246}{20}$
Step 3: Calculation and Simplification
We perform the division to find the exact decimal value of the height. We can simplify the fraction by dividing the numerator and the denominator by $2$ first:
$h = \frac{123}{10}$
Dividing by $10$ shifts the decimal point one place to the left:
$h = 12.3 \text{ cm}$
Final Solution: The missing height of the parallelogram is $12.3 \text{ cm}$.
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(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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