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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 $PQRS$.
$PQRS$ 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 $PQRS$.
Solution :
Given Variables & Initial Setup
We are presented with a parallelogram $PQRS$ in Euclidean space, along with specific orthogonal heights (altitudes) corresponding to its bases. The given parameters are structured as follows:
| Geometric Entity | Symbolic Representation | Magnitude / Value |
|---|---|---|
| Base of the Parallelogram | $SR$ | $12 \text{ cm}$ |
| Altitude to Base $SR$ | $QM$ | $7.6 \text{ cm}$ |
| Altitude to Base $PS$ | $QN$ | Unknown (Not required for Part A) |
Step 1: Theoretical Foundation for the Area of a Parallelogram
By the fundamental theorems of planar geometry, a parallelogram can be transformed into a rectangle of equal area by translating a right-angled triangle from one side to the other. Therefore, the area $A$ of any parallelogram is strictly equal to the product of the length of any chosen base ($b$) and its corresponding orthogonal height ($h$).
[Per the Area Axioms of Euclidean Geometry]:
$A = b \times h$
In the context of parallelogram $PQRS$, if we select $SR$ as the base, the corresponding height is the perpendicular distance from the opposite parallel side ($PQ$) to $SR$. This orthogonal segment is explicitly given as $QM$.
$A = SR \times QM$
Step 2: High-Precision Visual Representation
Below is the geometrically scaled representation of parallelogram $PQRS$, mapping the given dimensions and orthogonal projections. The coordinates are calculated to preserve the exact ratio of the base to the height ($12 : 7.6$).
Step 3: Algebraic Substitution and Calculation
We now substitute the known scalar quantities into our established area formula. Ensure that the units are consistent (both are in centimeters), which will yield an area in square centimeters ($\text{cm}^2$).
- Base ($SR$) = $12 \text{ cm}$
- Height ($QM$) = $7.6 \text{ cm}$
Executing the multiplication:
$A = 12 \text{ cm} \times 7.6 \text{ cm}$
To compute this precisely without a calculator, we can decompose the decimal:
$A = 12 \times \left(7 + 0.6\right)$
$A = (12 \times 7) + (12 \times 0.6)$
$A = 84 + 7.2$
$A = 91.2 \text{ cm}^2$
Final Solution: The area of the parallelogram $PQRS$ is $91.2 \text{ cm}^2$.
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(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$^2$.
- 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(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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