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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.
$PQRS$ 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.
Solution :
Given Variables & Initial Setup
We are analyzing the parallelogram $PQRS$ with two distinct base-height pairings. The geometric properties provided are:
- Base $SR = 12 \text{ cm}$
- Height corresponding to base $SR$, denoted as $QM = 7.6 \text{ cm}$
- Base $PS = 8 \text{ cm}$
- Height corresponding to base $PS$, denoted as $QN$, which is the unknown variable to be determined.
Step 1: Calculating the Area of Parallelogram $PQRS$
The area of a parallelogram is defined as the product of any chosen base and its corresponding perpendicular height. [Per the geometric theorem for the area of a parallelogram: $\text{Area} = \text{base} \times \text{height}$].
Using the given base $SR$ and its corresponding height $QM$:
$\text{Area of } PQRS = SR \times QM$
$\text{Area of } PQRS = 12 \text{ cm} \times 7.6 \text{ cm}$
$\text{Area of } PQRS = 91.2 \text{ cm}^2$
Step 2: Formulating the Equation for Height $QN$
The area of a specific polygon is an invariant scalar quantity; it remains constant regardless of which side is designated as the base. [By the principle of area invariance]. Therefore, calculating the area using base $PS$ and its corresponding height $QN$ must yield the exact same result.
$\text{Area of } PQRS = PS \times QN$
Substituting the known area ($91.2 \text{ cm}^2$) and the length of base $PS$ ($8 \text{ cm}$) into the equation:
$91.2 = 8 \times QN$
Step 3: Solving for $QN$
To isolate $QN$, we divide both sides of the equation by $8$:
$QN = \frac{91.2}{8}$
Performing the division:
$QN = 11.4 \text{ cm}$
Final Solution: The length of the height $QN$ is $11.4 \text{ cm}$.
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$^2$.
- 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 $PQRS$.
- 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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