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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 $DL$.
$DL$ 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 $DL$.
Solution :
Initial Setup & Given Variables
Let us define the geometric properties and given parameters of the parallelogram $ABCD$:
- Area of parallelogram $ABCD = 1470 \text{ cm}^2$
- Length of base $AB = 35 \text{ cm}$
- Length of base $AD = 49 \text{ cm}$
- $DL \perp AB$, establishing $DL$ as the corresponding altitude (height) to the base $AB$.
- $BM \perp AD$, establishing $BM$ as the corresponding altitude (height) to the base $AD$.
Geometric Visualization
The following high-precision diagram models the parallelogram to scale, where $10 \text{ units} = 1 \text{ cm}$. The exact coordinates are calculated using trigonometric projections to ensure spatial accuracy ($\angle DAB \approx 59^\circ$).
Step 1: Calculating the Length of Altitude $DL$
The area of a parallelogram is defined by the product of any chosen base and its corresponding perpendicular altitude [Per the Euclidean geometric principle of area equivalence].
The fundamental formula is:
$\text{Area} = \text{Base} \times \text{Corresponding Height}$
Taking $AB$ as the base, the corresponding height is the perpendicular segment $DL$. Substituting the known values into the area equation:
$\text{Area}(ABCD) = AB \times DL$
$1470 = 35 \times DL$
Isolating $DL$ by dividing both sides by $35$:
$DL = \frac{1470}{35}$
$DL = 42 \text{ cm}$
Step 2: Calculating the Length of Altitude $BM$
The area of the parallelogram remains constant regardless of which base is chosen for the calculation [By the invariant property of 2D geometric areas]. We now apply the area formula using $AD$ as the base. The corresponding height for base $AD$ is the perpendicular segment $BM$.
$\text{Area}(ABCD) = AD \times BM$
Substituting the known area and the length of base $AD$:
$1470 = 49 \times BM$
Isolating $BM$ by dividing both sides by $49$:
$BM = \frac{1470}{49}$
$BM = 30 \text{ cm}$
Final Conclusion
Final Solution: The length of the altitude $DL$ is $42 \text{ cm}$ and the length of the altitude $BM$ is $30 \text{ cm}$.
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 $PQRS$.
- 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.
- 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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