Find the best tutors and institutes for Class 10 Tuition
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
Top Tutors who teach Coordinate Geometry
I am a strong believer in concept-based gain of knowledge. I start from scratch while teaching the very basic concepts and take my students to a level where they have a deep understanding of the subject. Once they understand the concept, the next step is that practice a lot of questions/numerical/problem in the class itself as practice is the key to retain knowledge. I always refer to prescribed books by CBSE/ICSE board but always encourage students to refer additional 1 or 2 books based on their speed to practice. In last I provide them enough homework to practice at home and send me the doubts before the next class. Also for 10th students, we prepare right from beginning by covering the previous year questions and this approach till date have been fruitful as most of my students excel in the board exams
My son is taking Math and Science Tuition with Rakesh sir for last 2 weeks. I am very much satisfied with his teaching techniques. He uses visual aids which helps the student in understanding the concepts in a better way.
I am a dedicated mathematics tutor with seven years of experience teaching students from grades 8 to 10 across IB, IGCSE, ICSE, and CBSE boards. I understand that many students struggle with math due to fear, lack of conceptual clarity, or difficulty in applying formulas. My approach focuses on simplifying complex concepts through real-life examples, step-by-step problem-solving, and interactive techniques to make learning engaging. I emphasize building a strong foundation and boosting students' confidence, ensuring they not only understand math but also enjoy it as a subject. I have found that students sometimes lack skills and sometimes they are not very well focused , so I can easily point these out
I have a Science background completed my electrical engineering.... I am giving home tuition for 2 years.... I am also preparing for UPSC civil service exam.
Have taught in school for 10th grade students.
This is Prita Chatterjee with 27 years of teaching in the Engineering College and Schools .I teach Maths for all the boards - IB AA(SL& HL) ,IB - AI(HL&SL),Integrated Maths ,MYP ,Cambridge A Level , Additional Maths, IGCSE O Level, ISC, ICSE , CBSE .Presently, I am tutoring students from U.S. France, Austrailia , Netherland , Japan, Singapore, Malaysia , Thailand , Phillipines , Dubai , Riyadh , Doha, and pan India . I would love to teach from basics . .If you would like me to teach your child , kindly let me know ,so we can schedule a demo ,as per suitability.
She is a good teacher with very well and skilled explanations and making sure that students do each and every sum from text book.Hoping that she teaches out of the book&concepts too.:) my only concern is I can’t expect the reply for my messages as quick.
Find more Tutor for Coordinate Geometry in your City
- Bangalore Mathematics Tutors
- Delhi Mathematics Tutors
- Chennai Mathematics Tutors
- Gurgaon Mathematics Tutors
- Noida Mathematics Tutors
- Hyderabad Mathematics Tutors
- Mumbai Mathematics Tutors
- Ghaziabad Mathematics Tutors
- Chandigarh Mathematics Tutors
- Pune Mathematics Tutors
- Jaipur Mathematics Tutors
- Surat Mathematics Tutors
Download free CBSE - Class 9 Mathematics Coordinate Geometry EXERCISE 9.1 worksheets
Download Now