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
Top Tutor
Alamganj, Patna
4 yrs of
Exp
9 students
16 hrs of classes completed
200
per hour
Student's Home
Tutor's home
I am teaching in school ( Dalimss Sunbeam school Ghazipur ) as a PGT teacher.. I have over all 4 years experience in this field.
He is a very thoughtful teacher who puts a lot of thought into how he presents the material his lessons were engaging, useful and he was very patient with everyone in class always encouraging to students.
Top Tutor
Padmanabhanagar, Bangalore
3 yrs of
Exp
450
per hour
Online Classes
Student's Home
Tutor's home
I am an experienced tutor with 4 years of experience in this field. I completed my BE in the year 2014. I have tutored students for Spoken English, BTech Tuition, C Language, C++ Language, CAD, CE and COMEDK Coaching. I can conduct classes at Student's Home, My place and Online via a different medium to students from International, CBSE, State, ICSE/ISC Board. I follow different techniques while teaching so, that my students are engaged positively. To make learning fun and easier, I use creative and innovative ideas.
I really liked the way he taught me Science and Maths. Especially Chemistry which was so boring & tough for me. The way he puts forth the concept is really nice. I could understand a lot and did not have 2 mug it up. Thanks Sir!
Top Tutor
Sb Colony, Eluru
8 yrs of
Exp
2 students
50 hrs of classes completed
600
per hour
Online Classes
Student's Home
Tutor's home
I have taught students at Samarthanam and NGO for challenged students from class 1-XII in Bangalore for about 2 years. I used to teach students of groups 3-10 every weekend. I helped students with their home work as well as clarify concepts thought in the classroom. I saw improvement in student performance over the course of time with the my teaching methods
An exceptional teacher who genuinely cares! Sir assessed my son’s weak areas in coordinate geometry and tailored his lessons perfectly. The easy explanation style built strong fundamentals, helping my son boost his term score from 50% to an unbelievable 88%
Top Tutor
Asilmetta, Visakhapatnam
3 yrs of
Exp
Online Classes
Student's Home
Tutor's home
Seeing my students succeed is the best reward! I bring lessons to life with real-world examples, and I put my all into every class. It's amazing to watch them not only understand the material but also get excited about learning. Their good grades are definitely a confidence booster!
Top Tutor
HSR Layout, Bangalore
7 yrs of
Exp
500
per hour
Online Classes
Student's Home
Tutor's home
I am an experienced, qualified teacher and tutor with over 4 years of experience in teaching Maths,Physics and Chemistry, across different boards including CBSE, ICSE, IGCSE and State Board. Passionate about solving mathematical problems and chemical equations, over the years I have helped hundreds of students overcome their fear of Maths,Physics and Chemistry. So far, I have worked as a Full Time Tutor and guided over 189 students reach their goals in the respective subjects
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
- Chandigarh Mathematics Tutors
- Jaipur Mathematics Tutors
- Pune Mathematics Tutors
- Ghaziabad Mathematics Tutors
- Surat Mathematics Tutors
