Find the best tutors and institutes for Class 10 Tuition
Q1(e):
Find the area of each of the following parallelograms:
(e) 
Find the area of each of the following parallelograms:
(e) 
Solution :
Step 1: Given Variables & Initial Setup
Based on the standard geometric parameters provided in the referenced figure for this specific problem, we extract the following dimensions for the parallelogram:
- Base ($b$): $2\text{ cm}$
- Corresponding Height/Altitude ($h$): $4.4\text{ cm}$
[Note: In a parallelogram, any of the four sides can be chosen as the base. The corresponding height is the perpendicular distance from the chosen base to the opposite parallel side.]
Step 2: Geometric Visualization
Below is a mathematically scaled, high-precision vector representation of the parallelogram, illustrating the relationship between the base and its corresponding altitude.
Step 3: Mathematical Formulation
By the fundamental axioms of Euclidean geometry, a parallelogram can be transformed into a rectangle of equal area by translating the right-angled triangle formed by the altitude to the opposite side. Therefore, the area ($A$) of a parallelogram is strictly defined as the product of its base and its corresponding height.
The governing formula is:
$\text{Area} = \text{Base} \times \text{Height}$
$A = b \times h$
Step 4: Execution & Algebraic Calculation
Substitute the given scalar values into the area formula:
$A = 2\text{ cm} \times 4.4\text{ cm}$
To ensure absolute precision, we can perform the multiplication by converting the decimal to a fraction [Per standard arithmetic operations]:
$A = 2 \times \left(\frac{44}{10}\right)$
$A = \frac{88}{10}$
$A = 8.8$
Step 5: Dimensional Analysis
We must verify the units of the final result. Multiplying a one-dimensional length by another one-dimensional length yields a two-dimensional area:
$[\text{cm}] \times [\text{cm}] = [\text{cm}^2]$
The magnitude is $8.8$ and the unit is $\text{cm}^2$. The calculation is dimensionally consistent and mathematically sound.
Final Solution: The area of the parallelogram is $8.8\text{ cm}^2$.
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)
- 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.
- 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
Top Tutors who teach Coordinate Geometry
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.
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!
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%
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!
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
Download free CBSE - Class 9 Mathematics Coordinate Geometry EXERCISE 9.1 worksheets
Download Now