Find the best tutors and institutes for Class 10 Tuition
Q1(c):
Find the area of each of the following parallelograms:
(c) 
Find the area of each of the following parallelograms:
(c)
Solution :
Initial Setup & Given Variables
Based on the standard geometric configuration for this specific problem, we extract the primary dimensions of the given parallelogram from the provided visual data:
- Base ($b$): $2.5 \text{ cm}$
- Corresponding Altitude/Height ($h$): $3.5 \text{ cm}$
Step 1: Theoretical Foundation
The area of a parallelogram is determined by the product of its base and its corresponding altitude. [Per the geometric principle of area equivalence, a parallelogram can be orthogonally decomposed and rearranged into a rectangle of identical base and height without any loss of area].
The governing mathematical formula is:
$\text{Area} = b \times h$
Step 2: Substitution of Dimensional Values
We substitute the given scalar quantities into the area formula. It is critical to ensure that both measurements share the same unit ($\text{cm}$) before proceeding with the multiplication.
$\text{Area} = 2.5 \text{ cm} \times 3.5 \text{ cm}$
Step 3: Arithmetic Computation & Analytical Breakdown
To ensure absolute precision and avoid floating-point errors, we convert the decimal values into their fractional equivalents prior to multiplication:
$2.5 = \frac{25}{10}$
$3.5 = \frac{35}{10}$
Multiplying the fractions algebraically:
$\text{Area} = \left( \frac{25}{10} \right) \times \left( \frac{35}{10} \right)$
$\text{Area} = \frac{25 \times 35}{10 \times 10}$
Calculating the numerator ($25 \times 35$):
$25 \times 30 = 750$
$25 \times 5 = 125$
$750 + 125 = 875$
Substituting the numerator back into the fraction:
$\text{Area} = \frac{875}{100}$
Converting the fraction back to a standard decimal format by shifting the decimal point two places to the left:
$\text{Area} = 8.75 \text{ cm}^2$
Final Solution: The area of the parallelogram is $8.75 \text{ cm}^2$.
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(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.
- 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
Top Tutor
Flanders, Mount Olive
5 yrs of
Exp
2 students
22 hrs of classes completed
100
per hour
Online Classes
I have volunteer teacher for evidyaloka ..I thought higg school students math and science
Top Tutor
Jayanagar, Bangalore
18 yrs of
Exp
28 students
2317 hrs of classes completed
500
per hour
Online Classes
Private teacher. Classes will be taken for science also but only for Physics and Chemistry. I have been teaching for Class 10 for the past 15 years. Focus will be given to each student regarding their difficult concepts. Previous year question papers will be discussed, along with practice worksheets that will be given for each chapter. Coaching will be given to score good marks in respective board exams by giving them tricks to manage the timings
Wonderful teacher with enormous knowledge about the subject with a smile every time while teaching...
Top Tutor
Bhestan, Surat
2 yrs of
Exp
5 students
355 hrs of classes completed
600
per hour
Online Classes
Tutor's home
I have motivated many 10th std student who scored more than 85% I love teaching and I can make you understand what actually studies is!!! It's all about practise and applications
I highly recommend Shivam Sir for Class 10 Mathematics and Science. He taught my brother, and the improvement in his understanding and confidence was remarkable. Before joining his classes, my brother found several concepts difficult, but Shivam Sir explained every topic in a very simple, practical, and easy-to-understand manner. He always ensured that every concept was clear before moving on to the next chapter. His teaching style is interactive and student-friendly. He patiently answers every doubt, gives plenty of practice questions, conducts regular tests, and provides useful tips for board exams. He also shares short tricks, revision strategies, and focuses on strengthening concepts rather than just memorizing answers.
Top Tutor
SCB Nagar, Kothagudem
20 yrs of
Exp
31 students
2094 hrs of classes completed
450
per hour
Online Classes
22 Years Experience in Schools IN ICSE/CBSE/IGCSE IN DIFFERENT STATE BOARDS
Sir is very friendly teacher he will support us more he is the best teacher of the year his teaching skills are really good.
Top Tutor
Patel Nagar, Gurgaon
13 yrs of
Exp
300
per hour
Online Classes
Student's Home
Tutor's home
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
