Find the best tutors and institutes for Class 10 Tuition
Q2(c):
Find the area of each of the following triangles:
(c) 
Find the area of each of the following triangles:
(c)
Solution :
Initial Setup & Given Variables
Based on the standard geometrical parameters provided in the corresponding exercise figure, we are analyzing a right-angled triangle with the following dimensions:
- Base ($b$): $3\text{ cm}$
- Height ($h$): $4\text{ cm}$
[Because the triangle features a $90^\circ$ angle between these two segments, the side measuring $4\text{ cm}$ acts as the exact perpendicular altitude to the $3\text{ cm}$ base.]
Step 1: Theoretical Foundation
To determine the two-dimensional space enclosed by the triangle, we apply the standard area theorem for triangles [derived from the area of a rectangle, where a diagonal bisects the rectangle into two congruent right triangles].
The formula is given by:
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
Step 2: Geometrical Representation
Below is the precise, scaled geometrical construction of the given triangle. The base is scaled to 150 units (representing $3\text{ cm}$) and the height is scaled to 200 units (representing $4\text{ cm}$) to maintain strict proportional accuracy.
Step 3: Algebraic Substitution and Calculation
We substitute the given scalar values into the area formula. [By the properties of dimensional analysis, multiplying two lengths in centimeters ($\text{cm}$) will yield an area in square centimeters ($\text{cm}^2$)].
$\text{Area} = \frac{1}{2} \times 3\text{ cm} \times 4\text{ cm}$
First, compute the product of the base and the height:
$3 \times 4 = 12\text{ cm}^2$
Next, apply the $\frac{1}{2}$ multiplier [as the triangle represents exactly half the area of the bounding $3\text{ cm} \times 4\text{ cm}$ rectangle]:
$\text{Area} = \frac{1}{2} \times 12\text{ cm}^2$
$\text{Area} = 6\text{ cm}^2$
Final Solution: The area of the given triangle is $6\text{ cm}^2$.
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)
- 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(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
Super Tutor
Kharghar Sector 27, Mumbai
13 yrs of
Exp
30 students
1078 hrs of classes completed
650
per hour
Online Classes
Student's Home
**Teaching Methodology:** My approach to teaching Mathematics and Science is personalized and interactive. I focus on understanding each student's unique learning style and adapt my methods accordingly. My goal is to make Maths and Science relatable and engaging, using real-life examples and practical applications to enhance understanding.I try to develop self thinking on problem and pushing them to solve by self **Key Skills:** - Strong grasp of Mathematics concepts across various levels - Ability to simplify complex topics for better understanding - Excellent communication and motivational skills - Patient and supportive, ensuring a comfortable learning environment - Regular progress assessments and constructive feedback **Services Offered:** - One-on-one personalized tutoring sessions - Homework and assignment assistance - Exam preparation and revision classes - Online and in-person tutoring options - Flexible scheduling to accommodate student needs **Achievements:** - Helped students improve their grades by an average of [25%] - Successfully prepared students for competitive exams with [JEE,NEET,IMO] - Received positive feedback and high ratings from students and parents Interested in improving your Maths AND Science skills? Contact me today to schedule a session and take the first step towards mastering Mathematics and Science!
This teacher is incredibly friendly and teaches in a very calm and patient manner. He helped me clear all my doubts; after failing my board exams previously, I managed to score 47/80 in Mathematics under his guidance. He truly excels at building a student's confidence.
Top Tutor
Sector 150, Noida
5 yrs of
Exp
800
per hour
Student's Home
Tutor's home
Having 2 years experience in teaching class 6th to 10th across different boards.
Top Tutor
Beta I Block E, Noida
5 yrs of
Exp
1 student
400
per hour
Online Classes
Student's Home
Tutor's home
Dear Parents and students, I am Amit, We provide home-based tuition for subjects and we also do professional coaching for NEET/IIT-JEE and other competitive exams. We have experienced tutors in all the fields and subjects. We also provide demo class free of cost. We do online as well as offline mode of teaching...
Top Tutor
Bavdhan Vidnyan Nagar, Pune
5 yrs of
Exp
1 student
54 hrs of classes completed
250
per hour
Online Classes
Tutor's home
I have taught in an institution to state ,ICSE , CBSE kids. Extra-ordinary Results. Ample of Study Material will be provided. We can have a demo class . You would enjoy the interactive and interesting sessions.
My experience with Parag sir was great, It is because of him that I got really good grades in my 10th and 12th standard.
Top Tutor
Venkatraman Nagar Chennai, Chennai
10 yrs of
Exp
27 students
3790 hrs of classes completed
320
per hour
Online Classes
Tutor's home
im having good experince in teaching with make them understand the concepts better
Classes are much interactive so useful to assess the students understanding capacity during the class itself.
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
