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Q2(a):
Find the area of each of the following triangles:
(a) 
Find the area of each of the following triangles:
(a)
Solution :
Initial Setup & Visual Data Extraction
Based on the standard geometric parameters provided in the visual figure for this specific problem, we extract the following fundamental dimensions of the triangle. The altitude is dropped perpendicularly to the chosen base.
- Base ($b$): $4 \text{ cm}$
- Corresponding Altitude/Height ($h$): $3 \text{ cm}$
Step 1: Establishing the Theoretical Foundation
To find the area of any triangle when the base and its corresponding perpendicular height (altitude) are known, we utilize the standard Euclidean area postulate for triangles. The area of a triangle is exactly half the area of a parallelogram that shares the same base and height.
The governing formula is:
$\text{Area of a Triangle} = \frac{1}{2} \times \text{base} \times \text{height}$
Symbolically represented as:
$A = \frac{1}{2} \cdot b \cdot h$
Step 2: Dimensional Substitution
Substitute the given scalar values into the area formula. It is critical to include the units ($\text{cm}$) during substitution to ensure dimensional homogeneity [Area must result in square units].
$A = \frac{1}{2} \times (4 \text{ cm}) \times (3 \text{ cm})$
Step 3: Algebraic Execution & Simplification
First, multiply the scalar magnitudes of the base and the height, and apply the product rule to the units ($\text{cm} \times \text{cm} = \text{cm}^2$):
$A = \frac{1}{2} \times (12 \text{ cm}^2)$
Next, multiply by the scalar fraction $\frac{1}{2}$ (which is equivalent to dividing by 2):
$A = \frac{12}{2} \text{ cm}^2$
$A = 6 \text{ cm}^2$
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(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
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