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Q2(d):
Find the area of each of the following triangles:
(d) 
Find the area of each of the following triangles:
(d)
Solution :
Given Variables & Initial Setup
Based on the standard geometric parameters provided in the visual data for this specific problem, we extract the following dimensions for the triangle:
- Base ($b$): $3\text{ cm}$
- Height ($h$): $2\text{ cm}$
Below is the precise geometric reconstruction of the given figure. Note that for an obtuse-angled triangle, the altitude (height) dropped from the top vertex intersects the extended base outside the boundary of the triangle.
Step 1: Geometric Analysis of the Figure
The figure represents an obtuse-angled triangle $\triangle ABC$. [By definition, an obtuse triangle contains one interior angle strictly greater than $90^\circ$]. When calculating the area of an obtuse triangle using a base that forms one of the sides of the obtuse angle, the corresponding altitude (perpendicular height) must be drawn from the opposite vertex to the line containing the base. This altitude falls outside the triangle, meeting the extended base at a right angle (point $D$).
Step 2: Formulating the Area Equation
The area ($A$) of any triangle in Euclidean geometry is determined by the product of its base and its corresponding altitude, halved. [Per Euclidean geometry principles, the area of a triangle is exactly half the area of a parallelogram constructed on the same base and between the same parallels].
The governing formula is:
$A = \frac{1}{2} \times b \times h$
Where:
- $b$ is the length of the base segment ($BC$).
- $h$ is the length of the perpendicular altitude ($AD$).
Step 3: Substitution and Algebraic Calculation
We substitute the identified scalar values into the area formula. It is critical to include units during the calculation to ensure dimensional consistency.
$A = \frac{1}{2} \times (3\text{ cm}) \times (2\text{ cm})$
First, multiply the scalar magnitudes and the units:
$A = \frac{1}{2} \times 6\text{ cm}^2$
Next, apply the scalar multiplication by $\frac{1}{2}$:
$A = 3\text{ cm}^2$
Final Solution: The area of the given triangle is $3\text{ cm}^2$.
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(c): Find the area of each of the following triangles: (c)
- 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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