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Q2(b):
Find the area of each of the following triangles:
(b) 
Find the area of each of the following triangles:
(b) 
Solution :
Given Variables & Initial Setup
Based on the principles of Coordinate Geometry, when a geometric figure is presented on a Cartesian plane, its dimensions must be extracted by identifying the coordinates of its vertices. For this analysis, we extract the coordinates of the triangle's vertices from the provided two-dimensional grid.
- Vertex $A$ (Apex): $(0, 5)$
- Vertex $B$ (Left base endpoint): $(-4, 0)$
- Vertex $C$ (Right base endpoint): $(3, 0)$
[Theoretical Justification: In a Cartesian system, the distance between two points lying on the same horizontal or vertical axis can be determined using the absolute difference of their respective non-zero coordinates, per the 1D Distance Formula $d = |a_2 - a_1|$].
Step 1: Determining the Length of the Base ($b$)
The base of the triangle, segment $BC$, lies entirely on the x-axis. Because both points share the same y-coordinate ($y = 0$), the length of the base is the absolute difference between their x-coordinates.
Let $x_B = -4$ and $x_C = 3$.
$ \text{Base } (b) = |x_C - x_B| $
$ b = |3 - (-4)| $
$ b = |3 + 4| = 7 \text{ units} $
Step 2: Determining the Altitude / Height ($h$)
The height of a triangle is the perpendicular distance from the apex to the line containing the base. Since the base lies on the x-axis, the perpendicular distance from vertex $A(0, 5)$ to the x-axis is simply the absolute value of its y-coordinate.
Let $y_A = 5$ and the y-coordinate of the base $y_{base} = 0$.
$ \text{Height } (h) = |y_A - y_{base}| $
$ h = |5 - 0| = 5 \text{ units} $
Step 3: Calculating the Area of the Triangle
According to Euclidean geometry, the area ($A$) of a triangle is given by half the product of its base and its corresponding altitude.
$ A = \frac{1}{2} \times b \times h $
Substituting the derived values:
$ A = \frac{1}{2} \times 7 \times 5 $
$ A = \frac{1}{2} \times 35 $
$ A = 17.5 \text{ square units} $
Step 4: Rigorous Verification via Coordinate Geometry Formula
To ensure absolute precision, we verify the result using the determinant-based area formula for a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:
$ A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| $
Assigning the coordinates: $(x_1, y_1) = (0, 5)$, $(x_2, y_2) = (-4, 0)$, $(x_3, y_3) = (3, 0)$.
$ A = \frac{1}{2} |0(0 - 0) + (-4)(0 - 5) + 3(5 - 0)| $
$ A = \frac{1}{2} |0 + (-4)(-5) + 3(5)| $
$ A = \frac{1}{2} |0 + 20 + 15| $
$ A = \frac{1}{2} |35| = 17.5 \text{ square units} $
[The identical result confirms the geometric extraction is mathematically sound and verified across multiple theorems].
Final Solution: The area of the given triangle is 17.5 square units.
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(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
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