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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
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