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Q4(c):
Find the missing values: Base = $22$ cm, Height = ______, Area of Triangle = $170.5$ cm$^2$.
Solution :
Given Variables & Initial Setup
We are tasked with determining the missing perpendicular height of a triangle given its base dimension and total enclosed area. The known parameters are defined as follows:
- Base ($b$) = $22$ cm
- Area ($A$) = $170.5$ cm$^2$
- Height ($h$) = Unknown
Step 1: State the Governing Geometric Formula
The fundamental relationship between the area, base, and height of any triangle in Euclidean geometry is expressed by the formula:
$A = \frac{1}{2} \times b \times h$
[Per the standard area axiom for a Euclidean triangle, which establishes that the area of a triangle is exactly half the area of a parallelogram sharing the same base and height].
Step 2: Substitute the Known Values
By substituting the given numerical values for the area ($A$) and the base ($b$) into the governing equation, we establish the following algebraic relationship:
$170.5 = \frac{1}{2} \times 22 \times h$
Step 3: Algebraic Manipulation and Simplification
First, simplify the right side of the equation by evaluating the product of the constant terms:
$170.5 = 11 \times h$
[By applying the associative property of multiplication and simplifying $\frac{1}{2} \times 22 = 11$].
Next, isolate the unknown variable $h$ by dividing both sides of the equation by the coefficient $11$:
$h = \frac{170.5}{11}$
[Per the Division Property of Equality, ensuring the equation remains balanced].
Step 4: Final Calculation
Perform the arithmetic division to determine the exact numerical value of the height:
$h = 15.5$ cm
Geometric Visualization
Below is a strictly scaled geometric representation of the triangle. The coordinates are mapped such that the base spans exactly $220$ units and the altitude spans exactly $155$ units, preserving the $22 : 15.5$ ratio of the physical dimensions.
Final Solution: The missing height of the triangle is $15.5$ cm.
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)
- 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$^2$.
- 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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