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Q4(a):
Find the missing values: Base = $15$ cm, Height = ______, Area of Triangle = $87$ cm$^2$.
Solution :
Step 1: Given Variables & Initial Setup
We are tasked with determining the missing height of a triangle given its base and total area. The known parameters are defined as follows:
- Base ($b$) = $15 \text{ cm}$
- Area ($A$) = $87 \text{ cm}^2$
- Height ($h$) = ?
Step 2: Theoretical Foundation & Formula Application
The relationship between the area, base, and height of a two-dimensional triangle is governed by the fundamental geometric theorem for Euclidean triangles [Per the standard area postulate for polygons]. The formula is expressed as:
$A = \frac{1}{2} \times b \times h$
This formula dictates that the area of a triangle is exactly half the area of a parallelogram that shares the same base and height.
Step 3: Algebraic Manipulation & Isolation of the Variable
To find the unknown height ($h$), we substitute the given values into the area formula:
$87 = \frac{1}{2} \times 15 \times h$
Next, we isolate $h$ by applying the multiplication property of equality. Multiplying both sides of the equation by $2$ to eliminate the fractional coefficient [By the axiom of equality]:
$87 \times 2 = 15 \times h$
$174 = 15 \times h$
Step 4: Final Calculation & Verification
Divide both sides by $15$ to solve for $h$:
$h = \frac{174}{15}$
Performing the division yields:
$h = 11.6 \text{ cm}$
To verify the result, we can substitute the height back into the original formula: $\frac{1}{2} \times 15 \times 11.6 = 7.5 \times 11.6 = 87 \text{ cm}^2$. The calculation is mathematically sound.
Step 5: Visual Representation
Below is a scaled geometric representation of the triangle, demonstrating the proportional relationship between the base, the altitude (height), and the enclosed area. The dimensions in the SVG are strictly scaled to the ratio of $15 : 11.6$.
Final Solution: The missing Height is $11.6 \text{ cm}$.
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(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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