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Q4(b):
Find the missing values: Base = ______, Height = $31.4$ mm, Area of Triangle = $1256$ mm$^2$.
Solution :
Given Variables & Initial Setup
We are tasked with determining the missing base dimension of a triangle given its height and total area. The known parameters are defined as follows:
- Height ($h$): $31.4 \text{ mm}$
- Area ($A$): $1256 \text{ mm}^2$
- Base ($b$): Unknown
Step 1: Stating the Geometric Formula
The relationship between the area, base, and height of any planar triangle is governed by the standard area formula [Per the fundamental theorems of Euclidean geometry]:
$A = \frac{1}{2} \times b \times h$
Step 2: Algebraic Substitution and Simplification
Substitute the given numerical values into the area formula to establish the equation for the unknown base ($b$):
$1256 = \frac{1}{2} \times b \times 31.4$
Next, simplify the right side of the equation by dividing the height by $2$:
$1256 = b \times \left( \frac{31.4}{2} \right)$
$1256 = b \times 15.7$
Step 3: Isolating the Unknown Variable
To solve for $b$, isolate the variable by dividing both sides of the equation by $15.7$ [By the Division Property of Equality]:
$b = \frac{1256}{15.7}$
To perform the division with precision, eliminate the decimal in the denominator by multiplying both the numerator and the denominator by $10$:
$b = \frac{12560}{157}$
By analyzing the numbers, we can test multiples of $157$. Notice that $157 \times 8 = 1256$. Therefore, multiplying by $80$ yields:
$157 \times 80 = 12560$
Thus, the exact value of the base is:
$b = 80 \text{ mm}$
Geometric Visualization
Below is a strictly scaled geometric representation of the triangle. The SVG coordinates are mathematically calculated to maintain the exact ratio of the base ($80 \text{ mm}$) to the height ($31.4 \text{ mm}$), utilizing a scale factor of $4 \text{ pixels per mm}$ (Base $= 320\text{px}$, Height $= 125.6\text{px}$).
Final Solution: The missing base of the triangle is $80 \text{ mm}$.
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(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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