Step 1: Given Variables & Initial Setup

To determine the area of the remaining aluminium sheet, we must first establish the dimensions of the primary geometric figures involved in the system. Let $s$ represent the side length of the square aluminium sheet, and $r$ represent the radius of the circular cutout.

• Side of the square sheet, $s = 6 \text{ cm}$
• Radius of the circular cutout, $r = 2 \text{ cm}$
• Constant approximation, $\pi = 3.14$

Step 2: Geometric Visualization

The following scale diagram illustrates the spatial relationship between the square sheet and the circular cutout. The dimensions are mapped to a precise coordinate system where $1 \text{ cm} = 40 \text{ units}$.

Step 3: Calculating the Area of the Square Sheet

The total area of the initial aluminium sheet is calculated using the standard area formula for a regular quadrilateral [Per Euclidean Geometry principles].

$A_{\text{square}} = s^2$
$A_{\text{square}} = (6 \text{ cm})^2$
$A_{\text{square}} = 36 \text{ cm}^2$

Step 4: Calculating the Area of the Circular Cutout

The area of the removed material is defined by the area of the circle. We apply the area formula for a circle, substituting the given approximation for $\pi$.

$A_{\text{circle}} = \pi r^2$
$A_{\text{circle}} = 3.14 \times (2 \text{ cm})^2$
$A_{\text{circle}} = 3.14 \times 4 \text{ cm}^2$
$A_{\text{circle}} = 12.56 \text{ cm}^2$

Step 5: Determining the Area of the Leftover Sheet

To find the area of the remaining aluminium, we subtract the area of the circular cutout from the total area of the square sheet [By the Additive Property of Area, which states that the area of a whole is equal to the sum of the areas of its non-overlapping parts].

$A_{\text{leftover}} = A_{\text{square}} - A_{\text{circle}}$
$A_{\text{leftover}} = 36 \text{ cm}^2 - 12.56 \text{ cm}^2$
$A_{\text{leftover}} = 23.44 \text{ cm}^2$

Summary of Area Calculations

Final Solution: The area of the left over aluminium sheet is $23.44 \text{ cm}^2$.

Given Variables & Initial Setup

We are tasked with determining the missing base dimension of a parallelogram given its height and total area. The known parameters are defined as follows:

• Height ($h$): $15 \text{ cm}$
• Area ($A$): $154.5 \text{ cm}^2$
• Base ($b$): Unknown

Theoretical Foundation

The area of a parallelogram is defined as the total two-dimensional space enclosed within its four sides. Geometrically, a parallelogram can be rearranged into a rectangle of the same base and height. Therefore, the governing formula for the area of a parallelogram is:

$A = b \times h$

[Per the geometric postulate of area equivalence under translation, the area of a parallelogram is strictly the product of its base and the corresponding perpendicular height].

Step 1: Algebraic Substitution

Substitute the given numerical values into the standard area formula:

$154.5 = b \times 15$

Step 2: Isolation of the Variable

To solve for the unknown base ($b$), we must isolate it on one side of the equation. We achieve this by dividing both sides of the equation by the height ($15$).

$b = \frac{154.5}{15}$

[Applying the Division Property of Equality, which states that dividing both sides of an equation by the same non-zero number preserves the equality].

Step 3: Arithmetic Computation & Decimal Manipulation

To perform the division with precision, we can eliminate the decimal in the numerator by multiplying both the numerator and the denominator by $10$:

$b = \frac{154.5 \times 10}{15 \times 10}$

$b = \frac{1545}{150}$

Now, simplify the fraction by dividing both terms by their common factors. First, divide by $5$:

$\frac{1545 \div 5}{150 \div 5} = \frac{309}{30}$

Next, divide by $3$:

$\frac{309 \div 3}{30 \div 3} = \frac{103}{10}$

Converting the simplified fraction back into a decimal yields:

$b = 10.3 \text{ cm}$

Geometric Visualization

Below is a scaled, mathematically accurate representation of the parallelogram, demonstrating the relationship between the base, the perpendicular height, and the enclosed area.

Final Verification

To ensure absolute accuracy, we substitute the calculated base back into the original area formula:

$A = 10.3 \text{ cm} \times 15 \text{ cm}$

$A = 154.5 \text{ cm}^2$

The calculated area matches the given area perfectly, confirming the validity of the derived base.

Final Solution: The missing value for the Base is $10.3 \text{ cm}$.

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 $80 \text{ mm}$.

Step 1: Given Variables & Initial Setup

To determine the area of the remaining card sheet, we must first establish the dimensions of the original geometric figure and the components that are removed. The problem provides the following parameters:

• Radius of the original large circular sheet, $R = 14 \text{ cm}$
• Radius of the two smaller removed circles, $r = 3.5 \text{ cm}$
• Length of the removed rectangle, $l = 3 \text{ cm}$
• Breadth of the removed rectangle, $b = 1 \text{ cm}$
• Constant for Pi, $\pi = \frac{22}{7}$

Step 2: Visual Representation of the Geometric Configuration

The following scale-accurate diagram illustrates the original circular sheet with the two smaller circles and the rectangle removed. [The diagram uses a scale factor of $10\times$ for visual clarity, where $14 \text{ cm}$ is represented by $140 \text{ units}$].

Step 3: Calculating the Area of the Original Circular Sheet

The area of a circle is given by the formula $A = \pi r^2$. Applying this to the large circular sheet:

$A_{\text{total}} = \pi R^2$
$A_{\text{total}} = \frac{22}{7} \times (14)^2$
$A_{\text{total}} = \frac{22}{7} \times 196$
$A_{\text{total}} = 22 \times 28$
$A_{\text{total}} = 616 \text{ cm}^2$

Step 4: Calculating the Area of the Removed Components

[Per the geometric principle of area addition, we must calculate the individual areas of all removed disjoint regions and sum them].

1. Area of the Two Small Circles:
First, we find the area of one small circle. To simplify the calculation, we can express the radius $3.5 \text{ cm}$ as the fraction $\frac{7}{2} \text{ cm}$.

$A_{\text{small\_circle}} = \pi r^2$
$A_{\text{small\_circle}} = \frac{22}{7} \times \left(\frac{7}{2}\right)^2$
$A_{\text{small\_circle}} = \frac{22}{7} \times \frac{49}{4}$
$A_{\text{small\_circle}} = \frac{22 \times 7}{4} = \frac{154}{4} = 38.5 \text{ cm}^2$

Since two identical circles are removed, their combined area is:

$A_{\text{both\_circles}} = 2 \times 38.5 \text{ cm}^2 = 77 \text{ cm}^2$

2. Area of the Rectangle:
The area of a rectangle is the product of its length and breadth ($A = l \times b$).

$A_{\text{rectangle}} = 3 \text{ cm} \times 1 \text{ cm} = 3 \text{ cm}^2$

3. Total Removed Area:
Summing the areas of the removed components yields:

$A_{\text{removed}} = A_{\text{both\_circles}} + A_{\text{rectangle}}$
$A_{\text{removed}} = 77 \text{ cm}^2 + 3 \text{ cm}^2 = 80 \text{ cm}^2$

Step 5: Calculating the Area of the Remaining Sheet

[By the axiom of area subtraction, the area of a region remaining after removing sub-regions is equal to the area of the original region minus the total area of the removed sub-regions].

$A_{\text{remaining}} = A_{\text{total}} - A_{\text{removed}}$
$A_{\text{remaining}} = 616 \text{ cm}^2 - 80 \text{ cm}^2$
$A_{\text{remaining}} = 536 \text{ cm}^2$

Final Solution: The area of the remaining sheet is $536 \text{ cm}^2$.

Given Variables & Initial Setup

We are given an isosceles triangle, $\triangle ABC$, with the following dimensional properties:

• Length of equal sides: $AB = AC = 7.5 \text{ cm}$
• Length of the base: $BC = 9 \text{ cm}$
• Altitude (height) corresponding to base $BC$: $AD = 6 \text{ cm}$

We are tasked with determining two values:

1. The total area of $\triangle ABC$.
2. The length of the altitude $CE$, which is the perpendicular dropped from vertex $C$ to the side $AB$.

Step 1: Calculating the Area of $\triangle ABC$

The fundamental theorem of Euclidean geometry states that the area of any triangle can be calculated using the length of any chosen base and its corresponding perpendicular height (altitude). The formula is given by:

$ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} $

Taking $BC$ as the base, the corresponding height is the altitude dropped from vertex $A$, which is $AD$. Substituting the given values:

• $\text{Base } (b_1) = BC = 9 \text{ cm}$
• $\text{Height } (h_1) = AD = 6 \text{ cm}$

$ \text{Area of } \triangle ABC = \frac{1}{2} \times BC \times AD $

$ \text{Area of } \triangle ABC = \frac{1}{2} \times 9 \text{ cm} \times 6 \text{ cm} $

$ \text{Area of } \triangle ABC = 9 \times 3 \text{ cm}^2 = 27 \text{ cm}^2 $

Step 2: Establishing the Invariance of Area to Find $CE$

[Per the invariant property of the area of a polygon], the area of a rigid geometric figure remains constant regardless of the perspective or the specific base-height pair used for calculation. Therefore, if we reorient our perspective and treat side $AB$ as the base, the corresponding height is the altitude $CE$.

We set up the area equation using the new base-height pair:

• $\text{Base } (b_2) = AB = 7.5 \text{ cm}$
• $\text{Height } (h_2) = CE \text{ (Unknown)}$

$ \text{Area of } \triangle ABC = \frac{1}{2} \times AB \times CE $

Step 3: Algebraic Isolation and Calculation of $CE$

Since the area is strictly invariant, we equate the expression from Step 2 to the calculated area from Step 1 ($27 \text{ cm}^2$):

$ 27 = \frac{1}{2} \times 7.5 \times CE $

To isolate $CE$, we multiply both sides of the equation by $2$:

$ 54 = 7.5 \times CE $

Next, divide both sides by $7.5$:

$ CE = \frac{54}{7.5} $

To simplify the division, multiply the numerator and the denominator by $10$ to eliminate the decimal:

$ CE = \frac{540}{75} $

Divide both the numerator and the denominator by their greatest common divisor, which is $15$:

$ CE = \frac{540 \div 15}{75 \div 15} = \frac{36}{5} $

Converting the improper fraction to a decimal:

$ CE = 7.2 \text{ cm} $

Final Solution: The area of $\triangle ABC$ is exactly $27 \text{ cm}^2$, and the height from $C$ to $AB$ (altitude $CE$) is $7.2 \text{ cm}$.

Given Variables & Initial Setup

Let us define the fundamental parameters of the circle based on the provided data:

• Circumference ($C$) = $31.4 \text{ cm}$
• Mathematical constant ($\pi$) $\approx 3.14$
• Radius ($r$) = ?
• Area ($A$) = ?

Step 1: Deriving the Radius from the Circumference

[Per the geometric definition of a circle's circumference], the total arc length (perimeter) of a circle is directly proportional to its radius. The governing equation is:

$C = 2\pi r$

By substituting the given values into the equation, we can isolate the unknown variable, $r$:

$31.4 = 2 \times 3.14 \times r$

$31.4 = 6.28 \times r$

Dividing both sides by $6.28$ to solve for $r$:

$r = \frac{31.4}{6.28}$

$r = 5 \text{ cm}$

Step 2: Geometric Visualization

The following diagram illustrates the spatial relationship between the center, the calculated radius, and the given circumference of the circle.

Step 3: Calculating the Area of the Circle

[Per the fundamental theorem of circle geometry], the area $A$ enclosed by a circle is proportional to the square of its radius. The formula is expressed as:

$A = \pi r^2$

Substitute the derived radius ($r = 5 \text{ cm}$) and the given value of $\pi$ ($3.14$) into the area formula:

$A = 3.14 \times (5)^2$

First, evaluate the exponent:

$A = 3.14 \times 25$

Next, perform the multiplication to find the final area:

$A = 78.5 \text{ cm}^2$

Final Solution: The radius of the circle is $5 \text{ cm}$ and the area of the circle is $78.5 \text{ cm}^2$.

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 $11.6 \text{ cm}$.

Given Variables & Initial Setup

We are given a single continuous wire of a fixed length that is bent into two different geometric shapes (a circle and a square) in separate scenarios. The fundamental principle governing this problem is the Conservation of Perimeter, which states that the total boundary length of any shape formed by the wire must equal the total length of the wire.

• Total length of the wire, $L = 44 \text{ cm}$
• Mathematical constant, $\pi = \frac{22}{7}$

Step 1: Analyzing the Circular Shape and Finding the Radius

When the wire is bent into the shape of a circle, the entire length of the wire forms the boundary of the circle. [Per the geometric definition of circumference, the perimeter of a circle is its circumference].

Let $r$ be the radius of the circle. The formula for the circumference $C$ is:

$C = 2\pi r$

Equating the circumference to the total length of the wire:

$2\pi r = 44$

Substituting the given value of $\pi$:

$2 \left( \frac{22}{7} \right) r = 44$

$\frac{44}{7} r = 44$

Isolating $r$ by multiplying both sides by $\frac{7}{44}$:

$r = 44 \times \frac{7}{44}$

$r = 7 \text{ cm}$

Step 2: Calculating the Area of the Circle

The area $A_{circle}$ enclosed by a circle is given by the formula:

$A_{circle} = \pi r^2$

Substituting $r = 7 \text{ cm}$ and $\pi = \frac{22}{7}$:

$A_{circle} = \frac{22}{7} \times (7)^2$

$A_{circle} = \frac{22}{7} \times 49$

$A_{circle} = 22 \times 7$

$A_{circle} = 154 \text{ cm}^2$

Step 3: Analyzing the Square Shape and Finding the Side Length

When the same wire is bent into the shape of a square, the total length of the wire forms the perimeter of the square. [Per the properties of regular polygons, a square has four sides of equal length].

Let $s$ be the side length of the square. The formula for the perimeter $P$ of a square is:

$P = 4s$

Equating the perimeter to the total length of the wire:

$4s = 44$

Isolating $s$ by dividing both sides by $4$:

$s = \frac{44}{4}$

$s = 11 \text{ cm}$

Step 4: Calculating the Area of the Square

The area $A_{square}$ enclosed by a square is given by the formula:

$A_{square} = s^2$

Substituting $s = 11 \text{ cm}$:

$A_{square} = (11)^2$

$A_{square} = 121 \text{ cm}^2$

Step 5: Visualizing and Comparing the Enclosed Areas

We now compare the calculated areas to determine which geometric figure maximizes the enclosed space for a given perimeter (an application of the Isoperimetric Inequality).

• Area of the Circle: $154 \text{ cm}^2$
• Area of the Square: $121 \text{ cm}^2$

Since $154 \text{ cm}^2 > 121 \text{ cm}^2$, the circle encloses a significantly larger area than the square.

Final Solution: The radius of the circle is $7 \text{ cm}$ and its area is $154 \text{ cm}^2$. When bent into a square, the length of each side is $11 \text{ cm}$ and its area is $121 \text{ cm}^2$. Comparing the two, the circle encloses more area than the square.

Given Variables & Initial Setup

We are analyzing a composite two-dimensional geometric system where a smaller circle is concentrically removed from a larger circular sheet. The parameters defining this system are:

• Radius of the outer circular sheet, $R = 4 \text{ cm}$
• Radius of the inner removed circle, $r = 3 \text{ cm}$
• Approximation for Pi, $\pi = 3.14$

Geometric Visualization

The figure below illustrates the concentric nature of the two circles. The region between the outer boundary and the inner boundary represents the remaining sheet, geometrically defined as an annulus.

Step 1: Formulating the Area Equation

To find the area of the remaining sheet, we must subtract the area of the inner circle from the area of the outer circular sheet. Let $A$ represent the area of the remaining sheet.

$A = \text{Area of Outer Circle} - \text{Area of Inner Circle}$

$A = \pi R^2 - \pi r^2$ [Per the standard area formula for a circle, $Area = \pi \times \text{radius}^2$]

To optimize the calculation, we factor out the common constant $\pi$:

$A = \pi (R^2 - r^2)$ [By the distributive property of multiplication over subtraction]

Step 2: Algebraic Substitution

Substitute the given numerical values into the factored equation:

$A = 3.14 \times (4^2 - 3^2)$

Step 3: Evaluating the Expression

First, resolve the exponents inside the parentheses:

• $4^2 = 4 \times 4 = 16$
• $3^2 = 3 \times 3 = 9$

Substitute these squared values back into the equation:

$A = 3.14 \times (16 - 9)$

Perform the subtraction operation:

$A = 3.14 \times 7$

Step 4: Final Computation

Multiply the remaining terms to find the final area:

$A = 21.98$

Since the radii were provided in centimeters ($\text{cm}$), the resulting area must be expressed in square centimeters ($\text{cm}^2$) [Per the dimensional analysis of area, $L \times L = L^2$].

Final Solution: The area of the remaining sheet is $21.98 \text{ cm}^2$.

Given Variables & Initial Setup

We are given the following geometric parameters for a circle:

• Radius ($r$) = $14 \text{ mm}$
• Mathematical constant ($\pi$) = $\frac{22}{7}$

Theoretical Foundation & Formula

The area ($A$) of a circle represents the total two-dimensional space enclosed within its boundary (circumference). [Per the geometric principles of Euclidean space], the area of a circle is directly proportional to the square of its radius. The formula is given by:

$A = \pi r^2$

Step 1: Substituting the Values into the Equation

To find the area, we substitute the given radius $r = 14 \text{ mm}$ and the fractional approximation of $\pi = \frac{22}{7}$ into the standard area formula:

$A = \left(\frac{22}{7}\right) \times (14 \text{ mm})^2$

Step 2: Algebraic Manipulation and Simplification

First, expand the squared term to separate the numerical values from the units:

$A = \frac{22}{7} \times (14 \times 14) \text{ mm}^2$

$A = \frac{22}{7} \times 196 \text{ mm}^2$

To optimize the calculation without dealing with large numbers, we can simplify the expression by dividing one of the $14$ factors by the denominator $7$ [By the fundamental property of fractions and associative multiplication]:

$A = 22 \times \left(\frac{14}{7}\right) \times 14 \text{ mm}^2$

$A = 22 \times 2 \times 14 \text{ mm}^2$

Now, proceed with sequential multiplication:

$A = (22 \times 2) \times 14 \text{ mm}^2$

$A = 44 \times 14 \text{ mm}^2$

Perform the final multiplication step:

$A = 44 \times (10 + 4) \text{ mm}^2$

$A = 440 + 176 \text{ mm}^2$

$A = 616 \text{ mm}^2$

Step 3: Dimensional Analysis

The radius is given in millimeters ($\text{mm}$). When the radius is squared in the formula ($r^2$), the unit is also squared ($\text{mm} \times \text{mm} = \text{mm}^2$). This confirms that our final unit correctly represents a two-dimensional area.

Final Solution: The area of the circle is $616 \text{ mm}^2$.