Measuring Space: Perimeter and Area — Notes

Perimeter of a Shape

The perimeter of a shape is the total distance around its boundary or outline. Imagine an ant crawling along the edge of a shape until it returns to its starting point; the total distance the ant travels is the perimeter.

Key Concepts:

• Boundary: The outer edge or limit of a shape.
• Units: Perimeter is measured in linear units (e.g., metres, centimetres, inches, feet).

Perimeter of Polygons

For polygons (shapes with straight sides), the perimeter is the sum of the lengths of all its sides.

Perimeter of a Square

A square has four equal sides. If the length of one side is '$a$' units, then the perimeter of the square is:

Perimeter of a Rectangle

A rectangle has two pairs of equal sides: length ('$l$') and width ('$w$'). The perimeter of a rectangle is:

• Relationship between Square and Rectangle Perimeter: The formula for the perimeter of a square is a special case of the rectangle formula where the length equals the width ($l=w=a$).

Example 1: Perimeter of a Rectangle

Given: A rectangular garden has a length of 15 metres and a width of 8 metres. To find: The perimeter of the garden.

Solution:

1. Identify the formula for the perimeter of a rectangle: $P=2(l+w)$.
2. Substitute the given length ($l=15$ m) and width ($w=8$ m) into the formula: $P=2(15\text{ m}+8\text{ m})$.
3. Calculate the sum inside the parentheses: $P=2(23\text{ m})$.
4. Multiply by 2 to get the perimeter: $P=46\text{ m}$.

Answer: The perimeter of the garden is 46 metres.

Example 2: Perimeter of a Square

Given: A square park has a side length of 50 metres. To find: The perimeter of the park.

Solution:

1. Identify the formula for the perimeter of a square: $P=4a$.
2. Substitute the given side length ($a=50$ m) into the formula: $P=4\times 50\text{ m}$.
3. Calculate the perimeter: $P=200\text{ m}$.

Answer: The perimeter of the park is 200 metres.

Example 3: Word Problem

Given: The perimeter of a rectangular field is 100 metres. Its width is 20 metres. To find: The length of the field.

Solution:

1. Recall the formula for the perimeter of a rectangle: $P=2(l+w)$.
2. Substitute the given perimeter ($P=100$ m) and width ($w=20$ m): $100\text{ m}=2(l+20\text{ m})$.
3. Divide both sides by 2: $\frac{100\text{ m}}{2}=l+20\text{ m}$.
4. Simplify: $50\text{ m}=l+20\text{ m}$.
5. Subtract 20 m from both sides to find the length: $l=50\text{ m}-20\text{ m}$.
6. Calculate the length: $l=30\text{ m}$.

Answer: The length of the field is 30 metres.

Why Perimeter Matters

• Fencing: To calculate the amount of fencing needed to enclose a garden, field, or yard.
• Running Tracks: To design athletic tracks of specific lengths. The mention of a 400 m athletics track highlights how perimeter calculations are crucial for sports infrastructure, ensuring fairness for all athletes by managing the distance they run in different lanes (this is related to the "stagger" mentioned in the reference).
• Framing: To determine the length of material needed to frame a picture or a window.
• Borders: To calculate the length of decorative borders for walls or tablecloths.

Remember

Perimeter is a one-dimensional measurement (length). It measures only the distance around the outside, not the space enclosed within.

Exam-smart

You might be given the perimeter and one dimension of a rectangle and asked to find the other, or given the side of a square and asked for its perimeter. Be careful with units. Case studies might involve calculating fencing for a rectangular plot.

Perimeter of a Circle — The C / D Ratio

Ancient mathematicians observed that for any circle, the ratio of its circumference (the distance around the circle, often denoted by $C$) to its diameter (the distance across the circle through its center, denoted by $D$) is always constant, regardless of the circle's size. This constant ratio is fundamental to understanding circles.

The C/D Ratio: Introducing Pi ($\pi$)

This constant ratio, $C\mathrm{/}D$, is a special mathematical constant we represent with the Greek letter $\pi$ (pi). It is pronounced like "pie".

This means that the circumference of any circle can be calculated if you know its diameter, using the formula:

Why this matters: This relationship is crucial because it links the linear measurement of the diameter to the curved measurement of the circumference. It's a bridge between straight lines and curves, and it appears in many areas of mathematics and science.

Estimating Pi ($\pi$)

You can get a sense of the value of $\pi$ through simple experiments.

Method:

1. Take an object with a circular base (like a tin can or a cotton reel).
2. Measure its diameter ($D$) accurately using a ruler.
3. Take a piece of string and carefully wrap it around the circumference of the object exactly once.
4. Unwrap the string and measure its length ($C$).
5. Calculate the ratio $C\mathrm{/}D$.

You will find that this ratio is consistently between 3 and 4, and closer to 3.1.

Historical Approximations of Pi ($\pi$)

Mathematicians have sought to find increasingly accurate values for $\pi$ throughout history.

The discovery of infinite series by Mādhava was a revolutionary step. It moved from approximating $\pi$ geometrically to calculating it analytically, leading to the development of calculus.

Symbol Usage

The symbol $\pi$ was first used to denote the C/D ratio by William Jones in 1706. It was popularized by the Swiss German mathematician Leonhard Euler.

Diameter and Radius

The radius ($r$) of a circle is the distance from the center to any point on the circumference. The diameter is twice the radius: $D=2r$.

This gives us an alternative formula for the circumference:

Example 1: Given: A circle has a diameter of 14 cm. To find: The circumference of the circle. Solution: The formula for circumference is $C=\pi D$. We are given $D=14$ cm. So, $C=\pi \times 14$ cm. Using the approximation $\pi \approx \frac{22}{7}$: $C=\frac{22}{7}\times 14$ cm. $C=22\times \frac{14}{7}$ cm. $C=22\times 2$ cm. $C=44$ cm. Answer: The circumference is 44 cm.

Example 2: Given: The circumference of a circular park is 132 meters. To find: The radius of the park. Solution: The formula for circumference is $C=2\pi r$. We are given $C=132$ m. So, $132=2\pi r$. To find $r$, rearrange the formula: $r=\frac{C}{2\pi }$. Using the approximation $\pi \approx \frac{22}{7}$: $r=\frac{132}{2\times \frac{22}{7}}$ m. $r=\frac{132}{\frac{44}{7}}$ m. $r=132\times \frac{7}{44}$ m. $r=\frac{132}{44}\times 7$ m. Since $132=3\times 44$, $r=3\times 7$ m. $r=21$ m. Answer: The radius of the park is 21 meters.

Remember

The value of $\pi$ is an irrational number, meaning it cannot be expressed as a simple fraction and its decimal representation goes on forever without repeating. We use approximations like $\frac{22}{7}$ or $3.14$ for practical calculations.

Exam-smart

Questions might provide the diameter or radius and ask for the circumference, or provide the circumference and ask for the diameter or radius. Always check which value is given and which is asked for. You might also be asked to use a specific approximation for $\pi$ (like $\frac{22}{7}$ or $3.14$), or to leave the answer in terms of $\pi$.

Try this: The diameter of a bicycle wheel is 70 cm. Calculate its circumference. Answer: 220 cm.

π Is Irrational

You've learned that π (pi) is the ratio of a circle's circumference to its diameter ($\pi =\frac{C}{D}$). This ratio is a constant value for all circles. However, the true nature of π is that its decimal representation goes on forever without repeating, meaning it cannot be expressed as a simple fraction of two whole numbers.

Understanding Rational and Irrational Numbers

• A rational number can be expressed as a fraction $\frac{a}{b}$, where 'a' and 'b' are integers and $b\mathrm{\ne }0$. The decimal expansions of rational numbers either terminate (like $\frac{1}{4}=0.25$) or repeat in a predictable pattern (like $\frac{1}{3}=\mathrm{0.333...}$ or $\frac{1}{7}=\mathrm{0.142857142857...}$).

• An irrational number cannot be expressed as a simple fraction $\frac{a}{b}$. Its decimal expansion is non-terminating and non-repeating.

Key Fact: The number π is irrational. This means its decimal digits continue infinitely without any repeating sequence.

Historical Context

Mathematicians like Āryabhaṭa and Zu Chongzhi suspected π was irrational. Later, in 1761, Johann Lambert proved that π is indeed irrational using advanced mathematical techniques.

Approximations of π

Because π is irrational, we can only use approximations for practical calculations.

• The fraction $\frac{22}{7}$ is a common approximation for π. However, it's crucial to remember that π is not equal to $\frac{22}{7}$. We use the symbol $\approx$ to show it's an approximation: $\pi \approx \frac{22}{7}$.

• Another commonly used approximation is 3.14. Again, this is an approximation: $\pi \approx 3.14$.

• A more accurate approximation for π is $\frac{355}{113}$.

Example 1: Using Approximations

Given: The radius of a circle is 7 cm. To find: The circumference of the circle using the approximation $\pi \approx \frac{22}{7}$.

Solution: The formula for the circumference of a circle is $C=2\pi r$. Given radius, $r=7$ cm. Using the approximation $\pi \approx \frac{22}{7}$:

$C\approx 44$ cm. Answer: The circumference is approximately 44 cm.

Example 2: Using a Different Approximation

Given: The diameter of a circular garden is 14 metres. To find: The circumference of the garden using the approximation $\pi \approx 3.14$.

Solution: The formula for the circumference of a circle is $C=\pi D$. Given diameter, $D=14$ metres. Using the approximation $\pi \approx 3.14$:

To calculate $3.14\times 14$:

$C\approx 43.96$ metres. Answer: The circumference is approximately 43.96 metres.

Why it Matters

The irrational nature of π means that any calculation involving π will also result in an irrational number unless we use an approximation. For most everyday purposes, approximations like $\frac{22}{7}$ or 3.14 are sufficient. However, in advanced scientific and engineering fields, more precise approximations are used to ensure accuracy.

Fun Fact: The digits of π can be remembered using the phrase: "How I wish I could recollect pi." Counting the letters in each word gives you the first few digits: 3, 1, 4, 1, 5, 9, 2. This gives $\pi \approx 3.141592$.

Pi Day: March 14th (3-14) is celebrated as Pi Day worldwide, and July 22nd (22-7) is celebrated as Pi Approximation Day.

Remember

π is an irrational number, meaning it cannot be written as a simple fraction $\frac{a}{b}$ where a and b are integers. Its decimal expansion is infinite and non-repeating.

Exam-smart

You might be asked to identify whether a number is rational or irrational. If given a decimal that goes on forever without a repeating pattern, it's irrational. You might also be asked to calculate circumference or area using a given approximation for π (like 3.14 or $\frac{22}{7}$), and you'll need to use the correct formula and perform the calculation carefully.

Try this: Calculate the circumference of a circle with a radius of 3.5 cm, using the approximation $\pi \approx \frac{22}{7}$. Answer: 22 cm

Length of an Arc of a Circle

The circumference of a circle is the total distance around its boundary. It is given by the formula $C=\pi d$ or $C=2\pi r$, where $d$ is the diameter and $r$ is the radius of the circle.

An arc of a circle is a portion of its circumference.

Semicircle Arc Length

A semicircle is half of a circle, formed by cutting the circle along its diameter. If a circle has radius $r$, its full circumference is $2\pi r$. The length of a semicircle's arc is half of the full circumference.

Formula for semicircle arc length: Arc Length (semicircle) = $\frac{1}{2}\times 2\pi r=\pi r$

This can also be thought of as rotating the circle by ${180}^{\circ }$ to form a semicircle.

Example 1

Given: A circle with radius 7 cm. To find: The length of its semicircular arc. Solution: The radius of the circle is $r=7$ cm. The formula for the length of a semicircular arc is $\pi r$. We use the approximation $\pi \approx \frac{22}{7}$. Arc Length = $\frac{22}{7}\times 7$ cm Arc Length = $22$ cm Answer: The length of the semicircular arc is 22 cm.

Quarter Circle Arc Length

A quarter circle is one-fourth of a circle, formed by dividing the circle by two perpendicular diameters. The length of a quarter circle's arc is one-fourth of the full circumference.

Formula for quarter circle arc length: Arc Length (quarter circle) = $\frac{1}{4}\times 2\pi r=\frac{\pi r}{2}$

This can be visualized by rotating the circle by ${90}^{\circ }$.

Example 2

Given: A circle with diameter 28 cm. To find: The length of its quarter circular arc. Solution: The diameter of the circle is $d=28$ cm. The radius of the circle is $r=\frac{d}{2}=\frac{28}{2}=14$ cm. The formula for the length of a quarter circular arc is $\frac{\pi r}{2}$. We use the approximation $\pi \approx \frac{22}{7}$. Arc Length = $\frac{1}{2}\times \frac{22}{7}\times 14$ cm Arc Length = $\frac{1}{2}\times 22\times 2$ cm Arc Length = $22$ cm Answer: The length of the quarter circular arc is 22 cm.

General Formula for Arc Length

For any arc of a circle, its length depends on the angle it subtends at the centre of the circle. The angle subtended is the angle formed by the two radii connecting the endpoints of the arc to the centre.

Let the angle subtended by the arc at the centre be ${\theta }^{\circ }$. The full circle has an angle of ${360}^{\circ }$ at its centre. The arc length is a fraction of the total circumference, determined by the ratio of the subtended angle to ${360}^{\circ }$.

Formula for the length of an arc: Arc Length = $\frac{\theta }{{360}^{\circ }}\times 2\pi r$

Where:

• $\theta$ is the angle subtended by the arc at the centre in degrees.
• $r$ is the radius of the circle.
• $2\pi r$ is the circumference of the circle.

Example 3

Given: An arc of a circle with radius 10 cm subtends an angle of ${60}^{\circ }$ at the centre. To find: The length of the arc. Solution: Radius, $r=10$ cm. Angle subtended, $\theta ={60}^{\circ }$. Circumference of the circle = $2\pi r=2\times \pi \times 10=20\pi$ cm. Arc Length = $\frac{{60}^{\circ }}{{360}^{\circ }}\times 20\pi$ cm Arc Length = $\frac{1}{6}\times 20\pi$ cm Arc Length = $\frac{10\pi }{3}$ cm Using $\pi \approx 3.14$: Arc Length $\approx \frac{10\times 3.14}{3}$ cm Arc Length $\approx \frac{31.4}{3}$ cm Arc Length $\approx 10.47$ cm Answer: The length of the arc is $\frac{10\pi }{3}$ cm (or approximately 10.47 cm).

Example 4

Given: The length of an arc of a circle of radius 21 cm is 11 cm. To find: The angle subtended by the arc at the centre. Solution: Radius, $r=21$ cm. Arc Length = 11 cm. Let the angle subtended be ${\theta }^{\circ }$. Using the formula: Arc Length = $\frac{\theta }{{360}^{\circ }}\times 2\pi r$

Answer: The angle subtended by the arc at the centre is ${30}^{\circ }$.

Remember

The formula for arc length is essentially a fraction of the circumference, determined by the angle the arc makes at the centre. Ensure the angle is in degrees if using the $\frac{\theta }{{360}^{\circ }}$ ratio.

Exam-smart

You might be given the radius and angle and asked for the arc length, or given the radius and arc length and asked for the angle. Always check the units of the angle (degrees or radians) and use the appropriate formula. Sometimes, you might need to find the radius using the arc length and angle.

A Closer Look at a 400 m Athletics Track

An athletics track typically consists of two straight sections and two semicircular ends. A standard 400 m track is designed such that if a runner stays exactly 0.3 m away from the inner edge (lane line), they will cover a distance of 400 m in one lap.

• Straight sections: Each lane has two straight sections of length 84.39 m. Total length of straight sections = $2\times 84.39$ m = $168.78$ m.
• Curved sections: The curved ends are semicircles. The innermost lane has a radius of 36.5 m for its inner boundary. If a runner runs 0.3 m from this inner boundary, their running path has a radius of $36.5+0.3=36.8$ m. The two semicircular ends together form a full circle. Circumference of the two curved sections = $2\pi r=2\times \pi \times 36.8$ m. Using $\pi \approx 3.1416$: Circumference $\approx 2\times 3.1416\times 36.8$ m $\approx 231.22$ m.
• Total distance: The total distance covered by the runner in the innermost lane is the sum of the lengths of the straight and curved sections. Total distance = $168.78$ m + $231.22$ m = $400$ m.

Difference in radius between lanes: The width of each lane is 1.22 m. If the inner radius of lane 1 is $r_1=36.5$ m, the inner radius of lane 2 is $r_2=r_1+1.22=36.5+1.22=37.72$ m. The difference in radius between the first and second lanes is $1.22$ m.

Stagger: Runners in outer lanes have to cover a larger distance on the curves because their semicircles have a larger radius. To ensure fairness, the starting positions for runners in outer lanes are staggered. For the runner in the second lane, the radius of their running path (0.3 m from the inner boundary of lane 2) is $36.8+1.22=38.02$ m. The extra distance covered by the runner in the second lane compared to the first lane in one curved end (semicircle) is: Extra arc length = $\pi r_{lane2}-\pi r_{lane1}$ Extra arc length = $\pi (38.02)-\pi (36.8)=\pi (38.02-36.8)=\pi (1.22)$ m. This extra distance needs to be covered in the curved portion. The stagger is the extra distance the runner in the outer lane starts ahead of the runner in the inner lane. Stagger for lane 2 = Extra distance in one semicircle = $\frac{1}{2}\times \pi (1.22\times 2)$ m = $\pi \times 1.22$ m. Using $\pi \approx 3.1416$: Stagger $\approx 3.1416\times 1.22\approx 3.83$ m.

An equal stagger is needed between the third and second lanes, and so on for subsequent lanes, because the radius difference between consecutive lanes is constant.

Try this:

An arc of a circle of radius 14 cm subtends an angle of ${90}^{\circ }$ at the centre. Calculate the length of the arc. Answer: 22 cm

Problems, Puzzles, and Paradoxes on Perimeter

This section explores interesting problems and puzzles related to perimeter, demonstrating how seemingly complex shapes can have surprisingly simple perimeter calculations. It also introduces the concept of the area of a rectangle.

Area of a Rectangle

The area of a two-dimensional shape is the amount of space it occupies in a plane. The standard unit for measuring area is a square with sides of length 1 unit, referred to as 1 square unit (or $1{\text{unit}}^2$).

For a rectangle with length $l$ units and breadth $b$ units, its area is calculated by multiplying its length by its breadth.

Example 1:

A rectangular garden has a length of 20 meters and a breadth of 15 meters. Calculate its area.

Given: Length of the garden, $l=20\text{m}$ Breadth of the garden, $b=15\text{m}$

To find: Area of the garden

Solution:

Answer: The area of the rectangular garden is $300{\text{m}}^2$.

Example 2:

A square park has a side of 50 meters. What is its area?

Given: Side of the square park, $s=50\text{m}$

To find: Area of the square park

Solution: A square is a special type of rectangle where length and breadth are equal. So, for a square with side $s$, the area is $s\times s=s^2$.

Answer: The area of the square park is $2500{\text{m}}^2$.

Remember

The units of area are always squared units (e.g., $m^2$, $c{m}^2$, $f{t}^2$).

Exam-smart

You might be given the perimeter and one dimension of a rectangle and asked to find the area, or given the area and one dimension and asked to find the other. Always calculate the missing dimension first if needed.

Example 1: Perimeter of Intersecting Circles

Given: Two circles of equal radius $r$. Each circle passes through the center of the other. The shape formed by the two circles (excluding the overlapping dotted portions) is shown in Figure 6.12.

To find: The perimeter of the shape formed by the two circles in terms of $r$.

Solution: Let the centers of the two circles be A and B. Let the circles intersect at points C and D. Consider the triangle ABC. AB is the distance between the centers, which is $r$ (since each circle passes through the center of the other). AC is the radius of the first circle, so $AC=r$. BC is the radius of the second circle, so $BC=r$. Since all three sides of triangle ABC are equal to $r$, it is an equilateral triangle. Therefore, $\mathrm{\angle }CAB={60}^{\circ }$. Similarly, in triangle ABD, $\mathrm{\angle }DAB={60}^{\circ }$. The angle subtended by the arc CD at the center A is $\mathrm{\angle }CAD=\mathrm{\angle }CAB+\mathrm{\angle }DAB={60}^{\circ }+{60}^{\circ }={120}^{\circ }$. The portion of the circumference of the first circle that forms the perimeter of the shape is the arc CD. The length of an arc is given by $\frac{\theta }{{360}^{\circ }}\times 2\pi r$, where $\theta$ is the angle at the center. Length of arc CD (on the first circle) = $\frac{{120}^{\circ }}{{360}^{\circ }}\times 2\pi r=\frac{1}{3}\times 2\pi r=\frac{2\pi r}{3}$. Similarly, the angle subtended by the arc CD at the center B is $\mathrm{\angle }CBD={120}^{\circ }$. Length of arc CD (on the second circle) = $\frac{{120}^{\circ }}{{360}^{\circ }}\times 2\pi r=\frac{1}{3}\times 2\pi r=\frac{2\pi r}{3}$. The total perimeter of the shape is the sum of the lengths of these two arcs. Total Perimeter = $\frac{2\pi r}{3}+\frac{2\pi r}{3}=\frac{4\pi r}{3}$.

Answer: The perimeter of the shape formed by the two circles is $\frac{4\pi r}{3}$ units.

Remember

For shapes formed by intersecting circles, identify the relevant arcs and the angles they subtend at the centers.

Exam-smart

Questions may involve diagrams where you need to deduce that certain triangles are equilateral or isosceles to find central angles. Be careful to only include the specified arcs in the perimeter calculation.

Example 2: Comparing Paths of Equal Length

Given: Figure 6.13 shows two paths connecting points P and Q. Path 'a' is a single semicircle. Path 'b', 'c', and 'd' are three semicircles.

To find: Which path is longer: Path 'a' or Path 'b' + 'c' + 'd'.

Solution: Let the diameter of semicircle 'a' be $PQ$. Let its radius be $r_a$. The length of Path 'a' is the circumference of this semicircle: $L_a=\pi r_a$. The diameter $PQ$ is also equal to the sum of the diameters of semicircles 'b', 'c', and 'd'. Let the radii of semicircles 'b', 'c', and 'd' be $r_b$, $r_c$, and $r_d$, respectively. The diameter of 'b' is $2r_b$, of 'c' is $2r_c$, and of 'd' is $2r_d$. So, $PQ=2r_a=2r_b+2r_c+2r_d$. Dividing by 2, we get $r_a=r_b+r_c+r_d$.

The length of Path 'b' + 'c' + 'd' is the sum of the circumferences of these three semicircles: $L_{bcd}=\pi r_b+\pi r_c+\pi r_d=\pi (r_b+r_c+r_d)$.

Since $r_a=r_b+r_c+r_d$, we can substitute this into the equation for $L_a$: $L_a=\pi r_a=\pi (r_b+r_c+r_d)$.

Comparing $L_a$ and $L_{bcd}$:

Therefore, $L_a=L_{bcd}$.

Answer: The two paths have equal length.

Remember

This is an illustration of the Archimedes' twin paradox (or similar geometrical paradoxes), showing that a longer path composed of smaller segments can have the same length as a shorter, direct path. The key is that the sum of the diameters of the smaller semicircles equals the diameter of the larger semicircle.

Exam-smart

Such questions often involve visual illusions or counter-intuitive results. Focus on breaking down the shapes into their component parts and using the formulas for lengths of semicircles or full circles. Algebraic representation is crucial.

Area of a Parallelogram

The area of a parallelogram is the measure of the two-dimensional space it occupies. It is calculated using its base and its height.

A parallelogram is a quadrilateral with two pairs of parallel sides. In a parallelogram, opposite sides are equal in length, and opposite angles are equal.

The base of a parallelogram is any one of its sides. The height is the perpendicular distance from the base to the opposite side. It is crucial that the height is measured at a 90-degree angle to the base.

The formula for the area of a parallelogram is derived from the area of a rectangle. Imagine cutting a triangular piece from one side of a parallelogram and moving it to the other side to form a rectangle. This transformation results in a rectangle with the same base and the same height as the original parallelogram. Since the area of a shape is conserved when it is rearranged without overlapping, the area of the parallelogram is equal to the area of this equivalent rectangle.

The area of a rectangle is given by length × width. In the context of a parallelogram, this translates to base × height.

Where:

• $A$ is the area of the parallelogram.
• $b$ is the length of the base.
• $h$ is the perpendicular height.

Example 1: Finding the Area of a Parallelogram

Example

Given: A parallelogram has a base of 10 cm and a height of 7 cm. To find: The area of the parallelogram. Solution: The formula for the area of a parallelogram is $A=b\times h$. Substitute the given values into the formula:

Answer: The area of the parallelogram is $70{\text{cm}}^2$.

Example 2: Word Problem Involving Parallelogram Area

Example

Given: A farmer has a field in the shape of a parallelogram. The length of one side of the field is 50 meters, and the perpendicular distance to the opposite side is 30 meters. To find: The area of the farmer's field. Solution: The side of the field can be considered the base ($b$), and the perpendicular distance is the height ($h$).

The formula for the area of a parallelogram is $A=b\times h$. Substitute the values:

Answer: The area of the farmer's field is $1500{\text{m}}^2$.

Consideration for 'Thin' Parallelograms

Sometimes, the foot of the perpendicular from one vertex to the line containing the base might fall outside the base segment itself, especially in 'thin' or skewed parallelograms (Fig. 6.18). This occurs when the angle between the base and the adjacent side is obtuse. However, the formula $A=b\times h$ remains valid. The height $h$ is always the perpendicular distance, regardless of whether it falls within the base segment. The transformation to a rectangle can still be visualized by extending the base and using the perpendicular height.

The area of a parallelogram depends on its base and height, not solely on the lengths of its sides. If the lengths of the sides are fixed, changing the angle between them will alter the height and thus the area. A larger angle (closer to 90 degrees) will generally result in a larger height and area, while a smaller angle will result in a smaller height and area.

Remember

The height of a parallelogram must always be the perpendicular distance between the base and its opposite side. Using the length of the adjacent slanted side instead of the height will lead to an incorrect area.

Exam-smart

In a source-based question, you might be given a diagram of a parallelogram with its base and height clearly marked, or a word problem describing a scenario. You'll need to identify the correct base and height values from the given information and apply the formula $A=b\times h$. Be careful to distinguish the height from the length of the slanted side.

Try this: A parallelogram has a base of 15 cm and a height of 8 cm. Calculate its area. Answer: $120{\text{cm}}^2$

Area of a Triangle

The area of a triangle is the measure of the space enclosed within its three sides. We can derive the formula for the area of a triangle by understanding its relationship with a parallelogram.

Deriving the Area Formula

• Congruent Triangles Form a Parallelogram: If we take two identical (congruent) copies of a triangle, we can arrange them to form a parallelogram. For example, if triangle ABC is congruent to triangle A'B'C', they can be fitted together to create a parallelogram.

• Area of Parallelogram: We already know that the area of a parallelogram is given by the product of its base and its height:

where $b$ is the length of the base and $h$ is the perpendicular height from the base to the opposite vertex.

• Area of a Triangle: Since two congruent triangles form a parallelogram, the area of one triangle is exactly half the area of the parallelogram it forms.

Here, $b$ is the length of the base of the triangle, and $h$ is the perpendicular height of the triangle from the base to the opposite vertex.

Example 1

Given: A triangle with a base of 10 cm and a perpendicular height of 6 cm. To find: The area of the triangle. Solution: The formula for the area of a triangle is $Area=\frac{1}{2}\times base\times height$.

Answer: The area of the triangle is $30{\text{cm}}^2$.

Example 2

Given: A right-angled triangle with sides forming the right angle measuring 8 cm and 15 cm. To find: The area of the triangle. Solution: In a right-angled triangle, one of the sides forming the right angle can be considered the base, and the other side is the perpendicular height. Let base $b=15\text{cm}$ and height $h=8\text{cm}$. The formula for the area of a triangle is $Area=\frac{1}{2}\times b\times h$.

Answer: The area of the triangle is $60{\text{cm}}^2$.

Property of a Median

A median of a triangle is a line segment drawn from a vertex to the midpoint of the opposite side.

• Theorem: A median of a triangle divides it into two triangles of equal area.
• Explanation: Consider a triangle ABC, and let AD be its median. This means D is the midpoint of BC, so $BD=DC$. Both triangle ABD and triangle ACD share the same perpendicular height from vertex A to the base BC. Since their bases ($BD$ and $DC$) are equal and they have the same height, their areas are equal.

Since $BD=DC$, it follows that $Are{a}_{\mathrm{△}ABD}=Are{a}_{\mathrm{△}ACD}$.

Example 3

Given: In triangle PQR, PS is a median. The area of triangle PQS is $25{\text{cm}}^2$. To find: The area of triangle PRS. Solution: Since PS is a median, it divides triangle PQR into two triangles of equal area: triangle PQS and triangle PRS. Therefore, $Are{a}_{\mathrm{△}PQS}=Are{a}_{\mathrm{△}PRS}$. Given $Are{a}_{\mathrm{△}PQS}=25{\text{cm}}^2$. So, $Are{a}_{\mathrm{△}PRS}=25{\text{cm}}^2$. Answer: The area of triangle PRS is $25{\text{cm}}^2$.

Remember

The height of a triangle must always be the perpendicular distance from the base to the opposite vertex.

Exam-smart

Questions might involve finding the area given base and height, or finding one of them if the area is known. Also, problems may test the understanding that a median divides a triangle into two equal areas. You might be given the area of the whole triangle and asked for the area of one of the smaller triangles formed by a median, or vice-versa.

Heron’s Formula

Heron's formula allows us to calculate the area of a triangle when we only know the lengths of its three sides. This is particularly useful when we don't know the height of the triangle, which is often required for the basic area formula ($\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}$).

The formula was discovered by Heron of Alexandria.

The Formula:

For a triangle with side lengths $a$, $b$, and $c$:

1. Calculate the semi-perimeter ($s$): The semi-perimeter is half the perimeter of the triangle.

2. Apply Heron's Formula: The area (often denoted as $A$ or $\mathrm{△}$) is given by:

Why it Works (Conceptual Understanding):

Heron's formula is derived from the general area formula of a triangle by using trigonometry and algebraic manipulation, specifically the Pythagorean theorem and the difference of squares identity ($x^2-y^2=(x-y)(x+y)$). While the proof is complex and usually studied in higher classes, the formula itself provides a direct link between the side lengths and the area.

Key Terms:

• Semi-perimeter ($s$): Half the sum of the lengths of the sides of a polygon.
• Heron's Formula: A formula for the area of a triangle given its side lengths.

Example 1: Equilateral Triangle

Given: An equilateral triangle with side length $a$.

To find: The area of the equilateral triangle using Heron's formula.

Solution: The sides are $a$, $a$, and $a$. The semi-perimeter is $s=\frac{a+a+a}{2}=\frac{3a}{2}$. Now, apply Heron's formula: Area $=\sqrt{s(s-a)(s-b)(s-c)}$ Area $=\sqrt{\frac{3a}{2}(\frac{3a}{2}-a)(\frac{3a}{2}-a)(\frac{3a}{2}-a)}$ Area $=\sqrt{\frac{3a}{2}\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)}$ Area $=\sqrt{\frac{3a^4}{16}}$ Area $=\frac{\sqrt{3}{a}^2}{4}$

Answer: The area of an equilateral triangle with side $a$ is $\frac{\sqrt{3}{a}^2}{4}$ sq. units.

Example 2: Isosceles Triangle

Given: An isosceles triangle with equal sides of length $a$ and base of length $2b$.

To find: The area of the isosceles triangle using Heron's formula.

Solution: The sides are $a$, $a$, and $2b$. The semi-perimeter is $s=\frac{a+a+2b}{2}=\frac{2a+2b}{2}=a+b$. Now, apply Heron's formula: Area $=\sqrt{s(s-a)(s-b)(s-c)}$ Area $=\sqrt{(a+b)((a+b)-a)((a+b)-a)((a+b)-2b)}$ Area $=\sqrt{(a+b)(b)(b)(a-b)}$ Area $=\sqrt{b^2(a+b)(a-b)}$ Area $=\sqrt{b^2(a^2-b^2)}$ Area $=b\sqrt{a^2-b^2}$

Answer: The area of the isosceles triangle is $b\sqrt{a^2-b^2}$ sq. units.

Example 3: Right-Angled Triangle

Given: A triangle with sides 3 units, 4 units, and 5 units.

To find: The area of the triangle using Heron's formula.

Solution: The sides are $a=3$, $b=4$, and $c=5$. The semi-perimeter is $s=\frac{3+4+5}{2}=\frac{12}{2}=6$. Now, apply Heron's formula: Area $=\sqrt{s(s-a)(s-b)(s-c)}$ Area $=\sqrt{6(6-3)(6-4)(6-5)}$ Area $=\sqrt{6(3)(2)(1)}$ Area $=\sqrt{36}$ Area $=6$

Answer: The area of the triangle is 6 sq. units. (Note: Since $3^2+4^2=9+16=25=5^2$, this is a right-angled triangle. Its area can also be calculated as $\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 3\times 4=6$ sq. units, confirming Heron's formula.)

Remember

Heron's formula is most useful when the height of the triangle is not known or difficult to calculate. It only requires the lengths of the three sides.

Exam-smart

A question might provide the sides of a triangle and ask for its area directly, requiring Heron's formula. Alternatively, it might give a scenario where calculating the height is cumbersome (e.g., a triangle formed by three intersecting lines where equations are given, but not easily the height), making Heron's formula the efficient choice. You might also be asked to verify Heron's formula for a known triangle type (like equilateral or right-angled).

Area of a Circle

The area of a circle is the measure of the space enclosed within its boundary. It is directly proportional to the square of its radius.

Formula: The area ($A$) of a circle with radius ($r$) is given by:

where $\pi$ (pi) is a mathematical constant approximately equal to $\frac{22}{7}$ or $3.14159$.

Derivation Concept: Archimedes demonstrated that the area of a circle is equivalent to the area of a right-angled triangle whose sides are equal to the radius ($r$) and the circumference ($2\pi r$) of the circle. Area = $\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times (2\pi r)\times r=\pi r^2$. Another visual explanation involves dividing the circle into many sectors. When these sectors are rearranged, they form a shape resembling a parallelogram with a base of $\pi r$ (half the circumference) and a height of $r$ (the radius).

Example 4: Area of a Circle

Given: A circle with a radius of 7 cm.

To find: The area of the circle.

Solution: The radius is $r=7$ cm. Using the formula $A=\pi r^2$:

Using $\pi \approx \frac{22}{7}$:

Answer: The area of the circle is $154{\text{ cm}}^2$ (using $\pi \approx \frac{22}{7}$).

Remember

Always use the value of $\pi$ as specified in the question. If not specified, using $\frac{22}{7}$ or $3.14$ is common, but $\frac{22}{7}$ often simplifies calculations when the radius is a multiple of 7.

Exam-smart

Questions will typically provide the radius or diameter and ask for the area, or give the area and ask for the radius. Word problems might involve circular gardens, fields, or containers.

Area of Sector of a Circle

A sector of a circle is the portion of a circle enclosed by two radii and the arc between them.

Formula: The area of a sector with radius ($r$) and a central angle of $\theta$ degrees is a fraction of the total area of the circle. The fraction is $\frac{\theta }{{360}^{\circ }}$.

Conceptual Understanding: Imagine slicing a pizza. Each slice is a sector. The area of the slice depends on how big the slice is (the angle $\theta$) and the size of the pizza (the radius $r$). If $\theta ={360}^{\circ }$, the sector is the entire circle. If $\theta ={180}^{\circ }$, it's a semi-circle (half the area). If $\theta ={90}^{\circ }$, it's a quarter circle (one-fourth the area).

Example 5: Area of a Sector

Given: A sector of a circle with radius 7 cm and a central angle of ${60}^{\circ }$.

To find: The area of the sector.

Solution: Radius $r=7$ cm Central angle $\theta ={60}^{\circ }$ Using the formula for the area of a sector: Area of Sector $=\frac{\theta }{{360}^{\circ }}\times \pi r^2$ Area of Sector $=\frac{{60}^{\circ }}{{360}^{\circ }}\times \pi (7\text{ cm}{)}^2$ Area of Sector $=\frac{1}{6}\times \pi \times 49{\text{ cm}}^2$ Using $\pi \approx \frac{22}{7}$: Area of Sector $=\frac{1}{6}\times \frac{22}{7}\times 49{\text{ cm}}^2$ Area of Sector $=\frac{1}{6}\times 22\times 7{\text{ cm}}^2$ Area of Sector $=\frac{154}{6}{\text{ cm}}^2$ Area of Sector $=\frac{77}{3}{\text{ cm}}^2$

Answer: The area of the sector is $\frac{77}{3}{\text{ cm}}^2$ (or approximately $25.67{\text{ cm}}^2$).

Remember

Ensure the angle $\theta$ is in degrees when using the formula $\frac{\theta }{{360}^{\circ }}\times \pi r^2$. If the angle is in radians, the formula is $\frac{1}{2}{r}^2\theta$.

Exam-smart

These questions often involve clock hands sweeping an area, parts of circular fields, or designs. Be mindful of whether you need to find the area of a minor sector or a major sector (where the angle would be ${360}^{\circ }-\theta$).