Area

Rectangle and Squares

Chapter 14: Area

Page 1 of 5: Rectangles and Squares

Welcome to the world of shapes and spaces! In this chapter, we'll explore one of the most fundamental concepts in geometry: Area. It's the key to answering questions like "How much paint do I need for this wall?" or "Which field is larger for planting crops?".

{{FORMULA: expr=Area = length × width | symbols=Area: The amount of space inside a 2D shape, length: The longer side of a rectangle, width: The shorter side of a rectangle}}

Concept Introduction

Imagine you're helping decorate for a festival. You have two beautiful rectangular rangoli designs to fill with colored powder. One is long and narrow, and the other is shorter but wider. A friend asks, "Which one will need more powder?"

This is a question about area. Area is the measure of the amount of surface a two-dimensional shape covers. It’s not about the length of the boundary (that's perimeter!), but about the space inside the boundary. To figure out which rangoli needs more powder, we need to calculate and compare the space inside each one. We do this by seeing how many standard-sized squares can fit perfectly into the shape.

Definitions & Formulas

Before we start calculating, let's get our terms straight.

The most fundamental formulas for this section are:

• Area of a Rectangle: A = l × w
• Area of a Square: A = s × s = s²

Derivation / Logic: Why "Length × Width"?

Have you ever wondered why this simple multiplication works? It's all about counting squares. Let's break down the logic.

1. The Standard Unit: We first decide on a unit square. Let's say it's a square with sides of 1 cm each. Its area is 1 square centimeter, written as 1 cm². This is our basic building block for measuring area.

2. Visualizing the Rectangle: Imagine a rectangle with a length of 7 cm and a width of 4 cm.

   {{VISUAL: diagram: A 7cm by 4cm rectangle overlaid with a grid of 1cm by 1cm unit squares. The horizontal axis is labeled "length = 7 cm" and shows 7 squares. The vertical axis is labeled "width = 4 cm" and shows 4 squares.}}

3. Laying the First Row: If we line up our 1 cm² unit squares along the 7 cm length, we can fit exactly 7 squares in a single row.

4. Stacking the Rows: The width is 4 cm. This means we can stack 4 such rows of 7 squares on top of each other to completely fill the rectangle.

5. Counting the Total: We have 4 rows, and each row has 7 squares. The total number of unit squares is:

   Total Squares = (Squares per row) × (Number of rows)
   

6. The Formula Emerges: This translates directly to our dimensions. "Squares per row" is the length, and "Number of rows" is the width.

   Area = length × width
   

   Area = 7 cm × 4 cm = 28 cm²
   

So, the formula A = l × w is simply a fast way of counting all the unit squares that fit inside a rectangle!

{{KEY: type=concept | title=Perimeter is NOT a Measure of Area | text=Two shapes can have the same perimeter but wildly different areas. A long, skinny rectangle and a square can have the same boundary length, but the square will enclose a much larger area. Always use the area formula (l × w) to measure the space inside a shape, not its perimeter (2(l+w)).}}

Solved Examples

Let's put this knowledge into practice with a few examples.

Example 1: Basic Area Calculation (Easy)

Given: A rectangular mobile phone screen has a length of 15 cm and a width of 7 cm.

To Find: The area of the screen.

Solution:

1. Identify the given dimensions. The length l is 15 cm and the width w is 7 cm.

2. Use the formula for the area of a rectangle.

   Area = length × width
   

3. Substitute the values and calculate.

   Area = 15 cm × 7 cm
   

   Area = 105 cm²
   

Final Answer: The area of the mobile phone screen is 105 cm².

Example 2: Finding a Missing Side (Medium)

Given: A rectangular classroom floor has an area of 56 m². Its width is 7 m.

To Find: The length of the classroom.

Solution:

1. Write down the knowns. We have the Area (A) = 56 m² and the width (w) = 7 m.

2. Start with the area formula.

   Area = length × width
   

3. Substitute the known values into the equation.

   56 m² = length × 7 m
   

4. To find the length, rearrange the formula by dividing the area by the width.

   length = 56 m² ÷ 7 m
   

   length = 8 m
   

Final Answer: The length of the classroom is 8 m.

Example 3: Area of a Path (Hard)

Given: A rectangular park is 40 m long and 20 m wide. A path 2 m wide is built all around the outside of the park.

To Find: The area of the path.

Solution:

1. This problem has two rectangles: the inner park (let's call it EFGH) and the outer rectangle including the path (let's call it ABCD). The area of the path is the difference between the area of the outer rectangle and the area of the inner park.

   {{VISUAL: diagram: A rectangular park labeled EFGH with dimensions 40m x 20m. A larger rectangle ABCD is drawn around it, creating a shaded path of uniform width 2m between the two rectangles.}}

2. First, calculate the area of the inner park (EFGH).

   Area_inner = length_inner × width_inner
   

   Area_inner = 40 m × 20 m = 800 m²
   

3. Next, determine the dimensions of the outer rectangle (ABCD). The path is 2 m wide on all sides. So, 2 m is added to the top and bottom of the width, and 2 m is added to the left and right of the length.

   length_outer = 40 m + 2 m + 2 m = 44 m
   

   width_outer = 20 m + 2 m + 2 m = 24 m
   

4. Now, calculate the area of the outer rectangle.

   Area_outer = length_outer × width_outer
   

   Area_outer = 44 m × 24 m = 1056 m²
   

5. Finally, subtract the inner area from the outer area to find the area of the path.

   Area_path = Area_outer - Area_inner
   

   Area_path = 1056 m² - 800 m² = 256 m²
   

Final Answer: The area of the path is 256 m².

Example 4: Area vs. Perimeter (Tricky)

Given: Rectangle A has a length of 10 cm and a width of 4 cm. Rectangle B has a length of 8 cm and a width of 5 cm.

To Find: Compare their perimeters and areas. Which one has a larger perimeter? Which one has a larger area?

Solution:

1. Calculate for Rectangle A:

   • Find the perimeter: P = 2 × (l + w)

     Perimeter_A = 2 × (10 cm + 4 cm) = 2 × 14 cm = 28 cm
     

   • Find the area: A = l × w

     Area_A = 10 cm × 4 cm = 40 cm²
     

2. Calculate for Rectangle B:

   • Find the perimeter: P = 2 × (l + w)

     Perimeter_B = 2 × (8 cm + 5 cm) = 2 × 13 cm = 26 cm
     

   • Find the area: A = l × w

     Area_B = 8 cm × 5 cm = 40 cm²
     

3. Compare the results:

   • Perimeter: Perimeter_A (28 cm) > Perimeter_B (26 cm)
   • Area: Area_A (40 cm²) = Area_B (40 cm²)

   This shows that even though Rectangle A has a larger perimeter, both rectangles have the exact same area. This proves that perimeter is not a reliable indicator of area.

Final Answer: Rectangle A has a larger perimeter (28 cm vs 26 cm), but both rectangles have the same area (40 cm²).

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. A piece of wire is bent to form a square with an area of 81 cm². If the same wire is unbent and then re-bent to form a rectangle with a length of 10 cm, what is the area of the rectangle?

   💡 Answer: The area of the square is 81 cm², so its side is √81 = 9 cm. The total length of the wire (perimeter) is 4 × 9 = 36 cm. For the rectangle, the perimeter is also 36 cm. 2 × (10 + w) = 36, so 10 + w = 18, and the width w is 8 cm. The area of the rectangle is 10 cm × 8 cm = 80 cm².

2. A rectangular hall is 12 m long and 10 m wide. It is to be paved with square tiles of side 25 cm. How many tiles are required?

   💡 Answer: First, convert all units to cm. Hall length = 1200 cm, width = 1000 cm. Area of hall = 1200 × 1000 = 1,200,000 cm². Area of one tile = 25 × 25 = 625 cm². Number of tiles = (Area of hall) ÷ (Area of one tile) = 1,200,000 ÷ 625 = 1920 tiles.

3. A square has a side length of 8 cm. A rectangle has the same perimeter as the square. If the length of the rectangle is 10 cm, which shape has a larger area, and by how much?

   💡 Answer: The square's perimeter is 4 × 8 = 32 cm. Its area is 8 × 8 = 64 cm². The rectangle's perimeter is also 32 cm. 2 × (10 + w) = 32, so 10 + w = 16, and the width w is 6 cm. The rectangle's area is 10 × 6 = 60 cm². The square has a larger area by 64 cm² - 60 cm² = 4 cm².

Mini Cheatsheet

Triangles

Page 2: Area of Triangles

Welcome to the second part of our journey into the world of Area! We've mastered rectangles, but many shapes in the world aren't so straightforward. From the sail of a boat to a slice of pizza, triangles are everywhere. How do we measure the space they occupy?

This lesson will demystify the area of a triangle. We will not just learn the formula but understand why it works by connecting it back to the area of a rectangle. We will see how this simple formula can solve complex problems and discover some surprising properties of triangles within other shapes. Get ready to see rectangles and triangles in a whole new light!

{{FORMULA: expr=Area = ½ × base × height | symbols=Area:Total surface enclosed by the triangle, b:The length of the chosen base side, h:The perpendicular height from the base to the opposite vertex}}

Definitions & Formulas

Before we dive into solving problems, let's clearly define the terms we'll be using. For a triangle, the base can be any of its three sides. The height (or altitude) is the perpendicular distance from the chosen base to the opposite vertex.

The fundamental formula that connects these is:

A = ½ × b × h

Deriving the Area Formula

Why is there a ½ in the formula for a triangle's area? The answer lies in its relationship with the rectangle, a shape we already understand. Let's build the logic step-by-step.

1. Start with any triangle. Let's call it ΔABC. We will consider side BC as its base. Its height is the perpendicular distance from vertex A to the base BC.

2. Enclose it in a rectangle. We can draw a rectangle around ΔABC. Let's call it BCDE, where the base of the triangle BC is one side of the rectangle. The height of the rectangle CD is drawn to be exactly equal to the height of the triangle.

{{VISUAL: diagram: A triangle ABC enclosed in a rectangle BCDE. The base of the triangle BC is a side of the rectangle, and the height from A to BC is shown as equal to the side CD of the rectangle. This illustrates Area(ΔABC) = ½ Area(Rectangle BCDE).}}

3. Recognize the relationship. A key property of rectangles is that a diagonal divides it into two congruent (identical) triangles. In the simplest case of a right-angled triangle, the hypotenuse acts as this diagonal. For any triangle, we can see that it occupies exactly half the space of its "parent" rectangle.

4. Use the Rectangle's Area Formula. We know the area of the rectangle BCDE is simply its length times its width. In this case, that's BC × CD.

Area (Rectangle BCDE) = base × height

5. Connect Triangle Area to Rectangle Area. Since the triangle ΔABC occupies exactly half the area of the rectangle BCDE, we can conclude:

Area (ΔABC) = ½ × Area (Rectangle BCDE)

6. Final Formula. By substituting the rectangle's area formula from Step 4 into the equation from Step 5, we arrive at the universal formula for the area of a triangle:

Area (ΔABC) = ½ × base × height

This logic holds true even for obtuse-angled triangles, where the area can be calculated as the difference between the areas of two right-angled triangles. The final formula remains the same!

Solved Examples

Let's apply this formula to solve some problems, starting from easy and moving to more challenging ones.

Example 1: Basic Calculation (Easy)

Given: A triangle with a base of 10 cm and a corresponding height of 6 cm.

To Find: The area of the triangle.

Solution:

1. Identify the given values. Here, b = 10 cm and h = 6 cm.

2. Apply the area formula A = ½ × b × h.

A = ½ × 10 cm × 6 cm

3. Calculate the product.

A = ½ × 60 cm²

A = 30 cm²

Final Answer: The area of the triangle is 30 cm².

Example 2: Finding a Missing Dimension (Medium)

Given: The area of a triangle is 48 m². Its base is 12 m.

To Find: The height (altitude) of the triangle corresponding to the given base.

Solution:

1. List the known values: A = 48 m² and b = 12 m.

2. Start with the area formula.

A = ½ × b × h

3. Substitute the known values into the equation.

48 = ½ × 12 × h

4. Simplify the equation and solve for h.

48 = 6 × h

h = 48 / 6

h = 8 m

Final Answer: The height of the triangle is 8 m.

Example 3: Multiple Altitudes (Hard)

Given: In ΔPQR, the base QR = 15 cm and its corresponding altitude PM = 8 cm. The side PR = 12 cm.

To Find: The length of the altitude QN to the side PR.

Solution:

1. First, calculate the area of the triangle using the known base and altitude (QR and PM).

Area (ΔPQR) = ½ × QR × PM

Area (ΔPQR) = ½ × 15 cm × 8 cm

Area (ΔPQR) = 60 cm²

{{VISUAL: diagram: A scalene triangle PQR. Base QR = 15cm. Altitude from P to QR is PM = 8cm. Another side PR = 12cm. A second altitude from Q to PR is drawn, labeled QN. The length of QN is unknown.}}

2. The area of the triangle remains the same regardless of which side is chosen as the base. Now, consider PR as the base and QN as the height.

3. Use the area formula again with the new base and the unknown height QN.

Area (ΔPQR) = ½ × PR × QN

4. We already know the area is 60 cm². Substitute the values and solve for QN.

60 = ½ × 12 × QN

60 = 6 × QN

QN = 60 / 6

QN = 10 cm

Final Answer: The length of the altitude QN is 10 cm.

{{KEY: type=concept | title=Constant Area | text=The area of a triangle is a fixed value. You can calculate it using any base-altitude pair. If you know the area and a side length, you can always find the corresponding altitude to that side.}}

Example 4: Triangles within a Rectangle (Tricky)

Given: A rectangle ABCD has a length of 16 cm and a width of 10 cm. The diagonals AC and BD intersect at point O.

To Find: The area of ΔAOB.

Solution:

1. Understand the properties of a rectangle's diagonals. They bisect each other, meaning they cut each other into two equal halves. So, AO = OC and BO = OD.

2. The diagonals divide the rectangle into four smaller triangles: ΔAOB, ΔBOC, ΔCOD, and ΔDOA. Let's find the area of one of the larger triangles formed by a diagonal, say ΔABC.

Area (ΔABC) = ½ × base × height

Area (ΔABC) = ½ × AB × BC = ½ × 16 cm × 10 cm

Area (ΔABC) = 80 cm²

3. Now consider the triangle ΔABC. The line segment BO is a median to the side AC (since O is the midpoint of AC). A key property of triangles is that a median divides a triangle into two triangles of equal area.

{{VISUAL: diagram: A rectangle ABCD with length 16cm and width 10cm. Diagonals AC and BD intersect at O. Triangle AOB is shaded, and the median BO of triangle ABC is highlighted.}}

4. Therefore, the area of ΔAOB is exactly half the area of ΔABC.

Area (ΔAOB) = ½ × Area (ΔABC)

Area (ΔAOB) = ½ × 80 cm²

Area (ΔAOB) = 40 cm²

5. By the same logic, all four triangles (ΔAOB, ΔBOC, ΔCOD, ΔDOA) have equal areas. The total area of the rectangle is 16 × 10 = 160 cm². Each of the four triangles has an area of 160 / 4 = 40 cm².

Final Answer: The area of ΔAOB is 40 cm².

Tips & Tricks

Mastering triangle area calculations is easier with these shortcuts in mind.

Common Mistakes

Avoid these common pitfalls when calculating the area of triangles.

Brain-Teaser Questions

Test your understanding with these higher-order thinking problems.

1. If the base of a triangle is doubled and its height is tripled, by what factor does its area increase?

   💡 Answer: Let the original area be A₁ = ½ × b × h. The new base is b' = 2b and the new height is h' = 3h. The new area is A₂ = ½ × (2b) × (3h) = ½ × 6 × b × h = 6 × (½ × b × h) = 6A₁. The area increases by a factor of 6.

2. A square and an isosceles triangle have the same base length. If the area of the triangle is equal to the area of the square, what is the ratio of the square's side to the triangle's height?

   💡 Answer: Let the side of the square be s. Its area is s². The triangle has base b = s. Let its height be h. Its area is ½ × s × h. Given that the areas are equal: s² = ½ × s × h. We can divide both sides by s (since s cannot be zero): s = ½ × h. The ratio of the square's side (s) to the triangle's height (h) is s / h = ½. The ratio is 1:2.

3. In a rectangle ABCD, a point X is chosen anywhere on the side AB. Which triangle has a greater area: ΔXDC or ΔYBC, where Y is a point anywhere on side AD? Assume the rectangles are identical.

   💡 Answer: The area of ΔXDC is ½ × base × height = ½ × DC × (perpendicular distance from X to DC). Since X is on AB, the perpendicular distance to DC is simply the width of the rectangle, BC. So, Area(ΔXDC) = ½ × DC × BC. The area of ΔYBC is ½ × base × height = ½ × BC × (perpendicular distance from Y to BC). Since Y is on AD, the perpendicular distance to BC is the length of the rectangle, DC. So, Area(ΔYBC) = ½ × BC × DC. Both triangles have the exact same area, which is half the area of the rectangle.

Mini Cheatsheet

Here's a quick summary of the key formulas and concepts from this page. Screenshot this for your last-minute revisions!

Area of any Polygon & Parallelogram

{{FORMULA: expr=Area of Parallelogram = base × height | symbols=base: the length of any one side, height: the perpendicular distance from the base to the opposite side}}

Concept Introduction

Imagine you're an urban planner tasked with designing a new public park. The plot of land isn't a simple rectangle; it's an irregular five-sided shape, a pentagon. To order the right amount of grass turf, you need to calculate its exact area. How would you do it? You don't have a direct formula for the area of a pentagon.

This is where the power of simple shapes comes in. You can divide the complex pentagonal park into smaller, manageable triangles. By drawing a few lines from one corner to others, you can create three triangles. You already know how to find the area of a triangle! By calculating the area of each triangle and adding them up, you can find the total area of the park. This method, called triangulation, is a fundamental technique that allows us to find the area of any polygon, no matter how complex it looks.

Definitions & Formulas

To find the area of polygons, especially parallelograms, we use specific measurements. Understanding these terms is the first step to mastering the calculations.

The key formulas we will use on this page are:

• Area of a Triangle = ½ × base × height
• Area of a Parallelogram = base × height

Derivation: Area of a Parallelogram

Why is the area of a parallelogram simply base × height and not the product of its adjacent sides? The answer lies in a clever rearrangement. We can transform any parallelogram into a rectangle of the same area, and we already know the area of a rectangle is length × width.

Let's see how this works in a few steps.

{{VISUAL: diagram: A parallelogram ABCD is shown. A perpendicular AX is dropped from A to the base CD. The triangle ΔAXD is shown in a different color. An arrow indicates this triangle being cut and moved to the right side, fitting perfectly next to BC, to form the rectangle AXX'B.}}

1. Start with a parallelogram ABCD. Let's consider CD as the base (b).

2. Draw a height (h) from vertex A, perpendicular to the base CD. Let the point where it meets the base (or its extension) be X. This creates a right-angled triangle, ΔAXD.

3. Imagine you "cut" this triangle ΔAXD from the left side of the parallelogram. The remaining shape is a trapezium, ABCX.

4. Now, "slide" the cut triangle ΔAXD over to the right side of the trapezium. The side AD of the triangle will perfectly match the side BC of the parallelogram because opposite sides of a parallelogram are equal in length (AD = BC).

5. The new shape you have formed is a rectangle, AXX'B. The length of this rectangle is XX', which is the same as the parallelogram's base CD. The width of this rectangle is AX, which is the parallelogram's height h.

6. The area of this new rectangle is length × width, which is CD × AX or b × h. Since we only rearranged the parts without adding or removing anything, the area of the original parallelogram must be the same as the area of the rectangle.

Area of Parallelogram ABCD = Area of Rectangle AXX'B = base × height

{{KEY: type=concept | title=The Perpendicular is Key | text=The most critical part of finding a parallelogram's area is using the perpendicular height, not the slanted side length. The height must form a 90° angle with the base.}}

Solved Examples

Let's apply these concepts to solve some problems, starting from easy and moving to more challenging ones.

Example 1: Area of a Quadrilateral (Easy)

Given: A quadrilateral ABCD with diagonal AC = 22 cm. The lengths of perpendiculars from B and D to the diagonal AC are BM = 3 cm and DN = 3 cm, respectively.

To Find: The area of the quadrilateral ABCD.

{{VISUAL: diagram: A quadrilateral ABCD with diagonal AC = 22 cm. Perpendiculars BM = 3 cm and DN = 3 cm are drawn from vertices B and D to the diagonal AC, forming two triangles ΔABC and ΔADC.}}

Solution:

1. The diagonal AC divides the quadrilateral ABCD into two triangles: ΔABC and ΔADC.

2. The area of the quadrilateral is the sum of the areas of these two triangles. Area(ABCD) = Area(ΔABC) + Area(ΔADC)

3. Calculate the area of ΔABC. Here, the base is AC and the height is BM.

   Area(ΔABC) = ½ × base × height = ½ × AC × BM
   

   Area(ΔABC) = ½ × 22 cm × 3 cm = 33 cm²
   

4. Calculate the area of ΔADC. Here, the base is AC and the height is DN.

   Area(ΔADC) = ½ × base × height = ½ × AC × DN
   

   Area(ΔADC) = ½ × 22 cm × 3 cm = 33 cm²
   

5. Add the areas of the two triangles to get the total area of the quadrilateral.

   Area(ABCD) = 33 cm² + 33 cm² = 66 cm²
   

Final Answer: The area of the quadrilateral ABCD is 66 cm².

Example 2: Area of a Parallelogram (Medium)

Given: A parallelogram has a base of 18 cm and a corresponding height of 10 cm.

To Find: The area of the parallelogram.

Solution:

1. Identify the given values. The base b is 18 cm and the height h is 10 cm.

2. Use the formula for the area of a parallelogram.

   Area = base × height
   

3. Substitute the given values into the formula.

   Area = 18 cm × 10 cm
   

4. Calculate the final product.

   Area = 180 cm²
   

Final Answer: The area of the parallelogram is 180 cm².

Example 3: Finding the Other Height (Hard)

Given: A parallelogram has adjacent sides of length 12 cm and 9 cm. The altitude corresponding to the base of 12 cm is 6 cm.

To Find: The length of the altitude corresponding to the base of 9 cm.

Solution:

1. First, calculate the area of the parallelogram using the given base and its corresponding altitude. Let b₁ = 12 cm and h₁ = 6 cm.

   Area = b₁ × h₁
   

   Area = 12 cm × 6 cm = 72 cm²
   

2. The area of the parallelogram remains the same regardless of which side is chosen as the base.

3. Now, consider the other side as the base, b₂ = 9 cm. Let the unknown corresponding altitude be h₂. The area is still 72 cm².

   Area = b₂ × h₂
   

4. Substitute the known values and solve for h₂.

   72 cm² = 9 cm × h₂
   

5. Rearrange the equation to find h₂.

   h₂ = 72 cm² ÷ 9 cm = 8 cm
   

Final Answer: The length of the altitude corresponding to the base of 9 cm is 8 cm.

Example 4: Area of a Composite Shape (Tricky)

Given: A pentagon ABCDE is shown. It is divided into three parts: ΔABE, rectangle BCDE, and ΔCDF is outside, but we only need area of ABCDE. Let's assume the question meant a different shape. Let's use the one from the NCERT text: a field in the shape of a pentagon. To find its area, it is divided into a triangle at the top and a rectangle below. Let's use a different shape to make it more unique. A quadrilateral PQRS is inside a rectangle ABCD. A(2,8), B(10,8), C(10,2), D(2,2). The vertices of PQRS are P(4,6), Q(8,6), R(9,4), S(3,4).

To Find: Area of pentagon ABCDE, which is composed of ΔADE on top of rectangle ABCD. Given: AB = 10 cm, BC = 8 cm. The height of ΔADE from E to side AD is 6 cm. Note: AD is the top side of the rectangle.

{{VISUAL: diagram: A pentagon ABCDE, where ABCD is a rectangle and ADE is a triangle built on top of the side AD. Dimensions are labeled: AB=10cm, BC=8cm, and the altitude from E to AD is 6cm.}}

Solution:

1. The total area of the pentagon is the sum of the area of the rectangle ABCD and the area of the triangle ADE.

   Area(ABCDE) = Area(Rectangle ABCD) + Area(ΔADE)
   

2. First, calculate the area of the rectangle ABCD. The length is AB = 10 cm and the width is BC = 8 cm.

   Area(Rectangle ABCD) = length × width = 10 cm × 8 cm = 80 cm²
   

3. Next, calculate the area of the triangle ADE. The base of the triangle is the side AD. Since ABCD is a rectangle, AD = BC = 8 cm. The given height of the triangle is 6 cm.

   Area(ΔADE) = ½ × base × height = ½ × AD × 6 cm
   

   Area(ΔADE) = ½ × 8 cm × 6 cm = 24 cm²
   

4. Finally, add the two areas together to find the total area of the pentagon.

   Area(ABCDE) = 80 cm² + 24 cm² = 104 cm²
   

Final Answer: The area of the pentagon ABCDE is 104 cm².

Tips & Tricks

Here are some shortcuts and key ideas to help you solve area problems faster and more accurately.

Common Mistakes

Many students make similar errors when calculating the area of polygons. Here’s a guide to avoid them.

Brain-Teaser Questions

Test your understanding with these slightly more challenging problems.

1. If the base of a parallelogram is doubled and its height is halved, what happens to its area?

   💡 Answer: The area remains unchanged. Let the original area be A = b × h. The new area A' will be A' = (2b) × (h/2) = (2/2) × (b × h) = b × h = A.

2. A quadrilateral is drawn on a grid. You are told its diagonals are 10 units and 8 units long. Can you find its area with only this information? What else might you need?

   💡 Answer: No, you cannot. The lengths of the diagonals are not sufficient. You would also need the angle at which they intersect. If they intersect at 90°, it's a special case. For a general quadrilateral, the best way is to use triangulation, for which you'd need the lengths of the perpendiculars from the other vertices to one of the diagonals.

3. An L-shaped polygon has six sides. What is the minimum number of non-overlapping triangles you need to divide it into to find its area?

   💡 Answer: An L-shaped polygon is a hexagon (6 sides). Using the formula n-2, where n is the number of sides, the minimum number of triangles is 6 - 2 = 4. You can achieve this by drawing diagonals from one vertex.

Mini Cheatsheet

Here’s a quick summary of the key concepts from this page. Screenshot this for your last-minute revision!

Rhombus & Trapezium

Area of a Rhombus & Trapezium

Welcome back! In our previous lesson, we saw how to find the area of basic shapes like rectangles and triangles. But the world is filled with more interesting quadrilaterals. Think about the beautiful repeating patterns on a floor tile, the classic diamond shape of a kite, or the sloped sides of a bridge support. These are often in the shape of a rhombus or a trapezium.

While we could always divide these shapes into triangles, that's not very efficient. Just as we have a specific formula for rectangles, mathematicians have developed elegant, direct formulas for these shapes too. In this section, we will explore these special formulas. We'll learn how to derive them by cleverly cutting and rearranging the shapes, and then apply them to solve practical problems.

{{FORMULA: expr=Area = ½ × (a + b) × h | symbols=a:length of one parallel side, b:length of the other parallel side, h:perpendicular height}}

Definitions & Formulas

Before we jump into calculations, let's clearly define our terms. The formulas for these shapes rely on very specific measurements.

Deriving the Formulas

How do we know these formulas work? We don't just memorize them; we understand them by transforming these shapes into something we already know—a rectangle!

Area of a Rhombus

The key property of a rhombus is that its diagonals are perpendicular bisectors of each other. This allows for a clever rearrangement.

1. Start with a rhombus ABCD, with diagonals AC = d₁ and BD = d₂. The diagonals intersect at point O.

2. The rhombus is made up of four congruent right-angled triangles (∆AOB, ∆BOC, ∆COD, ∆DOA).

3. Let's cut the rhombus along its diagonals and rearrange these four triangles. We can place them together to form a rectangle.

{{VISUAL: diagram: A rhombus ABCD is shown. Arrows indicate it being cut into four right-angled triangles along its diagonals d₁ and d₂. These four triangles are then rearranged to form a rectangle with side lengths d₁ and d₂/2.}}

4. Wait, there's an even simpler way! Enclose the rhombus in a rectangle PQRS such that the sides of the rectangle are parallel to the diagonals of the rhombus.

5. The length of this new rectangle will be equal to one diagonal (d₁) and its width will be equal to the other diagonal (d₂). The area of this large rectangle is d₁ × d₂.

6. Notice that the area of the rhombus is exactly half the area of this enclosing rectangle. The four corner triangles outside the rhombus are identical to the four triangles inside it!

Therefore, the formula for the area of a rhombus is:

Area = ½ × d₁ × d₂

Area of a Trapezium

A trapezium has one pair of parallel sides. We can find its area by converting it into a parallelogram or a rectangle.

1. Start with a trapezium ABCD with parallel sides AB (a) and DC (b), and height h.

2. Make an identical copy of this trapezium, let's call it A'B'C'D'.

3. Rotate the copy by 180° and place it next to the original trapezium, joining side BC with B'C'.

{{VISUAL: diagram: A trapezium with parallel sides 'a' and 'b' and height 'h'. An identical, inverted trapezium is placed next to it. The combined shape is a large parallelogram with base (a+b) and height 'h'.}}

4. The combined shape is a parallelogram. The base of this new parallelogram is the sum of the two parallel sides of the trapezium (a + b).

5. The height of the parallelogram is the same as the height of the trapezium, h. The area of this large parallelogram is Base × Height = (a + b) × h.

6. Since the parallelogram was formed using two identical trapeziums, the area of one trapezium must be exactly half the area of the parallelogram.

Therefore, the formula for the area of a trapezium is:

Area = ½ × (a + b) × h

{{KEY: type=concept | title=Average of Parallel Sides | text=A great way to remember the trapezium formula is 'Area = Average of parallel sides × Height'. The term (a + b)/2 is simply the average length of the two parallel bases.}}

Solved Examples

Let's put these formulas into practice with some examples, moving from simple to more challenging.

Example 1: Area of a Kite (Easy)

Given: A kite, which has the shape of a rhombus, has diagonals of length 12 cm and 20 cm.

To Find: The area of the kite.

Solution:

1. Identify the given values. The diagonals are d₁ = 12 cm and d₂ = 20 cm.

2. Use the formula for the area of a rhombus.

   Area = ½ × d₁ × d₂
   

3. Substitute the given values into the formula.

   Area = ½ × 12 cm × 20 cm
   

4. Calculate the product.

   Area = ½ × 240 cm² = 120 cm²
   

Final Answer:

The area of the kite is 120 cm².

Example 2: Area of a Plot of Land (Medium)

Given: A plot of land is in the shape of a trapezium. Its parallel sides are 25 m and 45 m. The perpendicular distance between them is 18 m.

To Find: The area of the plot.

Solution:

1. Identify the given values. The parallel sides are a = 25 m and b = 45 m. The height is h = 18 m.

2. Use the formula for the area of a trapezium.

   Area = ½ × (a + b) × h
   

3. Substitute the values into the formula.

   Area = ½ × (25 m + 45 m) × 18 m
   

4. First, calculate the sum of the parallel sides.

   Area = ½ × (70 m) × 18 m
   

5. Calculate the final product.

   Area = 35 m × 18 m = 630 m²
   

Final Answer:

The area of the plot of land is 630 m².

Example 3: Finding a Missing Diagonal (Hard)

Given: The floor of a building consists of 2000 rhombus-shaped tiles. The area of the floor is 400 m². Each tile has one diagonal of length 20 cm.

To Find: The length of the other diagonal of a single tile.

{{VISUAL: diagram: A floor tiled with rhombus-shaped tiles. One tile is highlighted, showing its diagonals d₁ and d₂. The length d₁ is labeled '20 cm'.}}

Solution:

1. First, find the area of a single tile. The total area is 400 m² for 2000 tiles. It's crucial to work with consistent units. Let's convert the total area to cm². Since 1 m = 100 cm, 1 m² = 100 cm × 100 cm = 10000 cm².

   Total Area = 400 × 10000 cm² = 4,000,000 cm²
   

2. Now, calculate the area of one tile.

   Area of one tile = Total Area / Number of tiles
   

   Area of one tile = 4,000,000 cm² / 2000 = 2000 cm²
   

3. We know the area of one tile (2000 cm²) and one of its diagonals (d₁ = 20 cm). We can use the rhombus area formula to find the other diagonal, d₂.

   Area = ½ × d₁ × d₂
   

4. Substitute the known values and solve for d₂.

   2000 cm² = ½ × 20 cm × d₂
   

   2000 cm² = 10 cm × d₂
   

5. Isolate d₂ by dividing both sides by 10 cm.

   d₂ = 2000 cm² / 10 cm = 200 cm
   

Final Answer:

The length of the other diagonal of a single tile is 200 cm (or 2 m).

Example 4: Trapezium with a Twist (Tricky)

Given: The area of a trapezium is 384 cm². Its parallel sides are in the ratio 3:5 and the perpendicular distance between them is 12 cm.

To Find: The length of the longer of the parallel sides.

Solution:

1. Let the parallel sides be a and b. Since they are in the ratio 3:5, we can represent them using a common multiplier, x.

   a = 3x
   b = 5x
   

2. The height is given as h = 12 cm, and the area is 384 cm². Let's use the trapezium area formula.

   Area = ½ × (a + b) × h
   

3. Substitute the expressions for a and b, and the given values for Area and h.

   384 = ½ × (3x + 5x) × 12
   

4. Simplify the equation and solve for x.

   384 = ½ × (8x) × 12
   

   384 = 4x × 12
   

   384 = 48x
   

   x = 384 / 48 = 8
   

5. Now that we have the value of x, we can find the lengths of the two parallel sides.

   a = 3x = 3 × 8 = 24 cm
   b = 5x = 5 × 8 = 40 cm
   

6. The question asks for the length of the longer side, which is b.

Final Answer:

The length of the longer parallel side is 40 cm.

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. A rhombus and a square have the same perimeter of 40 cm. The rhombus has a diagonal of 12 cm. Which shape has a larger area, and by how much?

   💡 Answer: The square's side is 40/4 = 10 cm, so its Area = 10² = 100 cm². The rhombus's side is also 10 cm. Using Pythagoras on one of its four right-triangles (hypotenuse=10, one leg=12/2=6), the other leg is √(10²-6²) = √64 = 8 cm. This is half of the second diagonal. So, d₂ = 16 cm. Rhombus Area = ½ × 12 × 16 = 96 cm². The square has a larger area by 4 cm².

2. The cross-section of a swimming pool is a trapezium. The pool is 3 m deep at the shallow end and 8 m deep at the deep end. If the top width of the water is 50 m, what additional information do you absolutely need to find the area of this cross-section?

   💡 Answer: You have the parallel sides (a=3m, b=8m). The missing piece of information is the height (h), which in this context is the horizontal length of the pool from the shallow end to the deep end. The top width (50m) is not the height.

3. Can you divide any trapezium into two new trapeziums that have equal area? If so, how would you draw the dividing line?

   💡 Answer: Yes. The dividing line must be drawn parallel to the two original parallel bases. It won't be exactly halfway up the height. It will be positioned closer to the longer parallel side. Finding its exact length involves a more advanced formula, but the key is that the dividing line is parallel to the bases.

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Areas in Real Life & Summary

Areas in Real Life & Summary

Putting Area to Work

Have you ever helped paint a room or tile a floor? Before starting, you need to know how much paint or how many tiles to buy. Guessing could mean buying too little and having to rush back to the store, or buying too much and wasting money. This is where calculating area becomes a crucial real-life skill. By measuring the length and width of a wall, you can calculate its area and determine the exact amount of paint needed. Similarly, by finding the area of a floor, you can figure out the precise number of tiles required to cover it. Area helps us quantify two-dimensional space, turning abstract measurements into practical, actionable plans. From farming and construction to interior design and even mapping, understanding area is essential for planning, budgeting, and executing tasks efficiently in the world around us.

{{KEY: type=concept | title=What is Area? | text=Area is the measure of a two-dimensional space enclosed by a boundary. It tells us how much surface a shape covers and is measured in square units like square centimeters (cm²), square meters (m²), or square kilometers (km²).}}

Key Shapes: Definitions & Formulas

Here is a quick reference for the area formulas of the most common polygons you will encounter.

Derivation: Area of a Parallelogram

Why is the area of a parallelogram simply base × height, just like a rectangle? The answer lies in a clever bit of geometric transformation. We can show that any parallelogram can be rearranged to form a rectangle with the same base, height, and area.

{{VISUAL: diagram: A parallelogram ABCD. An altitude is dropped from vertex A to the base CD, meeting at point X. The triangle AXD on the left is shaded. An arrow shows this triangle being cut and moved to the right side, where side AD aligns with side BC, forming a rectangle.}}

1. Start with a parallelogram ABCD. Let the base be CD and the height be h. The height is the perpendicular distance between the parallel bases AB and CD.

2. Draw an altitude (height) from vertex A straight down to the base CD. Let's call the point where it meets the base X. This creates a right-angled triangle, ∆AXD.

3. Imagine cutting off this triangle ∆AXD from the left side of the parallelogram.

4. Now, slide this triangle across to the right side of the shape. The side AD of the triangle is equal in length to the side BC of the parallelogram.

5. Place the triangle so that its side AD perfectly aligns with the side BC. The point D will meet point C, and the point A will meet point B.

6. The new shape you've formed is a rectangle! Its length is the base CD of the original parallelogram, and its width is the height h (AX). The area of this new rectangle is base × height. Since we only rearranged the parts without adding or removing anything, the area of the original parallelogram must also be base × height.

Area(Parallelogram ABCD) = Area(Rectangle AX... ) = base × height

Solved Examples

Example 1: The Garden Path (Easy)

Given: A rectangular park is 50 m long and 30 m wide. A path 2 m wide is constructed outside the park.

To Find: The area of the path.

Solution:

1. This problem involves two rectangles: the inner park (let's call it EFGH) and the outer rectangle including the path (let's call it ABCD). The area of the path is the difference between the area of the outer rectangle and the inner rectangle.

2. First, calculate the area of the inner park.

   • Length (l₁) = 50 m
   • Width (w₁) = 30 m

   Area_park = l₁ × w₁ = 50 m × 30 m = 1500 m²
   

3. Next, determine the dimensions of the outer rectangle. The path is 2 m wide on all sides. So, 2 m is added to the length on both sides, and 2 m is added to the width on both sides.

   • Outer Length (l₂) = 50 m + 2 m + 2 m = 54 m
   • Outer Width (w₂) = 30 m + 2 m + 2 m = 34 m

4. Now, calculate the area of the outer rectangle.

   Area_total = l₂ × w₂ = 54 m × 34 m = 1836 m²
   

5. Finally, subtract the park's area from the total area to find the area of the path.

   Area_path = Area_total - Area_park = 1836 m² - 1500 m²
   

   Area_path = 336 m²
   

Final Answer: The area of the path is 336 m²

Example 2: Quadrilateral Field (Medium)

Given: A quadrilateral field ABCD where the diagonal AC is 22 m. The perpendiculars dropped from B and D to the diagonal AC are BM = 3 m and DN = 3 m.

To Find: The area of the quadrilateral field ABCD.

Solution:

1. A quadrilateral can be divided into two triangles by drawing a diagonal. Here, the diagonal AC divides the quadrilateral ABCD into two triangles: ∆ABC and ∆ADC.

{{VISUAL: diagram: A general quadrilateral ABCD. A diagonal is drawn from A to C. A perpendicular line is drawn from B to AC, labeled BM = 3m. Another perpendicular is drawn from D to AC, labeled DN = 3m. The diagonal AC is labeled 22m.}}

2. The area of the quadrilateral is the sum of the areas of these two triangles.

   Area(ABCD) = Area(∆ABC) + Area(∆ADC)
   

3. For ∆ABC, the base is AC and the height is BM.

   • Base (b) = 22 m
   • Height (h₁) = 3 m

   Area(∆ABC) = ½ × b × h₁ = ½ × 22 m × 3 m = 11 m × 3 m = 33 m²
   

4. For ∆ADC, the base is AC and the height is DN.

   • Base (b) = 22 m
   • Height (h₂) = 3 m

   Area(∆ADC) = ½ × b × h₂ = ½ × 22 m × 3 m = 11 m × 3 m = 33 m²
   

5. Add the areas of the two triangles to get the total area of the quadrilateral.

   Area(ABCD) = 33 m² + 33 m² = 66 m²
   

Final Answer: The area of the quadrilateral field is 66 m²

Example 3: Finding an Unknown Altitude (Hard)

Given: In triangle ABC, base BC = 8 units, and the corresponding altitude AX = 4 units. Another side AC = 6 units.

To Find: The length of the altitude BY from vertex B to the side AC.

Solution:

1. The area of a triangle can be calculated using any side as the base and the corresponding altitude. The area remains the same regardless of which base-altitude pair you choose.

2. First, calculate the area of ∆ABC using the given base BC and altitude AX.

   • Base (b₁) = BC = 8 units
   • Height (h₁) = AX = 4 units

   Area(∆ABC) = ½ × b₁ × h₁ = ½ × 8 × 4 = 16 sq. units
   

3. Now, we can express the area of the same triangle using the base AC and the unknown altitude BY.

   • Base (b₂) = AC = 6 units
   • Height (h₂) = BY = ?

   Area(∆ABC) = ½ × b₂ × h₂ = ½ × 6 × BY = 3 × BY
   

4. Since the area is the same, we can equate the two expressions for the area.

   3 × BY = 16
   

5. Solve for BY.

   BY = 16 / 3
   

Final Answer: The length of the altitude BY is 16/3 units (or approx. 5.33 units)

Example 4: The Midpoint Triangle (Tricky)

Given: In ∆XYZ, M is the midpoint of side XY and N is the midpoint of side XZ.

To Find: What fraction of the area of ∆XYZ is the area of ∆XMN?

Solution:

1. This problem requires understanding the properties of triangles formed by joining midpoints. Let's draw an altitude from X to the base YZ and call it h.

2. The line segment MN joining the midpoints of two sides of a triangle is parallel to the third side (YZ) and is half its length. This is a key property (Midpoint Theorem).

   • MN = ½ × YZ

{{VISUAL: diagram: A large triangle XYZ. M is the midpoint on side XY, N is the midpoint on side XZ. The line MN is drawn, creating a smaller triangle XMN at the top. An altitude is drawn from X down to YZ, intersecting MN at point P.}}

3. Now consider the altitude of the small triangle ∆XMN. If we draw an altitude from X to MN, let's call the intersection point P. Because N is the midpoint of XZ and M is the midpoint of XY, the line MN is exactly halfway up the triangle. Therefore, the altitude of ∆XMN (which is XP) is half the altitude of the big triangle ∆XYZ.

   • Altitude(∆XMN) = ½ × Altitude(∆XYZ) = h/2

4. Now we can write the area formulas for both triangles.

   • Area(∆XYZ) = ½ × Base × Height = ½ × YZ × h
   • Area(∆XMN) = ½ × Base × Height = ½ × MN × (h/2)

5. Substitute the relationships from steps 2 and 3 into the area formula for ∆XMN.

   Area(∆XMN) = ½ × (½ × YZ) × (h/2)
   

6. Simplify the expression.

   Area(∆XMN) = (1/8) × YZ × h
   

7. Now, compare this to the area of the large triangle. We know Area(∆XYZ) = ½ × YZ × h. Let's find the ratio.

   Ratio = Area(∆XMN) / Area(∆XYZ) = ((1/8) × YZ × h) / (½ × YZ × h)
   

   Ratio = (1/8) / (1/2) = (1/8) × 2 = 2/8 = 1/4
   

Final Answer: The area of ∆XMN is 1/4 of the area of ∆XYZ

Tips & Tricks

Common Mistakes to Avoid

Brain-Teaser Questions

1. Two identical squares of side length 10 cm overlap. The corner of one square is placed exactly at the center of the other square. What is the area of the overlapping (shaded) region?

   💡 Answer: The area is 25 cm². When a corner of one square is at the center of another identical square, the overlapping region will always be a smaller square with sides equal to half the side of the original square. So, the area is (10/2) × (10/2) = 5 × 5 = 25 cm².

2. You have 36 meters of fencing. What is the largest rectangular area you can enclose? What shape is it?

   💡 Answer: The largest area is 81 m², and the shape is a square. For a fixed perimeter, a square always encloses the maximum possible rectangular area. The perimeter is 36 m, so each side of the square would be 36/4 = 9 m. The area would be 9 m × 9 m = 81 m². Any other rectangle (e.g., 10×8, A=80; 12×6, A=72) will have a smaller area.

3. In a quadrilateral ABCD, the diagonals intersect at point O. If Area(∆AOB) = 12 cm², Area(∆BOC) = 18 cm², and Area(∆COD) = 24 cm², what is the area of ∆DOA?

   💡 Answer: The area of ∆DOA is 16 cm². In any convex quadrilateral, the product of the areas of opposite triangles formed by the intersection of diagonals is equal. Therefore, Area(∆AOB) × Area(∆COD) = Area(∆BOC) × Area(∆DOA). This gives us 12 × 24 = 18 × Area(∆DOA), so 288 = 18 × Area(∆DOA). Solving for the area gives 288 / 18 = 16 cm².

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