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Geometry and Measurement - Area of 2D Shapes (Trapeziums, Rhombuses, General Polygons)

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A trapezium (or trapezoid) is a quadrilateral with at least one pair of parallel sides. The area is found by multiplying the average of the parallel sides (aa and bb) by the perpendicular height (hh).

Diagram of a trapezium showing parallel sides a and b and perpendicular height h.
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A rhombus is a special type of parallelogram where all four sides are equal. Its area can be calculated using the lengths of its diagonals (d1d_1 and d2d_2), which bisect each other at 90∘90^\circ.

Rhombus showing two intersecting diagonals d1 and d2.
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The area of a general polygon can be found by decomposing it into simpler shapes like triangles and rectangles, or by using the 'offset' method in a field book where distances are measured along a main diagonal (base line).

General polygon divided into triangles and trapeziums using a central diagonal.
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For a rhombus, if the base (bb) and the vertical height (hh) are known, the area is simply Area=b×hArea = b \times h, the same as a parallelogram.

📐Formulae

Area of a Trapezium: A=frac12(a+b)hA = \\frac{1}{2}(a + b)h

Area of a Rhombus (Diagonal Method): A=frac12timesd1timesd2A = \\frac{1}{2} \\times d_1 \\times d_2

Area of a Rhombus/Parallelogram (Base/Height Method): A=btimeshA = b \\times h

Area of General Polygons: Atotal=A1+A2+...+AnA_{total} = A_1 + A_2 + ... + A_n

💡Examples

Problem 1:

Calculate the area of a trapezium where the two parallel sides measure 14textcm14\\text{ cm} and 22textcm22\\text{ cm}, and the perpendicular distance between them is 9textcm9\\text{ cm}.

Solution:

  1. Identify the given dimensions: a=14textcma = 14\\text{ cm}, b=22textcmb = 22\\text{ cm}, and h=9textcmh = 9\\text{ cm}.
  2. Write down the trapezium area formula: A=frac12(a+b)hA = \\frac{1}{2}(a + b)h.
  3. Substitute the values into the formula: A=frac12(14+22)times9A = \\frac{1}{2}(14 + 22) \\times 9.
  4. Add the bases inside the parentheses: A=frac12(36)times9A = \\frac{1}{2}(36) \\times 9.
  5. Multiply: A=18times9=162A = 18 \\times 9 = 162.

Explanation:

To find the area, we sum the parallel bases, multiply by the perpendicular height, and then take half of that product. The final area is 162textcm2162\\text{ cm}^2.

Problem 2:

A rhombus has a side length of 5textcm5\\text{ cm} and its two diagonals measure 6textcm6\\text{ cm} and 8textcm8\\text{ cm}. Find the area of the rhombus.

Solution:

  1. Identify the relevant dimensions for the diagonal formula: d1=6textcmd_1 = 6\\text{ cm} and d2=8textcmd_2 = 8\\text{ cm} (the side length of 5textcm5\\text{ cm} is not needed for this formula).
  2. State the formula: A=frac12timesd1timesd2A = \\frac{1}{2} \\times d_1 \\times d_2.
  3. Substitute the values: A=frac12times6times8A = \\frac{1}{2} \\times 6 \\times 8.
  4. Perform the multiplication: A=frac12times48A = \\frac{1}{2} \\times 48.
  5. Final calculation: A=24A = 24.

Explanation:

Using the diagonal method is the most direct way to find the area here. Note that the side length was extra information provided to test your ability to choose the correct dimensions. The area is 24textcm224\\text{ cm}^2.

Problem 3:

An irregular plot of land is shaped like a pentagon ABCDEABCDE. A diagonal ADAD is drawn as a base line of length 40 m40\text{ m}. Perpendiculars (offsets) are dropped from vertices BB, CC, and EE to the base line. BB is 10 m10\text{ m} away from ADAD at a point 10 m10\text{ m} from AA. CC is 15 m15\text{ m} away from ADAD at a point 30 m30\text{ m} from AA. EE is 12 m12\text{ m} away from ADAD on the opposite side, at a point 20 m20\text{ m} from AA. Calculate the total area of the plot.

Polygon ABCDE with diagonal AD used as base for offsets.

Solution:

The plot is divided into triangles and trapeziums.\text{The plot is divided into triangles and trapeziums.} Area of △ABF=12×10×10=50 m2\text{Area of } \triangle ABF = \frac{1}{2} \times 10 \times 10 = 50\text{ m}^2 Area of trapezium BCFG=12×(10+15)×(30−10)=12×25×20=250 m2\text{Area of trapezium } BCFG = \frac{1}{2} \times (10 + 15) \times (30 - 10) = \frac{1}{2} \times 25 \times 20 = 250\text{ m}^2 Area of △CDG=12×15×(40−30)=75 m2\text{Area of } \triangle CDG = \frac{1}{2} \times 15 \times (40 - 30) = 75\text{ m}^2 Area of △ADE=12×40×12=240 m2\text{Area of } \triangle ADE = \frac{1}{2} \times 40 \times 12 = 240\text{ m}^2 Total Area=50+250+75+240=615 m2\text{Total Area} = 50 + 250 + 75 + 240 = 615\text{ m}^2

Explanation:

We split the polygon into four parts along the diagonal ADAD: two triangles and one trapezium on the top side, and one large triangle on the bottom side. We calculate the area of each and sum them up.

Problem 4:

A window frame is in the shape of a trapezium. The top width is 80 cm80\text{ cm} and the bottom width is 120 cm120\text{ cm}. The area of the glass required is 5000 cm25000\text{ cm}^2. What is the height of the window frame?

Trapezium window with top 80cm and bottom 120cm.

Solution:

Given: a=80,b=120, Area =5000\text{Given: } a = 80, b = 120, \text{ Area } = 5000 Using Area formula: 5000=12(80+120)h\text{Using Area formula: } 5000 = \frac{1}{2}(80 + 120)h 5000=12(200)h5000 = \frac{1}{2}(200)h 5000=100h5000 = 100h h=5000100=50 cmh = \frac{5000}{100} = 50\text{ cm}

Explanation:

Substitute the known values into the trapezium area formula and solve for the unknown height hh.