Math Problem Statement

Calculate the area of a hexagonal park ABCDEF in which CMLAD, PLAD, BMLAD, and ENLAD are given with the respective dimensions.

Solution

The question is asking to calculate the area of a hexagonal park ABCDEFABCDEF by splitting it into triangles and trapeziums. The given measurements are:

  • FP=6cm,PL=1.5cm,CM=4.5cm,MD=2.25cmFP = 6 \, \text{cm}, PL = 1.5 \, \text{cm}, CM = 4.5 \, \text{cm}, MD = 2.25 \, \text{cm}
  • EN=9cm,LN=6cm,AP=4.5cm,BL=6cm,NM=1.5cmEN = 9 \, \text{cm}, LN = 6 \, \text{cm}, AP = 4.5 \, \text{cm}, BL = 6 \, \text{cm}, NM = 1.5 \, \text{cm}

Solution:

  1. Break the hexagon into parts:

    • CML,ENL,PLF\triangle CML, \triangle ENL, \triangle PLF (triangles)
    • Trapeziums: APLD and BMLD\text{Trapeziums: } \text{APLD and BMLD}
  2. Calculate areas of triangles: For each triangle, the area formula is: Area of triangle=12×base×height\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height}

    • PLF:base FP=6cm, height PL=1.5cm\triangle PLF: \text{base } FP = 6 \, \text{cm}, \text{ height } PL = 1.5 \, \text{cm} Area of PLF=12×6×1.5=4.5cm2\text{Area of } \triangle PLF = \frac{1}{2} \times 6 \times 1.5 = 4.5 \, \text{cm}^2

    • CML:base CM=4.5cm, height MD=2.25cm\triangle CML: \text{base } CM = 4.5 \, \text{cm}, \text{ height } MD = 2.25 \, \text{cm} Area of CML=12×4.5×2.25=5.0625cm2\text{Area of } \triangle CML = \frac{1}{2} \times 4.5 \times 2.25 = 5.0625 \, \text{cm}^2

    • ENL:base EN=9cm, height LN=6cm\triangle ENL: \text{base } EN = 9 \, \text{cm}, \text{ height } LN = 6 \, \text{cm} Area of ENL=12×9×6=27cm2\text{Area of } \triangle ENL = \frac{1}{2} \times 9 \times 6 = 27 \, \text{cm}^2

  3. Calculate areas of trapeziums: The area of a trapezium is given by: Area=12×(sum of parallel sides)×height\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}

    • Trapezium APLD:parallel sides AP=4.5cm,BL=6cm,height PL=1.5cm\text{Trapezium } APLD: \text{parallel sides } AP = 4.5 \, \text{cm}, BL = 6 \, \text{cm}, \text{height } PL = 1.5 \, \text{cm} Area of APLD=12×(4.5+6)×1.5=12×10.5×1.5=7.875cm2\text{Area of } APLD = \frac{1}{2} \times (4.5 + 6) \times 1.5 = \frac{1}{2} \times 10.5 \times 1.5 = 7.875 \, \text{cm}^2

    • Trapezium BMLD:parallel sides BL=6cm,NM=1.5cm,height MD=2.25cm\text{Trapezium } BMLD: \text{parallel sides } BL = 6 \, \text{cm}, NM = 1.5 \, \text{cm}, \text{height } MD = 2.25 \, \text{cm} Area of BMLD=12×(6+1.5)×2.25=12×7.5×2.25=8.4375cm2\text{Area of } BMLD = \frac{1}{2} \times (6 + 1.5) \times 2.25 = \frac{1}{2} \times 7.5 \times 2.25 = 8.4375 \, \text{cm}^2

  4. Sum up all the areas: Total Area=4.5+5.0625+27+7.875+8.4375=52.875cm2\text{Total Area} = 4.5 + 5.0625 + 27 + 7.875 + 8.4375 = 52.875 \, \text{cm}^2

Final Answer:

The area of the hexagonal park is 52.875cm252.875 \, \text{cm}^2.


Would you like to discuss details or have further questions?

5 Related Questions:

  1. How do you find the area of a hexagon without dividing it into simpler shapes?
  2. Can the properties of symmetry in hexagons simplify such calculations?
  3. How does this formula change if the hexagon is regular?
  4. Why are trapeziums useful in geometry for irregular polygons?
  5. What are other methods to calculate areas of composite shapes?

Tip:

Always verify the units of each dimension before performing area calculations to ensure consistency.

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Math Problem Analysis

Mathematical Concepts

Geometry
Area Calculation
Hexagons
Triangles
Trapeziums

Formulas

Area of a triangle = (1/2) × base × height
Area of a trapezium = (1/2) × (sum of parallel sides) × height

Theorems

Basic Area Formulas in Geometry

Suitable Grade Level

Grades 8-10