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Example 3 - There is a regular hexagon MNOPQR of side 5 cm

Example 3 - Chapter 11 Class 8 Mensuration - Part 2
Example 3 - Chapter 11 Class 8 Mensuration - Part 3
Example 3 - Chapter 11 Class 8 Mensuration - Part 4
Example 3 - Chapter 11 Class 8 Mensuration - Part 5
Example 3 - Chapter 11 Class 8 Mensuration - Part 6
Example 3 - Chapter 11 Class 8 Mensuration - Part 7
Example 3 - Chapter 11 Class 8 Mensuration - Part 8

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Transcript

Example 3 There is a regular hexagon MNOPQR of side 5 cm (Fig 11.20) . Aman and Ridhima divided it in two different ways. Find the area of this hexagon using both ways (Fig 11.21). Finding Area by Ridhima’s Method Area of hexagon MNOPQR = Area of triangle MNO + Area of rectangle MOPR + Area of triangle PQR By symmetry, Area of triangle MNO = Area of triangle PQR = 2 × Area of triangle MNO + Area of rectangle MOPR Area of triangle MNO In triangle MNO, Base = OM = 15 m & For height Height + 5 + Height = 11 2 × Height + 5 = 11 2 × Height = 11 – 5 2 × Height = 6 Height = 6/2 = 3 cm ∴ Area = 1/2 × Base × Height = 1/2 × 8 × 3 = 4 × 3 = 12 cm2 Area of rectangle MOPR MOPR is a rectangle with Length = MO = 8 cm Breadth = OP = 5 cm Area of rectangle MOPR = Length × Breadth = 8 × 5 = 40 cm2 Thus, Area of hexagon MNOPQR = 2 × Area of triangle MNO + Area of rectangle MOPR = 2 × 12 + 40 = 24 + 40 = 64 cm2 Finding Area by Aman’s Method Area of hexagon MNOPQR = Area of trapezium NOPQ + Area of trapezium NMRQ By symmetry, Area of trapezium NOPQ = Area of trapezium NMRQ = 2 × Area of trapezium NOPQ Area of trapezium NOPQ In trapezium NOPQ, OP & NQ are parallel sides Height is OA Here, OP = 5 cm NQ = 11 cm and Height = OA = 𝑂𝑀/2 = 8/2 = 4 cm Area of trapezium NOPQ = 1/2 × Sum of parallel sides × Height = 1/2 × (OP + NQ) × OA = 1/2 × (5 + 11) × 4 = 1/2 × 16 × 4 = 8 × 4 = 32 cm2 Thus, Area of Hexagon MNOPQR = 2 × Area of trapezium BCDF = 2 × 32 = 64 cm2 Therefore, area of hexagon by Ridhima’s method is equal to the area of hexagon by Aman’s method.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.