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.

Made by

Davneet Singh

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.

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