Ex 11.2, 10 There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways. Find the area of this park using both ways. Can you suggest some other way of finding its area?
Finding Area by Jyothi’s Diagram
Area of pentagon ABCDE
= Area of trapezium BCDF + Area of trapezium BAEF
By symmetry,
Area of trapezium BCDE = Area of trapezium BAFE
= 2 × Area of trapezium BCDF
Area of trapezium BCDF
In trapezium BCDF,
BF & CD are parallel sides
DF is height
Here,
BF = 30 m
CD = 15 m
and Height = DF = 𝐷𝐸/2 = 15/2 m
Area of trapezium BCDF = 1/2 × Sum of parallel sides × Height
= 1/2 × (BF + CD) × DF
= 1/2 × (30 + 15) × 15/2
= 1/2 × 45 × 15/2
= (45 × 15)/4
Thus,
Area of Pentagon ABCDE = 2 × Area of trapezium BCDF
= 2 × (45 × 15)/4
= (45 × 15)/2
= 675/2
= 337.5 m2
= 1/2 × (30 + 15) × 15/2
= 1/2 × 45 × 15/2
= (45 × 15)/4
Thus,
Area of Pentagon ABCDE = 2 × Area of trapezium BCDF
= 2 × (45 × 15)/4
= (45 × 15)/2
= 675/2
= 337.5 m2
Finding Area by Kavita’s Diagram
Area of pentagon ABCDE
= Area of triangle ABC + Area of Square ACED
Finding area of ∆ ABC
Area = 1/2×𝐵𝑎𝑠𝑒×𝐻𝑒𝑖𝑔ℎ𝑡
Putting
Putting
Base = AC = 15 m
& Height = BD – CD
= 30 – 15
= 15 m
∴ Area = 1/2 × Base × Height
= 1/2 × 15 × 15
= 225/2
= 112.5 m2
Area of square ACDE
By Symmetry,
ACDE is a square with side 15 m
Area of square ACDE = (Side)2
= 152
= 225 m2
Thus,
Area of pentagon ABCDE
= Area of triangle ABC + Area of Square ACED
= 112.5 + 225
= 337.5 m2
Another way to find Area of Pentagon
Area of pentagon = Area of rectangle FGDE
− Area of triangle AFB
− Area of triangle CGB
By symmetry,
Area of triangle AFB = Area of triangle GCB
= Area of rectangle FGDE − 2 × Area of triangle AFB
Area of rectangle FGDE
Area of rectangle FGDE = ED × GD
= 15 × 30
= 450 m2
Area of triangle AFB
In Δ AFB
AF is the height
BF is the base
AF = 30 – 15
= 15 m
and
BF = 1/2 × ED = 15/2 = 7.5 m
Area of triangle AFB = 1/2 × BF × AF
= 1/2 × 7.5 × 15
Now,
Area of pentagon = Area of rectangle FGDE − 2 × Area of triangle AFB
= 450 – 2 × 1/2 × 7.5 × 15
= 450 – 7.5 × 15
= 450 – 15/2 × 15
= 450 – 225/2
= 450 – 112.5
= 337.5 m2

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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