Question 1. Find the area of the pentagon shown in figure below, if AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm, BF = 5 cm, CG = 7 cm and EH = 3 cm.
Solution:
Given:
AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm BF = 5 cm, CG = 7 cm, EH = 3 cm
From given data,
FG = AG – AF = 8 – 5 = 3 cm
And,
GD = AD – AG = 10 – 8 = 2 cm
From given figure:
Area of Pentagon = (Area of triangle AFB) + (Area of trapezium FBCG) +
(Area of triangle CGD) + (Area of triangle ADE)
= (0.5 x AF x BF) + [0.5 x (BF + CG) x (FG)] + (0.5 x GD x CG) + (1/2 x AD x EH).
= (0.5 x 5 x 5) + [0.5 x (5 + 7)x (3) + (0.5 x 2 x 7) + (0.5 x 10 x 3)
= 12.5 + 18 + 7 + 15 = 52.5 cm2
Questions 2. Find the area enclosed by each of the following fig(Fig. (i)-(iii)J as the sum of the areas of a rectangle and a trapezium.
Solution:
(i) Figure can be divided into 2 parts a square and a trapezium as shown:
Area = (Area of square) + (Area of trapezium)
= (side)2 + (0.5 x (sum of parallel sides) x height
= (18 x 18) + 0.5 x (18 + 7) × (8)
= 324+100
= 424 cm2
(ii) Figure can be divided into 2 parts a rectangle and a trapezium as shown:
From the figure:
Height of trapezium = 28 – 20 = 8 cm
Area = (Area of rectangle) + (Area of trapezium)
= (length x breadth) + (0.5 x (sum of parallel sides) x height)
= (20 x 15) + [0.5 x (15 + 6) × (8)]
= 300 + 84
= 384 cm2
(iii) Figure can be divided into 2 parts one trapezium and one rectangle:
Pythagoras theorem in one of the right angle triangle:
52 = 42 + (base)2
base2 = 25 – 16
base = √9 = 3 cm
Therefore,
The height of trapezium = 3 cm
One side of trapezium = 6 + 4 + 6 = 14 cm
Area = (Area of rectangle) + (Area of trapezium)
= (length x breadth) + (0.5 x (sum of parallel sides) x height)
= (6 x 4) + (0.5 x (14 + 6) x (3))
= 24 + 30
= 54 cm2
Question 3. There is a pentagonal shaped park as shown in Fig. Jyoti and Kavita divided it in two different ways.
Find the area of this park using both ways. Can you suggest some another way of finding its area?
Solution:
Jyoti and Kavita divided the park in two different ways.
(i) Jyoti divided the park into two equal trapeziums:
From the figure the park divided into equal trapezium having height 7.5 m and sides 30 m and 15 m
Area of the park = 2 x (Area of a trapezium)
= 2 x (0.5 x (sum of parallel sides) x height)
= 2 x (0.5 x (30 + 15) x (7.5))
= 337.5 m2
(ii) Kavita divided the park into a rectangle and a triangle:
From diagram,
The height of the triangle = 30 – 15 = 15 m
Area of the park = (Area of square) + (Area of triangle)
= (15 x 15) + (0.5 x 15 x 15)
= 225 + 112.5
= 337.5 m2
Question 4. Find the area of the following polygon, if AL = 10 cm, AM = 20 cm, AN = 50 cm, AO = 60 cm and AD = 90 cm.
Solution:
Given:
AL = 10 cm, AM = 20 cm, AN = 50 cm
AO = 60 cm, AD = 90 cm
From given data,
MO = AO – AM = 60 – 20 = 40 cm
OD = AD – A0 = 90 – 60 = 30 cm
ND = AD – AN = 90 – 50 = 40 cm
LN = AN – AL = 50 – 10 = 40 cm
Area of Polygon = (Area of triangle AMF) + (Area of trapezium MOEF) +
(Area of triangle DNC) + (Area of trapezium NLBC) +
(Area of triangle ALB)
= (0.5 x AM x MF) + [0.5 x (MF + OE) x OM] + (0.5 x OD x OE) +
(0.5 x DN x NC) + [0.5 x (LB + NC) x NL] + (0.5 x AL x LB)
= (0.5 x 20 x 20) + [0.5 x (20 + 60) x (40)] +(0.5 x 30 x 60) +
(0.5 x 40 x 40) +[0.5 x (30 + 40) x (40)] + (0.5 x 10 x 30)
= 200 + 1600 + 900 + 800 + 1400 +150 = 5050 cm2
Question 5. Find the area of the following regular hexagon.
Solution:
As it is a regular Hexagon so all sides are 13 cm and AN = BQ
From the figure,
As the diagonal QN is 23 cm,
QB + BA + AN = QN
AN + 13 + AN = 23
2AN = 23 – 13 = 10
AN = 5 cm
Hence, AN = BQ = 5 cm
Applying Pythagoras theorem in triangle MAN:
MN2 = AN2 + AM2
169 = 25 + AM2
AM² = 169 – 25
AM = √144
AM = 12cm
From figure,
OM = RP = 2 × AM = 2 x 12 = 24 cm
This Hexagon can be divided into 3 parts 2 triangles and one rectangle therefore,
Area of the regular hexagon = (area of triangle MON) + (area of rectangle MOPR) +
(area of triangle RPQ)
= (0.5 x OM x AN) + (RP X PO) + (0.5 x RP x BQ)
= (0.5 x 24 x 5) + (24 x 13) + (0.5 x 24 x 5)
= 60 + 312 + 60
= 432 cm2