Page 278 - Start Up Mathematics_8 (Non CCE)

Solution: Area of octagon PQRSTUVW = (Area of trapezium PQRS) + (Area of rectangle PSTW) +
(Area of trapezium WTUV) = 2(Area of trapezium PQRS) + (Area of rectangle PSTW)
{ 1 }
= + (WT × ST) = 2 2 ¥ ( 511+ ) ¥ 4 + (11 × 5)

Ê 1 ˆ
= 2 Á Ë 2 ¥ 16 4¥ ˜ + (11 × 5) = 64 + 55 = 119 m 2
¯

MATHS LAB ACTIVITY

Objective: To find the area of an irregular polygon by breaking it into triangles and trapeziums
Material required: Chart paper, different coloured papers, scale, pencil, a pair of scissors, glue stick/fevicol

Step 1: Draw a line AB of length 30 cm on a white N
chart paper.
L
A 30 cm B

Step 2:. Take four points on AB, say, P, Q, R and S Q S B
and draw perpendiculars PL, QM, RN and A P R
ST of different lengths.
Step 3: Join AL, LN, NB, AM, MT and TB. T
Step 4: Measure and write the following lengths: M
AP = ____ cm PL = ____ cm N
AL = ____ cm PR = ____ cm L
RN = ____ cm LN = ____ cm
RB = ____ cm BN = ____ cm A Q S B
SB = ____ cm ST = ____ cm P R
QS = ____ cm TM = ____ cm
QM = ____ cm AM = ____ cm T
AQ = ____ cm M
Step 5: Now using different coloured papers, cut out ∆ ALP, trapezium LPRN, ∆ NRB, ∆ AQM,
trapezium MQST and ∆ TSB.
Step 6: Paste these cut-outs on their respective shapes on the figure.
Observations:

Area of ∆ ALP = ________________________ cm 2

Area of trapezium LPRN = ________________________ cm 2

Area of ∆ NRB = ________________________ cm 2

Area of ∆ AQM = ________________________ cm 2

Area of trapezium MQST = ________________________ cm 2

Area of ∆ TSB = ________________________ cm 2
Area of polygon ALNBTM Total = ____________________ cm 2

Conclusion: Area of any polygon can be found by breaking it into triangles and trapeziums.

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