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\ 06-Jan-2025 Bharat Arora Proof-7 Reader’s Sign _______________________ Date __________

Riddhi’s method:

Since it is a hexagon and NQ divides the hexagon into two congruent N
trapeziums. You can verify it by paper folding.
1 M O
2
Now, area of trapezium MNQR = × 4 × (11 + 5) cm = 32 cm 2 5 cm 4 cm
2 11 cm 4 cm 11 cm 5 cm
2
2
So, the area of hexagon MNOPQR = 2 × 32 cm = 64 cm . R P
Hence, the area calculated in both methods is the same. Q 4 cm
Example 13: The top surface of a raised platform is in the shape of a regular B 5 m C
octagon as shown in the given figure. Find the area of the octagonal surface. A D 4 m
Solution: Join AD and HE as shown in figure given below. Draw BN ⊥ AD 11 m 5 m
and GM ⊥ HE. H E

Area of the octagonal surface = Area of trapezium ABCD + Area of rectangle G F
ADEH + Area of trapezium HEFG
1 1
= (BC + AD) × BN + AD × AH + (GF + HE) × GM
2 2
1 1 B 5 m C
= (5 + 11) m × 4 m + 11 m × 5 m + (5 + 11) m × 4 m 4 m
2 2 A D
[Q GM = BN = 4 m] N 11 m 5 m
1 1 H M E
2
= × 16 m × 4 m + 55 m + × 16 m × 4 m
2 2 4 m
2
2
2
2
= 32 m + 55 m + 32 m = 119 m . G 5 m F
Father: Raman! now you can find the area of our agricultural land as given in the map.
In the map, AD = 200 m, AM = 30 m, MN = 50 m, NP = 60 m, PQ = 20 m.
E
QD = AD − AQ = 200 m – (30 + 50 + 60 + 20) m F
= 200 m − 160 m = 40 m 40 m 60 m 20 m

NQ = NP + PQ = 60 m + 20 m = 80 m A A M 50 m N 60 m P Q D
PD = PQ + QD = 20 m + 40 m = 60 m 30 m 50 m 40 m
C
MP = MN + NP = 50 m + 60 m = 110 m B
200 m
AN = AM + MN = 30 + 50 = 80 m

1 1
Area of DAMF = × AM × MF = × 30 m × 40 m
2 2
= 600 m 2

1 1
Area of DEPD = × PD × EP = × 60 m × 60 m
2 2
= 1800 m 2

1 1
Area of DQCD = × QD × CQ = × 40 m × 40 m
2 2
= 800 m 2

Mathematics-8 240