Question

K
(4) Find the area of hexagonal field KLMNOP which is in hexagonal shape that
LO = 35 m MN = 28 m. KM = 48 m. KQ = 20 m. PR = 16 m LR=45 m, QO = 40 m.
L
R
M
N​

Answer 1

Answer:

Method adopted by Suresh

Since it is a regular hexagon. So, NQ divides the hexagon into two congruent trapeziums. You can verify it by paper folding.

Now area of trapezium MNQR

=4×

2

11+5

=2×16=32 cm

2

So the area of hexagon MNOPQR=2×32=64 cm

2

Method adopted by Rushikas

△MNO and △RPQ are congruent triangles with altitude 3cm (fig). You can verify this by cutting off these two triangles and placing them on one another.

Area of MNO=×8×3=12cm

2

= Area of △RPQ

Area of rectangle MOPR=8 5=40cm

2

Now, area of hexagon MNOPQR=40+12+12=64 cm

2

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