The regular hexagon if side 5 cms. Find tge area of this hexagon by 2 methods
Answers
Answer:
1 method
area of hexagon MNOPQR
= area of traingle MNO + area of rectangle MOPR + area of traingle PQR
by symmetry
area of traingle MNO = area of triangle PQR
= 2×area of triangle MNO + area of rectangle MOPR
area of traingle MNO
in traingle MNO,
base=OM=15m
& for height
height +5+height=11
2×height +5=11
2×height =11-5
2×height =6
height =6/2=3cm
there fore area =1/2 ×base×height
= 1/2×8×3
=5×3
=12cm^2
Area of rectangle MOPR
MOPR is a rectangle with
length =MI=8cm
breadth =Op=5cm
area of rectangle MOPR =length ×breadth
=8×5= 40cm^2
thus,
area of hexagon MNOPQR
= 2÷area of traingle MNO+area of rectangle MOPR
=2×12+40
=24+40
=64cm^2
2 method
area of hexagon MNOPQR
= area of trapezium NOPQ +area of trapezium NMRQ
by symmetry
area of trapezium NOPQ =area of trapezium NMRQ
=2×area of trapezium NOPQ
area of trapezium NOPQ
in trapezium
OP& NQ are parallel sides
height is OA
here,
OP=5cm
NO=11cm
and height =OA=OM/2=8/2=4cm
area of trapezium NOPQ
=1/2 ×sum of parallel sides ×height
=1/2×(OP+NQ)×OA
=1/2 ×(5+11)×4
=1/2×16×4
=8×4
=32cm^2
thus,
area of hexagon MNOPQR
=2×area of trapezium BCDF
=2×32
=64cm^2