Question

MNOPQR is a hexagon of side 6 cm each. Find the area of the given hexagon in two different methods.
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Answer 1

QU€STION:

MNOPQR is a hexagon of side 6 cm each. Find the area of the given hexagon in two different methods.

ANSW€R:

Area of traepezium NOPQ

in traepezium NOPQ

OP & NQ are parallel sides

Height is OA

Here,

OP = 5 cm

NQ = 11cm

and height = OA = OM/2 = 8/2 = 4cm

Area of traepezium NOPQ = ½ × sum of parallel sides × height = ½ × (OP + NQ) × OA

= ½ × (5 + 11) × 4

= ½ × 16 ×4

= 8 × 4

= 32cm

Area of hexagon MNOPQ = 2 × area of traepezium BCDF

= 2 × 32

= 64cm

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@MყʂƚเƈαʅPσɯҽɾʂ❤️⚡

Answer 2

Answer:

Method 1 :-

divide the given hexagon into two similar trapezia by joining QN

area of the hexagon MNOPQR = 2× area of trapezium MNQR

= 2×½(6+11) ×4

=17×4

=68 cm²

Method 2 :-

The hexagon MNOPQR is divided into three parts two similar triangles and one rectangle by joining MO, RP

NS = 11cm - 6cm / 2

= 5/2 CM

= 2.5 CM

Area of hexagon MNOPQR = 2×area pop rectangle MRPO

= 2×[½×MO×NS] + [RP×MR]

=MO × NS +RP×MP

=8×25+8×6

=20+48

=68 CM

Hope It Helps My Dear Darling❤

By:-

Yeashwanth

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