Question

find the area of the regular hexagon and MNOPQR of side 5 cm​

Answer 1

Answer:

A≈64.95cm²

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Answer 2

Answer:

the area of the regular hexagon M N O P Q R of side 5 cm is 64 CM square

Step-by-step explanation:

1) Method adopted by Suresh:

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

We know that area of trapezium =21× sum of the parallel sides× distance between them

∴  area of trapezium MNQR=4×211+5

=2×16=32 cm2

So, the area of hexagon MNOPQR=2×32=64 cm2.

(2) Method adopted by Rushika:

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

We know that, area of a triangle =21× base× height

∴  Area of △MNO=21×8×3=12 cm2= Area of △RPQ

We know that, area of a rectangle =length× breadth

∴  Area of rectangle MOPR=8×5=40 cm2

Now, area of hexagon MNOPQR=40+12+12=64 cm2.

Hence, area of the given hexagon is 64 cm2, as calculated by both methods.

#please follow any one of the methods

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