Question

There is a hexagon MNOPQR of side 5 cm. Amana and Rithima divided it into different ways.
Find the area of this hexagon using ways?

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

Answer:

Aman's method:

Since it is a hexagon so NQ divides the hexagon into two congruent trapeziums.You can verify it by folding paper.(Attachment 3)

Now area of trapezium MNQR,

$\sf{4 \times \frac{(11 + 5)}{2} = 2 \times 16 = 32 \: {cm}^{2} }$

So the area of hexagon MNOPQR(By Aman's method)=2×32=64 cm².

Ridhima's method:

ΔMNO and ΔRPQ are congruent triangles with altitude 3 cm.(Attachment 2)

You can verify this by cutting off these two triangles and placing them on one another.

Now area of ΔMNO,

$\sf{ \frac{1}{2} \times 8 \times 3 = 12 \: {cm}^{2} = Area \: of \: \triangle RPQ}$

☆Area of rectangke MOPR=8×5=40 cm²

Now,

☆Area of hexagon MNOPQR=40+12+12=64 cm²