Question

Find the area of hexagon given below. You are me
given that MP = 10 cm, MD = 8 cm, MC = 7 cm,
MB = 5 cm, MA = 3 cm, AN = 4 cm, OC = 6 cm,
DQ = 3 cm and BR = 5 cm.
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Answer 1

Step-by-step explanation:

i hope the abovd images will help you

Answer 2

Answer:

To Solve this divide it into segments,

Right Triangle Segments are ΔMNA, ΔOCP, ΔDQP and ΔMBR

Area of Right Triangle = $\frac{1}{2}(b)(h)$

For ΔMNA,

MA -> Base

AN -> Height

$\frac{1}{2}(3)(4)\\= 2(3)\\= 6 cm^2$

For ΔOCP,

CP -> Base

OC -> Height

To calculate CP -> (MP - MD) + (MD - MC)

= (10-8) + (8-7)

= 2+1 = 3

$\frac{1}{2} (3)(6)\\= 3(3)\\= 9cm^2$

For ΔDQP,

DP -> Base

DQ -> Height

DP = (MP-MD) = 2

$\frac{1}{2} (2)(3)\\=1(3)\\= 3cm^2$

For ΔMBR,

MB -> Base

BR -> Height

$\frac{1}{2} (5)(5)\\= 2.5(5)\\= 12.5 cm^2$

Trapezium Segments are □NOCA and □BDQR

Area of Trapezium = $\frac{1}{2} (height)[Sum of Parallel Sides]$

For □NOCA,

NA and OC -> Parallel Sides

AC -> Height

AC = (MB - MA) + (MC - MB)

= (5-3) + (7 - 5) = 2 + 2

= 4

$\frac{1}{2} (4)(4+6)\\= \frac{1}{2} (4)(10)\\= 2(10)\\= 20cm^2$

For □BDQR

BR and DQ -> Parallel sides

BD -> Height

BD = MD - MB

= 8-5

= 3cm

$\frac{1}{2} (3)(3+5)\\= \frac{1}{2} (3)(8)\\= 3(4)\\= 12cm^2$

Total Area = 6 + 9 + 3 + 12.5 + 12 + 20

= 62.5 cm^2