Question

Find the area of the hexagon ABCDEF
By splitting into two congruent trapezium.​

Answer 1

Answer:

Plssss mark it as the brainliest answer plssss

Step-by-step explanation:

For making it a trapezium join C and F, resulting in the formation of trapezium CBAF and trapezium FCDE

For finding the length of CF consider a right angle triangle CBF in which ,

h = CF

p = BF = 10 cm

b = BC = 7 cm

$h^{2} = p^{2} + b^{2} \\\\CF^{2} = BF^{2} + BC^{2} \\\\CF^{2} = (10)^{2} + (7)^{2}\\\\CF^{2} = 100 + 49\\\\CF^{2} = 149\\\\CF = \sqrt{149}\\ \\CF = 12.2 cm$

To find the length of AB & AF

It is given that both of them are equal

&

angle A = 90

Therefore

h = BF

b = AF

p = AB

$h^{2} = p^{2} + b^{2} \\\\BF^{2} = AF^{2} + BA^{2} \\\\(100)^{2} = (a)^{2} + (a)^{2}\\\\(100) = 2a^{2} \\\\\frac{100}{2} = a^{2} \\\\ a = \sqrt{50}\\ \\a = 5\sqrt{2}\\ \\a = 5(1.41)\\\\a = 7.05 cm$

Therefore ,

AB = AF = DE = CD = 7.05cm

In trapezium CBAF ,

Height = AF

Parallel lines = BA & CF

Area of a trapezium = $\frac{1}{2} (height)(sum of parallel sides)$

                               = $\frac{1}{2} (AF)(AB + CF)$

                               $\frac{1}{2} (7.05)(7.05 + 12.2)\\\\\frac{1}{2} (7.05)(19.25)\\\\\frac{86.01}{2} \\\\= 43.005\\\\= 43.1 cm^{2}$

In trapezium FCDE ,

Height = CD

Parallel lines = ED & CF

Since it is given that AB = AF = DE = CD

Therefore in trapezium FCDE we can say that

Height = AF

Parallel lines = BA & CF

As the sides are same thus the measurements will be same and thus the area will also be same

Therefore area of trapezium FCDE = area of trapezium CFAB

                     area of trapezium FCDE = $43.1 cm^{2}$

As the area of hexagon =

                            area of trapezium FCDE + area of trapezium CFAB

                                  = 43.1 +43.1

                                 = $86.2 cm^{2}$

Answer 2

Answer:

well for a i got correct but in book backside its not given for b so i am not sure. mine b is coming 120 cm square for 1 trapezium is 60 cm square.