Question

(Fan and Medal will be given)!
Help with Area
Find the area of the image below:

Answer 1

hint:
the area of your geometrical shape, is given by the sum of the areas of these triangles:

Answer 2

so what exactly do I do ? Im not too well with this benchmark D: .

Answer 3

we have: area of triangle ABE:
\[A = \frac{{BE \times AK}}{2} = \frac{{8 \times 5}}{2} = ...\]

Answer 4

so 8 * 5 = 40 an d 40 divided by 2 is 20 .

Answer 5

ok!

Answer 6

area of triangle CDE:
\[A = \frac{{CE \times DK}}{2} = \frac{{5 \times 4}}{2} = ...\]

Answer 7

5 * 4 equals 20 and 20 divded by 2 is 10

Answer 8

ok!

Answer 9

finally:
area of triangle CBE:
\[A = \frac{{BE \times CE}}{2} = \frac{{8 \times 5}}{2} = ...\]

Answer 10

8*5 equals 40 again divided by two is 20 :D

Answer 11

ok!

Answer 12

so total area is:
20+10+20=...

Answer 13

50

Answer 14

yes!
nevertheless that answer is not an option of yours

Answer 15

wow , :/

Answer 16

my reasoning is right, I don't understand
maybe is there a scale factor, for your drawing?

Answer 17

namely, how many units are the area of the littlest square, of your drawing?

Answer 18

can u help me find the perimeter for it please michele , D: its not area , its perimeter. sorry ive made a mistake

Answer 19

ok!

Answer 20

then we have to apply the theorem of Pitagora repeatedly

Answer 21

so we can write this:
\[BC = \sqrt {C{E^2} + B{E^2}} = \sqrt {{5^2} + {8^2}} = \sqrt {25 + 64} = ...?\]

Answer 22

ok is that is 89

Answer 23

yes! and what is:sqrt(89)=...?

Answer 24

9.433

Answer 25

that's right!

Answer 26

now we have:
\[AB = \sqrt {A{H^2} + B{H^2}} = \sqrt {{5^2} + {6^2}} = \sqrt {25 + 36} = ...?\]

Answer 27

its 61

Answer 28

and what is:
sqrt(61)=...?

Answer 29

the square root of 61is 7.81

Answer 30

perfect!

Answer 31

:D

Answer 32

now we have:
\[AE = \sqrt {A{H^2} + E{H^2}} = \sqrt {{5^2} + {2^2}} = \sqrt {25 + 4} = ...?\]

Answer 33

25 plus 4 is 29 and the square root of 29 is 5.38

Answer 34

yes! correct!

Answer 35

next we have:
\[ED = \sqrt {K{D^2} + E{K^2}} = \sqrt {{4^2} + {2^2}} = \sqrt {16 + 4} = ...?\]

Answer 36

16 + 4 = 20 and square root of 20 is 4.47

Answer 37

ok!

Answer 38

finally, we have:
\[CD = \sqrt {K{D^2} + C{K^2}} = \sqrt {{4^2} + {3^2}} = \sqrt {16 + 9} = ...?\]

Answer 39

5 :D

Answer 40

perfect!

Answer 41

so the requested perimeter, is:
BC+AB+AE+ED+CD=

=9.43+7.81+5.39+4.47+5=...?

Answer 42

32.1 :D yipee

Answer 43

that's right! :)

Answer 44

Thank you sooo much :D i will be adding these into my notes thanks a lot I appreciate it ! God bless.

Answer 45

Thanks!! :) :)

Answer 46

Yw :)

Answer 47

:)