Question

I need helping finding Figure A,B,C,D,E
pleassee

Answer 1

ignore the top question (that answer is 1170 btw) just look at the figures

Answer 2

@Michele_Laino

Answer 3

here, we have a geometric shape, which is composed by a rectangle and two trapezoids.
Now the area of the rectangle is \(35 \cdot 18=...?\)

Answer 4

630

Answer 5

correct!

Answer 6

the area of one trapezoid is \(base \times height= 18 \cdot 15=...?\)

Answer 7

270

Answer 8

correct! So total area of the geometric shape is:
\(270+270+630=...?\)

Answer 9

1170

Answer 10

correct!

Answer 11

okay since that part is done how do we do the figures a b c d and e

Answer 12

sorry!, I go on parts a, b, c, d, e

Answer 13

huh?

Answer 14

if the perimeter of the rombus is \(20\) then each of its side measures \(20/4=...?\)

Answer 15

5

Answer 16

correct!

Answer 17

is that the answer or is there more to it?

Answer 18

no, we have to do more computations.
Now the situation is like below:

Answer 19

okay

Answer 20

ok

Answer 21

since the lenght of diagonal is \(8\) then the length of half diagonal is \(4\), and the length of the other half diagonal is \(x\)
we have to establish what is the value of \(x\)

Answer 22

yeah

Answer 23

If we apply the theorem of Pitagora, I can write this:
\[x = \sqrt {{5^2} - {4^2}} = \sqrt {25 - 16} = ...?\]

Answer 24

-11?

Answer 25

we have \(25-16=9\), so what is square root of \(9\)?

Answer 26

3

Answer 27

correct! so the length of second diagonal is: \(3 \cdot 2=6\)

Answer 28

alright

Answer 29

Now, the area \(A\) of the rombus is given by the subsequent formula:
\[\Large A = \frac{{{d_1}{d_2}}}{2} = \frac{{8 \cdot 6}}{2} = ...?\]

Answer 30

24

Answer 31

that's right!

Answer 32

is that the answer ?

Answer 33

yes! Please write the right unit of measure, so we have: \(A=24\,m^2\)
now I go to solve part B.

Answer 34

okay great

Answer 35

here we can find the length of each side of the rombus, so we get:
\[\Large side = \frac{{perimeter}}{4} = \frac{{52}}{4} = ...?\]

Answer 36

13

Answer 37

correct!

Answer 38

now we have this situation:

since the lenght of half diagonal is \(24/2=12\)

Answer 39

okay

Answer 40

again, I apply the theorem of Pitagora, in order to establish the value of the second half diagonal \(x\), so I can write:
\[\Large x = \sqrt {{{13}^2} - {{12}^2}} = \sqrt {169 - 144} = ...?\]

Answer 41

131

Answer 42

hint:
we have: \(169-144=25\), so what is the square root of \(25\)?

Answer 43

5

Answer 44

correct! So the length of second diagonal is: \(5 \cdot 2=10\)

Answer 45

Again I compute the area \(A\) of the rombus like below:
\[\Large A = \frac{{{d_1}{d_2}}}{2} = \frac{{24 \cdot 10}}{2} = ...?\]

Answer 46

120

Answer 47

that's right!

Answer 48

more precisely we have \(A=120\,m^2\)

Answer 49

yay ok

Answer 50

Now, I solve part C.

Answer 51

mhm

Answer 52

here we have an equilateral triangle, and the length of each side is \(15\,m\)
here is the corresponding drawing:

Answer 53

oh okay

Answer 54

CH is the height of the equilateral triangle with respect to the base AB, so, we can write:
\[BH = \frac{{AB}}{2} = \frac{{15}}{2} = 7.5\]

Answer 55

oh okay os BH and CH are the same

Answer 56

No, please CH is different from BH

Now I apply the theorem of Pitagora, so, I can write:
\[\Large x = \sqrt {{{15}^2} - {{7.5}^2}} = \sqrt {225 - 56.25} = ...?\]