anonymous
  • anonymous
area of a parallelogram did i do this right ?
Geometry
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
anonymous
  • anonymous
i got 105?
geerky42
  • geerky42
Yeah 105 is correct answer.

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geerky42
  • geerky42
what about unit?
anonymous
  • anonymous
thanks and um that would be cm
geerky42
  • geerky42
No. \(\text{cm}\times\text{cm} = \text{cm}^2\), right?
geerky42
  • geerky42
Since you did \(7~\text{cm}\times 15~\text{cm}\)
anonymous
  • anonymous
oh yes i forgot sorry
geerky42
  • geerky42
okay, so answer is \(105~\text{cm}^2\)
anonymous
  • anonymous
ok so @geerky42 is this right
geerky42
  • geerky42
Unit is wrong
geerky42
  • geerky42
You did \(\dfrac{1}{2} (8~\text{cm}\times10~\text{cm}) \)
anonymous
  • anonymous
yes
anonymous
  • anonymous
so cm3
geerky42
  • geerky42
How?
anonymous
  • anonymous
1/2 x 8 x 10
geerky42
  • geerky42
1/2 has no unit
anonymous
  • anonymous
oh wait cm2
geerky42
  • geerky42
yeah
anonymous
  • anonymous
ok
geerky42
  • geerky42
You can treat units like variable
geerky42
  • geerky42
\(\text{cm}\times\text{cm} = \text{cm}^2\) Just like how \(x\times x = x^2\)
anonymous
  • anonymous
ah ok that makes more sense
anonymous
  • anonymous
i see
anonymous
  • anonymous
what about this
anonymous
  • anonymous
geerky42
  • geerky42
Area should be \(3.14~\text{yd}^2\)
anonymous
  • anonymous
how is that
geerky42
  • geerky42
Because you have \(\pi r^2\) So you squared unit too.
geerky42
  • geerky42
\[\pi r^2~~~\rightarrow~~~\pi(1~\text{yd})^2\]
anonymous
  • anonymous
ohh
geerky42
  • geerky42
Yeah. What about circumference?
anonymous
  • anonymous
ok and my circumference is 6.28yd2
geerky42
  • geerky42
why \(\text{yd}^2\)?
anonymous
  • anonymous
2*3.14
geerky42
  • geerky42
\[2\pi r ~~~\rightarrow~~~2\pi(1~\text{yd})\]
geerky42
  • geerky42
that "2" has no unit.
anonymous
  • anonymous
ah ok so circumference is usually 1pi
geerky42
  • geerky42
\[C = 2\pi (1~\text{yd})\]
geerky42
  • geerky42
So 6.28 yd
anonymous
  • anonymous
yeah that's the formula i used
anonymous
  • anonymous
ok
anonymous
  • anonymous
im not sure what to do here
geerky42
  • geerky42
It's basically semicircle subtract triangle
anonymous
  • anonymous
if the radius is 6 that means the diameter is 12
anonymous
  • anonymous
huh? so how do I start to solve
geerky42
  • geerky42
find area of triangle and area of semicircle, then you do semicircle - triangle
anonymous
  • anonymous
ah ok onr sec
anonymous
  • anonymous
wait but only the height is given for the triangle how would i find the base
geerky42
  • geerky42
|dw:1433376534008:dw|
geerky42
  • geerky42
So base (aka diameter) is 12 yd.
anonymous
  • anonymous
ok i got that because diameter is twice as much as radius
geerky42
  • geerky42
yeah. Do you understand how I got base.
anonymous
  • anonymous
no how is the diameter the base
anonymous
  • anonymous
?
geerky42
  • geerky42
Try imagine rotating height into base, then you will see that height is half the base.
geerky42
  • geerky42
|dw:1433377025268:dw|
geerky42
  • geerky42
Height is also radius.
anonymous
  • anonymous
ok
geerky42
  • geerky42
Still don't understand?
anonymous
  • anonymous
i see
anonymous
  • anonymous
so what is my next step
geerky42
  • geerky42
Find areas
geerky42
  • geerky42
"find area of triangle and area of semicircle, then you do semicircle - triangle"
anonymous
  • anonymous
ok so then i have 36 for the triangle and 56.52
anonymous
  • anonymous
and you get 20.52
geerky42
  • geerky42
yeah
geerky42
  • geerky42
That's your answer.
geerky42
  • geerky42
unit is?
anonymous
  • anonymous
thanks that wasn't hard
geerky42
  • geerky42
yeah.
anonymous
  • anonymous
yd1
geerky42
  • geerky42
\(\text{yd}^2\)?
anonymous
  • anonymous
srry typo
geerky42
  • geerky42
oh okay
geerky42
  • geerky42
yeah 2
anonymous
  • anonymous
hey do you know about midpoint
geerky42
  • geerky42
yeah
anonymous
  • anonymous
anonymous
  • anonymous
i would do x1+x2/2 y1+y2/2 right
geerky42
  • geerky42
Yeah
geerky42
  • geerky42
\[\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\]
anonymous
  • anonymous
ok i think i got this wring
anonymous
  • anonymous
(4,-12/2) @geerky42
geerky42
  • geerky42
\((-4, -6)\)
geerky42
  • geerky42
\[\left(\dfrac{-6+-2}{2},\dfrac{-8+-4}{2}\right)\]
geerky42
  • geerky42
\[\left(\dfrac{-8}{2},\dfrac{-12}{2}\right)\]
anonymous
  • anonymous
8 divided by 2 is 4
geerky42
  • geerky42
yeah, so -8 divide by 2 is -4.
anonymous
  • anonymous
oh
anonymous
  • anonymous
what about this @geerky42
anonymous
  • anonymous
@geerky42

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