anonymous
  • anonymous
Please check my work http://prntscr.com/1owzv0 17) h=5 A= ½ (5) (6+9) A= ½ (20) A= 10 18) h=8 A= ½(8) (15 + 10) = ½ (8) (25) = ½ (200) A= 100 19) h=10 A= ½ (10) (12+6) A= ½ (180) A= 90 20) 8+8=16 2+8=10 ½(16)(10) ½(160) A=80in^ 21) ½(6)(6) ½ (36) A= 18m^ 22) 22) A = ½ (8) (6) A= ½ (48) A= 24ft^ 23) A= ½ (20) (30) A= ½ (600) A= 300 ft^ 24) A= ½ (10) (8) A= ½ (80) A= 40 in^ 25) A= ½ (6) (5) A= ½ (30) A= 15 m^
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
*** 17) h=5 A= ½ (5) (6+9) A= ½ (75) A= 37.5
anonymous
  • anonymous
Hold on, checking
anonymous
  • anonymous
okay thank you

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More answers

anonymous
  • anonymous
17 is not correct
anonymous
  • anonymous
h is 4, do you know why?
anonymous
  • anonymous
did you check if i did the areas and everything right
anonymous
  • anonymous
Pythagoras therorem, \[\sqrt{5^2+3^2}=4\]
anonymous
  • anonymous
17. \[width=\sqrt{5^{2}-3^{2}}=\sqrt{25-9}=\sqrt{16}=4\] \[area=\frac{ 6+9 }{2 }\times 4=30\]
anonymous
  • anonymous
Good!
anonymous
  • anonymous
I don't follow Surjithayer
anonymous
  • anonymous
You have a mistakes in all of your heights... the height you should use is |dw:1378140708677:dw|
anonymous
  • anonymous
mistake*
goformit100
  • goformit100
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anonymous
  • anonymous
Omg really -__-
anonymous
  • anonymous
Yep that's a pain xD
anonymous
  • anonymous
18. |dw:1378140558304:dw|
anonymous
  • anonymous
Wow thats confusing, so i did everything wrong
anonymous
  • anonymous
wait so for #18 i just use that height and do the same steps i did up there
anonymous
  • anonymous
take perpendicular distance
anonymous
  • anonymous
For #18, you should use h=6.93, which is what @surjithayer said
anonymous
  • anonymous
you are using slant height
anonymous
  • anonymous
Yeah, about it, but its 11, not 10 A= ½(8) (15 + 11)
anonymous
  • anonymous
|dw:1378141126637:dw|
anonymous
  • anonymous
@surjithayer did a good job there :)
anonymous
  • anonymous
\[AREA=\frac{ 15+11 }{2 }*4\sqrt{3}=52\sqrt{3}\]
anonymous
  • anonymous
why are you guys using trig to solve this, you're confusing me
anonymous
  • anonymous
Okay did I do 20 ~ 25 correct?
anonymous
  • anonymous
IN THIS QUSTION THEN HOW YOU FIND h and AE or DC
anonymous
  • anonymous
wait, i have to take my lunch.
anonymous
  • anonymous
Okay but can you tell me if i did 20 through 25 right
anonymous
  • anonymous
@Z3R0
anonymous
  • anonymous
Good
anonymous
  • anonymous
Wait
anonymous
  • anonymous
Are you sure, becuase you said 17,18,19 were also good, but ended up being wrong
anonymous
  • anonymous
Did I said that? I don't think so...
anonymous
  • anonymous
you only said 17 was incorrect and I asked you if i did the areas and everything right you said "Good"
anonymous
  • anonymous
But anyway, for #25, |dw:1378142242276:dw|
anonymous
  • anonymous
Aw sorry I thought you were saying "areas and everything" means only to #17...
anonymous
  • anonymous
ugh i don't understand any of these problems, i don't know which one is right and which one is wrong
anonymous
  • anonymous
#17 is right
anonymous
  • anonymous
how about 20
anonymous
  • anonymous
#18 and #19 is wrong, #20 is right
anonymous
  • anonymous
21, and 22 ?
anonymous
  • anonymous
Hold on...
anonymous
  • anonymous
Correct
anonymous
  • anonymous
OMG yay! :D
anonymous
  • anonymous
23, and 24
anonymous
  • anonymous
All good
anonymous
  • anonymous
According to http://www.wikihow.com/Calculate-the-Area-of-a-Rhombus
anonymous
  • anonymous
And yeah, as for #25, use the method I've showed you
anonymous
  • anonymous
23. |dw:1378144069265:dw|
anonymous
  • anonymous
25. |dw:1378144251590:dw|
anonymous
  • anonymous
BD=2*4=8 AREA=8*6=48
anonymous
  • anonymous
CORRECTION AREA=1/2*8*6=24
anonymous
  • anonymous
CORRECTION FOR 23' 1/2*60*40=1200 IF SEMI DIAGONAL IS GIVEN.
anonymous
  • anonymous
Can you please do 17 and 18 step by step please
anonymous
  • anonymous
are you here ?
anonymous
  • anonymous
18. |dw:1378175217644:dw| \[area of tapezoid= \frac{\sum of \parallel lines }{2 } \times perpendicular distance.\] \[=\frac{ 11+15 }{2 }\times 4\sqrt{3}=\frac{ 26 }{ 2 }\times 4\sqrt{3}=52\sqrt{3}\]
anonymous
  • anonymous
you can also solve by area of triangle and area of rectangle \[area=\frac{ 1 }{2 }\times 4\times 4\sqrt{3}+11\times4\sqrt{3}=8\sqrt{3}+44\sqrt{3}=52\sqrt{3}\]
anonymous
  • anonymous
oh okay can you please do 17 and 19, thats it. I have 13 more of the similar questions please do those 2

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