1. superraymond

can you help me first

2. LizzyLove<3

|dw:1364254756475:dw| Find the area of ABCD.

3. Hoa

let CO is the height of the triangle, and let F is the point between A and O. redraw it. I guide you then

4. LizzyLove<3

Ok like this??|dw:1364255189854:dw|

5. LizzyLove<3

@Hoa

6. Hoa

perfect, now, you have 2 regions. right? and you have formula to get the answer, try, the first region is parralelogram ADCF, the area formula is????

7. LizzyLove<3

b*h = Area for rectangle and parallelogram

8. jim_thompson5910

are you sure that BC = 8?

9. jim_thompson5910

it would make things a lot easier if BC = 8

10. jim_thompson5910

i meant, BC = 6

11. LizzyLove<3

Yea:) @jim_thompson5910 It does!!

12. jim_thompson5910

ok one sec

13. LizzyLove<3

Oh i know what ur saying... @jim_thompson5910

14. jim_thompson5910

anyways, we need CO, which will give us the height needed to find the area of the paralleogram but before we find CO, we need FB

15. jim_thompson5910

that length is 32 |dw:1364255489567:dw|

16. LizzyLove<3

Yea...

17. jim_thompson5910

this length is 22 |dw:1364255516091:dw| so this remaining length here |dw:1364255533268:dw| must be 10

18. jim_thompson5910

since 22+10 = 32

19. LizzyLove<3

Huh?

20. LizzyLove<3

Nvm i know what urr doing:)

21. jim_thompson5910

We now have this triangle |dw:1364255559257:dw|

22. LizzyLove<3

And so u have to do the pythagoreum theorm thing...

23. LizzyLove<3

to find 'h'

24. jim_thompson5910

it's not drawn to scale because this is actually a right triangle with hypotenuse of 10 so it really looks like this |dw:1364255624494:dw|

25. LizzyLove<3

But u cant do tht, if it doesnt say in my HW...

26. jim_thompson5910

27. jim_thompson5910

then 10-x must be this length here |dw:1364255697460:dw|

28. LizzyLove<3

Hold on there bro, let me see what ur doing, plz ... @jim_thompson5910

29. jim_thompson5910

so what this means is that we can say x^2 + h^2 = 6^2 and h^2 + (10-x)^2 = 8^2

30. jim_thompson5910

alright sorry

31. LizzyLove<3

|dw:1364255854635:dw| I dont get what u did here... @jim_thompson5910

32. jim_thompson5910

I broke up that length of 10 into two pieces

33. jim_thompson5910

one piece is x units long the other is 10-x units long

34. jim_thompson5910

both add back up to 10 (x) + (10-x) = x + 10-x = 10

35. LizzyLove<3

So y can t u just leave it as 10??

36. LizzyLove<3

@jim_thompson5910

37. jim_thompson5910

because breaking up this length of 10 will help us find the height h

38. jim_thompson5910

it's not the only way to find h, but it's probably one of the easier ways

39. LizzyLove<3

k hold on..

40. jim_thompson5910

alright

41. LizzyLove<3

|dw:1364256139846:dw|

42. jim_thompson5910

yes you break up the triangle like that and that will help you find h

43. LizzyLove<3

ok so now it makes sense...

44. jim_thompson5910

this first triangle here |dw:1364256244802:dw| has legs of x and h, with hypotenuse 6, so a^2 + b^2 = c^2 x^2 + h^2 = 6^2 x^2 + h^2 = 36

45. jim_thompson5910

this triangle here is a right triangle with leg1 = 10-x leg2 = h hypotenuse = 8 which means a^2 + b^2 = c^2 (10-x)^2 + h^2 = 8^2 (10-x)^2 + h^2 = 64

46. jim_thompson5910

with me so far?

47. LizzyLove<3

Kinda, i still dont get y u did 10 - x ?? lol sry

48. jim_thompson5910

well if you break up 10 into two pieces, then one piece is x units long the other must be 10-x units long

49. jim_thompson5910

example if x was say x = 2 inches, then the first piece would be 2 inches the remaining bit would be 10-x = 10-2 = 8 inches

50. LizzyLove<3

Yea i get it but y can u just split it in half??

51. jim_thompson5910

they must add back up to 10, and they do x + (10-x) 2 + (8) 10

52. jim_thompson5910

because we do NOT know whether the length h cuts the side 10 in half or not

53. jim_thompson5910

it could, but it could not (in this case, it doesn't) we just don't know

54. jim_thompson5910

so that's why we must go this route

55. LizzyLove<3

ok... sry for tht, its just i need to know how to do this, thxss :) so wats next?

56. jim_thompson5910

no need for sorries, questions are good

57. LizzyLove<3

lol

58. jim_thompson5910

ok we have these two equations x^2 + h^2 = 36 (10-x)^2 + h^2 = 64

59. jim_thompson5910

we can either use substitution or elimination to solve for x and h

60. jim_thompson5910

let's do substitution

61. LizzyLove<3

ok, i think there is a way simpler way... im pretty sure tht u can split tht in half, my teacher said tht we wouldn't need to use sub. or eli.

62. LizzyLove<3

Can we just assume, b/c it's a right angle??

63. jim_thompson5910

Start with equation 1 and isolate h^2 (not h, just h^2) x^2 + h^2 = 36 h^2 = 36 - x^2 ------------------------------------------------------- Now replace h^2 in equation 2 with 36 - x^2 (since h^2 = 36 - x^2 above) (10-x)^2 + h^2 = 64 (10-x)^2 + (36 - x^2) = 64 Now FOIL out, expand and simplify (10-x)^2 + (36 - x^2) = 64 100 - 20x + x^2 + 36 - x^2 = 64 100 - 20x + 36 = 64 136 - 20x = 64 -20x = 64-136 -20x = -72 x = -72/(-20) x = 18/5 x = 3.6

64. jim_thompson5910

This is why I asked if BC was 8 or not If BC = 6, then you could split the 10 into two equal pieces

65. LizzyLove<3

k let me get this down...

66. jim_thompson5910

alright

67. LizzyLove<3

So how did u get these two equations?? (10-x)^2 + h^2 = 64 x^2 + h^2 = 36

68. LizzyLove<3

@jim_thompson5910

69. jim_thompson5910

through the use of the pythagorean theorem

70. jim_thompson5910

a^2 + b^2 = c^2

71. jim_thompson5910

it's explained a bit more above

72. LizzyLove<3

where did u get 36 from?

73. jim_thompson5910

from squaring the 6

74. LizzyLove<3

So is it sub or elim tht ur doing ? @jim_thompson5910

75. LizzyLove<3

@jim_thompson5910 ??

76. jim_thompson5910

I'm doing substitution

77. LizzyLove<3

kk

78. jim_thompson5910

I isolated h^2 in the first equation and then replaced the other h^2 in the second equation with 36 - x^2

79. jim_thompson5910

notice how the h terms went away after that

80. LizzyLove<3

kk i need to write this

81. jim_thompson5910

alright

82. LizzyLove<3

Sry it took so long... but is this 3.6??

83. jim_thompson5910

good

84. jim_thompson5910

x = 3.6

85. jim_thompson5910

btw I showed how it's x = 3.6 above lol but I'm glad you got that on your own

86. LizzyLove<3

|dw:1364258514592:dw|

87. LizzyLove<3

Is tht 3.6 ??

88. jim_thompson5910

yep, that length you are pointing to is 3.6 units long

89. jim_thompson5910

so x^2 + h^2 = 6^2 x^2 + h^2 = 36 (3.6)^2 + h^2 = 36 12.96 + h^2 = 36 h^2 = 36 - 12.96 h^2 = 23.04 h = sqrt(23.04) h = 4.8

90. LizzyLove<3

kk

91. jim_thompson5910

So the height is 4.8

92. LizzyLove<3

So is tht for this one??

93. LizzyLove<3

|dw:1364258641203:dw|

94. jim_thompson5910

no this one here |dw:1364258674211:dw|

95. jim_thompson5910

that's what we wanted we really don't need x to find the area we only needed x to find h

96. LizzyLove<3

So, i thought u said that

97. LizzyLove<3

|dw:1364258753145:dw| trhis length is 10 - x ??

98. LizzyLove<3

*this

99. jim_thompson5910

yes

100. jim_thompson5910

but we found x, then used that to find h

101. jim_thompson5910

we don't need to know what 10-x is we just want h

102. LizzyLove<3

So the last equation u put was to find 'h' right??

103. LizzyLove<3

@jim_thompson5910

104. jim_thompson5910

exactly

105. LizzyLove<3

k hold on let me get tht down..

106. jim_thompson5910

I started with x^2 + h^2 = 36, plugged in x = 3.6, then solved for h

107. LizzyLove<3

OK!! lol i get what we did, so whats next?? Do we just do 32 * 4.8??

108. LizzyLove<3

@jim_thompson5910

109. jim_thompson5910

ok that's a lot of work to find h

110. jim_thompson5910

like I said, it would be a lot easier if BC = 6

111. jim_thompson5910

but it's still doable to find h even if BC wasn't 6

112. LizzyLove<3

so was i right?? or is there a different way to do this?

113. jim_thompson5910

anyways, this is a trapezoid with base1 = 32 base2 = 22 height = 4.8

114. jim_thompson5910

so we can find the area using this formula A = h*(b1+b2)/2 where in this case h = 4.8 b1 = 32 b2 = 22

115. jim_thompson5910

A = h*(b1+b2)/2 A = 4.8*(32+22)/2 A = 4.8*(54)/2 A = 4.8*27 A = 129.6

116. LizzyLove<3

kk hod it bro

117. jim_thompson5910

alright

118. LizzyLove<3

So thts it?? @jim_thompson5910

119. jim_thompson5910

yep pretty much, unless you want to convert that into a fraction

120. LizzyLove<3

No thank u, i perfer decimals:)

121. jim_thompson5910

ok great

122. jim_thompson5910

yeah decimals are more intuitive with area problems

123. LizzyLove<3

THANK THANK YOU SOOOOSOSOOOO MUCH:D:D:D:D: :):) UR THE BEST :D:D:D

124. jim_thompson5910

lol thx and glad to be of help