## Emily778 2 years ago Find the value of each variable. If your answer is not an integar, express it in simplest radical form.

1. Emily778

|dw:1362695099329:dw|

2. Emily778

I need to know step by step on how to solve this.

3. Emily778

4. jim_thompson5910

is this a right triangle?

5. Emily778

|dw:1362695278150:dw|

6. jim_thompson5910

ok thank you

7. jim_thompson5910

|dw:1362695556789:dw|

8. jim_thompson5910

this is because all angles must add to 180 notice how 45 + 45 + 90 = 180

9. jim_thompson5910

so we have a 45-45-90 triangle which means that the two legs are congruent

10. jim_thompson5910

|dw:1362695629297:dw|

11. jim_thompson5910

if the leg is x, then the hypotenuse is x*sqrt(2) So hypotenuse = x*sqrt(2) hypotenuse = sqrt(2)*sqrt(2) hypotenuse = (sqrt(2))^2 hypotenuse = 2

12. jim_thompson5910

|dw:1362695698593:dw|

13. Emily778

if you already figured out the hypotenuse's are the same, why'd you do all that work up there?

14. jim_thompson5910

the legs are the same, not the hypotenuses

15. Emily778

oh. Should I put down the equations you did to get a decent grade?

16. jim_thompson5910

and I did all that work based on the fact that if you have a 45-45-90 triangle, then you'll have this basic template |dw:1362695974272:dw|

17. jim_thompson5910

also, if you know the legs, you can find the hypotenuse using the pythagorean theorem, which is a^2 + b^2 = c^2

18. Emily778

you're kinda confusing me. Like, I'm understanding it more, but the way you're explaining it. You're just kind of throwing it out randomly. I need a step by step procedure.

19. jim_thompson5910

alright, did you see how I figured out how x = sqrt(2) ?

20. jim_thompson5910

the legs of a 45-45-90 triangle are congruent x is a leg and so is the sqrt(2) length, so that means x = sqrt(2)

21. jim_thompson5910

with me so far?

22. Emily778

can you write it down, like.."step 1, step 2" to make it more simple?

23. jim_thompson5910

the legs of a 45-45-90 triangle are congruent so x must be sqrt(2) units long since the other leg is sqrt(2) units long this means x = sqrt(2) --------------------------------------------- we know both legs are sqrt(2) units, so let's use them to find the hypotenuse y a^2 + b^2 = c^2 ( sqrt(2) )^2 + ( sqrt(2) )^2 = y^2 2 + 2 = y^2 4 = y^2 y^2 = 4 y = sqrt( 4 ) y = 2 So the hypotenuse is 2 units

24. jim_thompson5910

|dw:1362696451678:dw|

25. Emily778

would I write that down on a paper?

26. jim_thompson5910

yeah if you wanted to, you could also use the template above as a shortcut

27. jim_thompson5910

this template |dw:1362696580301:dw|

28. Emily778

Ugh..I'll show you what I have on my paper and you say if it's correct or not..

29. jim_thompson5910

ok

30. Emily778

|dw:1362696649159:dw|

31. jim_thompson5910

ok good so far, you just have shown that the missing angle is 45 degrees

32. jim_thompson5910

oh and you threw in x = sqrt(2)

33. jim_thompson5910

I would say why x is sqrt(2) and you can say that the legs are congruent (because it's a 45-45-90 triangle)

34. Emily778

okay

35. jim_thompson5910

the next thing you need to show is how to get y

36. jim_thompson5910

there are 2 ways to do it: use the template above or the pythagorean theorem

37. jim_thompson5910

both ways are shown in this post

38. Emily778

can you show them again. you can copy them

39. jim_thompson5910

here is the pythagorean theorem method

40. jim_thompson5910

the legs of a 45-45-90 triangle are congruent so x must be sqrt(2) units long since the other leg is sqrt(2) units long this means x = sqrt(2) --------------------------------------------- we know both legs are sqrt(2) units, so let's use them to find the hypotenuse y a^2 + b^2 = c^2 ( sqrt(2) )^2 + ( sqrt(2) )^2 = y^2 2 + 2 = y^2 4 = y^2 y^2 = 4 y = sqrt( 4 ) y = 2 So the hypotenuse is 2 units

41. jim_thompson5910

sound good?

42. Emily778

Yeah. Sorry, I was off the computer. Taking a break. :P

43. Emily778

Wait..Y=2? @jim_thompson5910

44. jim_thompson5910

yes, y = 2

45. jim_thompson5910

|dw:1362701799670:dw|

46. Emily778

could I make it look like this?|dw:1362701962485:dw|

47. Emily778

from the one thing you showed me? instead of putting down "sqrt"?

48. Emily778

@jim_thompson5910

49. jim_thompson5910

yes you could do it that way (it's a bit cleaner looking)

50. Emily778

just a question.. How does it equal y?

51. jim_thompson5910

what do you mean?

52. Emily778

nevermind I think I understand.

53. jim_thompson5910

are you sure?

54. jim_thompson5910

alright, I'm glad it's (sorta) making sense

55. Emily778

so, x= sqrt2? It can't be simplified?

56. jim_thompson5910

no it cannot

57. jim_thompson5910

that's as simple as it gets for x

58. Emily778

okay, can you help me with 2 more? Or no?

59. jim_thompson5910

sure I can help

60. jim_thompson5910

61. Emily778

the second one explains the same thing, but the problem is different.

62. Emily778

I'll draw it

63. jim_thompson5910

alright

64. Emily778

|dw:1362702888077:dw|

65. Emily778

:/

66. jim_thompson5910

the legs are equal (we have another 45-45-90 triangle)

67. jim_thompson5910

so y = x

68. Emily778

would I write down that y=x?

69. jim_thompson5910

use the pythagorean theorem to get |dw:1362703031520:dw|

70. jim_thompson5910

then you plug in y = x |dw:1362703073963:dw|

71. jim_thompson5910

and you solve for x

72. Emily778

is that " X^2 + X^2"?

73. Emily778

where did the y go?

74. jim_thompson5910

y = x so y and x are the same

75. Emily778

ohh. I get ya. How do I solve for X for this?

76. jim_thompson5910

so you can replace y with x

77. Emily778

How do i solve for x?

78. jim_thompson5910

well the two x^2 terms add up to 2x^2

79. jim_thompson5910

so you will have 2x^2 = 450 then what?

80. Emily778

well, I'd think you'd divide the 450 by 2 but it's squared..so, I don't know..

81. jim_thompson5910

yes, 2x^2 = 450 x^2 = 450/2 x^2 = 225 x = sqrt(225) x = 15

82. jim_thompson5910

so both x and y are 15 (since y = x)

83. Emily778

Oh, I get it. Thankyou. :)

84. Emily778

here's the 3rd one..

85. jim_thompson5910

yw

86. Emily778

|dw:1362704401057:dw|

87. jim_thompson5910

this is another 45-45-90 triangle (the legs are congruent)

88. Emily778

Okay

89. jim_thompson5910

so what does that mean for the value of y?

90. Emily778

it's bigger

91. jim_thompson5910

yes and what else

92. jim_thompson5910

how can you find the value of y?

93. Emily778

isn't it a^2 +b^2=c^2?

94. jim_thompson5910

very good

95. jim_thompson5910

replace a with sqrt(5) replace b with sqrt(5) and you want to solve for c

96. Emily778

I know. :)

97. jim_thompson5910

ok great

98. Emily778

Do I actually solve sqrt5^2 + sqrt5^2 in the calculator?

99. jim_thompson5910

well the square of the square roots cancel each other out

100. jim_thompson5910

so (sqrt(5))^2 = 5

101. jim_thompson5910

so you are left with 5 + 5 which turns into 10

102. jim_thompson5910

so c^2 = 10

103. Emily778

I'm confused. How did you get 5+5?

104. jim_thompson5910

|dw:1362706218096:dw|

105. jim_thompson5910

so that's how I got 2 five terms being added

106. Emily778

wouldn't I replace C with Y?

107. jim_thompson5910

you would

108. jim_thompson5910

so y^2 = 10

109. jim_thompson5910

making |dw:1362706377944:dw|

110. Emily778

is that it for that problem?

111. Emily778

@jim_thompson5910

112. jim_thompson5910

yes, you found y, so you're done

113. Emily778

Oh, I'm so sorry. but I don't get this other one. Can you help me with it? c:

114. jim_thompson5910

sure

115. Emily778

|dw:1362707591687:dw|

116. Emily778

@jim_thompson5910

117. Emily778

hullo?

118. jim_thompson5910

sry was afk for a sec

119. jim_thompson5910

were you able to get anywhere with this at all?

120. Emily778

No. :/

121. Emily778

I think Y = 2 sqrt 3?

122. Emily778

@jim_thompson5910

123. jim_thompson5910

sry i had to go afk, but I'm back

124. jim_thompson5910

y is actually 2 the legs are not congruent

125. jim_thompson5910

x = 2*y x = 2*2 x =4

126. jim_thompson5910

the reason how I know y is 2 is from this template |dw:1362709753724:dw|

127. Emily778

um, 60 degrees is down below and down below, its 2 sqrt3

128. Emily778

@jim_thompson5910

129. jim_thompson5910

well i flipped the triangle around, but if it were oriented the way the problem has it, then you would have |dw:1362710229246:dw|

130. Emily778

|dw:1362710294844:dw|

131. Emily778

you missed the 3?

132. Emily778

I mean 2

133. jim_thompson5910

the 2*sqrt(3) is above though isnt it?

134. Emily778

I just showed you the diagram. :P

135. jim_thompson5910

but in this one, the 2*sqrt(3) is above |dw:1362710579138:dw|

136. Emily778

Are you supposed to change what I showed you?

137. jim_thompson5910

no i just copied what you drew

138. Emily778

you're mixing up everything I just drew and you know that. ._.

139. Emily778

look up at the picture again..

140. jim_thompson5910

yeah i see the new one, but you drew a totally different one

141. jim_thompson5910

or is it a case the second one is the correct one and the first one was a typo?

142. Emily778

OH I'm sorry!

143. jim_thompson5910

yeah i was like...what in the world lol

144. Emily778

that was my fault.

145. jim_thompson5910

its fine, so which one is it?

146. Emily778

the second one is correct.

147. jim_thompson5910

ok one sec

148. Emily778

k

149. Emily778

are you gonna help? I don't have all day.

150. jim_thompson5910

|dw:1362712584316:dw|

151. jim_thompson5910

the value of x is exactly half that of the hypotenuse

152. Emily778

That's correct. xP Lol.

153. jim_thompson5910

yeah just wanted to make sure and put it up again

154. jim_thompson5910

so x = 2*sqrt(3)/2 = sqrt(3)

155. jim_thompson5910

which means |dw:1362712746325:dw|

156. Emily778

wait stop

157. jim_thompson5910

ok whats up

158. Emily778

is it 2 times sqrt 3/2?

159. jim_thompson5910

yes

160. jim_thompson5910

the 2 terms cancel

161. Emily778

so I write this down? x = 2*sqrt(3)/2 = sqrt(3)

162. jim_thompson5910

yes and that will give you the value of x

163. Emily778

Just to make sure we're on the same topic..|dw:1362713238955:dw|

164. jim_thompson5910

yep

165. Emily778

is that how it should be shown?

166. jim_thompson5910

yes you have it correct, so now that you have x, you can find y

167. jim_thompson5910

y = x*sqrt(3) y = sqrt(3)*sqrt(3) y = 3

168. Emily778

so, Y= 3 and X= Sqrt 3

169. Emily778

?

170. jim_thompson5910

yes good

171. Emily778

wanna help me with one more please? :D

172. jim_thompson5910

ok one more

173. Emily778

:) Thanks!

174. jim_thompson5910

sure thing

175. Emily778

|dw:1362714107413:dw|

176. jim_thompson5910

|dw:1362714258167:dw|

177. jim_thompson5910

x = 2*12 x = 24 since the hypotenuse is double the shorter leg in a 30-60-90 triangle

178. jim_thompson5910

y = 12*sqrt(3) because of the template given above

179. Emily778

where'd you get the 2 in the first one?

180. jim_thompson5910

in general, with any 30-60-90 triangle the hypotenuse is twice as long as the shorter leg

181. Emily778

oh, so that's why it's 2 x 24..Okay,

182. jim_thompson5910

2 * 12 = 24, but yeah you have it

183. Emily778

what's Y?

184. jim_thompson5910

y = 12*sqrt(3)

185. jim_thompson5910

sry mixed up the 30 and the 60

186. jim_thompson5910

|dw:1362714824453:dw|

187. Emily778

What is the answer for Y? I got 20..:/

188. jim_thompson5910

20 is incorect

189. jim_thompson5910

incorrect*

190. jim_thompson5910

|dw:1362714907609:dw|

191. Emily778

you can't simplify it?

192. jim_thompson5910

no

193. jim_thompson5910

that's as simplified as it gets

194. Emily778

how did you find y?

195. jim_thompson5910

in the template above, do you see how the longer leg is x*sqrt(3) ?

196. Emily778

ohhh.. now I get it.

197. Emily778

Thankyou ! :)

198. jim_thompson5910

ok I would write that template down, it's very handy

199. jim_thompson5910

also the 45-45-90 template as well

200. Emily778

thankyou for helping me

201. jim_thompson5910

you're welcome, sry for being afk a lot lol

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