If there's a hypotenuse, then you need to use the Pythagorean Theorem:
a²+b²=c²
a=leg
b=leg
c=hypotenuse
Plug in the information you have:
a²+60²=61²
a²+3600=3721
Subtract 3600 from both sides:
a²=121
You have to get the square root of both sides to get "a" alone
a=11
2006-09-18 06:19:42
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answer #1
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answered by Anonymous
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11
2006-09-18 06:20:28
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answer #2
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answered by spoiledprincess1269 1
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Pythagorean theorem:
A^2 + B^2 = C^2 where C is the hypotenuse
Let A = 60, and C = 61:
60^2 + B^2 = 61^2
3600 + B^2 = 3721
B^2 = 121
B = 11
Check:
60^2 + 11^2 = 61^2
3600 + 121 = 3721
3721 = 3721 (check!)
2006-09-18 06:20:27
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answer #3
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answered by ³√carthagebrujah 6
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11
the hypotenuse is 61 so using the pythagoreum therem, A^2 + B^2 = C^2, C is 61, C^2 is 3721
B is for the length 60 so B^2 is 3600
C^2 - B^2 is 3721 - 3600 = A^2 = 121
square root of 121 is 11
2006-09-18 06:23:07
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answer #4
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answered by Dennis K 4
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If the base is 60units and the hypotenuse is 61 units in length,
By the Pythagoras Theorem,
Perpendicular*Perpendicular=(Hypotenuse*Hypotenuse)-(Base*Base)
=61*61-60*60
=3721-3600
=121
Therefore,
Perpendicular=11 units
2006-09-18 06:22:41
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answer #5
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answered by Anonymous
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properly one way is which you would be waiting to apply the pythagorean theorum that's a squared + b squared = c squared and that interprets to 6 squared + b squared = 11 squared and then b = root 80 5 and that i think of via fact its a 30 60 ninety triangle the hypotenuse would would desire to be 12 and then h would equivalent 6 root 3
2016-12-18 12:31:21
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answer #6
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answered by shorb 4
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well don't forget the formula a^2 + b^2 = c^2 in this case its c^2 - a^2 = b^2. 60 x 60 = 3600 = a^2. 61x61 = 3721 = c^2. 3721- 3600 = 121. The square root of 121 is 11 which is the answer ^.^
2006-09-18 06:23:44
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answer #7
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answered by drE-drE 2
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61 squared - 60 squared + 121
square root of 121 = 11
11
2006-09-18 06:20:21
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answer #8
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answered by Anonymous
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People, people people -- the Pythagorean theorum is for RIGHT triangles. We do not know if that is the case here. Also -- why are you giving this person the answer? HELPING would be to provide the necessary formula or point the person in the direction of finding the correct formula.
2006-09-18 06:28:31
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answer #9
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answered by Anonymous
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I think there is a problem in this question. Though you have talked of base and hypotenuse, am not sure if you have in mind a right angled tiangle.
2006-09-18 06:24:49
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answer #10
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answered by Kinyua J 2
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