Triangle problems in math
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Triangle problems in math

[From: ] [author: ] [Date: 11-07-08] [Hit: ]
you have a hypotenuse of a and a side length of b√(2). Lets set up a ratio of the hypotenuse to the leg; it is a / [b√(2)]Let y be the length of the leg of the other triangle, the one with hypotenuse of c. The ratio of the hypotenuse to the leg is c/ySince the two triangles are similar, these two ratios are equala / [b√(2)] = c/ySolve for y.y = c / {a / [b√(2)]} = c * b√(2)/aIf y is the leg of the triangle and we want the area of the triangle (remember that the lengths of the two legs are equal),......
4√(3) = s√(3)/2
8 = s
The perimeter is 3 times the side, thus it is 3*8 = 24

5) d) 27
Each side is equal, so pick any two side and set them equal to each other and solve for x.
x = (x/3)+6
x - (x/3) = 6
(2/3)x = 6
x = 6* (3/2) = 9
The perimeter is 3 * 9 = 27

6) c) (b^2•c^2)/a^2
This one is tough, so it will take a little explanation.
If the area is b², and the formula for area is .5 L*W, then that must mean that both the length and the width are equal since there is only one variable. Let the length and width be x. The area is then .5x². Set this equal to b² and solve for x in terms of b.
.5x² = b²
x²= 2b²
x = b√(2)
In this triangle, you have a hypotenuse of a and a side length of b√(2). Let's set up a ratio of the hypotenuse to the leg; it is a / [b√(2)]
Let y be the length of the leg of the other triangle, the one with hypotenuse of c. The ratio of the hypotenuse to the leg is c/y
Since the two triangles are similar, these two ratios are equal
a / [b√(2)] = c/y
Solve for y.
y = c / {a / [b√(2)]} = c * b√(2)/a
If y is the leg of the triangle and we want the area of the triangle (remember that the lengths of the two legs are equal), in terms of y, the area is .5y².
Substitute c * b√(2)/a for y into the formula
.5y² = .5 * {c * b√(2)/a}² = .5 * c² * b² * 2/ a² = c² * b²/ a² = (b^2•c^2)/a^2

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Just1 (6,8,10)

6 (each side is 2)

1 to 2 (each side is doubled)

24 (each side is 8)

x = x/3 + 6 --> 2x/3 = 6 --> 2x = 18 --> x = 9 So perimeter is 27

(ab^2)/c or (cb^2)/a (We don't which is larger)

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1.A the only pythagorean triples with sides less that small are 3-4-5 and its multiples, like 6-8-10

2.C. Area = s^2*sqr3/4, so s^2/4 would have to equal 1. So s = 2 and P= 6

3.A. The perimeters are in the same ratio as the sides

4. E. An equilateral triangle has 60 degree angles. The alt. Forms a 30-60-90 right triangle with
The side as hypotenuse. So the side would be 8.

5.D. The sides would be equal , so x = 9. the sides are 9, 9 , and 9

6. C the ratio of the areas = (ratio of sides)^2 For similar figures so (a/c)^2= b^2/x

Good luck!
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