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3 years ago

The sides of a triangle are x , x +1 , 2 x -1 and its area is x root of 10 .Find the value of x.

1 answer:

6 0

Ok, I'm going to start off saying there is probably an easier way of doing this that's right in front of my face, but I can't see it so I'm going to use Heron's formula, which is A=√[s(s-a)(s-b)(s-c)] where A is the area, s is the semiperimeter (half of the perimeter), and a, b, and c are the side lengths.

Substitute the known values into the formula:

x√10=√{[(x+x+1+2x-1)/2][({x+x+1+2x-1}/2)-x][({x+x+1+2x-1}/2)-(x+1)][({x+x+1+2x-1}/2)-(2x-1)]}

Simplify:

x√10=√{[4x/2][(4x/2)-x][(4x/2)-(x+1)][(4x/2)-(2x-1)]}

x√10=√[2x(2x-x)(2x-x-1)(2x-2x+1)]

x√10=√[2x(x)(x-1)(1)]

x√10=√[2x²(x-1)]

x√10=√(2x³-2x²)

10x²=2x³-2x²

2x³-12x²=0

2x²(x-6)=0

2x²=0 or x-6=0

x=0 or x=6

Therefore, x=6 (you can't have a length of 0).

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Please EXPLAIN how to get this answer so I can do it with other problems. A straight answer would be great, but I have to do abo

OLEGan [10]

Answer:

PQ = 3.58, and RQ = 10.4

Step-by-step explanation:

We are given the hypotenuse of the triangle, and an angle. Use sin and cos to solve.

Hypotenuse = 11,

Opposite side is PQ

Adjacent side is RQ

x = 19

Sin x = (opposite side)/(hypotenuse)

Cos x = (adjacent side)/(hypotenuse)

For PQ, this is the side opposite to the angle, so use sin,

Sin 19 = x/11

11(Sin 19) = x

3.58 = x (rounded to the nearest hundredth)

For RQ, this is the side adjacent to the angle, so use cos,

Cos 19 = x/11

11(Cos 19) = x

10.4 = x (rounded to the nearest hundredth)

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DENIUS [597]

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3 years ago

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igor_vitrenko [27]

Answer:

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Step-by-step explanation:

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denis23 [38]

Answer:

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Step-by-step explanation:

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1 - Remember that two lines perpendicular to the same segment are parallel.

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Masteriza [31]

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