Answer:
3
Step-by-step explanation:
See picture for solution steps and answer.
Triangle = Equilateral Triangle
Therefore, all sides are equivalent to each other.
WORKING OUT:
5x - 22 = 4x - 10
5x - 4x = - 10 + 22
x = 12
CHECKING USING OTHER EQUATION COMBINATIONS:
4x - 10 = 3x + 2
4x - 3x = 2 + 10
x = 12
5x - 22 = 3x + 2
5x - 3x = 2 + 22
2x = 24
x = 24 / 2
x = 12
ANSWER:
x = 12
Answer:
x = √(a(a+b))
Step-by-step explanation:
We can also assume a > 0 and b > 0 without loss of generality. (If a and a+b have opposite signs, the maximum angle is 180° at x=0.)
We choose to define tan(α) = -(b+a)/x and tan(β) = -a/x. Then the tangent of ∠APB is ...
tan(∠APB) = (tan(α) -tan(β))/(1 +tan(α)tan(β))
= ((-(a+b)/x) -(-a/x))/(1 +(-(a+b)/x)(-a/x))
= (-bx)/(x^2 +ab +a^2)
This will be maximized when its derivative is zero.
d(tan(∠APB))/dx = ((x^2 +ab +a^2)(-b) -(-bx)(2x))/(x^2 +ab +a^2)^2
The derivative will be zero when the numerator is zero, so we want ...
bx^2 -ab^2 -a^2b = 0
b(x^2 -(a(a+b))) = 0
This has solutions ...
b = 0
x = √(a(a+b))
The former case is the degenerate case where ∠APB is 0, and the value of x can be anything.
The latter case is the one of interest:
x = √(a(a+b)) . . . . . . the geometric mean of A and B rotated to the x-axis.
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<em>Comment on the result</em>
This result is validated by experiments using a geometry program. The location of P can be constructed in a few simple steps: Construct a semicircle through the origin and B. Find the intersection point of that semicircle with a line through A parallel to the x-axis. The distance from the origin to that intersection point is x.
The total area of the room is 37.6376 and the no. of cans required to paint the wall is 3 cans.
The measurement of two of the walls is 2.86 metres and 3.16 metre
Area of the two walls = 2(length x breadth)
Area = 2(2.86 x 3.16) = 18.0752 m²
The measurement of the other two walls is 2.86 metres and 3.42 metres
Area of the two walls = 2(length × breadth)
Area = 2(2.86 × 3.42) = 19.5624 m²
Total area = 18.0752 + 19.5624 = 37.6376 m²
If one can of paint can cover 15 m², the no. of cans required to paint the bedroom will be
No. of cans = Total area/Area covered by one can of paint
No. of cans = 37.6376/15 = 2.5091 = 3 cans (approx.)
