NotThatSmart

We can use the law of cosines to solve this problem. 

First,let's say

\(AB = a = x\\ AC = b + 2x+1 \\ BC = c = 3x+5 \\ ∠BAC = 120°\)

The law of cosine states that 

From the problem, we can write \(cos(120) = \frac{(3x+5)^2+(2x+1)^2 - x^2}{2(2x+1)(3x+5)}\)

Expanding out everything and simplifying, we get \(18x^{2}+47x+31=0\)

Using the quadratic formula, we have \(x=\frac{-{47}\pm \sqrt{{47}^{2}-4\cdot {18}\cdot {31}}}{2\cdot{18}}\)

We get \(x=-\frac{47}{36}\pm \sqrt{23}\cdot (\frac{1}{36}i)\)

So \(x=-\frac{47}{36}+\sqrt{23}\cdot (\frac{1}{36}i)\)

Thanks! :)

Edit: THIS IS MY 200th answer! Not flexing...but noice :)

Very similar question

First up, let's isolate x for \(\frac{xy}{x+y}=2:\\ \)

We get \(x=\frac{2y}{y-2}\)

Subbing this in, we get 

\(\frac{\frac{2y}{y-2}z}{\frac{2y}{y-2}+z}=2\\ \frac{yz}{y+z}=2\)

\(\frac{2yz}{2y+z\left(y-2\right)}=2\\ \frac{yz}{y+z}=2\)

Now, we isolate y. 

\(\frac{2yz}{2y+z\left(y-2\right)}=2 \\ y=z\)

Subsituting in y=z, we get

\(\frac{zz}{z+z}=2\)

\(z=4\)

We get z=4 as our answer. 

Thanks!:)

Let's let the tangent point between Semicircle AB and the red circle as D, the tangent point between arc BC and the red circle as E, and the tangent point between the red circle and semicircle AC as F.  Label the center of semicircle AB to C1 the center of semicircle of BC to C2 and the radius of the red circle to C3.Also, let the radius be x. 

We know from the problem that \(BF=2\) so that means \(BC_3 = 2-x\)

We also know that \(C_1C_3 = 1+x.\)

Now, draw \(C_1C_3B.\) Notice this is a right triangle. 

Now, we can use the pythaogrean thereom to finish this poblem. 

\({1}^{2}+{(2-x)}^{2}={(1+x)}^{2}\) The x in the equation is cancelled out to get the radius of 2/3. 

2/3 is our answer. 

Thanks! :)

I don't see a graph posted for this problem. 

I'm not sure if there is supposed to be one, but this problem is impsossible without a graph. 

Thanks! :)

The diameter of AB is \(9 + 16 + 9 = 34.\) This means the radius is \(34/2 = 17\)

However, let's note something. 

\(QD = \sqrt{9(9 + 16)} = 15\), but in the problem, it says the square sidelenghth is 16. This means that PQRS is NOT a square, but a rectange with sidelengths 16 and 15. 

Like I said, the diameter is 34. We already know the height of the trapezoid, which we got earlier as 15. DC is 16, since PQRS is a rectangle. 

We have everything we need to solve for the area of the trapezoid! 

The area of a trapezoid is \(\frac{(b1+b2)h}{2}\) where b1 and b2 are the bases and h is the height. 

We have \(\frac{(25+16)15}{2} = (31)15/2 = 232.5\)

so the area of the trapezoid is 232.5. 

Thanks! :)

We want to put \( x^2+y^2=2x-6y+16+14x-4y+20\) in the form of a circle's equation. 

Moving all terms to one side, we get \(x^2-16x+y^2+10y=36\)

Completing the square for both x and y, we get \((x-8)^2+(y+5)^2=125\)

\(\left(x-8\right)^2+\left(y+5\right)^2=\left(5\sqrt{5}\right)^2\)

From this, we get that \(5\sqrt5\) is the radius. 

The area is \(\pi r^2 = \pi (125) = 125\pi\)

125pi is our answer

Thanks! :)