Question

3c) find both x and y.
sc 4 i can apply the pythagorean theorem and its converse, the distance formula
4a) how wide is this a - frame ski lodge at the bottom?
4b) how far apart are these points? (17, 29.5) and (-32.5, 80)
4c) what is the length of the side marked with “x”?

Explanation:

Step1: Solve for \(x\) and \(y\) in the angle - related problem

We know that vertical angles are equal. So, \(6x - 11=2x - 9\).
Subtract \(2x\) from both sides: \(6x-2x - 11=2x-2x - 9\), which gives \(4x-11=-9\).
Add 11 to both sides: \(4x-11 + 11=-9 + 11\), so \(4x = 2\).
Divide both sides by 4: \(x=\frac{2}{4}=\frac{1}{2}\).
Also, since \((6x - 11)+(20y + 19)=180\) (linear - pair of angles), substitute \(x=\frac{1}{2}\) into it.
\(6\times\frac{1}{2}-11+20y + 19 = 180\).
\(3-11+20y + 19 = 180\).
\(20y+(3 + 19-11)=180\).
\(20y + 11 = 180\).
Subtract 11 from both sides: \(20y=180 - 11=169\).
Divide both sides by 20: \(y=\frac{169}{20}=8.45\).

Step2: Solve 4a (width of A - frame ski lodge)

No relevant dimensions are given in the image for 4a, so we cannot solve it.

Step3: Solve 4b (distance between two points)

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
Here, \(x_1 = 17,y_1 = 29.5,x_2=-32.5,y_2 = 80\).
\(d=\sqrt{(-32.5 - 17)^2+(80 - 29.5)^2}\).
\(d=\sqrt{(-49.5)^2+(50.5)^2}\).
\(d=\sqrt{2450.25 + 2550.25}\).
\(d=\sqrt{5000.5}\approx70.71\).

Step4: Solve 4c (length of side marked \(x\))

Using the Pythagorean theorem in the right - triangle.
The two legs of the right - triangle are \(12\) and \(6\), and the hypotenuse of the smaller right - triangle is \(\sqrt{12^{2}+6^{2}}=\sqrt{144 + 36}=\sqrt{180}=6\sqrt{5}\).
The other right - triangle has one leg \(6\sqrt{5}\) and hypotenuse \(14\).
By the Pythagorean theorem \(x=\sqrt{14^{2}-(6\sqrt{5})^{2}}=\sqrt{196 - 180}=\sqrt{16}=4\).

Answer:

\(x=\frac{1}{2},y = 8.45\) (for the angle problem); Cannot solve 4a; Distance in 4b is approximately \(70.71\); \(x = 4\) in 4c.