Question

in the diagram, jl = 120. what are jk and kl? 4x + 6 7x + 15
q is the midpoint of pr. what are pq, qr, and pr? 6x - 7 5x + 1
if ∠rqt = 155, what are m∠rqs and m∠tqs? (4x - 20)° (3x + 14)°
the diameter of a circle is 22cm. what is the area of the circle?

Explanation:

Step1: Solve for \(x\) in the first - part (segment \(JL\))

Since \(JL = JK+KL\) and \(JL = 120\), \(JK = 4x + 6\), \(KL=7x + 15\), we have the equation \((4x + 6)+(7x + 15)=120\).
Combining like - terms: \(4x+7x+6 + 15=120\), which simplifies to \(11x+21 = 120\).
Subtract 21 from both sides: \(11x=120 - 21=99\).
Divide both sides by 11: \(x = 9\).
Then \(JK=4x + 6=4\times9+6=36 + 6=42\) and \(KL=7x + 15=7\times9+15=63 + 15=78\).

Step2: Solve for \(x\) in the second - part (segment \(PR\))

Since \(Q\) is the mid - point of \(PR\), \(PQ = QR\). So \(6x-7=5x + 1\).
Subtract \(5x\) from both sides: \(6x-5x-7=5x-5x + 1\), which gives \(x-7 = 1\).
Add 7 to both sides: \(x=8\).
Then \(PQ=6x-7=6\times8-7=48 - 7=41\), \(QR = 5x + 1=5\times8+1=40 + 1=41\), and \(PR=PQ + QR=41+41 = 82\).

Step3: Solve for \(x\) in the third - part (angle \(\angle RQT\))

Since \(\angle RQT=\angle RQS+\angle TQS\) and \(\angle RQT = 155^{\circ}\), \(\angle RQS=(4x - 20)^{\circ}\), \(\angle TQS=(3x + 14)^{\circ}\), we have the equation \((4x-20)+(3x + 14)=155\).
Combining like - terms: \(4x+3x-20 + 14=155\), which simplifies to \(7x-6 = 155\).
Add 6 to both sides: \(7x=155 + 6=161\).
Divide both sides by 7: \(x = 23\).
Then \(\angle RQS=4x-20=4\times23-20=92 - 20=72^{\circ}\) and \(\angle TQS=3x + 14=3\times23+14=69 + 14=83^{\circ}\).

Step4: Solve for the area of the circle

The formula for the area of a circle is \(A=\pi r^{2}\), and the diameter \(d = 22\mathrm{cm}\), so the radius \(r=\frac{d}{2}=\frac{22}{2}=11\mathrm{cm}\).
Then \(A=\pi\times(11)^{2}=121\pi\mathrm{cm}^{2}\approx121\times3.14 = 379.94\mathrm{cm}^{2}\).

Answer:

For \(JL\): \(JK = 42\), \(KL = 78\).
For \(PR\): \(PQ = 41\), \(QR = 41\), \(PR = 82\).
For \(\angle RQT\): \(\angle RQS = 72^{\circ}\), \(\angle TQS = 83^{\circ}\).
For the circle: \(A = 121\pi\mathrm{cm}^{2}\approx379.94\mathrm{cm}^{2}\)