Question

In the diagram, JK = LM = 24, NP= 3x, and NQ = 7x- 12. Find the radius of ON.

Answer 1

The radius of the circle is 15

Step-by-step explanation

Problem Statement

The question gives us a circle with two chords LM and JK that are equal in length and line segments NQ = 7x - 12 and NP = 3x. We are asked to find the radius of the circle.

Solution

- We have chord LM intersecting with line NQ at a right angle and chord JK intersecting with line NP at a right angle as well. This means we can apply the Chord and line segment theorem.

- This theorem states that "If a chord intersects a line drawn from the center of the circle, at a right angle, then the line bisects the chord into two equal parts."

- The implication of this theorem is that:

`\begin{gathered} \text{ Implication 1:} \\ LQ=QM \\ \\ \text{ Implication 2:} \\ LQ+QM=LM \\ \text{ Since }LQ=QM \\ 2LQ=LM \\ \therefore LQ=QM=(LM)/(2) \\ \\ \text{But we know that }LM=24 \\ \therefore LQ=QM=(24)/(2)=12 \end{gathered}`

- The same theorem can be applied to line NP and chord JK. And we shall reach similar conclusions. That is:

`\begin{gathered} \text{ Implication 1:} \\ JP=PK \\ \\ \text{ Implication 2:} \\ JP=PK=(JK)/(2)=(24)/(2) \\ JP=PK=12 \end{gathered}`

- We can redraw the circle to get a better understanding as shown below:

- Observe that NLQ and NJP form right-angle triangles. The hypotenuse of both triangles is indicated as r, which is also the radius of the circle. This means that we can apply the Pythagoras' theorem.

`\begin{gathered} \text{ Triangle NLQ:} \\ NQ^2+LQ^2=LN^2 \\ NQ=7x-12 \\ LQ=12 \\ LN=r \\ \\ \therefore(7x-12)^2+12^2=r^2\text{ (Equation 1)} \\ \\ \\ \text{ Triangle NJP:} \\ JP^2+NP^2=JN^2 \\ NP=3x \\ JP=12 \\ JN=r \\ \\ \therefore(3x)^2+12^2=r^2\text{ (Equation 1)} \end{gathered}`

- Notice that the two equations above, equate to r-squared. This means that they are equal to each other. When we equate them, we can get the value of x.

`\begin{gathered} (7x-12)^2+12^2=r^2 \\ (3x)^2+12^2=r^2 \\ \\ \therefore(7x-12)^2+12^2=(3x)^2+12^2 \\ \text{Subtract }12^2\text{ from both sides} \\ (7x-12)^2=(3x)^2 \\ \text{Take the squre root of both sides} \\ \sqrt[]{(7x-12)^2}=\sqrt[]{(3x)^2^{}} \\ \\ 7x-12=3x \\ \text{subtract 3x from both sides and add 12 to both sides} \\ 7x-3x=12 \\ 4x=12 \\ \text{Divide both sides by 4} \\ (4x)/(4)=(12)/(4) \\ \\ \therefore x=3 \end{gathered}`

- Now that we have the value of x, we can find the value of the radius (r) using any of the two equations from above.

- The value of radius (r) is gotten as follows:

`\begin{gathered} (3x)^2+12^2=r^2 \\ x=3 \\ (3*3)^2+12^2=r^2 \\ 9^2+12^2=r^2 \\ 81+144=r^2 \\ r^2=225 \\ \text{Take the square root of both sides} \\ r=15 \end{gathered}`

Final Answer

The radius of the circle is 15

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