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In the figure below, QT is a tangent to the circle at Q. AT is a straight line and QD is a chord, given that AX = 8cm, PT = 12cm, QX = 6.8cm and XD = 1cm.

Mathematics

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Question 11:
The problem provides a diagram with lengths and a description. There is an inconsistency between the lengths provided in the diagram for the two secants, which means the power of a point theorem (TP⋅TA=TX⋅TDTP \cdot TA = TX \cdot TD) cannot hold true for all given values simultaneously. We will proceed by using the values from one secant for calculations.

a) Find the length of XP.
The segment XP is a chord connecting points P and X on the circle. With the given information (lengths of secant segments and tangent), there is no direct theorem or formula to calculate the length of chord XP. Therefore, XP cannot be determined from the given data.

b) Find the length of QT.
We use the tangent-secant theorem, which states that the square of the length of the tangent segment from an external point is equal to the product of the length of the secant segment from the external point to the farther intersection point and the length of the secant segment from the external point to the nearer intersection point.
From the diagram, we have:
External point: T
Tangent segment: TQ
Secant segment TPA: TP = 12 cm, PA = 8 cm.
So, the whole secant length TA=TP+PA=12+8=20TA = TP + PA = 12 + 8 = 20 cm.

  1. Apply the tangent-secant theorem.

    TQ2=TP⋅TATQ^2 = TP \cdot TA

  2. Substitute the known values.

    TQ2=12 cm⋅(12+8) cmTQ^2 = 12 \text{ cm} \cdot (12 + 8) \text{ cm} TQ2=12 cm⋅20 cmTQ^2 = 12 \text{ cm} \cdot 20 \text{ cm} TQ2=240 cm2TQ^2 = 240 \text{ cm}^2

  3. Solve for TQ.

    TQ=240 cmTQ = \sqrt{240} \text{ cm} TQ=16×15 cmTQ = \sqrt{16 \times 15} \text{ cm} TQ=415 cmTQ = 4\sqrt{15} \text{ cm} TQ≈15.49 cmTQ \approx 15.49 \text{ cm}

    Since the problem has inconsistent data (using the other secant TXD: TX=6.8TX = 6.8 cm, XD=7XD = 7 cm, so TD=6.8+7=13.8TD = 6.8 + 7 = 13.8 cm, which would give TQ2=6.8⋅13.8=93.84TQ^2 = 6.8 \cdot 13.8 = 93.84, so TQ=93.84≈9.69TQ = \sqrt{93.84} \approx 9.69 cm), we have chosen to use the secant TPA.

    a) XP cannot be determined with the given information.

    b) QT≈15.49 cm\boxed{\text{QT} \approx \textbf{15.49 cm}}

    Question 12: Make rr the subject of the formula m=ht2r2+2wm = \sqrt{\frac{ht^2}{r^2+2w}}.

  4. Square both sides of the equation to remove the square root.

    m2=ht2r2+2wm^2 = \frac{ht^2}{r^2+2w}

  5. Multiply both sides by (r2+2w)(r^2+2w) to clear the denominator.

    m2(r2+2w)=ht2m^2(r^2+2w) = ht^2

  6. Divide both sides by m2m^2.

    r2+2w=ht2m2r^2+2w = \frac{ht^2}{m^2}

  7. Subtract $2w$ from both sides.

    r2=ht2m2−2wr^2 = \frac{ht^2}{m^2} - 2w

  8. Take the square root of both sides.

    r=±ht2m2−2wr = \pm \sqrt{\frac{ht^2}{m^2} - 2w}

    The subject of the formula is r=±ht2m2−2w\boxed{r = \pm \sqrt{\frac{ht^2}{m^2} - 2w}}.

    Question 13: Find the value of xx given that (5x23x)\begin{pmatrix} 5x & 2 \\ 3 & x \end{pmatrix} is a singular matrix.

    A matrix is singular if its determinant is equal to zero. For a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, the determinant is ad−bcad - bc.

  9. Calculate the determinant of the given matrix.

    det(5x23x)=(5x)(x)−(2)(3)\text{det}\begin{pmatrix} 5x & 2 \\ 3 & x \end{pmatrix} = (5x)(x) - (2)(3) det=5x2−6\text{det} = 5x^2 - 6

  10. Set the determinant to zero for a singular matrix.

    5x2−6=05x^2 - 6 = 0

  11. Solve for xx.

    5x2=65x^2 = 6 x2=65x^2 = \frac{6}{5} x=±65x = \pm \sqrt{\frac{6}{5}} x=±1.2x = \pm \sqrt{1.2} x≈±1.095x \approx \pm 1.095

    The value of xx is x=±65\boxed{x = \pm \sqrt{\frac{6}{5}}}.

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