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Two chords AB and CD intersect internally at X. Given that AX = 10cm, XB = 8cm and CD = 21cm, determine the length XD

Mathematics

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14. Determine the length XD

  1. Apply the Intersecting Chords Theorem.

    When two chords AB and CD intersect internally at a point X, the product of the segments of one chord equals the product of the segments of the other chord.

    AX⋅XB=CX⋅XDAX \cdot XB = CX \cdot XD

    Given AX=10 cmAX = 10 \text{ cm}, XB=8 cmXB = 8 \text{ cm}, and CD=21 cmCD = 21 \text{ cm}.
    Let XD=xXD = x. Then CX=CD−XD=21−xCX = CD - XD = 21 - x.

  2. Substitute the values into the theorem and form a quadratic equation.

    10⋅8=(21−x)⋅x10 \cdot 8 = (21 - x) \cdot x 80=21x−x280 = 21x - x^2

    Rearrange the equation into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0:

    x2−21x+80=0x^2 - 21x + 80 = 0
  3. Solve the quadratic equation for xx.

    We can factor the quadratic equation. We need two numbers that multiply to 80 and add up to -21. These numbers are -5 and -16.

    (x−5)(x−16)=0(x - 5)(x - 16) = 0

    This gives two possible values for xx:

    x−5=0  ⟹  x=5x - 5 = 0 \implies x = 5 x−16=0  ⟹  x=16x - 16 = 0 \implies x = 16
  4. State the possible lengths for XD.

    Both values are positive and physically possible lengths.
    If XD=5 cmXD = 5 \text{ cm}, then CX=21−5=16 cmCX = 21 - 5 = 16 \text{ cm}.
    If XD=16 cmXD = 16 \text{ cm}, then CX=21−16=5 cmCX = 21 - 16 = 5 \text{ cm}.
    Both solutions satisfy the theorem (10⋅8=8010 \cdot 8 = 80 and 5⋅16=805 \cdot 16 = 80).

    The length XD can be either 5 cm\boxed{\text{5 cm}} or 16 cm\boxed{\text{16 cm}}.

    15. Calculate the area bounded by the curve y=x2+2xy = x^2 + 2x, the lines x=1,x=3x = 1, x = 3 and the x-axis.

  5. Set up the definite integral for the area.

    The area AA bounded by the curve y=f(x)y = f(x), the x-axis, and the lines x=ax=a and x=bx=b is given by ∫abf(x) dx\int_{a}^{b} f(x)\,dx.
    For y=x2+2xy = x^2 + 2x in the interval [1,3][1, 3], the function is positive, so the area is directly given by the integral.

    A=∫13(x2+2x) dxA = \int_{1}^{3} (x^2 + 2x)\,dx
  6. Integrate the function.

    ∫(x2+2x) dx=x2+12+1+2x1+11+1+C=x33+2x22+C=x33+x2+C\int (x^2 + 2x)\,dx = \frac{x^{2+1}}{2+1} + \frac{2x^{1+1}}{1+1} + C = \frac{x^3}{3} + \frac{2x^2}{2} + C = \frac{x^3}{3} + x^2 + C
  7. Evaluate the definite integral using the limits of integration.

    A=[x33+x2]13A = \left[ \frac{x^3}{3} + x^2 \right]_{1}^{3} A=(333+32)−(133+12)A = \left( \frac{3^3}{3} + 3^2 \right) - \left( \frac{1^3}{3} + 1^2 \right) A=(273+9)−(13+1)A = \left( \frac{27}{3} + 9 \right) - \left( \frac{1}{3} + 1 \right) A=(9+9)−(13+33)A = (9 + 9) - \left( \frac{1}{3} + \frac{3}{3} \right) A=18−43A = 18 - \frac{4}{3} A=543−43A = \frac{54}{3} - \frac{4}{3} A=503A = \frac{50}{3}

    The area is 503 square units\boxed{\frac{50}{3} \text{ square units}}.

    16. Determine the coordinate of point A.

  8. Write down the given information in vector form.

    Point P has coordinates (8,4,−1)(8, 4, -1), so its position vector is p⃗=8i⃗+4j⃗−k⃗\vec{p} = 8\vec{i} + 4\vec{j} - \vec{k}.
    Point P divides line AB internally in the ratio 4:1, meaning AP:PB=4:1AP:PB = 4:1. So, m=4m=4 and n=1n=1.
    The position vector of point B is b⃗=i⃗+3j⃗−2k⃗\vec{b} = \vec{i} + 3\vec{j} - 2\vec{k}.
    Let the position vector of point A be a⃗=xi⃗+yj⃗+zk⃗\vec{a} = x\vec{i} + y\vec{j} + z\vec{k}.

  9. Apply the section formula for internal division.

    The formula for a point P dividing a line segment AB in the ratio m:nm:n is:

    p⃗=na⃗+mb⃗m+n\vec{p} = \frac{n\vec{a} + m\vec{b}}{m+n}

    Substitute the known values:

    8i⃗+4j⃗−k⃗=1⋅a⃗+4⋅(i⃗+3j⃗−2k⃗)1+48\vec{i} + 4\vec{j} - \vec{k} = \frac{1 \cdot \vec{a} + 4 \cdot (\vec{i} + 3\vec{j} - 2\vec{k})}{1+4} 8i⃗+4j⃗−k⃗=a⃗+4i⃗+12j⃗−8k⃗58\vec{i} + 4\vec{j} - \vec{k} = \frac{\vec{a} + 4\vec{i} + 12\vec{j} - 8\vec{k}}{5}
  10. Solve for the position vector a⃗\vec{a}.

    Multiply both sides by 5:

    5(8i⃗+4j⃗−k⃗)=a⃗+4i⃗+12j⃗−8k⃗5(8\vec{i} + 4\vec{j} - \vec{k}) = \vec{a} + 4\vec{i} + 12\vec{j} - 8\vec{k} 40i⃗+20j⃗−5k⃗=a⃗+4i⃗+12j⃗−8k⃗40\vec{i} + 20\vec{j} - 5\vec{k} = \vec{a} + 4\vec{i} + 12\vec{j} - 8\vec{k}

    Isolate a⃗\vec{a}:

    a⃗=(40i⃗+20j⃗−5k⃗)−(4i⃗+12j⃗−8k⃗)\vec{a} = (40\vec{i} + 20\vec{j} - 5\vec{k}) - (4\vec{i} + 12\vec{j} - 8\vec{k}) a⃗=(40−4)i⃗+(20−12)j⃗+(−5−(−8))k⃗\vec{a} = (40 - 4)\vec{i} + (20 - 12)\vec{j} + (-5 - (-8))\vec{k} a⃗=36i⃗+8j⃗+(−5+8)k⃗\vec{a} = 36\vec{i} + 8\vec{j} + (-5 + 8)\vec{k} a⃗=36i⃗+8j⃗+3k⃗\vec{a} = 36\vec{i} + 8\vec{j} + 3\vec{k}
  11. State the coordinate of point A.

    The position vector a⃗=36i⃗+8j⃗+3k⃗\vec{a} = 36\vec{i} + 8\vec{j} + 3\vec{k} corresponds to the coordinate (36,8,3)(36, 8, 3).

    The coordinate of point A is (36,8,3)\boxed{(36, 8, 3)}.

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