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In the given figure, AB||CD. Find the value of x, y and z.

Mathematics

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Answer

150π m2 or approximately 471.24 m2150\pi \text{ m}^2 \text{ or approximately } 471.24 \text{ m}^2

1:8

x=75∘,y=55∘,z=50∘x = 75^\circ, y = 55^\circ, z = 50^\circ

Step-by-step solution

5 steps
  1. Address Question 28 (Part 1).

    The question "a circle of radius 17 cm. Find the distance between the chords." is incomplete. To find the distance between chords, their lengths or other properties must be provided.

  2. Address Question 28 (Part 2).

    Find the square metres of canvas required for a conical tent with height h=3.5h = 3.5 m and base radius r=12r = 12 m.
    The canvas required is the lateral surface area of the cone, given by A=πrlA = \pi r l, where ll is the slant height.
    First, calculate the slant height ll:

    l=r2+h2l = \sqrt{r^2 + h^2} l=(12)2+(3.5)2l = \sqrt{(12)^2 + (3.5)^2} l=144+12.25l = \sqrt{144 + 12.25} l=156.25l = \sqrt{156.25} l=12.5 ml = 12.5 \text{ m}

    Now, calculate the lateral surface area:

    A=π×12 m×12.5 mA = \pi \times 12 \text{ m} \times 12.5 \text{ m} A=150π m2A = 150\pi \text{ m}^2

    Using π≈3.14159\pi \approx 3.14159:

    A≈150×3.14159 m2A \approx 150 \times 3.14159 \text{ m}^2 A≈471.2385 m2A \approx 471.2385 \text{ m}^2

    The canvas required is 150π m2 or approximately 471.24 m2\boxed{150\pi \text{ m}^2 \text{ or approximately } 471.24 \text{ m}^2}.

  3. Address Question 29 (Part 1).

    The surface areas of two spheres are in the ratio 1:41:4. Find the ratio of their volumes.
    Let the radii of the two spheres be r1r_1 and r2r_2.
    The surface area of a sphere is A=4πr2A = 4\pi r^2.
    Given A1A2=4πr124πr22=14\frac{A_1}{A_2} = \frac{4\pi r_1^2}{4\pi r_2^2} = \frac{1}{4}.

    r12r22=14\frac{r_1^2}{r_2^2} = \frac{1}{4} (r1r2)2=14\left(\frac{r_1}{r_2}\right)^2 = \frac{1}{4}

    Taking the square root of both sides:

    r1r2=14=12\frac{r_1}{r_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}

    The volume of a sphere is V=43πr3V = \frac{4}{3}\pi r^3.
    The ratio of their volumes is:

    V1V2=43πr1343πr23=r13r23=(r1r2)3\frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \frac{r_1^3}{r_2^3} = \left(\frac{r_1}{r_2}\right)^3

    Substitute the ratio of radii:

    V1V2=(12)3=18\frac{V_1}{V_2} = \left(\frac{1}{2}\right)^3 = \frac{1}{8}

    The ratio of their volumes is 1:8\boxed{1:8}.

  4. Address Question 29 (Part 2).

    In the given figure, AB∥CDAB \parallel CD. Find the value of x,yx, y and zz.
    Given:

    • Line ABAB is parallel to line CDCD.
    • ∠AEF=75∘\angle AEF = 75^\circ.
    • ∠EGD=125∘\angle EGD = 125^\circ.
    • x=∠EFGx = \angle EFG, y=∠FGEy = \angle FGE, z=∠FEGz = \angle FEG.
    1. Find xx: Since AB∥CDAB \parallel CD, and EF is a transversal, ∠AEF\angle AEF and ∠EFG\angle EFG are alternate interior angles. Therefore, ∠EFG=∠AEF\angle EFG = \angle AEF.

      x=75∘x = 75^\circ
    2. Find yy: Since AB∥CDAB \parallel CD, and EG is a transversal, ∠EGD\angle EGD and ∠FGE\angle FGE are angles on a straight line with the angle adjacent to ∠EGD\angle EGD on line CD. The angle ∠FGE\angle FGE and ∠EGD\angle EGD are not directly related as alternate interior or corresponding. However, ∠FGE\angle FGE and ∠EGC\angle EGC are supplementary if G is on CD. The angle ∠FGE\angle FGE is part of the triangle. The angle ∠EGC\angle EGC is 180∘−∠EGD180^\circ - \angle EGD (angles on a straight line). ∠EGC=180∘−125∘=55∘\angle EGC = 180^\circ - 125^\circ = 55^\circ. So, y=∠FGE=55∘y = \angle FGE = 55^\circ.

    3. Find zz: The sum of angles in a triangle is 180∘180^\circ. In △EFG\triangle EFG:

      ∠FEG+∠EFG+∠FGE=180∘\angle FEG + \angle EFG + \angle FGE = 180^\circ z+x+y=180∘z + x + y = 180^\circ z+75∘+55∘=180∘z + 75^\circ + 55^\circ = 180^\circ z+130∘=180∘z + 130^\circ = 180^\circ z=180∘−130∘z = 180^\circ - 130^\circ z=50∘z = 50^\circ

    The values are x=75∘,y=55∘,z=50∘\boxed{x = 75^\circ, y = 55^\circ, z = 50^\circ}.

  5. Address Question 30.

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