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In the adjoining figure, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to chord AB at point E. Show that triangle ABC is an isosceles triangle.

Mathematics

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Q. Cost of making a closed cone

  1. Identify the given values.

    Radius of base, r=6r = 6 m
    Slant height, l=8l = 8 m
    Rate of making = Rs. 10 per sq.m

  2. Calculate the total surface area (TSA) of the closed cone.

    The formula for the total surface area of a closed cone is TSA=πr(r+l)TSA = \pi r (r+l).

    TSA=π(6 m)(6 m+8 m)TSA = \pi (6 \text{ m}) (6 \text{ m} + 8 \text{ m}) TSA=π(6 m)(14 m)TSA = \pi (6 \text{ m}) (14 \text{ m}) TSA=84π m2TSA = 84\pi \text{ m}^2

  3. Calculate the total cost.

    Total cost = TSA ×\times Rate

    Total cost=84π m2×Rs. 10/m2Total \text{ cost} = 84\pi \text{ m}^2 \times \text{Rs. } 10/\text{m}^2 Total cost=840π Rs.Total \text{ cost} = 840\pi \text{ Rs.}

    Using π≈3.14159\pi \approx 3.14159:

    Total cost≈840×3.14159 Rs.Total \text{ cost} \approx 840 \times 3.14159 \text{ Rs.} Total cost≈2638.9356 Rs.Total \text{ cost} \approx 2638.9356 \text{ Rs.}

    The cost of making the closed cone is 840π Rs.\boxed{840\pi \text{ Rs.}} or approximately 2638.94 Rs.\boxed{2638.94 \text{ Rs.}}.

    Q.5. Solve any 1 out of 2.

    • i) If sin⁡ϕ=12\sin \phi = \frac{1}{2} then find cos⁡ϕ\cos \phi.

  4. Use the fundamental trigonometric identity.

    $$
\sin^2 \phi + \cos^2 \phi = 1
$$

  5. Substitute the given value of sin⁡ϕ\sin \phi.

    $$
\left(\frac{1}{2}\right)^2 + \cos^2 \phi = 1
$$
$$
\frac{1}{4} + \cos^2 \phi = 1
$$

  6. Solve for cos⁡2ϕ\cos^2 \phi.

    $$
\cos^2 \phi = 1 - \frac{1}{4}
$$
$$
\cos^2 \phi = \frac{3}{4}
$$

  7. Solve for cos⁡ϕ\cos \phi.

    $$
\cos \phi = \pm \sqrt{\frac{3}{4}}
$$
$$
\cos \phi = \pm \frac{\sqrt{3}}{2}
$$
The value of $\cos \phi$ is $\boxed{\pm \frac{\sqrt{3}}{2}}$.
    • ii) In the adjoining figure, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to chord AB at point E. Show that △ABC\triangle ABC is an isosceles triangle.

  8. Identify given information.

    • CD is a diameter of the circle with centre O.
    • CD is perpendicular to chord AB at point E. This means CE⊥ABCE \perp AB.

  9. Apply circle theorems.

    A perpendicular from the centre of a circle to a chord bisects the chord. Since CD passes through the centre O and is perpendicular to chord AB at E, E must be the midpoint of AB.  
Therefore, $AE = EB$.

  10. Consider triangles △CAE\triangle CAE and △CBE\triangle CBE.

    • AE=EBAE = EB (from Step 2).
    • ∠CEA=∠CEB=90∘\angle CEA = \angle CEB = 90^\circ (given that CD ⊥\perp AB).
    • CECE is a common side to both triangles.

  11. Prove congruence.

    By the Side-Angle-Side (SAS) congruence criterion, $\triangle CAE \cong \triangle CBE$.

  12. Conclude that △ABC\triangle ABC is isosceles.

    Since $\triangle CAE \cong \triangle CBE$, their corresponding parts are equal (CPCTC).  
Therefore, $CA = CB$.  
Since two sides of $\triangle ABC$ are equal ($CA = CB$), $\triangle ABC$ is an isosceles triangle.
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