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.

Q. Cost of making a closed cone

Step 1: Identify the given values. Radius of base, $r = 6$ m Slant height, $l = 8$ m Rate of making = Rs. 10 per sq.m

Step 2: Calculate the total surface area (TSA) of the closed cone. The formula for the total surface area of a closed cone is $TSA = \pi r (r+l)$. $TSA = \pi (6 m) (6 m + 8 m)$ $TSA = \pi (6 m) (14 m)$ $TSA = 84\pi m^2$

Step 3: Calculate the total cost. Total cost = TSA $\times$ Rate $Total cost = 84\pi m^2 \times Rs. 10/m^2$ $Total cost = 840\pi \text{ Rs.}$ Using $\pi \approx 3.14159$: $Total cost \approx 840 \times 3.14159 \text{ Rs.}$ $Total cost \approx 2638.9356 \text{ Rs.}$ The cost of making the closed cone is $\boxed{840\pi Rs.}$ or approximately $\boxed{2638.94 Rs.}$.

Q.5. Solve any 1 out of 2.

• i) If $\sin \phi = \frac{1}{2}$ then find $\cos \phi$. Step 1: Use the fundamental trigonometric identity. $\sin^2 \phi + \cos^2 \phi = 1$ Step 2: Substitute the given value of $\sin \phi$. $\left(\frac{1}{2}\right)^2 + \cos^2 \phi = 1$ $\frac{1}{4} + \cos^2 \phi = 1$ Step 3: Solve for $\cos^2 \phi$. $\cos^2 \phi = 1 - \frac{1}{4}$ $\cos^2 \phi = \frac{3}{4}$ Step 4: Solve for $\cos \phi$. $\cos \phi = \pm \sqrt{\frac{3}{4}}$ $\cos \phi = \pm \frac{\sqrt{3}}{2}$ 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 $\triangle ABC$ is an isosceles triangle. Step 1: Identify given information. • CD is a diameter of the circle with centre O. • CD is perpendicular to chord AB at point E. This means $CE \perp AB$.

  Step 2: 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, $AE = EB$.

  Step 3: Consider triangles $\triangle CAE$ and $\triangle CBE$. • $AE = EB$ (from Step 2). • $\angle CEA = \angle CEB = 90^\circ$ (given that CD $\perp$ AB). • $CE$ is a common side to both triangles.

  Step 4: Prove congruence. By the Side-Angle-Side (SAS) congruence criterion, $\triangle CAE \cong \triangle CBE$.

  Step 5: Conclude that $\triangle ABC$ 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.