The radius of a circle is 4cm, then find the diameter?

Q1 A. Choose the correct alternative for the following questions:

• i) The circle which passes through all the vertices of a triangle is called A circle that passes through all the vertices of a triangle is called a circumcircle. The correct option is (A) circumcircle.

• ii) Any point on the line $y=x$ is of the form For any point on the line $y=x$, its x-coordinate and y-coordinate are equal. Therefore, a point on this line is of the form $(a, a)$. The correct option is (A) $(a,a)$.

• iii) Which of the following is the value of $\sin 90^\circ$? The value of $\sin 90^\circ$ is 1. The correct option is (D) 1.

• iv) $2 \tan 45^\circ + \cos 45^\circ - \sin 45^\circ$ Step 1: Recall the values of the trigonometric functions. $\tan 45^\circ = 1$ $\cos 45^\circ = \frac{1}{\sqrt{2}}$ $\sin 45^\circ = \frac{1}{\sqrt{2}}$

  Step 2: Substitute the values into the expression. $2 \tan 45^\circ + \cos 45^\circ - \sin 45^\circ = 2(1) + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}$

  Step 3: Simplify the expression. $2(1) + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 2 + 0 = 2$ The correct option is (C) 2.

Q1 B. Solve the following.

• i) If the radius of a circle is 4cm, then find the diameter? Step 1: The relationship between radius ($r$) and diameter ($d$) is $d = 2r$. Step 2: Substitute the given radius $r = 4$ cm. $d = 2 \times 4 \text{ cm}$ $d = 8 \text{ cm}$ The diameter is $\boxed{8 cm}$.

• ii) What is the equation of Y-axis? The Y-axis is the vertical line where the x-coordinate of every point is 0. The equation of the Y-axis is $\boxed{x=0}$.

• iii) Find the value of $\tan 45^\circ$? The value of $\tan 45^\circ$ is 1. The value is $\boxed{1}$.

• iv) Find the volume of a cube if length = 1cm? Step 1: The formula for the volume ($V$) of a cube with side length ($s$) is $V = s^3$. Step 2: Substitute the given length $s = 1$ cm. $V = (1 cm)^3$ $V = 1 cm^3$ The volume of the cube is $\boxed{1 cm^3}$.

Q.2(A) Complete 2 activities out of 3. (4marks)

• i) Radius of a circle is 20cm. Distance of a chord from the centre of a circle is 12cm. Find the length of the chord.

  Solution: Let the centre of the circle be O. Radius = OD = 20 cm. Distance of the chord CD from O is 12 cm. seg OP $\perp$ seg CD.

  OP = $\boxed{12}$ cm

  Now CP = $\boxed{16}$ cm (perpendicular drawn from the centre bisects the chord).

  In the right angled $\triangle$ OPD, using Pythagoras' theorem $OP^2 + PD^2 = OD^2$. Step 1: Substitute the known values into the Pythagorean theorem. $12^2 + PD^2 = 20^2$ Step 2: Calculate the squares. $144 + PD^2 = 400$ Step 3: Solve for $PD^2$. $PD^2 = 400 - 144$ $PD^2 = 256$ Step 4: Find PD by taking the square root. $PD = \sqrt{256}$ $PD = 16 \text{ cm}$ Step 5: Since the perpendicular from the center bisects the chord, $CD = 2 \times PD$. $CD = 2 \times 16 \text{ cm}$ $CD = 32 \text{ cm}$ The length of the chord is $\boxed{32 cm}$.