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

Mathematics

ScanSolve AI Answer

Answer

A

circumcircle

Step-by-step solution

12 steps

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

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

  • iv) 2tan⁡45∘+cos⁡45∘−sin⁡45∘2 \tan 45^\circ + \cos 45^\circ - \sin 45^\circ

  1. Recall the values of the trigonometric functions.

    $\tan 45^\circ = 1$  
$\cos 45^\circ = \frac{1}{\sqrt{2}}$  
$\sin 45^\circ = \frac{1}{\sqrt{2}}$

  2. Substitute the values into the expression.

    $$
2 \tan 45^\circ + \cos 45^\circ - \sin 45^\circ = 2(1) + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}
$$

  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?

  4. The relationship between radius (rr) and diameter (dd) is d=2rd = 2r.

  5. Substitute the given radius r=4r = 4 cm.

    $$
d = 2 \times 4 \text{ cm}
$$
$$
d = 8 \text{ cm}
$$
The diameter is $\boxed{\text{8 cm}}$.

    • 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 x=0\boxed{x=0}.

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

    • iv) Find the volume of a cube if length = 1cm?

  6. The formula for the volume (VV) of a cube with side length (ss) is V=s3V = s^3.

  7. Substitute the given length s=1s = 1 cm.

    $$
V = (1 \text{ cm})^3
$$
$$
V = 1 \text{ cm}^3
$$
The volume of the cube is $\boxed{\text{1 cm}^3}$.

    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 = 12\boxed{12} cm

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

      In the right angled △\triangle OPD, using Pythagoras' theorem OP2+PD2=OD2OP^2 + PD^2 = OD^2.

  8. Substitute the known values into the Pythagorean theorem.

    $$
12^2 + PD^2 = 20^2
$$

  9. Calculate the squares.

    $$
144 + PD^2 = 400
$$

  10. Solve for PD2PD^2.

    $$
PD^2 = 400 - 144
$$
$$
PD^2 = 256
$$

  11. Find PD by taking the square root.

    $$
PD = \sqrt{256}
$$
$$
PD = 16 \text{ cm}
$$

  12. Since the perpendicular from the center bisects the chord, CD=2×PDCD = 2 \times PD.

    $$
CD = 2 \times 16 \text{ cm}
$$
$$
CD = 32 \text{ cm}
$$
The length of the chord is $\boxed{\text{32 cm}}$.
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