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A) State the special name of the triangle AOB.

Mathematics

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Answers

3 parts
b

180∘180^\circ

c

60∘60^\circ

d

7π3 cm\frac{7\pi}{3} \text{ cm}

Step-by-step solution

4 steps

Here is the solution for question 14:

Part a

State the special name of the triangle AOB.

Since AB is a diameter and O is the center, points A, O, and B are collinear. OA and OB are both radii, so OA = OB. A triangle with two equal sides is isosceles. Since the points are collinear, it is a degenerate isosceles triangle.

Part b

180∘180^\circ

Find the value of angle AOB.

Since AB is a diameter and O is the center, A, O, and B lie on a straight line. The angle on a straight line is 180∘180^\circ.
Therefore, ∠AOB=180∘\angle AOB = \boxed{180^\circ}.

Part c

60∘60^\circ

Find the value of angle AOC.

The angle subtended by an arc at the center is twice the angle subtended by the same arc at any point on the circumference.
Arc AC subtends ∠AOC\angle AOC at the center and ∠ABC\angle ABC at the circumference.
Given ∠ABC=30∘\angle ABC = 30^\circ.

∠AOC=2×∠ABC\angle AOC = 2 \times \angle ABC ∠AOC=2×30∘\angle AOC = 2 \times 30^\circ ∠AOC=60∘\angle AOC = \boxed{60^\circ}

Part d

7π3 cm\frac{7\pi}{3} \text{ cm}

If the radius of the circle is 7cm, calculate the length of the arc AC.

  1. Identify the given values.

    Radius r=7 cmr = 7 \text{ cm}.
    Angle subtended by arc AC at the center, ∠AOC=60∘\angle AOC = 60^\circ.

  2. Use the formula for arc length.

    The length of an arc LL is given by the formula:

    L=θ360∘×2πrL = \frac{\theta}{360^\circ} \times 2\pi r

    where θ\theta is the central angle in degrees.

  3. Substitute the values into the formula.

    LAC=60∘360∘×2π(7 cm)L_{AC} = \frac{60^\circ}{360^\circ} \times 2\pi (7 \text{ cm})
  4. Simplify the expression.

    LAC=16×14π cmL_{AC} = \frac{1}{6} \times 14\pi \text{ cm} LAC=14π6 cmL_{AC} = \frac{14\pi}{6} \text{ cm} LAC=7π3 cmL_{AC} = \boxed{\frac{7\pi}{3} \text{ cm}}
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