**Math Labs with Activity – Angle Subtended by an Arc at the Centre of a Circle is Double**

**OBJECTIVE**

To verify that the angle subtended by an arc at the centre of a circle is double the angle subtended by it at any point on the remaining part of the circle

**Materials Required**

- A sheet of white paper
- A piece of cardboard
- A sheet of tracing paper
- A sheet of colored paper
- A geometry box
- A tube of glue

**Theory**

The theorem may be proved as follows.

Consider a circle with centre O and radius r. Let AB be any arc of this circle. Join AB. Join CO and produce it to a point D outside the circle (see Figure 27.1).

Then, ∠AOD = ∠OAC + ∠ACO and ∠BOD = ∠OBC + ∠BCO.

[When a side of a triangle is produced, the exterior angle is equal to the sum of the opposite interior angles.]

But ∠OAC = ∠ACO [OC=OA=r]

and ∠OBC = ∠BCO. [OC=OB=r]

**∴** ∠AOD=2 ∠ACO and ∠BOD = 2∠BCO.

**∴** ∠AOB = ∠AOD + ∠BOD = 2 ∠ACO + 2 ∠BCO.

= 2 (∠ACO + ∠BCO) = 2 ∠ACB.

**Procedure**

**Step 1:** Paste the sheet of white paper on the cardboard. Mark a point O on this paper. With O as the centre, draw a circle of any radius.

**Step 2:** Mark two points A and B on this circle. Then, AB is an arc of the circle. Mark any point C on the remaining part of the circle. Join OA, OB, AC and BC. Then, ∠AOB is the angle subtended by AB at the centre of the circle and ∠ACB is the angle subtended by AB at any point on the remaining part of the circle (see Figure 27.2).

**Step 3:** Trace ∠ACB on the tracing paper and use it to make two exact copies of ∠ACB on the coloured paper as shown in Figure 27.3.

Cut these two copies of ∠ACB. Place them adjacent to each other over ∠AOB as shown in Figure 27.4.

**Observations**

We observe that the two copies of ∠ACB when placed adjacent to each other exactly cover ∠AOB.

Therefore, ∠AOB = 2 ∠ACB.

**Result**

It is verified that the angle subtended by an arc at the centre of a circle is double the angle subtended by it at any point on the remaining part of the circle.

**Remarks:**

If AB is a semicircle then ∠AOB = 180° and so, by the above theorem, ∠ACB = 90°, as shown in Figure 27.5.

Clearly, ∠ACB is an angle in a semicircle.

We, therefore, conclude that the angle in a semicircle is a right angle.

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