Question

the measure of a central angle equals one half of the measure of the arc it intercepts

true or false?​

Answer 1

Answer:

Inscribed Angle Theorem:

The measure of an inscribed angle is half the measure of the intercepted arc. That is, m∠ABC=12m∠AOC. This leads to the corollary that in a circle any two inscribed angles with the same intercepted arcs are congruent. Here, ∠ADC≅∠ABC≅∠AFC.

Answer 2

• False .
• The measure of a central angle equals same as the measure of the arc it intercepts .

To Find :- The measure of a central angle equals one half of the measure of the arc it intercepts .

• True or False ?

Solution :-

Central angle :- Central angle is the angle formed at the centre by the vertex . It is equal to the the arc it intercepts .

Example :- Let Length of arc it intercepts is equal to 60° . Then, measure of central angle is also equal to 60° . Or if central angle is 50° , then, length of arc is also equal to 50° .

Therefore, we can conclude that, given statement is False .

Extra knowledge :-

→ Inscribed angle :- Angle formed at the circumference . The measure of the inscribed angle is one half of the measure of the arc it intercepts .

Example :- If length of arc is equal to 70°, then inscribed angle is equal to 35° .

With this we can conclude that, the central angle is double of the inscribed angle when both intercept the same arc .

Proof :-

In ∆APO we have,

→ OP = OA { Radius of circle }

So,

→ ∠OPA = ∠OAP { Angle opposite to equal sides are equal in measure } ------ Equation (1)

also,

→ ∠POB = ∠OPA + ∠OAP { Exterior angle is equal to sum of opposite interior angles }

using Equation (1) in RHS,

→ ∠POB = ∠OAP + ∠OAP

→ ∠POB = 2•∠OAP ------------- Equation (2)

Similarly, In ∆AQO we have,

→ OQ = OA { Radius of circle }

So,

→ ∠OQA = ∠OAQ { Angle opposite to equal sides are equal in measure } ------ Equation (3)

also,

→ ∠QOB = ∠OQA + ∠OAQ { Exterior angle is equal to sum of opposite interior angles }

using Equation (3) in RHS,

→ ∠QOB = ∠OAQ + ∠OAQ

→ ∠QOB = 2•∠OAQ ------------- Equation (4)

finally, adding Equation (2) and Equation (4) we get,

→ ∠POB + ∠QOB = 2•∠OAP + 2•∠OAQ

→ ∠POQ = 2[∠OAP + ∠OAQ]

→ ∠POQ = 2•∠PAQ (Proved)

Therefore, we can conclude that, the measure of a central angle equals double of the inscribed angle and inscribed angle is half of the central angle .

Learn more :-

In the figure along side, BP and CP are the angular bisectors of the exterior angles BCD and CBE of triangle ABC. Prove ∠BOC = 90° - (1/2)∠A .

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