Revision: Term - 2 >> Measurements Mathematics SSLC (English Medium) Class 7 Tamil Nadu Board of Secondary Education

A circle is a closed curve where all points on the boundary (called the circumference) are at the same distance from a fixed point inside it.

• The fixed point inside the circle is called the center (O)

The radius is a straight line segment that connects the center of the circle to any point on its circumference.

Characteristics:

• Symbol: Usually represented as r

• All radii of a circle have the same length

• A circle has infinite radii (one to every point on the circumference)

• The radius is always half the diameter

• Radius = `"Diameter"/"2"`

 The diameter is a straight line segment that passes through the center of the circle and has both endpoints on the circumference.

Characteristics:

• The diameter passes through the center

• A circle has infinite diameters

• The diameter is the longest possible chord of a circle

• The diameter is twice the radius

• Diameter = 2 × Radius and

A chord is a straight line segment that connects any two points on the circumference of the circle.

Characteristics:

• A circle has infinite chords

• The diameter is the longest chord in any circle

• Chords closer to the centre are longer than chords farther from the center

Area of a circle: The area of a circle is the region occupied by the circle in a two-dimensional plane.

Area of the circle = πr²

Given: A circle with centre O and an external point T from which tangents TP and TQ are drawn to touch the circle at P and Q.

To prove: ∠PTQ = 2∠OPQ.

Proof: Let ∠PTQ = xº.

Then, ∠TQP + ∠TPQ + ∠PTQ = 180º   ...[∵ Sum of the ∠s of a triangle is 180º]

⇒ ∠TQP + ∠TPQ = (180º – x)   ...(i)

We know that the lengths of tangent drawn from an external point to a circle are equal.

So, TP = TQ.

Now, TP = TQ

⇒ ∠TQP = ∠TPQ

`= \frac{1}{2}(180^\text{o} - x)`

`= ( 90^\text{o} - \frac{x}{2})`

∴ ∠OPQ = (∠OPT – ∠TPQ)

`= 90^\text{o} - ( 90^\text{o} - \frac{x}{2})`

`= \frac{x}{2} `

`⇒ ∠OPQ = \frac { 1 }{ 2 } ∠PTQ`

⇒ 2∠OPQ = ∠PTQ

Given: TP and TQ are two tangents of a circle with centre O and P and Q are points of contact.

To prove: ∠PTQ = 2∠OPQ

Suppose ∠PTQ = θ.

Now by theorem, “The lengths of a tangents drawn from an external point to a circle are equal”.

So, TPQ is an isoceles triangle.

Therefore, ∠TPQ = ∠TQP

`= 1/2 (180^circ - θ)`

`= 90^circ - θ/2`

Also by theorem “The tangents at any point of a circle is perpendicular to the radius through the point of contact” ∠OPT = 90°.

Therefore, ∠OPQ = ∠OPT – ∠TPQ

`= 90^@ - (90^@ -  1/2theta)`

`= 1/2 theta`

= `1/2` ∠PTQ

Hence, 2∠OPQ = ∠PTQ.

We have to prove that

AQ = `1/2` (perimeter of ΔABC)

Perimeter of ΔABC = AB + BC + CA

= AB + BP + PC + CA

= AB + BQ + CR + CA

(∵ Length of tangents from an external point to a circle are equal ∴ BP = BQ and PC = CR)

= AQ + AR  ...(∵ AB + BQ = AQ and CR + CA = AR)

= AQ + AQ  ...(∵ Length of tangents from an external point are equal)

= 2AQ

⇒ AQ = `1/2` (Perimeter of ΔABC)

Hence proved.