Revision: Measurements Mathematics SSLC (English Medium) Class 8 Tamil Nadu Board of Secondary Education

The standard metre is defined in terms of the speed of light, according to which one metre is the distance travelled by light in `1/(299,792,458)` of a second in the air (or vacuum).

The quantity of matter contained in a body is known as its mass.

Mass is the measure of the amount of matter in an object.

It is defined as 1/24 the part of the mean solar day.

The average of the varying solar days, when the earth completes one revolution around the sun, is called mean solar day.

It is defined as the 1/1440 part of the mean solar day.

One year is defined as the time in which earth completes one complete revolution around the sun.

Measurement is the process of comparison of the given physical quantity with the known standard quantity of the same nature.

A standard metreis equal to 1650763.73 wavelengths in vacuum, of the radiation from krypton isotope of mass 86.

The physical quantities like mass, length and time which do not depend on each other are known as fundamental quantities.

The time taken by the earth to complete one rotation about its own axis is called solar day.

“A second is defined as 1/86400 the part of a mean solar day.”
OR
Second may also be defined “as to be equal to the duration of9,192,631,770 vibrations corresponding to the transition between two hyperfine levels of cesium – 133 atoms in the ground state.”

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

Central Angle: An angle whose vertex is the centre of the circle is called a central angle.

Sector of a circle
Region enclosed by two radii and the corresponding arc.

Minor sector
Sector with angle < 180°.

Major sector
Sector with angle > 180°.
Angle of major sector = 360° − angle of minor sector.

A net is a two-dimensional (2D) pattern that, when folded along its edges, forms a three-dimensional (3D) solid shape.

\[\text{Length of arc}=\frac{\theta}{360}\times2\pi r\]

\[\text{Area of sector}=\frac{\theta}{360}\times\pi r^2\]

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.

Suppose CD and AB are two parallel tangents of a circle with center O
Construction: Draw a line parallel to CD passing through O i.e. OP
We know that the radius and tangent are perpendicular at their point of contact.
∠OQC = ∠ORA = 90°
Now, ∠OQC + ∠POQ = 180°          (co-interior angles)
⇒ ∠POQ = 180° - 90° = 90°
Similarly, Now, ∠ORA +∠POR =180°       (co-interior angles)

⇒ ∠POQ = 180° - 90° = 90°
Now,∠POR + ∠POQ = 90° + 90°  =180°
Since, ∠POR and ∠POQare linear pair angles whose sum is 180°
Hence, QR is a straight line passing through center O.

Construction Join PO and OQ
In ΔPOR and ΔQOR
OP = OQ(Radii)
RP = RQ(Tangents from the external point are congruent)
OR = OR (Common)
By SSS congruency, ΔPOR ≅ ΔQOR
∠PRO = ∠QRO(C.P.C.T)

Now,∠PRO+ ∠QRO= ∠PRQ
⇒ 2 ∠PRO = 120°
⇒ ∠PRO = 60°
Now. In ΔPOR
cos 60° `=(PR)/(OR)`

⇒ `1/2 =(PR)/(OR)`
⇒ OR =  2PR
⇒  OR = PR + PR
⇒  OR = PR +RQ

Let ABCD be the quadrilateral circumscribing the circle.
Let E, F, G and H be the points of contact of the quadrilateral to the circle.

To Prove: AB + DC = AD + BC
Proof:
AB = AE + EB
AD = AH + HD
DC = DG + GC
BC = BF + FC

We have:
AE = AH   (Tangents drawn from an external point to the circle are equal.)
Similarly, we have:
BE = BF
DH = DG
CG = CF
Now, we have:
AB + DC = AE + EB + DG + GC
                = AH + BF + DH + CF
               = (AH + DH) + (BF + CF)
                = AD + BC
⇒ AB + DC = AD + BC

Thus, if a quadrilateral is drawn to circumscribe a circle, the sums of opposite sides are equal.
Hence, proved.

We know that tangent segments to a circle from the same external point are congruent
Now, we have
AR = AO, BR = BP and CP = CQ
Now, AB = AC
⇒ AR+ RB= AQ+ QC
⇒ AR + RB = AR + OC
⇒ RB  = QC
⇒  BP = CP
Hence, P bisects BC at P.

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.

Given: PQ is the diameter of the circle. Point P, Q, C are points of contact of the respective tangents.

To prove: ∠AOB = 90°

Construction: Draw seg OC

Proof:

In ∆OPA and ∆OCA,

side OP ≅ side OC   ...[Radii of the same circle]

side OA ≅ side OA   ...[Common side]

side PA ≅ side CA   ...[Tangent segment theorem]

∴ ∆OPA ≅ ∠OCA   ...[SSS test of congruency]

∴ ∠AOP ≅ ∠AOC   ...[C.A.C.T.]

Let m∠AOP = m∠AOC = x   ...(i)

Similarly, we can prove that ∠BOC ≅ ∠BOQ.

Let m∠BOC = m∠BOQ = y   ...(ii)

m∠AOP + m∠AOC + m∠BOC + m∠BOQ = 180°   ...[Linear angles]

∴ x + x + y + y = 180°   ...[From (i) and (ii)]

∴ 2x + 2y = 180°

∴ 2(x + y) = 180°

∴ x + y = 90°   ...(iii)

Now ∠AOB = ∠AOC + ∠BOC

= x + y   ...[From (i) and (ii)]

∴ ∠AOB = ∠AOC + ∠BOC

= x + y    

∴ ∠AOB = 90°   ...[From (iii)] 

In given figure,

`{:(AF = AE),(FB = BD),(EC = DC):}}`   ...(i) [Tangent Segment theorem]

In ▢ODCE,

∠ECD = 90°   ...[∠ACB = 90°, A–E–C, B–D–C]

`{:(∠ODC = 90^circ),(∠OEC = 90^circ):}}`   ...[Tangent theorem]

∴ ∠EOD = 90°  ...[Remaining angle of ▢ODCE]

∴ ▢ODCE is a rectangle.

Also, OE = OD = r   ...[Radii of the same circle]

∴ ▢ODCE is a square   ...`[("A Rectangle is square if it's"),("adjcent sides are congruent")]`

∴ OE = OD = CD = CE = r   ...(ii) [Sides of the square]

Consider R.H.S. = a + b – c

= BC + AC – AB

= (BD + DC) + (AE + EC) – (AF + FB)   ...[B–D–C, A–E–C, A–F–B]

= (FB + r) + (AF + r) – (AF + FB)   ...[From (i) and (ii)]

= FB + r + AF + r – AF – FB

= 2r

= L.H.S.

∴ 2r = a + b – c

Construction: Join OA, OC and OB

We know that the radius and tangent are perpendicular at their point of contact
∴ ∠OCA =  ∠OCB = 90°
Now, In Δ OCA and ΔOCB
∠OCA = ∠OCB = 90°
OA = OB (Radii of the larger circle)
OC = OC (Common)
By RHS congruency
Δ OCA ≅  Δ OCB
∴ CA =CB

We know that tangent segments to a circle from the same external point are congruent.
So, we have
EA = EC for the circle having center O₁
and
ED = EB for the circle having center O₁ 
Now, Adding ED on both sides in EA = EC. we get
EA+ ED  =  EC + ED
⇒  EA + EB = EC + ED
⇒  AB = CD

Since tangents from an exterior point to a circle are equal in length.

∴ AD = AF [Tangents from A]

BD = BE [Tangents from B]

CE = CF [Tangents from C]

Now,

AB = AC

⇒ AB – AD = AC – AD [Subtracting AD from both sides]

⇒ AB – AD = AC – AF [Using (i)]

⇒ BD = CF ⇒ BE = CF [Using (ii)]

⇒ BE = CE [Using (iii)]

Definition: A segment is a region of a circle bounded by a chord and its arc

Two Main Types:

• Minor Segment = smaller piece

• Major Segment = larger piece

Semicircle Special Case:

• Formed when chord = diameter

• Creates two perfectly equal segments

• Each semicircle = half the circle's area

1. Arc Definition: An arc is a curved portion of a circle's circumference between two points.

2. Two Types: Minor arc (< 180°) and Major arc (> 180°).

3. Semicircle: When the arc angle is exactly 180°, it's called a semicircle.

4. Complete Circle: Minor arc + Major arc = 360° (complete circumference).