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It is an exclusive e-Learning Blog that has been dedicated to help keen learn students to boost their knowledge in different subjects.
Definition of Parabola
: A parabola is a curve that looks like the one on the right. Its open end can point up, down, left or right. A curve of this shape is called 'parabolic', meaning 'like a parabola'.Quick navigation: Geometric Figure
: A u-shaped curve with certain specific properties. Formally, a parabola is defined as follows: For a given point, called the focus, and a given line not through the focus, called the directrix, a parabola is the locus of points such that the distance to the focus equals the distance to the directrix. [Adopted from MathWords]
Note: For a parabolic mirror, all rays of light emitting from the focus reflect off the parabola and travel parallel to each other (parallel to the axis of symmetry as well).
: A parabola is the set of all points in a plane that are equidistant from the focus and the directrix of the parabola. [Adopted from iCoachMath]
Focus:
Focus of a parabola lies on the axis of symmetry.Directrix:
Directrix is a line that is perpendicular to the axis of symmetry of a parabola.Geometric figure and Formulas:
Large image below:
Short Note:
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A few Geometric Figure
Geometric Figure 01:
Geometric Figure 02:
Geometric Figure 03:
Geometric Figure 04:
Geometric Figure 05:
Geometric Figure 06:
Quick navigation: Basic Rules of Trigonometry
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Proposition:
Let O be the centre of the circle ABCD. Let AB be the diameter and CD be a chord other than diameter of the circle. It is required to prove that AB > CD.Quick Navigation: Trigonometry Unit Circle
Construction:
Join O,C and O,D.Proof:
OA = OB = OC = OD (radius of the same circle)Now, in ΔOCD,
OC + OD > CD
or, OA + OB > CD
Therefore, AB > CD. (Proved)
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Chords equidistant from the centre of a circle are equal
Proposition:Let AB and CD be two chords of a circle with centre O. OE and OF are the perpendiculars from O to the chords AB and CD respectively. Then OE and OF represent the distances from centre to the chords AB and CD respectively.If OE = OF, it is to be proved that AB = CD.
Construction:
Join O,A and O,C.Step-1: Since OE⊥AB and OF⊥CD (right angles)
Therefore, ∠OEA = ∠OFC = 1 right angle.
Step-2: Now, between the right-angled
ΔOAE and ΔOCF
hypotenuse OA = hypotenuse OC (radius of same circle)
and OE = OF. (supposition)
∴ ΔOAE ≅ ΔOCF
∴ AE = CF.
Step-3: AE = 1/2 AB and CF = 1/2 CD. (Perpendicular from the centre)
Step-4: Therefore, 1/2 AB = 1/2 CD
i.e., AB = CD. (Proved)
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Proposition:
Let AB and CD be two equal chords of a circle with centre O. It is to be proved that the chords AB and CD are equidistant from the centre.Construction:
Draw from O, the perpendiculars OE and OF to the chords AB and CD respectively. Join O,A and O,C.Step-1: OE⊥AB and OF⊥CD (Perpendicular from the centre bisects the chord)
Therefore, AE = BE and CF = DF.
∴ AE = 1/2 AB
and CF = 1/2 CD
Step-2: But AB = DC (supposition)
∴ AE = CF.
Step-3: Now b etween the right-angled ΔOAE and ΔOCF (radius of same circle)
hypotenuse OA = hypotenuse OC and AE = CF.
∴ ΔOAE ≅ ΔOCF
∴ OE = OF.
Step-4: But OE and OF are the distances from O to the chords AB and CD respectively.
Therefore, the chords AB and CD are equidistant from the centre of the circle. (Proved)
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Two diagonals of a rhombus bisect each other at right angles
Proposition:Let the diagonals AC and BD of the rhombus ABCD intersect at O. It is required to prove that,(i) ∠AOB = ∠BOC = ∠COD = ∠DOA = 1 right angle
(ii) AO = CO, BO = DO.
Proof:Step-1: A rhombus is a parallelogram. Therefore, AO = CO, BO = DO. (Diagonals of a parallelogram bisect each other)
Step-2: Now in ΔAOB and ΔBOC,
AB = BC (sides of a rhombus are equal)
AO = CO
and OB = OB. (common side)
So ΔAOB = ΔBOC.
Therefore, ∠AOB = ∠BOC.
∠AOB + ∠BOC = 1 straight angle = 2 right angles.
∠AOB = ∠BOC = 1 right angle.
Similarly, it can be proved that, ∠COD = ∠DOA = 1 right angle. (Proved)
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The diagonals of a parallelogram bisect each other.
Proposition:Let the diagonals AC and BD of the parallelogram ABCD intersect at O. It is required to prove that AO = CO, BO = DO.
Proof: Step-1: The lines AB and DC are parallel and AC is their transversal.
Therefore, ∠BAC = alternate ∠ACD. (Alternate angles are equal)
Step-2: The lines BC and AD are parallel and BD is their transversal
Therefore, ∠BDC = alternate ∠ABD. (Alternate angles are equal)
Step-3: Now, between ΔAOB and ΔCOD
∠OAB = ∠OCD, ∠OBA = ∠ODC and AB = DC .
So ΔAOB ≅ ΔCOD.
Therefore, AO = CO and BO = DO. (Proved)
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If the square of a side of any triangle is equal to the sum of the squares of other two sides, the angle between the latter two sides is a right angle.
Proposition:
Let in ΔABC, AB2 = AC2 + BC2
It is required to prove that ∠C is a right angle.
Construction:Draw a triangle ΔDEF so that ∠F = 1 right angle.
EF = BC and DF = AC.
Proof:DE2 = EF2 + DF2 [Since in ΔDEF, ∠F is aright angle]
= BC2 + AC2 = AB2
∴ DE = AB
Now, in ΔABC and ΔDEF , BC = EF, AC = DF and AB = DE. [supposition]
∴ ΔABC ≅ ΔDEF; ∴ ∠C = ∠F
∴ ∠F =1 right angle.
∴ ∠C= 1 right angle. (Proved)
Proposition:
Let in ΔABC, AB2 = AC2 + BC2
It is required to prove that ∠C is a right angle.
Construction:Draw a triangle ΔDEF so that ∠F = 1 right angle.
EF = BC and DF = AC.
Proof:DE2 = EF2 + DF2 [Since in ΔDEF, ∠F is aright angle]
= BC2 + AC2 = AB2
∴ DE = AB
Now, in ΔABC and ΔDEF , BC = EF, AC = DF and AB = DE. [supposition]
∴ ΔABC ≅ ΔDEF; ∴ ∠C = ∠F
∴ ∠F =1 right angle.
∴ ∠C= 1 right angle. (Proved)
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In a right-angled triangle the square on the hypotenuse is equal to the sum of the squares on the two other sides.
Proposition:Let in the triangle ABC, B = 90°, hypotenuse AC = b, AB = c and BC = a. It is required to prove that, AC2 = AB2 + BC 2, i.e. b2 = c2 + a2.
Construction:Produce BC up to D such that CD = AB = c. Also draw perpendicular DE at D on BC produced, so that DE = BC = a. Join C, E and A, E.
Proof:Steps-1: In ΔABC and ΔCDE, AB = CD = c, BC = DE = a and included ∠ABC = included ∠CDE. [each right angle]
Hence, ΔABC ≅ ΔCDE.
∴ AC = CE = b and ∠BAC = ∠ECD.
Steps-2: Again, since AB⊥BD and ED⊥BD, AB ll ED.
Therefore, ABDE is a trapezium.
Steps-3: Moreover, ∠ACB + ∠BAC = ∠ACB + ∠ECD = 1 right angle.
∴ ∠ACE = 1 right angle.
Now, area of the trapezium ABDE = area of (Δ region ABC + Δ region CDE + Δ region ACE)
N.B:
1. Area of trapezium =1/2 Χ (sum of parallel sides Χ distance between parallel sides).
2. Pythagoras Theorem: In 6th century B.C. Greek philosopher Pythagoras discovered an important property of right-angled triangle. This property of right-angled triangle is known as Pythagorean property. It is believed that before the birth of Pythagoras, in Egyptian and Greek era, this special property of right-angled triangle was in use.
Proposition:Let in the triangle ABC, B = 90°, hypotenuse AC = b, AB = c and BC = a. It is required to prove that, AC2 = AB2 + BC 2, i.e. b2 = c2 + a2.
Construction:Produce BC up to D such that CD = AB = c. Also draw perpendicular DE at D on BC produced, so that DE = BC = a. Join C, E and A, E.
Proof:Steps-1: In ΔABC and ΔCDE, AB = CD = c, BC = DE = a and included ∠ABC = included ∠CDE. [each right angle]
Hence, ΔABC ≅ ΔCDE.
∴ AC = CE = b and ∠BAC = ∠ECD.
Steps-2: Again, since AB⊥BD and ED⊥BD, AB ll ED.
Therefore, ABDE is a trapezium.
Steps-3: Moreover, ∠ACB + ∠BAC = ∠ACB + ∠ECD = 1 right angle.
∴ ∠ACE = 1 right angle.
Now, area of the trapezium ABDE = area of (Δ region ABC + Δ region CDE + Δ region ACE)
N.B:
1. Area of trapezium =1/2 Χ (sum of parallel sides Χ distance between parallel sides).
2. Pythagoras Theorem: In 6th century B.C. Greek philosopher Pythagoras discovered an important property of right-angled triangle. This property of right-angled triangle is known as Pythagorean property. It is believed that before the birth of Pythagoras, in Egyptian and Greek era, this special property of right-angled triangle was in use.
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