Hindigk: ज्यामिति (Geometry) फार्मूला और ट्रिक

ज्यामिति (Geometry) फार्मूला और ट्रिक

L.A. (Line and Angle)

→ A line has no end points on either side

→ A line segment has two end points.

→ A Ray has one end point

► Types of Angle:-

(i) Acute angle : – 0° < θ < 90°

(ii) Right Angle :– θ = 90°

(iii) Obtuse Angle : - 90° < θ < 180°

(iv) Straight Angle : - θ = 180°

(v) Reflex Angle : - 180° < θ < 360°

(vi) Complete Angle : - θ = 360°

► Relation between Angles:-

(i) Complementary Angle:-

(ii) Supplementary Angle : -

(iii) Adjacent Angle:-

→ The have a common vertex

→ They have a common arm

→ The non-common arms are on either side of the common arm.

(iv) Linear pair Angles : - in last fig. If ∠1 + ∠2 = 180°

(v) Vertically opposite Angles : -

► Angle made by a transversal line :-

(i) Interior Angles : - ∠3, ∠4, ∠5, ∠6

(ii) External Angles : - ∠1, ∠2, ∠7, ∠8

(iii) Pair of corresponding Angles : -

∠1 and ∠5, ∠2 and ∠6, ∠4 and ∠7, ∠3 and ∠8

► 1. B → Triangle :-

→ perimeter = (a + b + c)

sub perimeter=(a+b+c)/2

→ ∠A + ∠B + ∠C = 180°

→ a + b > c, c + a > b, c + b > a

→

→ or. A =

► Centres Triangle :-

(i) In-centre : - The point of intersection of all the three angle bisectors.

→

→ ∠BIC = 90 + 1/2 ∠A

(ii) circum - centre of triangle-

The points of intersection of perpendicular bisectors of three side.

→ ∠ BOC = 2.∠A

→ circumcentre in Right angle

(iii) Centroid : - It is the point of intersection of all the three medians.

→ The centroid of triangle divides a median in ratio 2 : 1

→ A median divides area of a ∆ in exactly two parts.

→ A centroid divides area of a ∆ in exactly three parts.

(iv) Ortho centre : - Point of intersection of all three altitude.

(iii) ∆ABC is a right -angle triangle and ∠B = 90°, then

(a) BD =

(b) AD = AB2/AC

(d)

(iv) The ratio of the areas of two triangles is equal to the ratio of products of Base and its corresponding height

► I.C. Quadrilateral

→ ∠A + ∠B + ∠C + ∠D = 360°

→ Area = 1/2 ×BD × (AP + CQ)

→ Pair of adjacent sides = (AB, BC), (BC, CD), (CD, DA), (DA, AB)

► Types of Quadrilateral

(i) Parallelogram (11gm):-

→ AB = CD, AD = BC

→ AB || CD, AD || BC

→ ∠A = ∠C, ∠B = ∠D

→ ∠A + ∠D = ∠A + ∠B = ∠B + ∠C = ∠C + ∠D = 180°

→ AO = OC, OB = OD

→ Bisectors of angles of a 11gm form a rectangle.

→AC2 + BD2 = (AB2 + BC2 + CD2 + AD2) = 2 (AB2 + BC2)

→ Area = (Base × Height) = AB × h = AB × AD sin θ

(ii) Rhombus:-

→ AB = BC = CD = AD

→ ∠A = ∠C, ∠B = ∠D

→ AO = OC, OB = OD

→ d12 + d22 = (AB2 + BC2 + CD2 + DA2) = 4a2

→ If d1 = d2, then ABCD is a square

→ Area =

→ Perimeter = 4a

(iii) Trapezium:-

→ ∠A + ∠D = ∠B + ∠C = 180°

→ AB || CD

→ Median (EF) = 1/2 (AB + CD)

→ Area =

→ (AC2 + BD2) = (BC2 + AD2 + 2 × AB × CD)

(iv) Rectangle :-

→ AB = CD, AD = BC

→ AB || CD, AD||BC

→ AC = BD =

→ ∠A = ∠B = ∠C = ∠D = 90°

→ Area = (a × b)

→ Perimetre = 2 (a + b)

→ For the given perimenter of rectangle, a square has maximum area.

→ If 'p' is a point in rectangle, then (PA2 + PC2) = (PB2 + PD2)

(v) Square:-

→ AB = BC = CD = AD = a

→ ∠A = ∠B = ∠C = ∠D = 90°

→ AB || BC, CD || AD

→ AC = BD, AC ⊥ BD

→ AC = BD = a

→ Area = a2 = (side)2

→ Perimetre = 4(a)

► I.D. Polygon

(For Regular Polygon)

→ Sum of interior angle = (n – 2) × 180°

→ Each exterior angle =

→ Sum of all exterior angle = 360°

→ Numbers of diagonals =

► I.E. circle

→ Perimeter = 2πr

→ Radius (r) = OA

→ Area = πr2

→ PQ → SECANT

→ Tangent → PT

→ Diameter = AB = 2r

► Properties of circle

(i)

→ If OM ⊥ AB, then AM = MB

If AM = MB, then OM ⊥ AB

(ii)

→ If AB = CD, then ∠AOB = ∠COD

If ∠AOB = ∠COD, then AB = CD

(iii)

→ If OX = OY, then AB = CD

If AB = CD, then OX = OY

(iv)

→ ∠AOB = ∠ACB

(v)

→ ∠ADB = ∠ACB

(vi)

→ ∠ACD = 90° (The Angles in a semicircle is a Right angle)

(vii)

→ (∠A + ∠C = ∠B + ∠D = 180°)

(viii)

(ix)

→ Length of tangent

Here XY = d

then PQ = RS =

PS = RQ =

Concept 2

► 2.A → Cube and cuboid

→ Cube

→ Volume = a3

→ Area = 6a2

→ Diagonal = a

→ Cuboid

→ Volume = l × b × h cubic unit

→ Area of four walls = 2 (l + b) × h

→ Total surface area = 2 (lb + bh + hl)

→ Diagonal of cuboid =

► 2.B. Cone and Cylinder

→ Right circular cylinder

r → radius of base, h → height

→ V = Area of base × height

= π r2h cubic unit

→ Area of curved surface

= circumference of the base x height

= 2πrh sq. unit

→ Total surface Area = (curved surface Area) + (Area of two ends)

= 2πr (h + r)

→ Right circular cone

l = slant height =

→ V = × area of base × height = πr2h

→ curved surface area = πrl sq. unit.

→ Total surface Area = (curved surface area + base area) = πr(l + r)

→ Volume of cone = (volume of cylinder)

→ Volume of cylinder = 3 (Volume of cone)

{when r (cylinder) = r (cone) and h (cylinder) = h (cone)}

► 2.C. Sphere and Hemisphere

→ Sphere

(i) V = cubic units

(ii) Surface area = 4πr2 sq. units

→ Spherical shell

If r = inner radius of spherical shell, and R = outer radius, then

(i) V = 4/3 π (R3 – r3)

(ii) Total surface area = 4π (R2 + r2)

→ Hemisphere

→ V = 2/3 πr3 cubic unit

→ curved surface area = 2πr2 sq. unit

→ Total surface area = 3πr2 sq. unit.

► 2.D. Frustum of a right circular cone

→ Frustum

R → Radius of Base

r → Radius of top.

h → height of frustum

l → Slant height

(i) l = units

(ii) v = h cubic units

(iii) curved surface area = π (R + r) l + π (R2 + r2) sq. unit

= π [(R + r) l + (R2 + r2)] sq. unit

→ Right prism

→ Number of vertices = 2n

→ Number of faces = (n + 2)

→ Volume of the prism = (Area of the base × height)

→ Lateral surface area of the prism = (perimeter of base × height)

→ Total surface Area = (Lateral surface area + 2 × Base Area)

Concept 3

Theorem 1 → If two lines intersect each other, then vertical opposite angles are equal.

∠1 = ∠2 and ∠3 = ∠4

Parallel lines and transversal → A line intersects two or more lines at distinct points is called a transversal.

Theorem 2 → The sum of the angles of a triangle is 180°

∠A + ∠B + ∠C = 180

Triangles : -

(i) Congruence of triangle → There are four rules.

a) SAS rule → Two triangles are congruent if two sides and included angle of one triangle are equal to sides and included angle of other triangle.

Now if OA = OB and OD = OC

then ∆AOD ≅ ∆BOC

b) ASA rule → Two triangle are congruent if two angles and included side of one are equal to two angles and included side of other triangle.

If OA = OD and ∠AOB = ∠COD, ∠ABO = ∠DCO

then ∆COD ≅ AOB

c) SSC rule → If three sides of one triangles are equal to three sides of another triangle then both triangle will be congruent.

d) RHS rule → If in two right angle triangles the hypotenuse and one side of a triangle are equal to hypotenuse and one side of another triangle, then two triangles are opposite to each other.

If AP = BP, AQ = BQ and PQ = PQ

∆PAQ ≅ ∆PBQ

Theorem 3 → If two sides of a triangle are unequal, then angle opposite to the longer side is larger.

Theorem 4 → The sum of any two sides of a triangle is always greater than the third side.

Similarity of triangle →

Theorem 5 → If a line is drawn parallel to one side of a triangle of a triangle to intersect the other two sides in distinct points, the other sides are divided in the same ratio.

The vice-versa is also true.

If DE || BC, then

or or

Theorem 6 → If in two triangles, sides of one triangle are proportional to the sides of another triangle then their corresponding angles are equal and hence the two triangles are similar.