Geometry

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L.A. (Line and Angle)

→ A line has no end points on either side left right arrow with space space space space space space space straight A space space space straight B space space space space space space space space on top

→ A line segment has two end points. straight A space begin inline style fraction numerator space space space space space space space space space space space space space space space over denominator space space end fraction end style straight B

→ A Ray has one end point begin inline style rightwards arrow from Ray to straight O space space space space space space space space space space space space space space space space space space straight P space space space space space space space space space of end style

► 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:-

    box enclose straight alpha space plus space straight beta space equals space 90 degree end enclose                  

   

(ii) Supplementary Angle : -

    box enclose straight alpha space plus space straight beta space equals space 180 degree end enclose                 

   

(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 : -

    open table attributes columnalign right end attributes row cell angle AOD comma space angle BOC end cell row cell angle AOC comma space angle BOD end cell end table close curly brackets rightwards arrow space Are space vertically space opposite space Angle  

     

► 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

box enclose Area space equals space 1 half space cross times space Base space cross times space Height end enclose

→ or. A = square root of straight s left parenthesis straight s minus straight a right parenthesis left parenthesis straight s minus straight b right parenthesis left parenthesis straight s minus straight c right parenthesis end root

► Centres Triangle :-

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

box enclose In space radius space left parenthesis straight r right parenthesis space equals space open parentheses fraction numerator Area space of space straight a space increment space straight A over denominator straight s end fraction close parentheses end enclose

→ ∠BIC = 90 + 1 half∠A

(ii) circum - centre of triangle-

The points of intersection of perpendicular bisectors of three side.

→ ∠ BOC = 2.∠A

→ circumcentre in Right angle

     box enclose OA space equals space OB space equals space OC space equals space straight R end enclose                                   

   

(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.

box enclose angle DAE space equals space 1 half left parenthesis angle ABC space �

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

(a) BD = fraction numerator AB cross times BC over denominator AC end fraction

(b) AD = AB2/AC

(c) CD = BC squared over AC

(d) 1 over BD squared space equals space open parentheses 1 over AB squared plus 1 over BC squared close parentheses

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

box enclose fraction numerator Ar left parenthesis increment ABC right parenthesis over denominator Ar space left parenthesis increment PQR right parenthesis end fraction space equals space fraction numerator straight H subscript 1 cross times straight B subscript 1 over denominator straight H subscript 2 cross times straight B subscript 2 end fraction end enclose

► I.C. Quadrilateral

→ ∠A + ∠B + ∠C + ∠D = 360°

→ Area = 1 half×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

→ AO = OC, OB = OD

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

→ If d1 = d2, then ABCD is a square

→ Area = 1 half cross times straight d subscript 1 cross times straight d subscript 2

→ Perimeter = 4a

(iii) Trapezium:-

→ ∠A + ∠D = ∠B + ∠C = 180°

→ AB || CD

→ Median (EF) = 1 half (AB + CD)

→ Area = open curly brackets 1 half left parenthesis AB space plus space CD right parenthesis space cross times space straight H close curly brackets

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

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

(iv) Rectangle :-

→ AB = CD, AD = BC

→ AB || CD, AD||BC

→ AC = BD = open parentheses square root of straight a squared plus straight b squared end root close parentheses

→ ∠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 = square root of 2a

→ Area = a2 = (side)2

→ Perimetre = 4(a)

I.D. Polygon

(For Regular Polygon)

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

→ Each exterior angle = open parentheses fraction numerator 360 degree over denominator straight n end fraction close parentheses

→ Sum of all exterior angle = 360°

→ Numbers of diagonals = fraction numerator straight n left parenthesis straight n minus 3 right parenthesis over denominator 2 end fraction

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

or

If AM = MB, then OM ⊥ AB

(ii)

→ If AB = CD, then ∠AOB = ∠COD

or

If ∠AOB = ∠COD, then AB = CD

(iii)

→ If OX = OY, then AB = CD

or

If AB = CD, then OX = OY

(iv)

→ ∠AOB = 1 half ∠ACB

(v)

→ ∠ADB =  ∠ACB

(vi)

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

(vii)

→ (∠A + ∠C = ∠B + ∠D = 180°)

(viii)

      box enclose PA space cross times space PB space equals space PC space cross times space PD end enclose

(ix)

box enclose PT squared space equals space space PA space cross times space PB end enclose

→ Length of tangent

Here XY = d

then PQ = RS = square root of straight d squared minus left parenthesis straight r subscript 1 plus straight r subscript 2 right parenthesis squared end root

PS = RQ = square root of straight d squared minus left parenthesis straight r subscript 1 plus straight r subscript 2 right parenthesis squared end root

2.A → Cube and cuboid

→ Cube

→ Volume = a3

→ Area = 6a2

→ Diagonal = square root of 3a

→ 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 = square root of straight l squared space plus space straight b squared space plus space straight h squared end root

► 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 = square root of straight h squared space plus space straight r squared end root

→ V = 1 third× area of base × height = 1 thirdπr2h

→ curved surface area =  πrl sq. unit.

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

→ Volume of cone = 1 third (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 = 4 over 3 πr cubed spacecubic units

(ii) Surface area = 4πr2 sq. units

→ Spherical shell

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

(i) V = 4 over 3 π (R3 – r3)

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

→ Hemisphere

→ V = 2 over 3πr3 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 = square root of straight h squared plus left parenthesis straight R squared minus straight r squared right parenthesis end root units

(ii) v = 1 third straight pi space left parenthesis straight R squared plus straight r squared plus Rr right parenthesis 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)

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

AD over DB space equals space AE over EC or DB over AD space equals space EC over AE or AD over AB space equals space AE over AC

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.

If PQ || RS, ∠P = ∠S and ∠Q = ∠R

then, ∆POQ ~ ∆SOR [Similar triangles]

Theorem 7 → The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

If XY || BA, ∠X = ∠Y and ∠A = ∠B

then ∆ABC ~ ∆XBY [Similar figures]

So, fraction numerator Area space of space increment ABC over denominator Area space of space increment XBY end fraction space equals space open parentheses AB over YX close parentheses squared

Theorem 8 →

If 'O' is any point inside a rectangle then OB2 + OD2 = OA2 + OC2

Theorem 9 → Area of triangle by heron's formula

If a, b, c are 3 side of a triangle and semiperimeter S = fraction numerator straight a space plus space straight b space plus space straight c over denominator 2 end fraction

then area of triangle = square root of straight s space left parenthesis straight s space �

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