Lines, Angles and Triangles

Conditions of similarity:-

 

  • AAA Similarity:- If in two triangles, corresponding angles are equal, then triangles are similar.

 

 

 

 

 

 

If ∠A = ∠P , ∠B = ∠Q , ∠C = ∠R

Then △ABC ~ △ PQR

  • SSS Similarity:- If the corresponding sides of two triangles are proportional then they are similar.

  

 

If AB/PQ = BC/QR = AC/PR

                     Then   △ABC ~ △ PQR

  • SAS Similarity: If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar.

      

If AB/PQ = BC/QR  &  ∠B = ∠Q

Then △ABC ~ △ PQR

The similarity of Triangles:- Similarity of triangles is a special case where if either of the conditions of similarity holds, The other will hold automatically.

Congruency of Triangles:-If two triangles are congruent, then corresponding angles are equal.

Conditions of Congruency:-

1.SAS Congruency:- If two sides and an included angle of one triangle are equal to two sides and an included angle of another, the two triangles are congruent.

     

In  △ABC and △ PQR

IF AB=PQ,

BC=QR,

∠ABC = ∠PQR

Then △ABC~△ PQR

2.ASA Congruency:- If two angles and the included side of one triangle is equal to two angles and the included side of another, The two triangles are congruent.

In  △ABC and △ DEF

If  ∠A =∠D

AB = DE

∠B = ∠E, Then △ABC ≅ △ PQR

3. AAS Congruency:- If two angles and the side opposite to one of the angles is equal to the corresponding angles and the side of another triangle, the triangles are congruent.

If  In △ABC and △ DEF

∠A =∠D

∠B=∠E

AC=DF

Then △ABC≅△ DEF

4.SSS Congruency:- If three sides if one triangle is equal to three sides of another triangle, the two triangles are congruent.

In △ABC and △ DEF

If AB=DE, BC=EF and AC=DF

THEN  △ABC ≅ △ DEF

5.SSA Congruency:- If two sides and the angle opposite to any of the sides are equal to corresponding sides and the angle opposite to any of the sides, Then the triangles are congruent.

      

 

 

 

 

 

In △ABC and △ DEF

AB=DF

BC= EF

∠BAC = ∠EDF

Then △ABC ≅ △ DEF

Right Angled Triangle:

Pythagoras theorem: In the case of a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In right △ABC, AB2 + BC2 = AC2

Special figure:

Note: A lot of questions are based on the figure

In this figure, △ABC ~△ ADB ~ △BDC

  • △ABC ~△ ADB

 

AB/AD   =  BC/DB  =  AC/AB

 

  • △ADB ~△ BDC

AD/BD   =  DB/DC  = AB/BC

 

  • △ABC ~△ BDC

AB/BD = BC/DC = AC/BC

Area Theorem:

According to area theorem:The ratio of areas of two similar triangles is the square of the ratio of corresponding sides.

If △ABC ~ △ DEF

Ar(ABC): Ar(DEF) = (AB/DE)2 = (BC/EF)2 = (AC/DF)2

Q) For similar triangles, the ratio of their corresponding sides is 2:3. What is the ratio of their areas?

Ans) According to area theorem :The ratio of areas of two similar triangles is the square of the ratio of corresponding sides.

Ratio of areas= (2/3)2  =  4/9

Q) What is the area of the triangle shown?

 

 

 

 

 

 

Ans:-

Consruction : Draw AD⊥BC

∠B = ∠C = 60

∠A= 60

(Sum of angles of a triangle is equal to 180°)

 

∠A = ∠B = ∠C = 60°

(Angles of an equilateral triangle are equal to 60° & triangle having each angle equal to 60° is an equilateral triangle )

:. △ABC is an equilateral △

In △ABD

BD2 +AD2 = AB2 (Pythagoras theorem)

AD2 =75, ad = 5√3

Area of Triangle = ½ * Base * Height

=1/2 * 10*5√3

= 25√3  sq. unit

 

Ques:- What is the area of the triangle Shown?

Ans:-

x2 +( x + 1)2 =25 (by Pythagoras theorem)

2x2 +2x =24

2x2 + 2x -24 =0

x2 + x – 12 =0

x = 3 or -4

x= 3 (Because length cannot be negative)

Area = ½ (base) x (height) = ½(7)(8)   = 28 sq. units

Ques:- What is the value of x in the triangle shown?

Ans)      In △ABC

4 + 3 = AC  (BY Pythagoras theorem)

AC = 5

△ABC~ △BDC

 

AB/BD =BC/DC = AC/BC

 

4/x  =  5/3

12 = 5x          ,      x =(12/5)  = Ans

 

Polygons

A polygon is a closed figure whose sides are straight line segments.

The perimeter of the polygon is the sum of the lengths of the sides. A vertex of a polygon is the point where two adjacent sides meet.

A diagonal of a polygon is a line segment connecting two nonadjacent vertices.

A regular polygon has sides of equal length and interior angles of equal measure.

 

Sum of all angles of a polygon with n sides

= (n-2) π radians = (n-2) (180°)

 Area of Regular polygon = (ns2/4) x Cot (180/n)   

Where     s = length of side

      n = No. of sides

 

Parallelogram (IIgm

A quadrilateral with two pairs of parallel sides.

Properties of parallelogram:

Area of Parallelogram = Base x Height

Diagonals of Parallelogram bisect each other.

The opposite angles in a Parallelogram are equal.

 

Rectangle

A Parallelogram with four equal angles, each at right angle is a rectangle.

 

 AB = CD, BC =AD

∠ABC = ∠BCD = ∠CDA = ∠BAD

 Properties of a Rectangle:

 Diagonals of a rectangle are equal and bisect each other.

 

Rhombus

A Parallelogram having all the sides equal is a rhombus.

 

 Area of Rhombus = ½ x product of diagonals

 

Properties:-

Diagonals of a rhombus bisect each other at right angle.

Square

 A Square is a rectangle with adjacent sides equal or rhombus with angles equal to 90°.

 Properties:

Diagonals of square are equal and bisect each other.

Area = ½ x (diagonal) 2

 

Diameter of circle circumscribing a square also acts as a diagonal of square.

 

BD=Diameter of Circle

= Diagonal of Square = a √2

 

Trapezium

A Trapezium is a quadrilateral with only two sides parallel to each other.

 

 

 Area of Trapezium = ½ (Sum of Parallel sides)x height

Rectangular Hexagon

A Rectangular Hexagon is a actually a combination of 6 equilateral triangles all of side “a”

 

Area of hexagon = 6(3/4 a2) = 3√3 / 2 a2

 Q) Right △ABC and rectangle EFGH have the same perimeter. What is the value of x?

Ans) In △ABC

AB2 + BC2 = AC2 (BY Pythagoras theorem)

AC=5

Perimeter of △ABC = 3+4+5 = 12

Perimeter of rectangle EFGH =2(2+x)

12 = 2 (2+x)

2 + x = 6

x = 4

 

Q) What is the area of the square in the figure above?


Ans) (Side2 + Side2 = Diagonal2) (by Pythagoras theorem)

2 Side2  = 10= 100

Side = 5 √2

Area = Side2 = 50 sq units

 

Q) The area of a rectangle can be represented by the expression 2x2+9x+10. The length is 2x+5 and the width is 6. What is the value of the area of the rectangle?

Ans) Area = 2x2 +9x 10 = Length x Breadth

2x2 +9x 10 = (2x +5)6

2x2  – 3x – 20 = 0

x = 4 or -2.5

If x = -2.5, then length = 2(-2.5) + 5 = 0 (Length can’ t be zero,∴ Negative value of x is discarded)

x = 4

Area = 2 (16) +9 (4) +10

= 32 +36+ 10   = 78 sq Units

Q) In the figure shown, ABCD is a rectangle. AB=8 and BC=6. R,S,T, AND Q are midpoints of the sides of rectangle ABCD. What is the perimeter of RSTQ?

 

Ans) AB = CD = 8, BC = AD = 6

AR = RB = DT = CT

AQ = QC = BS = SD = 3

AR2 + AQ2 = RQ2  (By Pythagoras theorem)

32 + 42 = RQ2

RQ = 5

RQ=RS=ST=QT=5

Perimeter of RSTQ = 20

 

Q) In the figure shown ABCD is a square with a side length of 16. R,S,T AND Q are midpoints of the sides of ABCD .

What is the area of RSTQ?

 

Ans) AB=16

AR=RB=BS=SC=TC=DT=QD=AQ=8

AR2 + AQ2 = RQ2  (By Pythagoras Theorem)

RQ=8 √2

Similarly RQ=RS=QT=ST= 8 √2

In △ARQ, ∠QAR = 90°, AR = AQ

∠AQR = ∠DQT (Angles opposite to equal sides are equal)

∠AQR = ∠DQT = 45°

∠RQT=90(Linear pair)

Therefore, RSTQ is square

Area (8 √2)2= 128 sq units

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