## 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 ~△ 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

Problem 1) For similar triangles, the ratio of their corresponding sides is 2:3. What is the ratio of their areas?

Sol. 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

Problem 2) What is the area of the triangle shown?

Sol.

∠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)

Area of Triangle = ½ × Base × Height

=1/2 × 10 × 5√3

= 25√3  sq. unit

Problem 3) What is the area of the triangle Shown?

Sol.

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

Problem 4) What is the value of x in the triangle shown?

Sol.      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.

∠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

Problem 5) Right △ABC and rectangle EFGH have the same perimeter. What is the value of x?

Sol. 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

Problem 6) What is the area of the square in the figure above?

Sol. (Side2 + Side2 = Diagonal2) (by Pythagoras theorem)

2 Side2  = 10= 100

Side = 5 √2

Area = Side2 = 50 sq units

Problem 7) 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?

Sol. 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

Problem 8) 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?

Sol. 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

Problem 9) 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?

Sol. 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

Problem 10) In the given figure, AD||BC. Find the value of x.

a) x=8, $$\frac { 15 }{ 3 }$$

b) x=8, $$\frac { 11 }{ 3 }$$

c) x=8, $$\frac { 13 }{ 3 }$$

d) x=7, $$\frac { 13 }{ 3 }$$

Sol.

$$In\quad \triangle ODA\quad and\quad \triangle OBC,\quad AD\parallel BC\\ \angle ODA=\angle OBC$$ (Alternate interior Angles are equal)

$$\angle OAD=\angle OCB$$ (Alternate interior angles are equal)

$$\triangle ODA\sim \triangle OBC\\ \frac { OD }{ OB } =\frac { DA }{ BC } =\frac { OA }{ OC } \\ \frac { OD }{ OB } =\frac { AO }{ OC } \\ \frac { x-5 }{ x-3 } =\frac { 3 }{ 3x-19 } \\ (x-5)(3x-19)=3x-9\\ 3{ x }^{ 2 }-19x-15x+95-3x+9=0\\ 3{ x }^{ 2 }-37x+104=0\\$$

$$x=\frac { 37\pm \sqrt { { 37 }^{ 2 }-4(3)(104) } }{ 6 } =\frac { 37\pm 11 }{ 6 } \\ =8,\frac { 11 }{ 3 }$$

c) is correct.

Problem 11) In the right angle $$\triangle PQR$$, find RS?

Sol.

$$In\quad \triangle QRP\quad and\quad \triangle RSP\\ \angle QRP=\angle RSP\quad (both\quad are\quad right\quad angles)\\ \angle QPR=\angle SPR\quad (common)\\ Therefore,\quad \triangle QRP\sim \triangle RSP\\ \frac { QR }{ RS } =\frac { RP }{ SP } =\frac { QP }{ RP } \\ \frac { 5 }{ RS } =\frac { QP }{ 12 } \\ In\quad \triangle QRP\\ { QR }^{ 2 }+{ RP }^{ 2 }={ QP }^{ 2 }(By\quad pythagoras\quad theorem)\\ { 12 }^{ 2 }+{ 5 }^{ 2 }={ QP }^{ 2 }\\ QP=13\\ \frac { 5 }{ RS } =\frac { 13 }{ 12 } \\ RS={ (\frac { 60 }{ 13 } ) }^{ Ans }$$

Problem 12)

The area of semicircle O is $$16\pi$$.

$$\angle CDE=\angle ABE={ 30 }^{ 0 }\\ AC=\sqrt { 2 }$$

Quantity A:  DE

Quantity B:  $$3\sqrt { 5 }$$

Sol.

$$In\quad \triangle EDC\quad and\quad \triangle EBA\\ \angle EDC=\angle EBA\\ \angle CED=\angle AEB={ 90 }^{ 0 }\\ \triangle EDC\sim EBA\\ \frac { EC }{ AE } =\frac { DC }{ BA } \\ Let\quad EC\quad be\quad x\\ In\quad \triangle EDC\\ \angle CED={ 90 }^{ 0 },\quad \angle CDE={ 30 }^{ 0 },\quad \angle DCE={ 60 }^{ 0 }$$

Recall that the sides of a 30-60-90 triangle are in the proportion

$$x:x\sqrt { 3 } :2x$$.

.Or if you are familiar with trigonometry, just recall that

$$\sin { { 30 }^{ 0 } } =\frac { 1 }{ 2 } =\frac { perpendicular }{ hypotenuse } \\ Therefore,\quad \frac { CE }{ CD } =\frac { 1 }{ 2 } =\frac { x }{ CD } ,\quad CD=2x\\ \frac { EC }{ AE } =\frac { DC }{ BA } \\ \frac { x }{ x+\sqrt { 2 } } =\frac { 2x }{ BA }$$

Area of semicircle = $$\frac { \pi { r }^{ 2 } }{ 2 } =16\pi$$ $${ r }^{ 2 }=36,\quad r=4\sqrt { 2 } =OB\\ \frac { x }{ x+\sqrt { 2 } } =\frac { 2x }{ 8\sqrt { 2 } } \\ x+\sqrt { 2 } =4\sqrt { 2 } \\ x=3\sqrt { 2 } \\ Therefore,\quad CE=3\sqrt { 2 } ,\quad CD=6\sqrt { 2 }$$

Side of $$\triangle CDE$$ are in ratio $$x:x\sqrt { 3 } :2x$$ $$CE=3\sqrt { 2 } ,\quad CD=6\sqrt { 2 } \\ ED=3\sqrt { 6 } \\ Therefore,\quad DE=3\sqrt { 6 }$$

(Quantity A is greater than Quantity B).

Problem 13) What is the area of a triangle that has two sides that each have a length of 10, & whose perimeter is equal to that of a square whose area is 81?

a) 30

b) 48

c) 36

d) 60

e) 42

Sol.

Area of square = 81

Side of square = $$\sqrt { 81 } =9$$

Perimeter of square = 4×9 = 36

Perimeter of triangle = 36

Two sides of triangle have a length of 10.

Third side of triangle = 36-10-10 = 16

Any triangle that has equal sides is an isosceles triangle. Drop a perpendicular to divide it into two individual right triangles.

$$In\quad \triangle ABC,\quad AD\bot BC\\ In\quad \triangle ABD,\quad { AB }^{ 2 }={ BD }^{ 2 }+{ AD }^{ 2 }\quad (by\quad pythagoras\quad theorem)\\ { 10 }^{ 2 }={ 8 }^{ 2 }+{ AD }^{ 2 }\\ AD=6\\ Area\quad of\quad \triangle ABC=\frac { 1 }{ 2 } \times Base\times Height=\frac { 1 }{ 2 } \times 16\times 16=48\quad Sq.units$$

Problem 14) Michael wants to split his rhombus-shaped garden into two triangular plots, one for planting strawberries & one for planting vegetables, by erecting a fence from one corner of the garden to the opposite corner. If one corner of the garden measures $${ 60 }^{ 0 }$$ & one side of the garden measures $$x$$ meters, which of the following could be the area of the vegetable plot?

Indicate all such value

a) $$\frac { { x }^{ 2 }\sqrt { 3 } }{ 2 }$$

b) $$\frac { { x }^{ 2 }\sqrt { 3 } }{ 4 }$$

c) $$\frac { { x }^{ 2 } }{ 2 }$$

d) $$\frac { { x }^{ 2 }\sqrt { 3 } }{ 4 }$$

e) $$\frac { { x }^{ 2 } }{ 4 }$$

Sol.  Whenever a geometry problem comes without a figure, start by drawing one yourself:

Michael wants to split this garden into two triangles.

Try it this way first:

$$\triangle ACD$$ is an equilateral. (Because all angles are equal to $${ 60 }^{ 0 }$$)

Area of triangle ACD = Area of vegetable plot =

$$=\frac { \sqrt { 3 } { x }^{ 2 } }{ 4 }$$

This is one of the choices, but the garden might be split the other way:

$$Draw\quad AE\bot BD,\quad In\triangle AEB\\ \angle AEB={ 90 }^{ 0 },\quad \angle ABE={ 30 }^{ 0 },\quad \angle EAB={ 60 }^{ 0 }$$

Recall that the sides of a 30-60-90 triangle are in the proportion $$x:\sqrt { 3 } x:2x$$

Or if you are familiar with basic trigonometry

$$In\quad \triangle AEB,\quad \sin { { 30 }^{ 0 } } =\frac { 1 }{ 2 } =\frac { AE }{ AB } =\frac { AE }{ x } \\ AE=\frac { x }{ 2 } ,\quad EB=\frac { x\sqrt { 3 } }{ 2 }$$

Therefore, we can easily derive that

$$AE=\frac { x }{ 2 } ,\quad EB=\frac { x\sqrt { 3 } }{ 2 }$$ (by using properties of triangle or by using basic trigonometry)

Similarly, we can find that $$ED=\frac { \sqrt { 3 } x }{ 2 } ,\quad BD=\sqrt { 3 } x$$ $$Ar(\triangle ABD)=\frac { 1 }{ 2 } \times base\times height\\ =\frac { 1 }{ 2 } \times \sqrt { 3 } x\times \frac { x }{ 2 } \\ =\frac { \sqrt { 3 } { x }^{ 2 } }{ 4 }$$

This is same as the previous area, so only one answer choice is B.

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