Triangles

Given: Length of sides of triangle = 10,6.5,a

Formula Used/Theory:-

Sum of 2 sides of triangles is always greater than 3^rd side of triangle

6.5 + a>10

a>10 - 6.5

a>3.5

‘a’ must be greater than 3.5

∴ the minimum value above 3.5 can be taken from option is 4

Given: Δ DEF is a right triangle with ∠E = 90°

Formula Used/Theory:-

Sum of all angles of triangle is 180°

In Δ DEF

∠D + ∠E + ∠F = 180°

∠D + 90° + ∠F = 180°

∠D + ∠F = 180° - 90°

∠D + ∠F = 90°

∴ Angle D and F are complimentary angles

As their sum is 90°

Given: In ΔPQS

PQ = PS, Exterior angle with Q is 110°

In ΔPSR

∠PRS = 25°, ∠RPS = x

Formula Used/Theory:-

Exterior angle of triangle is equal to sum of 2 opposite interior angles .

As RQ is straight line

∠PQS + 110° = 180°

∠PQS = 180° - 110°

∠PQS = 70°

As ΔPQS is isosceles triangle and PQ = PS

∴ ∠PQS = ∠PSQ

⇒ ∠PSQ = 70°

In ΔPSR

With exterior angle ∠PSQ equal to sum of opposite interior angles

∠PSQ = ∠SPR + ∠PRS

70° = x + 25°

x = 70° - 25°

x = 45°

In right angled triangle

One angle is 90° and sum of other 2 angles is 90°

Then;

Both angles must be smaller than 90°

∴ Both angles must be acute

Given: An isosceles triangle with one angle is 70°

Formula Used/Theory:-

In an isosceles triangle 2 angles are equal corresponding with 2 equal sides

⇒ If 70° is 3^rd angle of triangle

Then;

70° + x + x = 180°

2x = 180° - 70°

x =

x = 55°

Both angles are 55°

⇒ If 70° is Base angle of isosceles triangle

Then;

70° + 70° + x = 180°

x = 180° - 140°

x = 40°

one angle is 40° and another is 70°

∴ the angles can be either 55°, 55° or 70°, 40°

Given: In triangle one angle is 90°

Formula Used/Theory:-

Sum of angles of triangle is 180°

⇒ Moving to 1^st option

The other two angles are of 45° each

This can be happen only if triangle is isosceles right angle Δ

∴ this can be but not necessary every time

⇒ Moving to 2^nd option

In remaining two angles, one angle is 90° and other is 45°

This cannot be possible because angle sum of triangle is beyond 180° which is not possible

⇒ Moving to 3^rd option

Remaining two angles are complementary

As one angle is 90° and sum of angles of triangle is 180°

∴ the sum of both angle is always 90°

∴ only 3^rd statement is correct.

Given: Lengths of sides of a triangle are 3 cm, 4 cm and 5 cm

Formula Used/Theory:-

Pythagoras theorem:-

Base² + Height² = Hypotenuse²

As Lengths of sides of a triangle are 3 cm, 4 cm and 5 cm

Squaring all sides

9,16,25

⇒ 9 + 16 = 25

Which is equal to square of 3^rd side

∴ Pythagoras theorem get proved

Hence;

It is a right angled triangle

Exterior angle of triangle is equal to sum of 2 opposite onterior angles.

As BC is straight line

∠PBD + 120° = 180°

∠PBD = 180° - 120°

∠PBD = 60°

As ΔPBD is isosceles triangle and PB = PD

∴ ∠PBD = ∠PDB

⇒ ∠PDB = 60°

In ΔPSR

With exterior angle ∠PSQ equal to sum of opposite inetrior angles

∠PDB = ∠DPC + ∠PCD

60° = x + 35°

x = 60° - 35°

x = 25°

Difference of 2 sides of triangle is always smaller than the 3^rd side of triangle

And

Sum of 2 sides of triangle is always greater than 3^rd side of triangle

Difference of 2 sides of triangle is always smaller than the 3^rd side of triangle

And

Sum of 2 sides of triangle is always greater than 3^rd side of triangle

Given: Height of broken tree = 5m

Base distance of tree touch the ground = 12cm

Formula Used/Theory:-

Pythagoras theorem:-

Base² + Height² = Hypotenuse²

Length of broken tree is :-

Base² + Height² = Hypotenuse²

12² + 5² = Hypotenuse²

144 + 25 = Hypotenuse²

169 = Hypotenuse²

Height of broken tree = √ 169 = 13m

Height of tree = length of broken tree + 5m

= 13m + 5m = 18m

As none of side is equal

∴ it cannot be isosceles triangle

Square of all sides are

16,25,64

16 + 25 = 41≠64(square of 3^rd side)

∴ it cannot be right angle triangle

Hence;

It is a scalene triangle

Given: Two trees 7 m and 4 m high;bases (roots) are 4 m apart

Formula Used/Theory:-

Pythagoras theorem:-

Base² + Height² = Hypotenuse²

Difference in height of both trees is 7m – 4m = 3m

∴ difference in height is 3m

And distance between their roots is 4m

⇒ it makes a right angled triangle

With height 3m

Base 4m

Hypotenuse = distance between their tops

Base² + Height² = Hypotenuse²

4² + 3² = Hypotenuse²

Hypotenuse² = 16 + 9 = 25

Hypotenuse = √25 = 5m

Distance between tops of trees is 5m

If 2 angles are 40°

Then;

Sum of all angles is 180°

40° + 40° + x = 180°

x = 100°

Hence;

If one angle is obtuse then it will form obtuse angled triangle

If 2 angles of triangle is 60°

Then;

60° + 60° + x = 180°

120° + x = 180°

x = 180° - 120°

x = 60°

∴ All angles are equal to 60°

Hence it is a equilateral triangle

Given: Length of rectangle 60cm

Diagonal of rectangle 61cm

Formula Used/Theory:-

Pythagoras theorem:-

Base² + Height² = Hypotenuse²

In rectangle every angle is 90°

∴ Length, Breadth, Diagonal forms a right angled triangle

With diagonal as hypotenuse

Base² + Height² = Hypotenuse²

Breadth² + Length² = Hypotenuse²

Breadth² + 60² = 61²

Breadth² = 61² – 60²

Breadth² = 3721 – 3600

Breadth² = 121

Breadth = √121 = 11cm

Perimeter = 2(length + breadth)

= 2(60 + 11)

= 142cm

In ∆PQR

if PQ = QR

Then PQR is a isosceles triangle

Having, ∠P = ∠R

∠Q = 100°

As sum of all angles of triangle is 180°

Then;

∠P + ∠R + ∠Q = 180°

∠R + ∠R + 100° = 180°

2∠R = 80°

∠R = = 40°

Difference of any two sides of a triangle is always smaller than the third side

Not greater than 3^rd side of triangle

Exterior angle is equal to sum of opposite interior angles

∠ACD = ∠A + ∠B

As ΔACB is isosceles triangle with AC = BC

Then ∠A must be equal to ∠B

∴ ∠ACD = 40° + 40° = 80°

Difference of 2 sides of triangle is always smaller than the 3^rd side of triangle

Difference = 9cm – 7cm = 2cm

∴ 3^rd side must be greater than 2cm

And

Sum of 2 sides of triangle is always greater than 3^rd side of triangle

Sum = 9cm + 7cm = 16cm

∴ 3^rd side must be smaller than 16cm

Hence;

Appropriate option comes out to be (C). 3 cm and 16 cm

As exterior angle is equal to sum of opposite interior angle

∴ ∠ACD = ∠A + ∠B

∠ACD = 25° + 35° = 60°

In another triangle

x is exterior angle

∴ x = 60° + ∠ACD

x = 60° + 60° = 120°

As we can there are 2 triangles ΔABC and ΔDEF

∴ By angle sum property of triangle

∠A + ∠B + ∠C = 180°

∠D + ∠E + ∠F = 180°

∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 180° + 180°

= 360°

Given: PQ = PR, RS = RQ and ST || QR. If the exterior angle RPU is 140°

Formula Used/Theory:-

Base angle are equal is isosceles triangle

Alternate angles are equal if lines are parallel

Exterior angle is equal to sum opposite interior angles

In ΔPQR

∠UPR is exterior angle

∴ ∠UPR = ∠Q + ∠R

As ΔQPR is isosceles triangle because PQ = PR

Hence;

∠Q = ∠R

140° = ∠Q + ∠R

2∠Q = 140°

∠Q = = 70°

As ΔQRS is isosceles triangle because QR = RS

∴ ∠Q = ∠QSR

∠QSR = 70°

Then;

∠Q + ∠QSR + ∠SRQ = 180°

70° + 70° + ∠SRQ = 180°

∠SRQ = 180° - 140° = 40°

As ST || QR

And SR is transverse

∠SRQ = ∠TSR ∵ Alternate interior angles.

∴ ∠TSR = 40°

Given: ∠BAC = 90°, AD ⊥ BC and ∠BAD = 50°

Formula Used/Theory:-

Sum of all angles of triangle is 180°

As ∠BAC = ∠BAD + ∠CAD

90° = 50° + ∠CAD

∠CAD = 90° - 50° = 40°

In Δ ADC

∠ADC = 90° ∵ AD is perpendicular to BC

∠ADC + ∠CAD + ∠ACD = 180°

90° + 40° + ∠ACD = 180°

130° + ∠ACD = 180°

∠ACD = 180° - 130° = 50°

In right angle triangle one angle is 90° and other 2 angles are always complementary.

Hence;

Only in right angle triangle one angle and sum of other 2 angle are always equal to 90°

Exterior angle is equal to the sum of opposite interior angles

Then;

Is both interior opposite angles are equal

Let both the interior angle be X

∴ 130° = X + X

2X = 130°

X = = 55°

As by the angle sum property

Sum of all angles is 180°

∴ if one angle of triangle is 110°

Then sum of other 2 angles is 180° -110° = 70°

If the bisector of both angles are made…

Then both angled get halved

If both angled get halved then their sum will also get halved

= 35°

∴ it forms a new triangle if the bisectors are joined

Though sum of bisected angles will be half of sum of angle of triangle

∴ the 3^rd angle will be

180° - 35°

= 145°

As E is midpoint of AC.

It divide AC in 2 equal parts

Joining midpoint of AC to B gives out median.

∴ BE is median

Exterior angle is sum of opposite interior angles

When QR is produced means exterior angle of ∠R

Having opposite angles ∠P and ∠Q

∴ exterior angle = ∠P + ∠Q

= 60° + 40° = 100°

Angle Sum property gives

Sum of angles of triangle is 180°

But in

40°, 132°, 18°

40° + 132° + 18° = 190≠180

∴ triangle cannot be possible with these 3 angles

Difference of 2 sides of triangle is always smaller than 3^rd side of triangle

Sum of 2 sides of triangle is always greater than 3^rd side of triangle

3^rd side<18cm + 14cm

3ed side<32cm

3^rd side>18cm – 14cm

3^rd side>4cm

Then side must above 4cm and below 32cm

∴ 5cm is between them

Altitude is perpendicular drawn from vertex to opposite side

As there are 3 vertex in triangle

∴ 3 altitude can be formed

Median is line joining vertex to mid-point of opposite side

Exterior angle is sum of interior opposite angles

∴ 120° = x + 50°

x = 120° - 50° = 70°

Angle sum property

Sum of all angles of triangle is 180°

50° + x + y = 180°

50° + 70° + y = 180°

120° + y = 180°

y = 180° - 120° = 60°

Difference of 2 sides of triangle is always smaller than 3^rd side of triangle

Sum of 2 sides of triangle is always greater than 3^rd side of triangle

3^rd side<10cm + 6cm

3^rd side<16cm

3^rd side>10cm – 6cm

3^rd side>4cm

Then side must above 4cm and below 16cm

∴ 6cm is between them

Given: angle B = 90°, BC = 3 cm and AC = 5 cm

Formula Used/Theory:-

Pythagoras theorem:-

Base² + Height² = Hypotenuse²

Base² + Height² = Hypotenuse²

3cm² + Height² = 5²

9 + Height² = 25

Height² = 25 – 9 = 16

Height = √16 = 4cm

∴ AB = 4cm

If in Δ ABC

∠B = 90° then its opposite side must be AC[hypotenuse]

And

AB and BC are 2 other sides of triangle

Obtuse angles have their altitude outside the triangle

Exterior angle is equal to the sum of opposite interior angles

Then;

In Δ ABC

As BC is extended

∠A + ∠B = ∠3 + ∠4

∴ ∠1 + ∠2 = ∠3 + ∠4

Given:

In Δ ABC, ∠A = 100°, AD bisects ∠A and AD ⊥ BC

Formula Used/Theory:-

Pythagoras theorem:-

Base² + Height² = Hypotenuse²

When ∠A is bisected

Then;

∠DAB = 50° and ∠DAC = 50°

Then;

In Δ ABD

As AD ⊥ BC

∠ADB = 90°

Sum of all angles is 180°

∠ADB + ∠DAB + ∠B = 180°

90° + 50° + ∠B = 180°

∴ ∠B = 180° - 140° = 40°

Given: ∠A = 50°, ∠B = 70° and bisector of ∠C meets AB in D

Formula Used/Theory:-

Angle Sum property

Sum of angles of triangle is 180°

In Δ ABC

∠A + ∠B + ∠C = 180°

50° + 70° + ∠C = 180°

∠C = 180° - 120° = 60°

If CD is bisector of ∠C

Then;

∠DCA = ∠DCB = = 30°

In Δ ADC

∠A + ∠DCA + ∠ADC = 180°

50° + 30° + ∠ADC = 180°

∠ADC = 180° - 80° = 100°

Because it does not following congruency criteria CAB ↔ EDF

AB must be equal to DF

In Δ ABM and Δ DCM

AM = MC ∵ M is midpoint of AC

DM = BM ∵ M is midpoint of BD

∠AMB = ∠DMC ∵ Vertically opposite angles

AMB ↔ CMD

Hence, Δ ABM ≅ Δ DCM

By congruency criteria AMB ↔ CMD

∠1 = ∠4;and ∠2 = ∠3

In isosceles triangle altitude and median are the same

As joining line from mid-point of side to 3^rd point we get median

So, D is midpoint of BC joining from point A gives AD as median

As In isosceles triangle altitude and median are the same

It possess 90° angle on BC

∴ ∠ADC = 90°

RHS congruency

It states that if in 2 right angle triangles Hypotenuse and any 1 side are equal then both triangles are congruent by RHS

ASA congruency

It states that if in 2 triangles 2 angles and a corresponding side between them are equal then both triangles are congruent by ASA

SAS congruency

It states that if in 2 triangles 2 sides and a corresponding angle between them are equal then both triangles are congruent by SAS

SSS congruency

It states that if in 2 triangles if all 3 sides are equal then both triangles are congruent by SSS

In Δ PRQ and Δ PSQ

PR = PS (both are length of a cm)

RQ = SQ (both are length of b cm)

PQ = PQ (common in both triangles)

PRQ ↔ PSQ

Δ PRQ ≅ Δ PSQ (By SSS criterion)

ASA congruency

It states that if in 2 triangles 2 angles and a corresponding side between them are equal then both triangles are congruent by ASA

SAS congruency

It states that if in 2 triangles 2 sides and a corresponding angle between them are equal then both triangles are congruent by SAS

SSS congruency

It states that if in 2 triangles if all 3 sides are equal then both triangles are congruent by SSS

But there is no congruency in AAA as having same angles cannot make sides equal of the triangle

Δ PQR ≅ Δ STU

PQR ↔ STU

∴ QR = TU

⇒ TU = 6cm

In Δ ABC and Δ DBC

AB = DC (given)

AC = DB (given)

BC = BC (common in both triangles)

ABC ↔ DCB

Δ ABC ≅ Δ DCB (By SSS criterion)

Obtuse angled

As obtuse angled is more than 90°

Then if we have a obtuse angled triangle

Drawing their perpendicular they get out of the triangles

Acute angled triangle always have their altitudes inside the triangle

180°

As making exterior angle line of triangle is produced

It is a straight line

And a straight line possess 180° when it measures from any point on line

Hypotenuse

Longest side in right angled triangle is hypotenuse

As by Pythagoras theorem also

We get sum of square of both sides of triangle equal to square of its hypotenuse.

Altitude

As in equilateral triangle

60°

As in equilateral triangle all sides are equal and all angles are equal

∴ sum of all angles is 180° by angle sum property

Hence; each angle would be = 60°

Equal

Two angle are equal are the one which are made by equal sides

As in isosceles triangle 2 sides are equal in length

∴ 3^rd side is made by joining both sides last point

Hence it possess same angle

Equal

As in isosceles triangle 2 sides are equal in length

∴ 3^rd side is made by joining both sides last point

Hence it possess same angle

90°

When one angle is equal to sum of other two

Means 2 quantities are equal

If by angle sum property all angles sum is 180°

∴ 2 quantities can only be equal if they get halved

Hence; angle = = 90°

Two

There are 3 types of triangle

⇒ Obtuse angled triangle

In obtuse angle triangle one angle is obtuse angle

Hence; sum of both remaining angles is below 90

∴ both angles are acute

⇒ Right angled triangle

In Right angle triangle one angle is 90°

Hence; sum of both remaining angles is 90°

∴ both angles are acute

⇒ Acute angled triangle

In Acute angle triangle all angles are acute

Same

As line segment is 1-dimension figure

Only length can make both lines equal

∴ having same length can make 2 line segment congruent

Congruent

As in angle there are rays on both arms which can extend its length

∴ the measure of angle makes them congruent

length ; breadth

As in rectangle all angles are 90°

∴ congruency depends on its length and breadth(dimensions)

Hence; if length and breadth are equal to other rectangle both are congruent

Side

As in square all angles are 90°

∴ congruency depends on its length of side(dimension)

Hence; if side of 1 square is equal to other square both are congruent

If both ∆PQR and ∆XYZ are congruent under the correspondence QPR ↔ XYZ

Then

All angle and side will be equal is respect to given congruency QPR ↔ XYZ

As line PQ = XZ

Line RQ = YZ

And ∠Q = ∠Z

Then PQR ↔ XZY by SAS congruency

As line PS = QR

Line PR = PR

And ∠SPR = ∠PRQ

Then PRS ↔ RPQ by SAS congruency

As line RQ = RQ

∠DQR = ∠PRQ

And ∠Q = ∠R

Then PQR ↔ DRQ by ASA congruency

When 2 angle are equal

∠AOR = ∠POQ (vertically opposite angle)

∠R = ∠Q (given 55°)

Third angle also be to other because the all have sum 180°

Hence;

∠A = ∠P

As line AO = OP

∠AOR = ∠POQ

And ∠A = ∠P

Then ARO ↔ PQO by SAS congruency

As line AB = AD

AC = AC

And ∠BAC = ∠DAC

Then ABC ↔ ADC by SAS congruency

Exterior angle property

Exterior angle of triangle is equal to sum of opposite interior angles

As ∠TPQ is exterior angle of ∠P it will be equal to ∠R + ∠Q of ΔPQR

As ∠UQR is exterior angle of ∠Q it will be equal to ∠P + ∠R of ΔPQR

As ∠PRS is exterior angle of ∠R it will be equal to ∠P + ∠Q of ΔPQR

False

It is valid only on Right angled triangle

Where sum of squares of two sides is equal to the square of the third side by Pythagoras theorem

False

Sum of two sides of a triangle is always greater to the third side

Sum never get equal to 3^rd side

True

The difference between the lengths of any two sides of a triangle is smaller than the length of third side

It cannot get equal because if it gets equal then their will be a line not a triangle.

False

If ∆ABC ≅ ∆PQR

In ABC ↔ PQR

Where AB is equal to PQ

Which is not true

As AB = 3.5cm and PQ = 5cm

False

In Obtuse angled triangle Sum of 2 angles is always smaller than 3^rd angle

Because of angle is above 90° and sum of all angles is always 180°

False

As by angle sum property

Sum of all angles of triangle is always equal to180°

False

No, it is not possible

Because in right angled triangle Pythagoras states that

Sum of Square of both sides is always equal to square of 3^rd side

But in equilateral all sides must be equal

True

As M is midpoint of AB

AM should be equal to MB

And 2 line segment can only be congruent if their lengths are equal

False

It is not possible

Because by angle sum property

Sum of angles of triangle is equal to 180°

If both angles get 90° then 3^rd angle must be 0 which is not possible

False

No, it is not possible

Because by angle sum property

Sum of angles of triangle is equal to 180°

If both angles get above 90° then sum of angles exceed 180° which is not possible

True

Every triangles have at least 2 acute angles

So it is a necessary condition to make a triangle

False

No, it is not possible

Because by angle sum property

Sum of angles of triangle is equal to 180°

If all angles get below 60° then sum of all angles must be less than 180° which is not possible

False

No, it is not possible

Because by angle sum property

Sum of angles of triangle is equal to 180°

If all angles get above 60° then sum of all angles must be greater than 180° which is not possible

True

Yes, it is possible

As in equilateral triangle all angles are 60°

And by angle sum property

Sum of angles of triangle is equal to 180°

Hence we can construct triangles with all angles 60°

False

No, it is not possible

Because in right angled triangle Pythagoras states that

Sum of Square of both sides is always equal to square of 3^rd side

But all sides never get equal

False

No, it is not possible

If two angles of a triangle are equal, the third angle is also equal to each of the other two angles

Gives us an equilateral triangle

⇒ But not in case of isosceles triangle

In which only 2 angles are equal

False

No they are not congruent by RHS

But they can be congruent by SAS

As in RHS congruency Hypotenuse of triangle must be equal in both the triangles

True

Yes it is true

AS congruent figure means equal in dimensions and

Both get compLet ely overlap each other

True

Yes they are congruent

Because they have same radii circles

Even if their money value gets different

They overlap each other

False

No, they are not congruent

Because in Kaleidoscope the lenses get bigger

So the radii of circle also get bigger

Hence; they can’t overlap each other

False

No it is not true

2 angles can only be congruent if they have equal measures

2 acute angles can only be congruent if they have equal measures

True

Yes it is true

Because

Right angles can only get their measures 90°

And 2 angles can only be congruent if they have equal measures

False

No this is not true

Because 2 figures with same shape are called similar

For 2 figures to be congruent they must be having same shape and size

True

As the square has only one dimension of side

And if the Area of 2 squares gets equal

Then, their side must be equal

As area = side²

False

As rectangle gets 2 dimensions (length and breadth)

If the area is equal that doesn’t mean its length and breadth gets equal

EG-May be length get double and breadth get half the area of both rectangle will be same but not congruent

True

As the circle has only one dimension of radius

And if the Area of 2 circles gets equal

Then, their radius must be equal

As area = π × radius²

True

As the square has only one dimension of side

And if the Perimeter of 2 squares gets equal

Then, their side must be equal

As Perimeter of square = 4×side

True

As the circle has only one dimension of radius

And if the Circumference of 2 circles gets equal

Then, their Radius must be equal

As Perimeter of circle = 2π×Radius

False

No they cannot be congruent

As when 2 angles of triangles are equal

Then;

It doesn’t make the sides of triangle equal

As same angled triangle can have length of sides to different

Hence, they are not of same size

∴ triangles cannot be congruent by2 angles equal

True

Yes it is true

If 2 legs of right angled triangle are equal

And the angle between both the sides is 90°

Then;

Both triangles are congruent by ASA congruency

False

No it is not true

If 2 sides of both the triangle are equal

And one angle is equal

Then;

It cannot be congruent until the angle should be in between 2 sides which is equal in both the triangles to get congruent by SAS congruency

True

If 2 triangles are congruent

Then,

Both triangles get equal in shape and size

∴ the corresponding angles and corresponding sides are equal of both the triangles

False

No it is not true

If 2 angles of both the triangle are equal

And one side is equal

Then;

It cannot be congruent until the side should be in between 2 angles which is equal in both the triangles to get congruent by ASA congruency

False

No it is not true

If only hypotenuse of both the triangle are equal

Then;

It cannot be congruent until with hypotenuse there should a side of triangle rather than hypotenuse which is equal in both the triangles to get congruent by RHS congruency

False

No it is not true

If hypotenuse of both the triangle are equal

And one angle is equal in both right angles triangle

Then;

It cannot be congruent until with hypotenuse there should a side of triangle rather than hypotenuse which is equal in both the triangles to get congruent by RHS congruency

False

AAS congruence criterion states that any 2 angles and with any 1 side are equal in both the triangles then these are congruent by AAS congruence criterion

ASA congruence criterion states that any 2 angles and with 1 side between 2angles are equal in both the triangles then these are congruent by ASA congruence criterion

False

In ∆ABD and ∆ACD

AD = AD (common)

∠BAD = ∠CAD

But AB≠AC

Hence hypotenuse is not equal in both triangles then

Both triangles cannot be congruent by RHS congruence criterion

Given:

Measure of three angles of a triangle are in the ratio 5 : 3 : 1

Formula Used/Theory:-

Angle sum property

Sum of all angles of triangle is 180°

If the ratio is 5:3:1

Then;

The 3 angles will be 5x,3x,x

Sum of angles will be 180°

5x + 3x + x = 9x

9x = 180°

⇒ x = = 20°

∴ angles are 5×20°, 3× 20°, 1×20°

100°, 60°, 20°

Result:- the angles of triangle are 100°,60°,20°

Formula Used/Theory:-

Angle sum property

Sum of all angles of triangle is 180°

In the right angled triangle where x lies

One angle is 55°

Other angle is x

And central angle is 90°

Then by angle sum property

55° + x + 90° = 180°

x + 145° = 180°

x = 180° - 145° = 35°

Result:- the value of x comes out to be 35°

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

→ Exterior angle is equal to sum of interior opposite angles

In Figure (i)

⇒ b = 60° + 70° = 130° (Exterior angle property)

⇒ c + 60° + 70° = 180° (Angle sum property)

c = 180° - 130° = 50°

⇒ b + a + 30° = 180° (Angle sum property)

a + 130° + 30° = 180°

a = 180° - 160° = 20°

In Figure (ii)

⇒ b = 60° + 55° = 115° (Exterior angle property)

⇒ a + 60° + 55° = 180° (Angle sum property)

a = 180° - 115° = 65°

⇒ b + c + 40° = 180° (Angle sum property)

c + 115° + 40° = 180°

c = 180° - 155° = 25°

Result:- In figure (i) a = 20°; b = 130°;c = 50°

In figure (ii) a = 65°; b = 115°;c = 25°

Given: ∠Z = 90°

∠X is 30° greater than ∠Y

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

Let ∠Y be ‘a’

Then;

∠X = ∠Y + 30°

∠X = a + 30°

∠Z = 90°

By angle sum property

∠X + ∠Y + ∠Z = 180°

(a + 30°) + a + 90° = 180°

2a + 120° = 180°

2a = 180° - 120° = 60°

a = = 30°

∠Y = a

∴ ∠Y = 30°

Result:- Measure of ∠Y comes to be 30°

Given: ∠A is 40° less than ∠B

∠A is 50° less than ∠C

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

Let ∠A be ‘x’

Then;

⇒ ∠A = ∠B - 40°

x = ∠B - 40°

∠B = x + 40°

⇒ ∠A = ∠C - 50°

x = ∠C - 50°

∠C = x + 50°

By angle sum property

∠A + ∠B + ∠C = 180°

x + (x + 40°) + (x + 50°) = 180°

3x + 90° = 180°

3x = 180° - 90° = 90°

x = = 30°

∠A = x

∴ ∠A = 30°

Result:- Measure of ∠A comes out to be 30°

Given:

Polygon have 3 sides

One angle is 15°

Another angle is 60°

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

If polygon have 3 sides than

Polygon must be a triangle .

By angle sum property

Sum of all angles of triangle is 180°

If;

One angle is 15°

Another angle is 60°

15° + 60° + x = 180°

x = 180° - 75° = 105°

if all angles are different then triangle cannot be equilateral or isosceles triangle and none of the angle is 90°

∴ triangle cannot be right angled triangle

∴ triangle is scalene triangle

Result:- polygon is scalene triangle

Given: 6km in east

8km in north

Formula Used/Theory:-

When jiya moved from east to north it possess 90° angle

And the distance of position to his home will be hypotenuse

And form an right angled triangle

⇒ Pythagoras theorem:-

side² + side² = Hypotenuse²

As 8km and 6km are 2 sides of right angled triangle

Then;

8² + 6² = Hypotenuse²

Hypotenuse² = 64 + 36 = 100

Hypotenuse = √100

Hypotenuse = 10km

∴ jiya is 10 km from his house

Result:- Jiya is 10 km from his house

Given: The park measures 60 meters × 80 meters

Formula Used/Theory:-

⇒ Pythagoras theorem:-

side² + side² = Hypotenuse²

As when we cut rectangular park diagonally

We get 2 right angled triangle

With length and breadth as 2 sides

And diagonal as hypotenuse

⇒ length² + breadth² = diagonal²

60² + 80² = diagonal²

3600 + 6400 = diagonal²

Diagonal² = 10000

Diagonal = √10000

= 100 meters

⇒ if she moves by edges she will cover

Distance of length + breadth

= 80 + 60 = 140 meters

⇒ 140 meters – 100 meters = 40 meters

She travel 40 meters less when she travel diagonally

Result:- Jayanti travel 40 meters less when she travel

diagonally

Given: ∠P = 30° ; PQ = PR

Formula Used/Theory:-

If 2 sides of triangle are equal

Then it will isosceles triangle

And its base angle will also be equal

→ Angle sum property

Sum of all angles of triangle is 180°

In Δ PQR

As PQ = PR

Hence;

Δ PQR is isosceles triangle

∴ ∠Q = ∠R

By angle sum property

Sum of all angles of triangle is 180°

∠P + ∠Q + ∠R = 180°

30° + ∠Q + ∠Q = 180°

2∠Q = 180° - 30°

2∠Q = 150°

∠Q = = 75°

As ∠Q = ∠R

∠R = 75°

Result:- ∠Q = ∠R = 75°

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

→ Exterior angle is equal to sum of interior opposite angles

By angle sum property

x + 45° + 60° = 180°

x + 105° = 180°

x = 180°-105°

x = 75°

By exterior angle property

y = 60° + x

y = 60° + 75°

y = 135°

Result:- x = 75°; y = 135°

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

→ Exterior angle is equal to sum of interior opposite angles

In Δ LOM

∠LOP = ∠MLO + ∠LMO

= 70° + 20° = 90°

In Δ PON

∠LOP = ∠NPO + ∠PNO

90° = ∠NPO + 70°

∠NPO = 90° - 70° = 20°

∠NPO + ∠PNO + ∠PON = 180° (Angle sum property)

20° + 70° + ∠PON = 180°

∠PON = 180° - 90°

∠PON = 90°

Result:- ∠PON = 90°; ∠NPO = 20°

Given: QP || RT

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

→ if lines are parallel then alternate angles are equal

AS QP || RT

And PR is transverse

x = 70° ∵ Alternate angles

By angle sum property

x + y + 30° = 180°

70° + y + 30° = 180°

Y + 100° = 180°

y = 180° - 100° = 80°

Result:- x = 70°; y = 80°

Formula Used/Theory:-

→ Exterior angle is equal to sum of interior opposite angles

In Δ ABC

Exterior angle producing BC is 115°

115° = ∠A + 65°

∠A = 115° - 65°

∠A = 50°

Result:- the value of ∠A comes out to be 50°

Given: In right angle triangle one angle is 35°

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

In a right angled triangle

One angle is always 90°

Another given is 35°

Let the 3^rd angle be X

Hence;

By angle sum property

90° + 35° + X = 180°

X = 180° - 125°

= 55°

Result:- The 3^rd angle comes out to be 55°

Given: Each of the two equal angles of an isosceles triangle is four

times the third angle

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

Let the 3^rd angle be X

The other 2 angle will be

4X

⇒ By angle sum property

X + 4X + 4X = 180°

9X = 180°

X = = 20°

The angles of triangle will be 4×20°;4×20°;1×20°

The angles of triangle is 80°;80°;20°

Result:- The angles of triangle is 80°;80°;20°

Given: The angles of a triangle are in the ratio 2 : 3 : 5

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

If angles of triangle are in ratio 2:3:5

The angles of triangles will be 2X;3X;5X

⇒ By angle sum property

2X + 3X + 5X = 180°

10X = 180°

X = = 18°

∴ The angles of triangles will be 2×18°;3×18°;5×18°

The angles of triangle is 36°;54°;90°

Result:- The angles of triangle is 36°;54°;90°

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

→ Exterior angle is equal to sum of interior opposite angles

Let 3 angle of triangle be x;y;z

Exterior angle opposite of x will be y + z

Exterior angle opposite of y will be x + z

Exterior angle opposite of z will be x + y

∴ sum of all 3 exterior angles will be

(y + z) + (x + z) + (x + y)

= 2x + 2y + 2z

= 2(x + y + z)

By angle sum property of triangle

x + y + z = 180°

sum of exterior angles is 2(x + y + z)

= 2× 180° = 360°

Result:- sum of exterior angles is 360°

Given: ∠A = ∠C, and exterior angle ABX = 140°

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

→ Exterior angle is equal to sum of interior opposite angles

If exterior angle ∠ABX = 140°

That means exterior angle is opposite of ∠B

∴ ∠ABX = ∠A + ∠C

140° = 2∠A

∠A = = 70°

∠C = 70°

⇒ By angle sum property

∠A + ∠B + ∠C = 180°

70° + ∠B + 70° = 180°

∠B = 180° - 140° = 40°

Result:- The angles of triangle are 70°;70°;40°

Formula Used/Theory:-

→ Exterior angle is equal to sum of interior opposite angles

In the bigger triangle

By exterior angle property

y = 30° + 45° = 75°

In smaller triangle

By exterior angle property

x = 50° + 30° = 80°

Result:- The value of x comes out to be 80°

The value of y comes out to be 75°

Formula Used/Theory:-

→ Exterior angle is equal to sum of interior opposite angles

In Δ ABC

When BC extended to D

Exterior angle ACD = ∠CAB + ∠ABC

∠ACD = 80° + 30° = 110°

∠ACD = ∠ACE + ∠ECD (As given in figure)

∠ACD = x + 90°

⇒ 110° = x + 90°

x = 110° - 90°

= 20°

Result:- The value of x comes out to be 20°

Given: Angle of triangle are in descending order

difference between two consecutive angles is 10°

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

Let the largest angle be x

Then the 2^nd angle will be x-10°

And 3^rd angle will be (x-10°)-10°

= x-20°

By angle sum property

x + (x-10°) + (x-20°) = 180°

3x-30° = 180°

3x = 180° + 30° = 210°

x = = 70°

the angle will be 70°;(70° -10°);(70°- 20°)

The angles of triangle will be 70°;60°;50°

Result:- The angles of triangle will be 70°;60°;50°

Given: DE || BC; ∠B = 30° ; ∠C = 40°

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

→ if 2 lines are parallel then their corresponding angles will be equal

As DE || BC

And AB is transverse

x = 30° ∵ Corresponding angles

As DE || BC

And AC is transverse

y = 40° ∵ Corresponding angles

By angle sum property in Δ ADE

x + y + z = 180°

30° + 40° + z = 180°

z = 180° -70°

z = 110°

Result:- The value of x;y;z is 30°;40°;110° respectively

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

→ Exterior angle is equal to sum of interior opposite angles

In Δ ABC

By angle sum property

∠A + ∠B + ∠C = 180°

90° + 60° + z = 180°

z = 180° - 150°

z = 30°

By exterior angle property

∠y = 60° + 60° = 120°

By exterior angle property

∠x = 30° + z

∠x = 30° + 30° = 60°

Result:- The value of x;y;z is 60°;120°;30°

Given: one angle of triangle is 60°

Other 2 angles are in ratio 1:2

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

1 angle is 60°

And other 2 angles are in ratio 1:2

The;

Let 2 angles x,2x

By angle sum property

60° + x + 2x = 180°

3x = 180° -60° = 120°

X = = 40°

The 2 angles of triangle are 40°;80°

Result:- The all 3 angles of triangle are 40°;80°;60°

Given: 3∠P = 4∠Q = 6∠R

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

Solutions:-

If 3∠P = 4∠Q = 6∠R

Taking LCM of 3,4,6

We get 12

Then;

Dividing LCM by magnitude of each angle gives ratio of all 3 angles

∠P = 12/3 = 4

∠Q = 12/4 = 3

∠R = 12/6 = 2

Means angles are in ratio 4:3:2

Then;

Let all 3 angles of triangle be 4x;3x;2x

By angle sum property

4x + 3x + 2x = 180°

9x = 180°

x = = 20°

The angles of triangle will be 4×20°;3×20°;2×20°

The angles of triangle are 80°;60°;40°

Result:- The angles of triangle are 80°;60°;40°

Given: ∠D = 60°, ∠E = 70°

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

In Δ DEF

∠D + ∠E + ∠F = 180°

60° + 70° + ∠F = 180°

∠F = 180° - 130° = 50°

Bisector of ∠E will be 35°

Bisector of ∠F will be 25°

Joining the bisector of ∠E and ∠F at O will make a Δ EOF

Where ∠OEF = 35°

And ∠OFE = 25°

By angle sum property

∠OEF + ∠OFE + ∠EOF = 180°

∠EOF + 35° + 25° = 180°

∠EOF = 180° - 60° = 120°

Result:- ∠EOF = 120°

∠F = 50°

Given: QP || ST ; US || RP; ∠P = 90°

Formula Used/Theory:-

If all sides of both the triangles are parallel than then they are similar triangles

And their angles will also be equal

In Δ PQR and Δ STU

PQ || ST (given)

PR ||SU (given)

QR || TU (both lies on same line)

∴ Δ PQR ~ Δ STU

Then;

∠P = ∠S; ∠Q = ∠T; ∠R = ∠U

If ∠P = 90°

Then; ∠S = 90°

Result:- angle S measured out to be 90°

Formula Used/Theory:-

ASA congruence criterion is in which 2 angles and a side between them are equal in both the triangles

(a) ∠A = ∠Q

But ∠B≠∠P

∴ Δ ABC and Δ PQR are not congruent

Result:- Δ ABC and Δ PQR are not congruent

(b) ∠ABD = ∠BDC

∠ADB = ∠DBC

BD = BD (common in both triangle)

∴ Δ ADB and Δ CBD are congruent by ASA

∆ADB ≅ ∆CBD

Result:- Δ ADB and Δ CBD are congruent by ASA

(c) ASA congruence criterion is in which 2 angles and a side between them are equal in both the triangles

∠X = ∠L

∠Y = ∠M

XY = ML

∴ Δ XYZ and Δ LMN are congruent by ASA

∆XYZ ≅ ∆LMN

Result:- Δ XYZ and Δ LMN are congruent by ASA

(d) → Angle sum property

Sum of all angles of triangle is 180°

By angle sum property

∠A + ∠B + ∠C = 180° ∠D + ∠E + ∠F = 180°

Equating both

We get;

∠A + ∠B + ∠C = ∠D + ∠E + ∠F

As ∠B = ∠F

∠A = ∠D

Cancelling out we get, ∠C = ∠E

∠C = ∠E

∠B = ∠F

BC = FE

∴ Δ ABC and Δ DFE are congruent by ASA

∆ABC ≅ ∆DFE

Result:- Δ ABC and Δ DFE are congruent by ASA

(e) In Δ PNO and Δ MNO

∠PNO = ∠MON

∠MNO≠∠PON

ON = ON (common in both triangles)

∴ Δ MNO and ΔPON are not congruent by ASA

Result:- ∴ Δ MNO and ΔPON are not congruent by ASA

(f) ∠D = ∠C

∠AOD = ∠COB

OD = CO

∴ Δ ADO and Δ BCO are congruent by ASA

∆ADO ≅ ∆BCO

Result:- Δ ADO and Δ BCO are congruent by ASA

Formula Used/Theory:-

RHS congruence criterion is in which hypotenuse and one side are equal in both the triangles

(a) AC = AB (Hypotenuse)

AD = AD (common in both triangles)

∴ Δ ADB and Δ ADC are congruent by RHS

∆ADB ≅ ∆ADC

Result:- Δ ADB and Δ ADC are congruent by RHS

(b) XZ = YU (Hypotenuse)

YZ = YZ (common in both triangles)

∴ Δ XYZ and Δ UZY are congruent by RHS

∆XYZ ≅ ∆UZY

Result:- Δ XYZ and Δ UZY are congruent by RHS

(c) AE = EB (Hypotenuse)

CE = ED

∴ Δ ACE and Δ BDE are congruent by RHS

∆ACE ≅ ∆BDE

Result:- Δ ACE and Δ BDE are congruent by RHS

(d) ⇒ Pythagoras theorem:-

Base² + Height² = Hypotenuse²

In Δ ABC

AC² = 6² + 8²

AC² = 36 + 64

AC = √100

AC = 10cm

CD = BD – BC = 14 cm – 8 cm

CD = 6cm

AC = CE (Hypotenuse)

AB = CD

∴ Δ ABC and Δ CDE are congruent by RHS

∆ABC ≅ ∆CDE

Result:- Δ ABC and Δ CDE are congruent by RHS

(e) XY = XY (Common Hypotenuse)

XZ≠YU

XU≠YZ

∴ Δ XYZ and Δ XYU are not congruent by RHS

Result:- Triangles are not congruent

(f) LM = LN (Hypotenuse)

LO = LO

∴ Δ LOM and Δ LON are congruent by RHS

∆LOM ≅ ∆LON

Result:- Δ LOM and Δ LON are congruent by RHS

Given:

Formula Used/Theory:-

If 2 sides are equal then triangle is Isosceles triangle

In isosceles triangle base angles of triangles are also equal

In Δ PQR

As QR = PR

Then;

∠PQR = ∠QPR

∠QPR = 50° (By vertical opposite angle)

∠PQR = x = 50°

Result:- the value of x comes out to be 50°

Formula Used/Theory:-

If 2 sides are equal then triangle is Isosceles triangle

In isosceles triangle base angles of triangles are also equal

→ Angle sum property

Sum of all angles of triangle is 180°

→ exterior angle is sum of interior opposite angles

∠TSU = 78° (Vertically opposite angles )

By isosceles triangle property

∠STU = ∠SUT

By angle sum property

∠STU + ∠SUT + ∠TUS = 180°

y + y + 78° = 180°

2y = 180° - 78° = 102° =

y = 51°

By exterior angle property

x = 78° + y

x = 78° + 51° = 129°

Result:- The value of x and y are 51° and 102°

respectively

Given: length of sides are 1.5,3.6,3.9

Formula Used/Theory:-

⇒ Pythagoras theorem:-

Base² + Height² = Hypotenuse²

Largest side is 3.9cm

Other 2 sides are 1.5cm,3.6cm

As hypotenuse is largest side

Hypotenuse² = (3.9)² = 15.21

Sum of square of other 2 sides of triangle

(3.6)² + (1.5)² = 12.96 + 2.25

= 15.21

LHS = RHS

Hence Pythagoras theorem verified

∴ it is a right angled triangle

Result:- Triangle having sides of these lengths is right

angled triangle

Given: Height = 8m

Base = 6m

Formula Used/Theory:-

⇒ Pythagoras theorem:-

Base² + Height² = Hypotenuse²

Hypotenuse is length of rope tied on top of pole to ground

(Length of rope)² = 8² + 6²

(Length of rope)² = 64 + 36 = 100

Length of rope = √100

= 10cm

Result:- Length of rope is 10cm

Formula Used/Theory:-

→ Angle sum property

Sum of all angles of triangle is 180°

→ exterior angle is sum of interior opposite angles

If y = 5× x = 5x

By angle sum property

∠R + ∠S + ∠RQS = 180°

60° + y + x = 180°

y + x = 180° -60° = 120°

5x + x = 120°

6x = 120°

x = 20°

y = 5x = 5× 20° = 100°

By exterior angle property

z = 60° + y

z = 60° + 100° = 160°

Result:- the values of x;y;z are 20°;100°;160° resp.

Formula Used/Theory:-

Perimeter = sum of all sides of triangle

There can be 2 cases

(i) If equal sides in isosceles triangle is equal to 9cm

Then;

Perimeter = 9cm + 9cm + 20cm

= 38cm

(ii) If equal sides in isosceles triangle is equal to 20cm

Then;

Perimeter = 20cm + 20cm + 9cm

= 49cm

∴ perimeter of triangle can be either 38cm or 49cm

Result:- Perimeter of triangle is either 38cm or 49cm.

(a) ST = DE ∠S = ∠D

TU = EF ∠T = ∠E

SU = DF ∠U = ∠F

(b) AB = LM ∠A = ∠L

BC = MN ∠B = ∠M

AC = LM ∠C = ∠N

(c) YZ = PQ ∠Y = ∠P

ZX = QR ∠Z = ∠Q

XY = RP ∠X = ∠R

(d) XY = ML ∠X = ∠M

YZ = LN ∠Y = ∠L

ZX = NM ∠Z = ∠N

(a) AB = LN

BC = ML

AC = NM

Hence; both triangles are congruent by SSS criterion

∆ABC ≅ ∆NLM

(b) LM = GH

MN = HI

NL = IG

Hence; both triangles are congruent by SSS criterion

∆LMN ≅ ∆GHI

(c) LM = LO

MN = ON

LN = LN

Hence; both triangles are congruent by SSS criterion

∆LMN ≅ ∆LON

(d) WY = ZX

WX = ZY

XY = XY

Hence; both triangles are congruent by SSS criterion

∆XYW ≅ ∆YXZ

(e) AO = OD

BO = OE

AB = DE

Hence; both triangles are congruent by SSS criterion

∆AOB ≅ ∆DOE

(f) TU = UV

TS = VS

US = US

Hence; both triangles are congruent by SSS criterion

∆TUS ≅ ∆VUS

(g) SR = PQ

PS = QR

PR = PR

Hence; both triangles are congruent by SSS criterion

∆PQR ≅ ∆RSP

(h) SU = PR

ST = PQ

TU≠RQ

Hence triangle are not congruent

Given: AB = AC; D is midpoint of BC

As D is midpoint of BC

BD = DC

In Δ ADB and Δ ADC

AB = AC (given)

BD = DC (D is midpoint of BC)

AD = AD (common)

⇒ these are all 3 equal parts of triangles ABD and ACD

⇒ yes ∆ABD ≅ ∆ACD

Because it follows SSS congruence criterion

Given: LM = ON and NL = MO

In Δ LMN and Δ MNO

LM = ON (given)

LN = OM (given)

MN = MN (common)

⇒ these are all 3 equal parts of triangles LMN and MNO

⇒ yes ∆NOM ≅ ∆MLN

Because it follows SSS congruence criterion

Given: Triangles DEF and LMN are both isosceles

DE = DF and LM = LN, respectively.

If DE = LM and EF = MN

DE = DF and LM = LN

If DE = LM

Then putting value on both sides

We get

DF = LN

In Δ DEF and Δ LMN

DE = LM

DF = LN

EF = MN

⇒ Hence Δ DEF ≅ Δ LMN

Both triangles are congruent by SSS criterion

As both triangles are congruent

∠E = ∠M

∴ ∠M = 40°

As LMN is isosceles triangle having base angles are equal

∠M = ∠N

∴ ∠N = 40°

Given: ∆PQR and ∆SQR are both isosceles triangle on a common

base QR

Formula Used/Theory:-

→ 2 sides of triangle are equal gives a isosceles triangle

→ If all sides of triangle are equal to all other sides of triangle

Then both triangle are congruent by SSS criterion

As ∆PQR and ∆SQR are both isosceles triangle on a common base QR

Then;

PQ = PR and SQ = SR

In Δ PSQ and Δ PSR

SQ = SR (stated above)

PQ = PR (stated above)

PS = PS (common)

∴ Δ PSQ ≅ Δ PSR

Hence; both triangles PSQ and PSR are congruent by SSS criterion

(i) QR = SU

QP = TU

∠Q = ∠U

Hence; both triangles are congruent by SAS criterion

Δ PQR ≅ Δ TUS

(ii) JK = MN

JL≠NO

∠J = ∠N

Hence; both triangles are not congruent by SAS criterion

(iii) AB = BC

AE = CD

∠A = ∠C

Hence; both triangles are congruent by SAS criterion

Δ EAB ≅ Δ DCB

(iv) TS = XZ

TU = YZ

∠T = ∠Z

Hence; both triangles are congruent by SAS criterion

Δ STU ≅ Δ XZY

(v) OD = OH

OF = OC

∠FOD = ∠COH

Hence; both triangles are congruent by SAS criterion

Δ FOD ≅ Δ COH

(vi) AB = DE

AC≠DF

∠A = ∠D

Hence; both triangles are not congruent by SAS criterion

(vii) SP = QR

SQ = SQ

∠PSQ = ∠RQS

Hence; both triangles are congruent by SAS criterion

Δ PSQ ≅ Δ RQS

(viii) LM = OM

MN = MN

∠LMN = ∠OMN

Hence; both triangles are congruent by SAS criterion

Δ LMN ≅ Δ OMN

PQ = ST

QR = TU

∠Q = ∠T

Hence; both triangles are congruent by SAS criterion

Δ PQR ≅ Δ STU

(b) AB = XZ

∠A = ∠X

YZ≠BC (Hypotenuse )

AC≠XY

Hence; both triangles are not congruent neither by RHS nor by

SAS congruence criterion

Given: PQ = PS and ∠1 = ∠2

Formula Used/Theory:-

→ when 2 sides and angle between both the sides are equal in both the triangles then triangles are congruent by SAS congruence criterion

In Δ PQR and Δ PSR

PQ = PS (Given)

PR = PR (Common)

∠QPR = ∠SPR (Given)

∴ ∆PQR ≅ ∆PSR

Hence; Δ PQR and Δ PSR are congruent by SAS criterion

If ∆PQR ≅ ∆PSR then;

All 6 perimeters are equal.

PQ = PS; PR = PR; QR = SR

∠1 = ∠2; ∠Q = ∠S; ∠PRQ = ∠PRS

Given: DE = IH, EG = FI and ∠E = ∠I

Formula Used/Theory:-

→ when 2 sides and angle between both the sides are equal in both the triangles then triangles are congruent by SAS congruence criterion

In Δ DEF and Δ GIH

⇒ DE = HI (Given)

EG = FI

If we add length of GF both sides

EG + GF = FI + GF

⇒ EF = IG

⇒ ∠E = ∠I

∴ ∆DEF ≅ ∆HIG

Hence; Both triangles are congruent by SAS criterion

Given: ∠1 = ∠2 and ∠3 = ∠4

Formula Used/Theory:-

→ when 2 angles and side between both the angles are equal in both the triangles then triangles are congruent by ASA congruence criterion

In ∆ADC and ∆ABC

∠1 = ∠2 (Given)

∠3 = ∠4 (Given)

AC = AC (common)

∆ADC ≅ ∆ABC

⇒ Hence; both triangles are congruent by ASA congruence

If ∆ADC ≅ ∆ABC then;

If triangles are congruent then all 3 sides of triangle are equal to all 3 sides of other triangle

AC = AC

⇒ AD = AB

⇒ CD = CB

Formula Used/Theory:-

→ when 2 angles and side between both the angles are equal in both the triangles then triangles are congruent by ASA congruence criterion

In ∆ABC and ∆DCB

∠B = ∠C (Both 70° )

∠ACB = ∠DCB (Both 30° )

BC = BC (common)

∆ABC ≅ ∆DCB

⇒ Hence; both triangles are congruent by ASA congruence

If ∆ABC ≅ ∆DCB then;

If triangles are congruent then all 3 sides of triangle are equal to all 3 sides of other triangle

BC = BC

⇒ AB = DC

⇒ AC = DB

Given: QS ⊥ PR, RT ⊥ PQ and QS = RT

Formula Used/Theory:-

⇒ If hypotenuse and 1 sides of Right angled triangle are equal in both the triangles then both triangles are congruent by RHS congruence criterion

In ∆QSR and ∆RTQ

As ∆QSR, ∆RTQ both are right angle triangle

Right angled at ∠QSR and ∠RTQ

QR = QR (Hypotenuse)

QS = TR (Given)

∆QSR ≅ ∆RTQ

Hence; both triangles are congruent by RHS criterion

If ∆QSR ≅ ∆RTQ then;

All 3 angles of one triangle will be equal to all 3 angles of another triangles

⇒ ∠Q = ∠R

∠QTR = ∠QSR

∠SQR = ∠QRT

In the given triangle

AC as distance of pound

AC² = 30² + 40²

AC² = 900 + 1600

AC² = 2500

AC = √2500

= 50m

AC = AB + BC

50m = AB + 12m

AB = 38m

Given:

Two poles of 10 m and 15 m

Distance between the tops is 13 m

Formula Used/Theory:-

Pythagoras theorem:-

Base² + Height² = Hypotenuse²

Difference between height of poles(Height) = 15m-10m = 5m

Distance between their tops is (Hypotenuse) = 13m

Distance by which poles are separated(Base)

Base² + Height² = Hypotenuse²

Base² + 5² = 13²

Base² + 25 = 169

Base² = 169-25 = 144

Base = √144 = 12m

Hence; poles are separated by 12m

Result:- poles are separated by 12m

Given: Height of window = 8m

Distance of ladder to wall = 6m

Formula Used/Theory:-

Pythagoras theorem:-

Base² + Height² = Hypotenuse²

Height of ladder (Hypotenuse)

Height of wall (Height) = 8m

Distance of wall and foot of ladder = 6m

Base² + Height² = Hypotenuse²

6² + 8² = Hypotenuse²

Hypotenuse² = 36 + 64 = 100

Hypotenuse = √100 = 10m

(i) Height of ladder is 10m

(ii) If ladder is shifted 8m from wall

Then;

Base = 8m

Height of ladder will remain constant(Hypotenuse) = 10m

Distance of wall till ladder reach(Height)

Base² + Height² = Hypotenuse²

8² + Height² = 10²

Height² + 64 = 100

Height² = 100-64 = 36

Height = √36 = 6m

Ladder reach 6m above the ground

Formula Used/Theory:-

⇒ If hypotenuse and 1 sides of Right angled triangle are equal in both the triangles then both triangles are congruent by RHS congruence criterion

In ∆ABC and ∆EOD

As ∆ABC, ∆EOD both are right angled triangle

Right angled at ∠ABC and ∠EOD

∠B = ∠O = 90°

AC = DE (Hypotenuse Given)

AB = EO (Given)

∆ABC ≅ ∆EOD

Hence; both triangles are congruent by RHS criterion