Question

# Arrange in descending order of magnitude and

Hint:

### Use LCM property.

## The correct answer is: descending order

### Complete step by step solution:

Here we can write, and

On taking the LCM of 3,6,9 we have LCM as 18.

So,

Likewise,

Likewise,

Now, we have and

So, here descending order is 64 > 27 > 16

### Related Questions to study

### In △ABC, AD⊥BC Also, Prove that

As BD = 3 CD ;We get BC = BD +CD = 4 CD

Applying pythagoras theorem in ΔACD ,We get

Applying pythagoras theorem in ΔABD ,We get

—Eq2

Substitute Eq2 in Eq1 ,

As BD = 3 CD ;We get

Multiplying with 2 on both sides

As BC = 4 CD

∴

Hence proved

### In △ABC, AD⊥BC Also, Prove that

As BD = 3 CD ;We get BC = BD +CD = 4 CD

Applying pythagoras theorem in ΔACD ,We get

Applying pythagoras theorem in ΔABD ,We get

—Eq2

Substitute Eq2 in Eq1 ,

As BD = 3 CD ;We get

Multiplying with 2 on both sides

As BC = 4 CD

∴

Hence proved

### In the figure, AD is a median to BC in and . and DE =

Aim :- Prove that

Hint :- apply in pythagoras theorem on triangles AEB , AEC and AED Find and values .Now, add them and do necessary modification to get in terms of p and a.

Explanation(proof ) :-

As D is mid point BD = DC = BC/2 = a/2

Applying pythagoras theorem to triangle AEB we get ,

Applying pythagoras theorem to triangle AEC we get ,

Applying pythagoras theorem to triangle AED we get ,

Adding Eq 1 and Eq 2

We get

[By applying

Substitute Eq3 in the above condition

Hence proved

### In the figure, AD is a median to BC in and . and DE =

Aim :- Prove that

Hint :- apply in pythagoras theorem on triangles AEB , AEC and AED Find and values .Now, add them and do necessary modification to get in terms of p and a.

Explanation(proof ) :-

As D is mid point BD = DC = BC/2 = a/2

Applying pythagoras theorem to triangle AEB we get ,

Applying pythagoras theorem to triangle AEC we get ,

Applying pythagoras theorem to triangle AED we get ,

Adding Eq 1 and Eq 2

We get

[By applying

Substitute Eq3 in the above condition

Hence proved

Given angles are in the ratio 5:6:7

Step 1:- Find angles in terms of k and find k

The angles are 5k ,6k and 7 k respectively

We know Sum of angles of triangle is 180°

5k +6k+7k = 180

18 k = 180

K = 10°

Step 2:- Find the actual angles of triangle

5k = 5×10 = 50°

6k = 6×10 = 60°

7k = 7×10 = 70°

Therefore 50°,60° and 70° are the angles of the triangle which satisfy the condition .

Given angles are in the ratio 5:6:7

Step 1:- Find angles in terms of k and find k

The angles are 5k ,6k and 7 k respectively

We know Sum of angles of triangle is 180°

5k +6k+7k = 180

18 k = 180

K = 10°

Step 2:- Find the actual angles of triangle

5k = 5×10 = 50°

6k = 6×10 = 60°

7k = 7×10 = 70°

Therefore 50°,60° and 70° are the angles of the triangle which satisfy the condition .

Explanation(proof ) :-

Applying pythagoras theorem in triangle ABC,we get

—- Eq1

Applying pythagoras theorem in triangle ABD ,we get

— Eq2

As AD is median , D is mid point then BC = 2BD

Subtract Eq1 - Eq2

We get ,

As D is mid point BD = CD ,

substitute BD = CD in the above equation

∴

Hence proved

Explanation(proof ) :-

Applying pythagoras theorem in triangle ABC,we get

—- Eq1

Applying pythagoras theorem in triangle ABD ,we get

— Eq2

As AD is median , D is mid point then BC = 2BD

Subtract Eq1 - Eq2

We get ,

As D is mid point BD = CD ,

substitute BD = CD in the above equation

∴

Hence proved

### The Quadrilateral PQRS has angles at S,Q right angles and the diagonals PR, QS are perpendicular. Prove that SR = QR.

Aim :- Prove that SR = QR

Explanation(proof ) :-

We know that if a quadrilateral has a sum of opposite angles 180°

∠S + ∠ Q = 90 + 90 = 180°(S and Q are opposite angle)

Here PQRS is cyclic and ∠S = 90°

As the angle angle in a semicircle is 90° then PSR lies in a semi circle

and PR is diameter.

As PR passes through centre and perpendicular to chord SQ

The perpendicular drawn from centre to chord bisects the chord

So, OS = OQ as PR bisects chord SQ

In OQR and OSR

OS = OQ (side)

∠SOR = ∠QOR =90° (Angle)

OP =OP (common side )

By SAS rule OQR ≅ OSR

By congruence we get , SR= QR

Hence proved

### The Quadrilateral PQRS has angles at S,Q right angles and the diagonals PR, QS are perpendicular. Prove that SR = QR.

Aim :- Prove that SR = QR

Explanation(proof ) :-

We know that if a quadrilateral has a sum of opposite angles 180°

∠S + ∠ Q = 90 + 90 = 180°(S and Q are opposite angle)

Here PQRS is cyclic and ∠S = 90°

As the angle angle in a semicircle is 90° then PSR lies in a semi circle

and PR is diameter.

As PR passes through centre and perpendicular to chord SQ

The perpendicular drawn from centre to chord bisects the chord

So, OS = OQ as PR bisects chord SQ

In OQR and OSR

OS = OQ (side)

∠SOR = ∠QOR =90° (Angle)

OP =OP (common side )

By SAS rule OQR ≅ OSR

By congruence we get , SR= QR

Hence proved

### P and Q are points on the sides CA and CB respectively of a right angled at c. Prove that

Hint :- Applying pythagoras theorem to both triangles ABC, ACQ,PCQ and PCB

find the equation, add them and substitute proper conditions to prove the condition

Hint :- Applying pythagoras theorem to both triangles ABC, ACQ,PCQ and PCB

find the equation, add them and substitute proper conditions to prove the condition

Explanation(proof ) :-

Applying pythagoras theorem to triangle AQC we get ,

Applying pythagoras theorem to triangle PBC we get ,

Applying pythagoras theorem to triangle ABC we get ,

Applying pythagoras theorem to triangle PCQ we get ,

Adding Eq1 and Eq2 we get ,

Hence proved

### P and Q are points on the sides CA and CB respectively of a right angled at c. Prove that

Hint :- Applying pythagoras theorem to both triangles ABC, ACQ,PCQ and PCB

find the equation, add them and substitute proper conditions to prove the condition

Hint :- Applying pythagoras theorem to both triangles ABC, ACQ,PCQ and PCB

find the equation, add them and substitute proper conditions to prove the condition

Explanation(proof ) :-

Applying pythagoras theorem to triangle AQC we get ,

Applying pythagoras theorem to triangle PBC we get ,

Applying pythagoras theorem to triangle ABC we get ,

Applying pythagoras theorem to triangle PCQ we get ,

Adding Eq1 and Eq2 we get ,

Hence proved

### In △ABC, ∠B=90^{∘} and is the mid point of BC. Prove that

Hint :- use pythagoras theorem in triangle ABD and ABC

We get and subtract them and substitute BC = 2BD

In the equation to get the result.

Explanation(proof ) :-

Applying pythagoras theorem in triangle ABC,we get

—- Eq1

Applying pythagoras theorem in triangle ABD ,we get

Eq2

As AD is median , D is mid point then BC = 2BD

Subtract Eq1 - Eq2

We get,

Hence proved

### In △ABC, ∠B=90^{∘} and is the mid point of BC. Prove that

Hint :- use pythagoras theorem in triangle ABD and ABC

We get and subtract them and substitute BC = 2BD

In the equation to get the result.

Explanation(proof ) :-

Applying pythagoras theorem in triangle ABC,we get

—- Eq1

Applying pythagoras theorem in triangle ABD ,we get

Eq2

As AD is median , D is mid point then BC = 2BD

Subtract Eq1 - Eq2

We get,

Hence proved

Explanation(proof ) :-

BK:KC = 2:1 KC :( KC+BK) = 1:(1+2) KC :CB = 1:3

In equilateral triangle all sides are equal

Let , AB = BC = CA = x

Given CK = BC = ( BC is trisected at K)

In equilateral triangle ,the altitudes and medians coincide

so,AE is median also i.e E is midpoint of BC

CE =

From diagram ,KE = CE - CK =

We get height of equilateral triangle when side x is

Applying pythagoras theorem on triangle ADE

We get,

Substitute x by AC as x = AC

Hence proved

Explanation(proof ) :-

BK:KC = 2:1 KC :( KC+BK) = 1:(1+2) KC :CB = 1:3

In equilateral triangle all sides are equal

Let , AB = BC = CA = x

Given CK = BC = ( BC is trisected at K)

In equilateral triangle ,the altitudes and medians coincide

so,AE is median also i.e E is midpoint of BC

CE =

From diagram ,KE = CE - CK =

We get height of equilateral triangle when side x is

Applying pythagoras theorem on triangle ADE

We get,

Substitute x by AC as x = AC

Hence proved

### The perpendicular AD on the base BC of Triangle ABC intersects BC in D such that BD = 3CD. Prove that

Explanation(proof ) :-

As BD = 3 CD ;We get BC = BD +CD = 4 CD

Applying pythagoras theorem in ΔACD ,We get

— Eq1

Applying pythagoras theorem in ΔABD ,We get

—Eq2

Substitute Eq2 in Eq1 ,

As BD = 3 CD ;We get

Multiplying with 2 on both sides

Hence proved

### The perpendicular AD on the base BC of Triangle ABC intersects BC in D such that BD = 3CD. Prove that

Explanation(proof ) :-

As BD = 3 CD ;We get BC = BD +CD = 4 CD

Applying pythagoras theorem in ΔACD ,We get

— Eq1

Applying pythagoras theorem in ΔABD ,We get

—Eq2

Substitute Eq2 in Eq1 ,

As BD = 3 CD ;We get

Multiplying with 2 on both sides

Hence proved

### In an equilateral the side BC is trisected at D . Prove that

.

Aim :- Prove that

Hint :- let the side of equilateral triangle be x , then we get height of equilateral

triangle as .Now , Find length of DE and apply pythagoras theorem in

Right angle triangle ADE to find length of AD in terms of AB.

Explanation(proof ) :-

In equilateral triangle all sides are equal

Let , AB = BC = CA = x

Given BD = ⅓ BC = x/3 ( BC is trisected at D)

n equilateral triangle ,the altitudes and medians coincide

so,AE is median also i.e E is midpoint of BC

BE = x/2

From diagram ,DE = BE - BD

We get height of equilateral triangle when side x is

Applying pythagoras theorem on triangle ADE

We get,

Substitute x by AB as x = AB

Hence proved

### In an equilateral the side BC is trisected at D . Prove that

.

Aim :- Prove that

Hint :- let the side of equilateral triangle be x , then we get height of equilateral

triangle as .Now , Find length of DE and apply pythagoras theorem in

Right angle triangle ADE to find length of AD in terms of AB.

Explanation(proof ) :-

In equilateral triangle all sides are equal

Let , AB = BC = CA = x

Given BD = ⅓ BC = x/3 ( BC is trisected at D)

n equilateral triangle ,the altitudes and medians coincide

so,AE is median also i.e E is midpoint of BC

BE = x/2

From diagram ,DE = BE - BD

We get height of equilateral triangle when side x is

Applying pythagoras theorem on triangle ADE

We get,

Substitute x by AB as x = AB

Hence proved

ABC and DBC are two isosceles triangles on the same base BC. Show

that

Explanation :-

In triangle ABC, AB = AC (isosceles triangle )

so, ∠ ABC = ∠ ACB

In triangle BDC, BD = DC (isosceles triangle )

∠ DBC =∠ DCB

Adding above results we get,

∠ABC + ∠ DBC = ∠ ACB + ∠ DCB

[ from diagram we get ∠ ABC + ∠ DBC = ∠ ABD and ∠ ACB + ∠ DCB = ACD

∠ ABD = ∠ACD

Hence proved

ABC and DBC are two isosceles triangles on the same base BC. Show

that

Explanation :-

In triangle ABC, AB = AC (isosceles triangle )

so, ∠ ABC = ∠ ACB

In triangle BDC, BD = DC (isosceles triangle )

∠ DBC =∠ DCB

Adding above results we get,

∠ABC + ∠ DBC = ∠ ACB + ∠ DCB

[ from diagram we get ∠ ABC + ∠ DBC = ∠ ABD and ∠ ACB + ∠ DCB = ACD

∠ ABD = ∠ACD

Hence proved

### PQ and RS are respectively the smallest and longest sides of a quadrilateral PQRS. Show that ∠ P > ∠ R

In triangle PQR,

PQ < QR (PQ is smallest)

so, ∠PRQ ∠QPR

In triangle PSR,

PS < SR (RS is longest side)

∠ PRS ∠ RPS

Adding above results we get,

∠PRQ + ∠PRS ∠QPR + ∠RPS

∠R ∠P

Hence proved

### PQ and RS are respectively the smallest and longest sides of a quadrilateral PQRS. Show that ∠ P > ∠ R

In triangle PQR,

PQ < QR (PQ is smallest)

so, ∠PRQ ∠QPR

In triangle PSR,

PS < SR (RS is longest side)

∠ PRS ∠ RPS

Adding above results we get,

∠PRQ + ∠PRS ∠QPR + ∠RPS

∠R ∠P

Hence proved

### Classify whether the following Surd is rational or irrational , justify. 3√24− 5√5

(It is an irrational number)

(It is an irrational number)

Here both the numbers are irrational.

On subtraction, we have , we get an irrational number.

This cant be written in the form , where p and q are integers.

∴ is an irrational number.

### Classify whether the following Surd is rational or irrational , justify. 3√24− 5√5

(It is an irrational number)

(It is an irrational number)

Here both the numbers are irrational.

On subtraction, we have , we get an irrational number.

This cant be written in the form , where p and q are integers.

∴ is an irrational number.

### In △ABC, AD and are the bisectors of ∠A and ∠C respectively. AD and CE intersect at . If ∠ABC = 90^∘, then find ∠AOC.

Explanation :-

Step 1:- find∠OAC and ∠OCA

In ABC ,

∠A+∠B+∠C = 180°(sum of angles in the triangle)

∠A+∠C = 180°- 90°

∠A+∠C = 90°

∠OAC = ½ ∠A (AD is angular bisector )

∠OCA = ½ ∠C (CE is angular bisector )

Step 2 :- Find ∠AOC

In AOC ,

∠AOC + ∠OAC + ∠OCA = 180°(sum of angles in the triangle)

∠AOC + ½ ∠A + ½ ∠C = 180°

∠AOC + ½ (∠A + ∠C) = 180° (Substitute ∠A+∠C = 90°)

∠AOC + ½ (90°) = 180° ∠AOC = 180°- 45° = 135°.

∴ ∠AOC = 135°

### In △ABC, AD and are the bisectors of ∠A and ∠C respectively. AD and CE intersect at . If ∠ABC = 90^∘, then find ∠AOC.

Explanation :-

Step 1:- find∠OAC and ∠OCA

In ABC ,

∠A+∠B+∠C = 180°(sum of angles in the triangle)

∠A+∠C = 180°- 90°

∠A+∠C = 90°

∠OAC = ½ ∠A (AD is angular bisector )

∠OCA = ½ ∠C (CE is angular bisector )

Step 2 :- Find ∠AOC

In AOC ,

∠AOC + ∠OAC + ∠OCA = 180°(sum of angles in the triangle)

∠AOC + ½ ∠A + ½ ∠C = 180°

∠AOC + ½ (∠A + ∠C) = 180° (Substitute ∠A+∠C = 90°)

∠AOC + ½ (90°) = 180° ∠AOC = 180°- 45° = 135°.

∴ ∠AOC = 135°

Aim :- Prove that

Hint :- Applying pythagoras theorem to both triangles ACM and BLA. find the BL

^{2}

and CM

^{2}add them and substitute proper conditions to prove the condition .

Explanation(proof ) :-

Given, BL and CM are the medians of a triangle ABC

M is midpoint of AB i.e AM = ½ AB

L is mid point of AC i.e AL = ½ AC

Applying pythagoras theorem to triangle BLA we get ,

As we know AL = ½ AC we get

Applying pythagoras theorem to triangle ACM we get ,

As we know AM = ½ AB we get — Eq2

Applying pythagoras theorem to triangle ABC we get ,

Adding Eq1 and Eq2 we get ,

Substitute Eq 3 in the above , we get

Hence proved

Aim :- Prove that

Hint :- Applying pythagoras theorem to both triangles ACM and BLA. find the BL

^{2}

and CM

^{2}add them and substitute proper conditions to prove the condition .

Explanation(proof ) :-

Given, BL and CM are the medians of a triangle ABC

M is midpoint of AB i.e AM = ½ AB

L is mid point of AC i.e AL = ½ AC

Applying pythagoras theorem to triangle BLA we get ,

As we know AL = ½ AC we get

Applying pythagoras theorem to triangle ACM we get ,

As we know AM = ½ AB we get — Eq2

Applying pythagoras theorem to triangle ABC we get ,

Adding Eq1 and Eq2 we get ,

Substitute Eq 3 in the above , we get

Hence proved