Question

# Simplify 4√12+5√27 - 3√75 +√300

Hint:

### Use prime factorization.

## The correct answer is: 18√3

### Complete step by step solution:

On prime factorization, we can write

and

This can be written as and respectively.

So,

### Related Questions to study

### Simplify √45 - 3√20 +4√5

On prime factorization, we can write

This can be written as and respectively.

So,

### Simplify √45 - 3√20 +4√5

On prime factorization, we can write

This can be written as and respectively.

So,

### Arrange in ascending order of magnitude and

Here we can write,

On taking the LCM of 2,3,6 we have LCM as 6.

So,

Likewise,

Likewise,

Now, we have and

So, here ascending order is 12 < 121 < 125

### Arrange in ascending order of magnitude and

Here we can write,

On taking the LCM of 2,3,6 we have LCM as 6.

So,

Likewise,

Likewise,

Now, we have and

So, here ascending order is 12 < 121 < 125

### The Diagonals of a square ABCD meet at O. Prove that

Aim :- Prove that

Let length of side of square be a

Then ,applying using pythagoras theorem in triangle ADC

We get

As diagonals bisect each other AC = 2OA

where a =AB

We get

Hence proved

### The Diagonals of a square ABCD meet at O. Prove that

Aim :- Prove that

Let length of side of square be a

Then ,applying using pythagoras theorem in triangle ADC

We get

As diagonals bisect each other AC = 2OA

where a =AB

We get

Hence proved

### △ PQR is a right triangle with ∠Q=90^{∘} M is the midpoint of QR. Prove that

As M is mid point of QR then, QM = ½ RQ

We get —Eq1

Applying Pythagoras theorem For triangle PQM

—Eq2

Substitute Eq2 in Eq1

We get

Hence proved

### △ PQR is a right triangle with ∠Q=90^{∘} M is the midpoint of QR. Prove that

As M is mid point of QR then, QM = ½ RQ

We get —Eq1

Applying Pythagoras theorem For triangle PQM

—Eq2

Substitute Eq2 in Eq1

We get

Hence proved

### In Triangle . Prove that

Explanation(proof ) :-Applying pythagoras theorem in Δ ACD ,We get

— Eq1

Applying pythagoras theorem in Δ ABD ,We get

Substitute Eq1 in Eq2 ,

Hence proved

### In Triangle . Prove that

Explanation(proof ) :-Applying pythagoras theorem in Δ ACD ,We get

— Eq1

Applying pythagoras theorem in Δ ABD ,We get

Substitute Eq1 in Eq2 ,

Hence proved

### In and prove that

Aim :- Prove that

Hint :- using inverse of pythagoras theorem, we get right angle at Q,

Now using angles prove the similarity of QMR and PMQ triangles. And find the

ratios to get the desired property

Explanation(proof ) :-

Given ,

Here ,PR is hypotenuse and angle Q is right angle.

Then

We know

By AA similarity

By similarity we get

Hence proved.

### In and prove that

Aim :- Prove that

Hint :- using inverse of pythagoras theorem, we get right angle at Q,

Now using angles prove the similarity of QMR and PMQ triangles. And find the

ratios to get the desired property

Explanation(proof ) :-

Given ,

Here ,PR is hypotenuse and angle Q is right angle.

Then

We know

By AA similarity

By similarity we get

Hence proved.

### In the figure,then find the value of x and y.

Step 1:-Find ∠QSR

We know that in a triangle, the exterior angle is always equal to the sum of the interior opposite angles.

∠QSR +∠SQR = ∠QRT

∠QSR +28° = 65°

∠QSR = 65° -28° = 37°

Step 2:- Find value of x

PQ || SR and QS intersects both the lines

We get angle x = ∠QSR ( alternate interior angles)

x =37°

Step 3:- Find value of y

In PSQ, ∠P + x + y = 180 °(sum of angles in triangles)

90°+37° +y =180 °

y = 180-127

y = 53 °

### In the figure,then find the value of x and y.

Step 1:-Find ∠QSR

We know that in a triangle, the exterior angle is always equal to the sum of the interior opposite angles.

∠QSR +∠SQR = ∠QRT

∠QSR +28° = 65°

∠QSR = 65° -28° = 37°

Step 2:- Find value of x

PQ || SR and QS intersects both the lines

We get angle x = ∠QSR ( alternate interior angles)

x =37°

Step 3:- Find value of y

In PSQ, ∠P + x + y = 180 °(sum of angles in triangles)

90°+37° +y =180 °

y = 180-127

y = 53 °

### Arrange in descending order of magnitude and

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

### Arrange in descending order of magnitude and

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

### 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