Maths-

General

Easy

Question

# The perpendicular bisectors of the sides of a triangle ABC meet at point I. prove that IA=IB=IC

Hint:

### Find the congruence rules used and thus find the equal sides.

## The correct answer is: IA = IB = IC

In ΔABC, in which AD is perpendicular bisector of BC

BE is perpendicular bisector of CA

CF is perpendicular bisector of AB

AD, BE and CF meet at I

To prove IA=IB=IC

In ΔBID and ΔCID

BD = DC (given)

(AD is perpendicular bisector of BC)

ID = ID (common)

Hence by SAS congruence rule, we have

ΔBID ⩭ ΔCID

So, IB = IC (Corresponding part of congruent triangle)

Similarly, In ΔCIE and ΔAIE

CE = AE (given)

(BE is perpendicular bisector of AC)

IE = IE (common)

By SAS congruence rule, ΔCIE ⩭ ΔAIE

IC = IA (Corresponding part of congruent triangle)

Thus, IA = IB = IC

### Related Questions to study

Maths-

### Use prime factorization method to find the square root of each of the following :

c) 1089

c) Let's find the prime factorization of the number.

1089

= 3 × 363

= 3 × 3 × 121

= 3 × 3 × 11 ×1 1

The square root of 1089 is 33.

1089

= 3 × 363

= 3 × 3 × 121

= 3 × 3 × 11 ×1 1

The square root of 1089 is 33.

### Use prime factorization method to find the square root of each of the following :

c) 1089

Maths-General

c) Let's find the prime factorization of the number.

1089

= 3 × 363

= 3 × 3 × 121

= 3 × 3 × 11 ×1 1

The square root of 1089 is 33.

1089

= 3 × 363

= 3 × 3 × 121

= 3 × 3 × 11 ×1 1

The square root of 1089 is 33.

Maths-

### Use prime factorization method to find the square root of each of the following :

b)175

Let's find the prime factorization of the number

30625 = 5 × 6125

= 5 × 5 × 1225

= 5 × 5 × 5 × 245

= 5 × 5 × 5 5 × 49

= 5 ×5 × 5 × 5 × 7 × 7

The square root of 30625 is 175.

30625 = 5 × 6125

= 5 × 5 × 1225

= 5 × 5 × 5 × 245

= 5 × 5 × 5 5 × 49

= 5 ×5 × 5 × 5 × 7 × 7

The square root of 30625 is 175.

### Use prime factorization method to find the square root of each of the following :

b)175

Maths-General

Let's find the prime factorization of the number

30625 = 5 × 6125

= 5 × 5 × 1225

= 5 × 5 × 5 × 245

= 5 × 5 × 5 5 × 49

= 5 ×5 × 5 × 5 × 7 × 7

The square root of 30625 is 175.

30625 = 5 × 6125

= 5 × 5 × 1225

= 5 × 5 × 5 × 245

= 5 × 5 × 5 5 × 49

= 5 ×5 × 5 × 5 × 7 × 7

The square root of 30625 is 175.

Maths-

### Use prime factorization method to find the square root of each of the following :

a)11025

Explanation :-

a) Let's find the prime factorization of the number.

11025 =3 × 3675

= 3 × 3 × 1225

= 3 × 3 × 5*245

= 3 × 3 × 5 × 5 × 49

= 3 × 3 × 5 × 5 × 7 × 7

=

The square root of 11236 is 105.

a) Let's find the prime factorization of the number.

11025 =3 × 3675

= 3 × 3 × 1225

= 3 × 3 × 5*245

= 3 × 3 × 5 × 5 × 49

= 3 × 3 × 5 × 5 × 7 × 7

=

The square root of 11236 is 105.

### Use prime factorization method to find the square root of each of the following :

a)11025

Maths-General

Explanation :-

a) Let's find the prime factorization of the number.

11025 =3 × 3675

= 3 × 3 × 1225

= 3 × 3 × 5*245

= 3 × 3 × 5 × 5 × 49

= 3 × 3 × 5 × 5 × 7 × 7

=

The square root of 11236 is 105.

a) Let's find the prime factorization of the number.

11025 =3 × 3675

= 3 × 3 × 1225

= 3 × 3 × 5*245

= 3 × 3 × 5 × 5 × 49

= 3 × 3 × 5 × 5 × 7 × 7

=

The square root of 11236 is 105.

Maths-

### In Triangle ABC , AB= AC, and D is a point on AB. Such that AD=DC=BC. Show that

In ΔABC, we have

CD = BC (given)

Let (since angles opposite to the equal sides are always equal)

Since AB = AC,

In ΔABC,

(By angle sum property of the triangle)

Since AD = CD (given),

So,

Now,

(angle sum property)

Now, .

### In Triangle ABC , AB= AC, and D is a point on AB. Such that AD=DC=BC. Show that

Maths-General

In ΔABC, we have

CD = BC (given)

Let (since angles opposite to the equal sides are always equal)

Since AB = AC,

In ΔABC,

(By angle sum property of the triangle)

Since AD = CD (given),

So,

Now,

(angle sum property)

Now, .

Maths-

### In the triangle ABC, the angles at A and C are 40° and 95° respectively. If N is the foot of the perpendicular from C to AB , prove that CN = BN.

In ΔABC, We have

(given)

(given)

(sum of interior angles of a triangle is 180)

We know that, ( CN ⊥ AB)

Hence,

So, In ΔCNB, the base angles are equal () hence sides

opposite to equal base angles are also equal. Hence we get, CN = NB

### In the triangle ABC, the angles at A and C are 40° and 95° respectively. If N is the foot of the perpendicular from C to AB , prove that CN = BN.

Maths-General

In ΔABC, We have

(given)

(given)

(sum of interior angles of a triangle is 180)

We know that, ( CN ⊥ AB)

Hence,

So, In ΔCNB, the base angles are equal () hence sides

opposite to equal base angles are also equal. Hence we get, CN = NB

Maths-

### Given XYZ is a triangle in which XW is the bisector of the angle x and XW ⊥ yz. Prove that xy= xz.

In ΔXYW and ΔXZW

(since XW is the bisector of angle X)

(XW ⊥ YZ)

XW = XW (common side)

Hence by ASA congruence rule, we have ΔXYW ⩭ ΔXZW

So, XY = XZ (Corresponding part of congruence triangles)

### Given XYZ is a triangle in which XW is the bisector of the angle x and XW ⊥ yz. Prove that xy= xz.

Maths-General

In ΔXYW and ΔXZW

(since XW is the bisector of angle X)

(XW ⊥ YZ)

XW = XW (common side)

Hence by ASA congruence rule, we have ΔXYW ⩭ ΔXZW

So, XY = XZ (Corresponding part of congruence triangles)

Maths-

### ABC is a triangle in which altitudes BE and CF to sides AC and AB respectively are equal. Show that

(i) ΔABE ⩭ ΔACF

(ii) AB = AC

(i) In ΔABE and ΔACF, we have

( Common angle)

(Since BE ⟂ AC and CF ⟂ AB)

BE = CF (given)

Hence, by AAS criterion ΔABE ⩭ ΔACF

(ii) Since ΔABE ⩭ ΔACF, we have

AB = AC (Corresponding part of congruent triangles)

### ABC is a triangle in which altitudes BE and CF to sides AC and AB respectively are equal. Show that

(i) ΔABE ⩭ ΔACF

(ii) AB = AC

Maths-General

(i) In ΔABE and ΔACF, we have

( Common angle)

(Since BE ⟂ AC and CF ⟂ AB)

BE = CF (given)

Hence, by AAS criterion ΔABE ⩭ ΔACF

(ii) Since ΔABE ⩭ ΔACF, we have

AB = AC (Corresponding part of congruent triangles)

Maths-

### In the figure, OA = OD and OB = OC. Show that:

(i) AOB ⩭ DOC

(ii) AB || CD

(i) In ΔAOB and ΔDOC, we have

AO = OD (given)

BO = OC (given)

Since CB and AD intersects, we have

(vertically opposite angles)

So by SAS congruence rule, we have

ΔAOB ⩭ ΔDOC

(ii)Since ΔAOB ⩭ ΔDOC, we have

( Corresponding part of congruent triangles)

But they form alternate interior angles and since they are equal , AB || CD.

AO = OD (given)

BO = OC (given)

Since CB and AD intersects, we have

(vertically opposite angles)

So by SAS congruence rule, we have

ΔAOB ⩭ ΔDOC

(ii)Since ΔAOB ⩭ ΔDOC, we have

( Corresponding part of congruent triangles)

But they form alternate interior angles and since they are equal , AB || CD.

### In the figure, OA = OD and OB = OC. Show that:

(i) AOB ⩭ DOC

(ii) AB || CD

Maths-General

(i) In ΔAOB and ΔDOC, we have

AO = OD (given)

BO = OC (given)

Since CB and AD intersects, we have

(vertically opposite angles)

So by SAS congruence rule, we have

ΔAOB ⩭ ΔDOC

(ii)Since ΔAOB ⩭ ΔDOC, we have

( Corresponding part of congruent triangles)

But they form alternate interior angles and since they are equal , AB || CD.

AO = OD (given)

BO = OC (given)

Since CB and AD intersects, we have

(vertically opposite angles)

So by SAS congruence rule, we have

ΔAOB ⩭ ΔDOC

(ii)Since ΔAOB ⩭ ΔDOC, we have

( Corresponding part of congruent triangles)

But they form alternate interior angles and since they are equal , AB || CD.

Maths-

### If AD ⟂ CD and BC ⟂ CD , AQ = PB and DP = CQ. Prove that

AD ⊥ CD (given)

and BC ⊥ CD (given)

AQ = BP and DP = CQ (given)

To prove : ∠ DAQ = ∠ CBP

AD ⊥ CD (given)

and BC ⊥ CD (given)

∴ ∠ D = ∠ C (each 90°)

∵ DP = CQ (given)

On adding PQ to both sides. we get

DP + PQ = PQ + CQ…(i)

⇒ DQ = CP

Now, in right angles ADQ and BCP

∴ AQ = BP (hypotenuse)

DQ = CP (from (i))

∴ Δ ADQ ≡ Δ BCP (Right angle hypotenuse side (RHS congruence rule))

∴ ∠ DAQ = ∠CBP (Corresponding part of congruent triangles)

and BC ⊥ CD (given)

AQ = BP and DP = CQ (given)

To prove : ∠ DAQ = ∠ CBP

AD ⊥ CD (given)

and BC ⊥ CD (given)

∴ ∠ D = ∠ C (each 90°)

∵ DP = CQ (given)

On adding PQ to both sides. we get

DP + PQ = PQ + CQ…(i)

⇒ DQ = CP

Now, in right angles ADQ and BCP

∴ AQ = BP (hypotenuse)

DQ = CP (from (i))

∴ Δ ADQ ≡ Δ BCP (Right angle hypotenuse side (RHS congruence rule))

∴ ∠ DAQ = ∠CBP (Corresponding part of congruent triangles)

### If AD ⟂ CD and BC ⟂ CD , AQ = PB and DP = CQ. Prove that

Maths-General

AD ⊥ CD (given)

and BC ⊥ CD (given)

AQ = BP and DP = CQ (given)

To prove : ∠ DAQ = ∠ CBP

AD ⊥ CD (given)

and BC ⊥ CD (given)

∴ ∠ D = ∠ C (each 90°)

∵ DP = CQ (given)

On adding PQ to both sides. we get

DP + PQ = PQ + CQ…(i)

⇒ DQ = CP

Now, in right angles ADQ and BCP

∴ AQ = BP (hypotenuse)

DQ = CP (from (i))

∴ Δ ADQ ≡ Δ BCP (Right angle hypotenuse side (RHS congruence rule))

∴ ∠ DAQ = ∠CBP (Corresponding part of congruent triangles)

and BC ⊥ CD (given)

AQ = BP and DP = CQ (given)

To prove : ∠ DAQ = ∠ CBP

AD ⊥ CD (given)

and BC ⊥ CD (given)

∴ ∠ D = ∠ C (each 90°)

∵ DP = CQ (given)

On adding PQ to both sides. we get

DP + PQ = PQ + CQ…(i)

⇒ DQ = CP

Now, in right angles ADQ and BCP

∴ AQ = BP (hypotenuse)

DQ = CP (from (i))

∴ Δ ADQ ≡ Δ BCP (Right angle hypotenuse side (RHS congruence rule))

∴ ∠ DAQ = ∠CBP (Corresponding part of congruent triangles)

Maths-

### Prove that the medians bisecting the equal sides of an isosceles triangle are also equal. If OA = OB and OD = OC, show that:

(i) AOD ⩭ BOC

(ii) AD || BC

(i) In ΔAOD and ΔBOC, we have

AO = OB (given)

OD = OC (given)

Since CD and AB intersects, we have

(vertically opposite angles)

So by SAS congruence rule, we have

ΔAOD ⩭ ΔBOC

(ii) SinceΔAOD ⩭ ΔBOC, we have

( Corresponding part of congruent triangles)

But they form alternate interior angles and since they are equal, AD || BC.

AO = OB (given)

OD = OC (given)

Since CD and AB intersects, we have

(vertically opposite angles)

So by SAS congruence rule, we have

ΔAOD ⩭ ΔBOC

(ii) SinceΔAOD ⩭ ΔBOC, we have

( Corresponding part of congruent triangles)

But they form alternate interior angles and since they are equal, AD || BC.

### Prove that the medians bisecting the equal sides of an isosceles triangle are also equal. If OA = OB and OD = OC, show that:

(i) AOD ⩭ BOC

(ii) AD || BC

Maths-General

(i) In ΔAOD and ΔBOC, we have

AO = OB (given)

OD = OC (given)

Since CD and AB intersects, we have

(vertically opposite angles)

So by SAS congruence rule, we have

ΔAOD ⩭ ΔBOC

(ii) SinceΔAOD ⩭ ΔBOC, we have

( Corresponding part of congruent triangles)

But they form alternate interior angles and since they are equal, AD || BC.

AO = OB (given)

OD = OC (given)

Since CD and AB intersects, we have

(vertically opposite angles)

So by SAS congruence rule, we have

ΔAOD ⩭ ΔBOC

(ii) SinceΔAOD ⩭ ΔBOC, we have

( Corresponding part of congruent triangles)

But they form alternate interior angles and since they are equal, AD || BC.

Maths-

### Kumar wants to spread a carpet on the floor of his classroom. The floor is a square

with an area of 4160.25 square feet. What is the length of carpet on each side?

Ans :- 64.5 feet

Explanation :-

Let the length of the classroom (square) be a cm .

The we get area of class room =

So we get ,

We know that the length of side of square is length of the carpet to be used to cover it

I.e length of side of square = the required length of square carpet.

Using the long division method to find value of square root of 4160.25

The length of carpet be on each side is .

Explanation :-

Let the length of the classroom (square) be a cm .

The we get area of class room =

So we get ,

We know that the length of side of square is length of the carpet to be used to cover it

I.e length of side of square = the required length of square carpet.

Using the long division method to find value of square root of 4160.25

The length of carpet be on each side is .

### Kumar wants to spread a carpet on the floor of his classroom. The floor is a square

with an area of 4160.25 square feet. What is the length of carpet on each side?

Maths-General

Ans :- 64.5 feet

Explanation :-

Let the length of the classroom (square) be a cm .

The we get area of class room =

So we get ,

We know that the length of side of square is length of the carpet to be used to cover it

I.e length of side of square = the required length of square carpet.

Using the long division method to find value of square root of 4160.25

The length of carpet be on each side is .

Explanation :-

Let the length of the classroom (square) be a cm .

The we get area of class room =

So we get ,

We know that the length of side of square is length of the carpet to be used to cover it

I.e length of side of square = the required length of square carpet.

Using the long division method to find value of square root of 4160.25

The length of carpet be on each side is .

Maths-

### The volume of a cubical box is 12167 cm^{3}. Find the side of the box and its 6 sides area?

Length of the side of the box is 23 cm. Area of each side is 529 .

Area of 6 sides combined = 3174 cm

Explanation :-

Step1:- find the length of side of the cube

Let the length of the side of the cube be a .

The volume of cube =

Then,

The length of the side of the square is 23 cm .

Step2:- find the side area of the cube

The side are of the cube =

Step 3:- find the side area of 6 sides .

In the cube the side area of all sides is equal so, we get as the area of 6 sides.

The side area of 6 sides = .

We get that Length of the side of the box is 23 cm. Area of 6 sides combined = 3174 cm

Area of 6 sides combined = 3174 cm

^{2}.Explanation :-

Step1:- find the length of side of the cube

Let the length of the side of the cube be a .

The volume of cube =

Then,

The length of the side of the square is 23 cm .

Step2:- find the side area of the cube

The side are of the cube =

Step 3:- find the side area of 6 sides .

In the cube the side area of all sides is equal so, we get as the area of 6 sides.

The side area of 6 sides = .

We get that Length of the side of the box is 23 cm. Area of 6 sides combined = 3174 cm

^{2}### The volume of a cubical box is 12167 cm^{3}. Find the side of the box and its 6 sides area?

Maths-General

Length of the side of the box is 23 cm. Area of each side is 529 .

Area of 6 sides combined = 3174 cm

Explanation :-

Step1:- find the length of side of the cube

Let the length of the side of the cube be a .

The volume of cube =

Then,

The length of the side of the square is 23 cm .

Step2:- find the side area of the cube

The side are of the cube =

Step 3:- find the side area of 6 sides .

In the cube the side area of all sides is equal so, we get as the area of 6 sides.

The side area of 6 sides = .

We get that Length of the side of the box is 23 cm. Area of 6 sides combined = 3174 cm

Area of 6 sides combined = 3174 cm

^{2}.Explanation :-

Step1:- find the length of side of the cube

Let the length of the side of the cube be a .

The volume of cube =

Then,

The length of the side of the square is 23 cm .

Step2:- find the side area of the cube

The side are of the cube =

Step 3:- find the side area of 6 sides .

In the cube the side area of all sides is equal so, we get as the area of 6 sides.

The side area of 6 sides = .

We get that Length of the side of the box is 23 cm. Area of 6 sides combined = 3174 cm

^{2}Maths-

### The area of a square is 289 m^{2}, find its perimeter?

Ans :- 68

Explanation :-

Step 1:-

Let the length of the square be a

The area of square = (length of side)

Then,

Step 2:- find the perimeter of square by using length

We find the perimeter of the square = 4a = 4 × 17 = 68m.

The perimeter of square whose area is 289 (m

Explanation :-

Step 1:-

Let the length of the square be a

The area of square = (length of side)

^{2 }Then,

Step 2:- find the perimeter of square by using length

We find the perimeter of the square = 4a = 4 × 17 = 68m.

The perimeter of square whose area is 289 (m

^{2}) is 68 m.### The area of a square is 289 m^{2}, find its perimeter?

Maths-General

Ans :- 68

Explanation :-

Step 1:-

Let the length of the square be a

The area of square = (length of side)

Then,

Step 2:- find the perimeter of square by using length

We find the perimeter of the square = 4a = 4 × 17 = 68m.

The perimeter of square whose area is 289 (m

Explanation :-

Step 1:-

Let the length of the square be a

The area of square = (length of side)

^{2 }Then,

Step 2:- find the perimeter of square by using length

We find the perimeter of the square = 4a = 4 × 17 = 68m.

The perimeter of square whose area is 289 (m

^{2}) is 68 m.Maths-

### What is the least seven digits which is a perfect square. Also find the square root of the number so obtained

Ans :- 1 00 00 00 (10

Explanation :-

The least seven digit number is 1 00 00 00(10

^{6})Explanation :-

The least seven digit number is 1 00 00 00(10

^{6}) which is already a perfect square as the given number is the perfect square of 1000### What is the least seven digits which is a perfect square. Also find the square root of the number so obtained

Maths-General

Ans :- 1 00 00 00 (10

Explanation :-

The least seven digit number is 1 00 00 00(10

^{6})Explanation :-

The least seven digit number is 1 00 00 00(10

^{6}) which is already a perfect square as the given number is the perfect square of 1000Maths-

### Find the least number which must be subtracted from 23416 to make it a perfect square.

Ans :- 7

Explanation :-

We need to subtract the 7 from 23416 to make it a perfect square.

Explanation :-

We need to subtract the 7 from 23416 to make it a perfect square.

### Find the least number which must be subtracted from 23416 to make it a perfect square.

Maths-General

Ans :- 7

Explanation :-

We need to subtract the 7 from 23416 to make it a perfect square.

Explanation :-

We need to subtract the 7 from 23416 to make it a perfect square.