Maths-

General

Easy

Question

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

Hint:

### Find the congruence rule used and thus find whether sides are equal.

## The correct answer is: 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)

### Related Questions to study

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.

Maths-

### Find the least number which must be added to 732733 to obtain a perfect square.

Ans :- 3

Explanation :-

Number = 732733

Square Root = 732733 + 3 = 732736 = (86)

We need to add 3 to the given number to make it a perfect square.

Explanation :-

Number = 732733

Square Root = 732733 + 3 = 732736 = (86)

^{2}.We need to add 3 to the given number to make it a perfect square.

### Find the least number which must be added to 732733 to obtain a perfect square.

Maths-General

Ans :- 3

Explanation :-

Number = 732733

Square Root = 732733 + 3 = 732736 = (86)

We need to add 3 to the given number to make it a perfect square.

Explanation :-

Number = 732733

Square Root = 732733 + 3 = 732736 = (86)

^{2}.We need to add 3 to the given number to make it a perfect square.

Maths-

### In the given figure, PQ || RS. Check whether ΔPOQ i congruent to ΔSOR Give reason for your answer:

In ΔPOQ and ΔSOR, we have

(Vertically opposite angles)

( since PQ || RS, alternate angles)

( since PQ || RS, alternate angles)

By, AAA similarity criterion we can say that ΔPOQ is similar to ΔSOR, but we can’t

conclude that ΔPOQ ⩭ ΔSOR

(Vertically opposite angles)

( since PQ || RS, alternate angles)

( since PQ || RS, alternate angles)

By, AAA similarity criterion we can say that ΔPOQ is similar to ΔSOR, but we can’t

conclude that ΔPOQ ⩭ ΔSOR

### In the given figure, PQ || RS. Check whether ΔPOQ i congruent to ΔSOR Give reason for your answer:

Maths-General

In ΔPOQ and ΔSOR, we have

(Vertically opposite angles)

( since PQ || RS, alternate angles)

( since PQ || RS, alternate angles)

By, AAA similarity criterion we can say that ΔPOQ is similar to ΔSOR, but we can’t

conclude that ΔPOQ ⩭ ΔSOR

(Vertically opposite angles)

( since PQ || RS, alternate angles)

( since PQ || RS, alternate angles)

By, AAA similarity criterion we can say that ΔPOQ is similar to ΔSOR, but we can’t

conclude that ΔPOQ ⩭ ΔSOR

Maths-

### Show that 64 is square as well as cube number

Explanation :-

64 = Here 6 is divisible by both 2 and 3 .

so,64 is both a square and cube number.

64 = Here 6 is divisible by both 2 and 3 .

so,64 is both a square and cube number.

### Show that 64 is square as well as cube number

Maths-General

Explanation :-

64 = Here 6 is divisible by both 2 and 3 .

so,64 is both a square and cube number.

64 = Here 6 is divisible by both 2 and 3 .

so,64 is both a square and cube number.

Maths-

### From the given figures, find by which criteria the two triangles ABC and PQR are congruent.

In the figure XY = PZ. Prove that XZ = PY.

In the figure XY = PZ. Prove that XZ = PY.

In ΔABC, We have

( Sum of interior angles of a triangle is 180)

Likewise, In ΔPQR, We have

( Sum of interior angles of a triangle is 180)

If we consider ΔABC and ΔPQR, we have

AB = RQ

(from the figure)

So, by ASA congruence rule we have

ΔABC⩭ ΔRQP

( Sum of interior angles of a triangle is 180)

Likewise, In ΔPQR, We have

( Sum of interior angles of a triangle is 180)

If we consider ΔABC and ΔPQR, we have

AB = RQ

(from the figure)

So, by ASA congruence rule we have

ΔABC⩭ ΔRQP

### From the given figures, find by which criteria the two triangles ABC and PQR are congruent.

In the figure XY = PZ. Prove that XZ = PY.

In the figure XY = PZ. Prove that XZ = PY.

Maths-General

In ΔABC, We have

( Sum of interior angles of a triangle is 180)

Likewise, In ΔPQR, We have

( Sum of interior angles of a triangle is 180)

If we consider ΔABC and ΔPQR, we have

AB = RQ

(from the figure)

So, by ASA congruence rule we have

ΔABC⩭ ΔRQP

( Sum of interior angles of a triangle is 180)

Likewise, In ΔPQR, We have

( Sum of interior angles of a triangle is 180)

If we consider ΔABC and ΔPQR, we have

AB = RQ

(from the figure)

So, by ASA congruence rule we have

ΔABC⩭ ΔRQP

Maths-

### Evaluate:

Explanation :-

3375 = 3 × 1125

= 3 × 3 × 375

= 3 ×3 × 3 × 125

= 3 × 3 × 3 × 5 × 25

= 3 × 3 × 3× 5 × 5 × 5

=

3375 = 3 × 1125

= 3 × 3 × 375

= 3 ×3 × 3 × 125

= 3 × 3 × 3 × 5 × 25

= 3 × 3 × 3× 5 × 5 × 5

=

### Evaluate:

Maths-General

Explanation :-

3375 = 3 × 1125

= 3 × 3 × 375

= 3 ×3 × 3 × 125

= 3 × 3 × 3 × 5 × 25

= 3 × 3 × 3× 5 × 5 × 5

=

3375 = 3 × 1125

= 3 × 3 × 375

= 3 ×3 × 3 × 125

= 3 × 3 × 3 × 5 × 25

= 3 × 3 × 3× 5 × 5 × 5

=

Maths-

### Evaluate:

Explanation :-

We know that

Substituting respective values we get

We know that

Substituting respective values we get

### Evaluate:

Maths-General

Explanation :-

We know that

Substituting respective values we get

We know that

Substituting respective values we get