NCERT Solutions for Exercise 8.1 Class 9 Maths Chapter 8 - Quadrilaterals

NCERT Solutions for Exercise 8.1 Class 9 Maths Chapter 8 - Quadrilaterals

Edited By Vishal kumar | Updated on Oct 07, 2023 10:16 AM IST

NCERT Solutions for class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1- Download Free PDF

NCERT Solutions for class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1- This class 9 Maths chapter 8 exercise 8.1 focuses on quadrilaterals, four-sided shapes. Exercise 8.1 helps you understand their basics, properties, and types. Our free PDF solutions simplify complex ideas, providing clear explanations and step-by-step guidance. These resources aid your understanding, exam preparation, and math success.

In NCERT solutions for Class 9 Maths chapter 8 exercise 8.1, the types of quadrilaterals are determined by the angles and lengths of their sides. Because the word "quad" means "four," all of these quadrilaterals have four sides, and the sum of their angles is 360 degrees.

Types of quadrilaterals are Squares, Trapezium, Parallelogram, Rectangle, Rhombus, and Kite.

Some points to be noted about the types of quadrilaterals are:

A square can likewise be considered a rectangle and furthermore, a rhombus, yet every rectangle or a rhombus is certainly not a square. A trapezium isn't a parallelogram (as just one set of inverse sides are equal in a trapezium and we require the two sets to be equal in a parallelogram)

Along with NCERT book Class 9 Maths, chapter 8 exercise 8.1 the following exercises are also present.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1

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Access Quadrilaterals Class 9 Chapter 8 Exercise: 8.1

Q1 The angles of quadrilateral are in the ratio \small 3:5:9:13 . Find all the angles of the quadrilateral.

Answer:

Given : The angles of a quadrilateral are in the ratio \small 3:5:9:13 .
Let the angles of quadrilateral be 3x,5x,9x ,13x .

Sum of all angles is 360.

Thus, 3x+5x+9x +13x=360 \degree

\Rightarrow 30x=360 \degree

\Rightarrow x=12 \degree

All four angles are : 3x=3\times 12=36\degree

5x=5\times 12=60 \degree

9x=9\times 12=108 \degree

13x=13\times 12=156 \degree

Q2 If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer:

Given: ABCD is a parallelogram with AC=BD.

To prove: ABCD is a rectangle.

Proof: In \triangle ABC and \triangle BAD,

1640170405939

BC= AD (Opposite sides of parallelogram)

AC=BD (Given )

AB=AB (common)

\triangle ABC \cong \triangle BAD (By SSS)

\angle ABC=\angle BAD (CPCT)

and \angle ABC+\angle BAD=180 \degree (co - interior angles)

2\angle BAD=180 \degree

\angle BAD= 90 \degree

Hence, it is a rectangle.

Q3 Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Answer:

Given: the diagonals of a quadrilateral bisect each other at right angles i.e.BO=OD.

To prove: ABCD is a rhombus.

Proof : In \triangle AOB and \triangle AOD,

1640170440375

\angle AOB=\angle AOD (Each 90 \degree )

BO=OD (Given )

AO=AO (common)

\triangle AOB \cong \triangle AOD (By SAS)

AB=AD (CPCT)

Similarly, AB=BC and BC=CD

Hence, it is a rhombus.

Q4 Show that the diagonals of a square are equal and bisect each other at right angles.

Answer:

Given : ABCD is a square i.e. AB=BC=CD=DA.

To prove : the diagonals of a square are equal and bisect each other at right angles i.e. AC=BD,AO=CO,BO=DO and \angle COD=90 \degree

Proof : In \triangle BAD and \triangle ABC,

\angle BAD = \angle ABC (Each 90 \degree )

AD=BC (Given )

AB=AB (common)

\triangle BAD \cong \triangle ABC (By SAS)

BD=AC (CPCT)

In \triangle AOB and \triangle COD,

\angle OAB= \angle OCD (Alternate angles)

AB=CD (Given )

\angle OBA= \angle ODC (Alternate angles)

\triangle AOB \cong \triangle COD (By AAS)

AO=OC ,BO=OD (CPCT)

In \triangle AOB and \triangle AOD,

OB=OD (proved above)

AB=AD (Given )

OA=OA (COMMON)

\triangle AOB \cong \triangle AOD (By SSS)

\angle AOB= \angle AOD (CPCT)

\angle AOB+ \angle AOD = 180 \degree

2. \angle AOB = 180 \degree

\angle AOB = 90 \degree

Hence, the diagonals of a square are equal and bisect each other at right angles.

Q5 Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Answer:

Given : ABCD is a quadrilateral with AC=BD,AO=CO,BO=DO, \angle COD = 90 \degree

To prove: ABCD is a square.

Proof: Since the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a rhombus.

Thus, AB=BC=CD=DA

In \triangle BAD and \triangle ABC,

AD=BC (proved above )

AB=AB (common)

BD=AC

\triangle BAD \cong \triangle ABC (By SSS)

\angle BAD = \angle ABC (CPCT)

\angle BAD+ \angle ABC = 180 \degree (Co-interior angles)

2. \angle ABC = 180 \degree

\angle ABC = 90 \degree

Hence, the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Q6 (i) Diagonal AC of a parallelogram ABCD bisects \small \angle A (see Fig. \small 8.19 ). Show that

it bisects \small \angle C also.

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Answer:

Given: \angle DAC= \angle BAC ................1

\angle DAC= \angle BCA.................2 (Alternate angles)

\angle BAC= \angle ACD .................3 (Alternate angles)

From equation 1,2 and 3, we get

\angle ACD= \angle BCA...................4

Hence, diagonal AC bisect angle C also.

Q6 (ii) Diagonal AC of a parallelogram ABCD bisects \small \angle A (see Fig. \small 8.19 ). Show that

ABCD is a rhombus.

1640170530307

Answer:

Given: \angle DAC= \angle BAC ................1

\angle DAC= \angle BCA.................2 (Alternate angles)

\angle BAC= \angle ACD .................3 (Alternate angles)

From equation 1,2 and 3, we get

\angle ACD= \angle BCA...................4

From 2 and 4, we get

\angle ACD= \angle DAC

In \triangle ADC,

\angle ACD= \angle DAC (proved above )

AD=DC (In a triangle,sides opposite to equal angle are equal)

A parallelogram whose adjacent sides are equal , is a rhombus.

Thus, ABCD is a rhombus.

Q7 ABCD is a rhombus. Show that diagonal AC bisects \small \angle A as well as \small \angle C and diagonal BD bisects \small \angle B as well as \small \angle D .

Answer:

1640170558677

In \triangle ADC,

AD = CD (ABCD is a rhombus)

\angle 3= \angle 1.................1(angles opposite to equal sides are equal )

\angle 3= \angle 2.................2 (alternate angles)

From 1 and 2, we have

\angle 1= \angle 2.................3

and \angle 1= \angle 4.................4 (alternate angles)

From 1 and 4, we get

\angle 3= \angle 4.................5

Hence, from 3 and 5, diagonal AC bisects angles A as well as angle C.

1640170587935

In \triangle ADB,

AD = AB (ABCD is a rhombus)

\angle 5= \angle 7.................6(angles opposite to equal sides are equal )

\angle 7= \angle 6.................7 (alternate angles)

From 6 and 7, we have

\angle 5= \angle 6.................8

and \angle 5= \angle 8.................9(alternate angles)

From 6 and 9, we get

\angle 7= \angle 8.................10

Hence, from 8 and 10, diagonal BD bisects angles B as well as angle D.

Q8 (i) ABCD is a rectangle in which diagonal AC bisects \small \angle A as well as \small \angle C . Show that:

ABCD is a square

Answer:

1640170617153

Given: ABCD is a rectangle with AB=CD and BC=AD \angle 1= \angle 2 and \angle 3= \angle 4.

To prove: ABCD is a square.

Proof : \angle 1= \angle 4 .............1(alternate angles)

\angle 3= \angle 4 ................2(given )

From 1 and 2, \angle 1= \angle 3.....................................3

In \triangle ADC,

\angle 1= \angle 3 (from 3 )

DC=AD (In a triangle, sides opposite to equal angle are equal )

A rectangle whose adjacent sides are equal is a square.

Hence, ABCD is a square.

Q8 (ii) ABCD is a rectangle in which diagonal AC bisects \small \angle A as well as \small \angle C . Show that:

diagonal BD bisects \small \angle B as well as \small \angle D .

Answer:

1640176768751

In \triangle ADB,

AD = AB (ABCD is a square)

\angle 5= \angle 7.................1(angles opposite to equal sides are equal )

\angle 5= \angle 8.................2 (alternate angles)

From 1 and 2, we have

\angle 7= \angle 8.................3

and \angle 7 = \angle 6.................4(alternate angles)

From 1 and 4, we get

\angle 5= \angle 6.................5

Hence, from 3 and 5, diagonal BD bisects angles B as well as angle D.

Q9 (i) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that \small DP=BQ (see Fig. \small 8.20 ). Show that: \small \Delta APD\cong \Delta CQB

1640170652426

Answer:

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that \small DP=BQ .

To prove : \small \Delta APD\cong \Delta CQB

Proof :

In \small \Delta APD\, \, and\, \, \Delta CQB,

DP=BQ (Given )

\angle ADP= \angle CBQ (alternate angles)

AD=BC (opposite sides of a parallelogram)

\small \Delta APD\cong \Delta CQB (By SAS)

Q9 (ii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that \small DP=BQ (see Fig. \small 8.20 ). Show that: \small AP=CQ

1640170672862

Answer:

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that \small DP=BQ .

To prove : \small AP=CQ

Proof :

In \small \Delta APD\, \, and\, \, \Delta CQB,

DP=BQ (Given )

\angle ADP= \angle CBQ (alternate angles)

AD=BC (opposite sides of a parallelogram)

\small \Delta APD\cong \Delta CQB (By SAS)

\small AP=CQ (CPCT)

Q9 (iii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that \small DP=BQ (see Fig. \small 8.20 ). Show that: \small \Delta AQB\cong \Delta CPD

1640170699729

Answer:

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that \small DP=BQ .

To prove : \small \Delta AQB\cong \Delta CPD

Proof :

In \small \Delta AQB\, \, and\, \, \Delta CPD,

DP=BQ (Given )

\angle ABQ= \angle CDP (alternate angles)

AB=CD (opposite sides of a parallelogram)

\small \Delta AQB\cong \Delta CPD (By SAS)

Q9 (iv) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that \small DP=BQ (see Fig. \small 8.20 ). Show that: \small AQ=CP

1640170719974

Answer:

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that \small DP=BQ .

To prove : \small AQ=CP

Proof :

In \small \Delta AQB\, \, and\, \, \Delta CPD,

DP=BQ (Given )

\angle ABQ= \angle CDP (alternate angles)

AB=CD (opposite sides of a parallelogram)

\small \Delta AQB\cong \Delta CPD (By SAS)

\small AQ=CP (CPCT)

Q9 (v) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that \small DP=BQ (see Fig. \small 8.20 ). Show that: APCQ is a parallelogram

1640170738363

Answer:

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that \small DP=BQ .

To prove: APCQ is a parallelogram

Proof :

In \small \Delta APD\, \, and\, \, \Delta CQB,

DP=BQ (Given )

\angle ADP= \angle CBQ (alternate angles)

AD=BC (opposite sides of a parallelogram)

\small \Delta APD\cong \Delta CQB (By SAS)

\small AP=CQ (CPCT)...............................................................1

Also,

In \small \Delta AQB\, \, and\, \, \Delta CPD,

DP=BQ (Given )

\angle ABQ= \angle CDP (alternate angles)

AB=CD (opposite sides of a parallelogram)

\small \Delta AQB\cong \Delta CPD (By SAS)

\small AQ=CP (CPCT)........................................2

From equation 1 and 2, we get

\small AP=CQ

\small AQ=CP

Thus, opposite sides of quadrilateral APCQ are equal so APCQ is a parallelogram.

Q10 (i) ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. \small 8.21 ). Show that \small \Delta APB\cong \Delta CQD

1640170770148

Answer:

Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.

To prove : \small \Delta APB\cong \Delta CQD

Proof: In \small \Delta APB\, \, and\, \, \Delta CQD ,

\angle APB= \angle CQD (Each 90 \degree )

\angle ABP= \angle CDQ (Alternate angles)

AB=CD (Opposite sides of a parallelogram )

Thus, \small \Delta APB\cong \Delta CQD (By SAS)

Q10 (ii) ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. \small 8.21 ). Show that \small AP=CQ

1640170798432

Answer:

Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.

To prove : \small AP=CQ

Proof: In \small \Delta APB\, \, and\, \, \Delta CQD ,

\angle APB= \angle CQD (Each 90 \degree )

\angle ABP= \angle CDQ (Alternate angles)

AB=CD (Opposite sides of a parallelogram )

Thus, \small \Delta APB\cong \Delta CQD (By SAS)

\small AP=CQ (CPCT)

Q11 (i) In \small \Delta ABC and \small \Delta DEF , \small AB=DE,AB\parallel DE,BC=EF and \small BC\parallel EF . Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. \small 8.22 ). Show that quadrilateral ABED is a parallelogram

1640170849147

Answer:

Given : In \small \Delta ABC and \small \Delta DEF , \small AB=DE,AB\parallel DE,BC=EF and \small BC\parallel EF .

To prove : quadrilateral ABED is a parallelogram

Proof : In ABED,

AB=DE (Given)

AB||DE (Given )

Hence, quadrilateral ABED is a parallelogram.

Q11 (ii) In \small \Delta ABC and \small \Delta DEF , \small AB=DE,AB\parallel DE,BC=EF and \small BC\parallel EF . Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. \small 8.22 ). Show that quadrilateral BEFC is a parallelogram.

1640170880705

Answer:

Given: In \small \Delta ABC and \small \Delta DEF , \small AB=DE,AB\parallel DE,BC=EF and \small BC\parallel EF .

To prove: quadrilateral BEFC is a parallelogram

Proof: In BEFC,

BC=EF (Given)

BC||EF (Given )

Hence, quadrilateral BEFC is a parallelogram.

Q11 (iii) In \small \Delta ABC and \small \Delta DEF , \small AB=DE,AB\parallel DE,BC=EF and \small BC\parallel EF . Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. \small 8.22 ). Show that \small AD\parallel CF and \small AD=CF

1640170909196

Answer:

To prove: \small AD\parallel CF and \small AD=CF

Proof :

In ABED,

AD=BE.................1(ABED is a parallelogram)

AD||BE .................2(ABED is a parallelogram)

In BEFC,

BE=CF.................3(BEFC is a parallelogram)

BE||CF .................4(BEFC is a parallelogram)

From 2 and 4 , we get

AD||CF

From 1 and 3, we get

AD=CF

Q11 (iv) In \small \Delta ABC and \small \Delta DEF , \small AB=DE,AB\parallel DE,BC=EF and \small BC\parallel EF. . Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. \small 8.22 ). Show that quadrilateral ACFD is a parallelogram.

1640170928287

Answer:

To prove : quadrilateral ACFD is a parallelogram

Proof :

In ABED,

AD=BE.................1(ABED is a parallelogram)

AD||BE .................2(ABED is a parallelogram)

In BEFC,

BE=CF.................3(BEFC is a parallelogram)

BE||CF .................4(BEFC is a parallelogram)

From 2 and 4 , we get

AD||CF...........................5

From 1 and 3, we get

AD=CF...........................6

From 5 and 6, we get

AD||CF and AD=CF

Thus, quadrilateral ACFD is a parallelogram

Q11 (v) In \small \Delta ABC and \small \Delta DEF , \small AB=DE,AB\parallel DE,BC = EF and \small BC\parallel EF . Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. \small 8.22 ). Show that \small AC=DF

1640170956375

Answer:

In ACFD,

AC=DF (Since, ACFD is a parallelogram which is prooved in part (iv) of the question)

Q11 (vi) In \small \Delta ABC and \small \Delta DEF , \small AB=DE,AB\parallel DE,BC=EF and \small BC\parallel EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. \small 8.22 ). Show that \small \Delta ABC\cong \Delta DEF

1640170982314

Answer:

In \small \Delta ABC and \small \Delta DEF ,

AB=DE (Given )

BC=EF (Given )

AC=DF ( proved in (v) part)

\small \Delta ABC\cong \Delta DEF (By SSS rule)

Q12 (i) ABCD is a trapezium in which \small AB\parallel CD and \small AD=BC (see Fig. \small 8.23 ). Show that \small \angle A=\angle B

[ Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

1640171017353

Answer:

Given: ABCD is a trapezium in which \small AB\parallel CD and \small AD=BC

To prove : \small \angle A=\angle B

Proof: Let \angle A be \angle 1, \angle ABC be \angle 2, \angle EBC be \angle 3, \angle BEC be \angle 4.

In AECD,

AE||DC (Given)

AD||CE (By cnstruction)

Hence, AECD is a parallelogram.

AD=CE...............1(opposite sides of a parallelogram)

AD=BC.................2(Given)

From 1 and 2, we get

CE=BC

In \triangle BCE,

\angle 3=\angle 4 .................3 (opposite angles of equal sides)

\angle 2+\angle 3=180 \degree ...................4(linear pairs)

\angle 1+\angle 4=180 \degree .....................5(Co-interior angles)

From 4 and 5, we get

\angle 2+\angle 3=\angle 1+\angle 4

\therefore \angle 2=\angle 1 \rightarrow \angle B=\angle A (Since, \angle 3=\angle 4 )

Q12 (ii) ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that \small \angle C=\angle D

[ Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

1640171044691

Answer:

Given: ABCD is a trapezium in which \small AB\parallel CD and \small AD=BC

To prove : \small \angle C=\angle D

Proof: Let \angle A be \angle 1, \angle ABC be \angle 2, \angle EBC be \angle 3, \angle BEC be \angle 4.

\angle 1+\angle D= 180 \degree (Co-interior angles)

\angle 2+\angle C= 180 \degree (Co-interior angles)

\therefore \angle 1+\angle D=\angle 2+\angle C

Thus, \small \angle C=\angle D (Since , \small \angle 1=\angle 2 )

Q12 (iii) ABCD is a trapezium in which \small AB\parallel CD and \small AD=BC (see Fig. \small 8.23 ). Show that \small \Delta ABC\cong \Delta BAD

[ Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

1640171072699

Answer:

Given: ABCD is a trapezium in which \small AB\parallel CD and \small AD=BC

To prove : \small \Delta ABC\cong \Delta BAD

Proof: In \small \Delta ABC\, \, and\, \, \, \Delta BAD ,

BC=AD (Given )

AB=AB (Common )

\angle ABC=\angle BAD (proved in (i) )

Thus, \small \Delta ABC\cong \Delta BAD (By SAS rule)

Q12 (iv) ABCD is a trapezium in which \small AB\parallel CD and \small AD=BC (see Fig. \small 8.23 ). Show that diagonal AC \small = diagonal BD

[ Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

1640171092180

Answer:

Given: ABCD is a trapezium in which \small AB\parallel CD and \small AD=BC

To prove: diagonal AC \small = diagonal BD

Proof: In \small \Delta ABC\, \, and\, \, \, \Delta BAD ,

BC=AD (Given )

AB=AB (Common )

\angle ABC=\angle BAD (proved in (i) )

Thus, \small \Delta ABC\cong \Delta BAD (By SAS rule)

diagonal AC \small = diagonal BD (CPCT)

More About NCERT Solutions for Class 9 Maths Exercise 8.1

NCERT syllabus Class 9 Maths exercise 8.1 includes some important theorems that are listed below:

A parallelogram diagonal divides it into two congruent triangles.

The opposite sides of a parallelogram are equal.

A parallelogram is formed when each pair of opposite sides of a quadrilateral is equal.

The opposite angles in a parallelogram are equal.

A parallelogram is formed when each pair of opposite angles in a quadrilateral is equal.

A parallelogram's diagonals cross each other.

A parallelogram is formed when the diagonals of a quadrilateral cross each other.

If two opposite sides of a quadrilateral are equal and parallel, the quadrilateral is a parallelogram.

Also Read| Quadrilaterals Class 9 Notes

Benefits of NCERT Solutions for Class 9 Maths Exercise 8.1

  • Exercise 8.1 Class 9 Maths, is based on the concepts of QUADRILATERALS.

  • From Class 9 Maths chapter 8 exercise 8.1 we learn about the angle property of quadrilaterals and the different types of quadrilaterals.

  • Understanding the concepts from Class 9 Maths chapter 8 exercise 8.1 will easily help us to comprehend the ideas identified with the various sorts of QUADRILATERALS like a parallelogram, square shape.

Key Features of 9th Class Maths Exercise 8.1 Answers

  1. Basic Understanding: The answers in this class 9 maths ex 8.1 provide a fundamental understanding of quadrilaterals, their properties, and classifications.

  2. Clear Explanations: Class 9 ex 8.1 offer clear and concise explanations for each question, making it easier for students to comprehend the concepts.

  3. Step-by-Step Solutions: The answers provide step-by-step solutions, helping students follow the problem-solving process easily.

  4. Varied Problem Types: Ex 8.1 class 9 cover a variety of problems related to quadrilaterals, ensuring that students are well-prepared for different question types.

  5. Practice Material: Exercise 8.1 answers serve as valuable practice material for students to reinforce their knowledge of quadrilaterals.

  6. Preparation for Assessments: These 9th class maths exercise 8.1 answers help students prepare for exams by offering a range of problems that may be encountered in tests and assessments.

Also, See

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. What is the main concept of NCERT solutions for Class 9 Maths exercise 8.1?

Quadrilateral, types of quadrilaterals and their properties and different theorem.

2. Define quadrilateral?

A quadrilateral is a closed four-sided shape formed by joining four points by straight lines.

3. What is the sum total of all angles of a quadrilateral?

The sum total of all angles of a quadrilateral is 360 degrees.

4. The minimum number of triangles that can be formed from a quadrilateral are?

We can get a minimum of two triangles to form a quadrilateral. 

5. What are the different types of quadrilaterals?

Different types of quadrilaterals are:

  1. Square

  2. Rectangle

  3. Rhombus

  4. Trapezium

  5. Parallelogram

  6. Kite

6. What are the similarities between square and rhombus?

The similarities between square and rhombus are:

  1. All the sides are equal to each other

  2. Their diagonal intersects at an angle of 90 degrees

7. The diagonals of a parallelogram separates it into how many congruent triangles?

The diagonals of a parallelogram separate it into two congruent triangles.

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0.34\; J

Option 2)

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Option 3)

1.00\; J

Option 4)

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Option 4)

12.89×10−3 kg

 

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Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

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Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

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Option 1)

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Option 2)

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Option 1)

0.02

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3.125 × 10-2

Option 3)

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Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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