Prasanna

HOTS Questions for Class 9 Science Chapter 6 Tissues

by

HOTS Questions for Class 9 Science Chapter 6 Tissues

These Solutions are part of HOTS Questions for Class 9 Science. Here we have given HOTS Questions for Class 9 Science Chapter 6 Tissues

Question 1.
What will happen if

  1. Apical meristem is damaged or cut ?
  2. Cork is not formed in older stems and roots ?
  3. Cells of epithelial tissue are not compactly packed.
  4. Lymph is not returned to blood ?

Answer.

  1. Apical Meristem Removed: Growth in length will stop.
  2. Cork is Not Formed: If cork is not formed in older stems and roots, the outer tissues will rupture with the increase in girth and expose the interior to desiccation and infection.
  3. Epithelial tissue will not be effective as protective impermeable layer.
  4. Lymph is Not Returned to Blood: Blood volume will decrease while passage of materials from tissues to blood and vice versa would be dislocated.

More Resources

Question 2.

  1. Identify figures : A,B and C.
    HOTS Questions for Class 9 Science Chapter 6 Tissues image - 1
  2. Which one of them provides both mechanical strength as well as flexibility ?
  3. Which one of them is commercially exploited to obtain Hemp and Jute ?
  4. Which one of them can be modified to form air cavities in aquatic plants ?
  5. Which one of them has heavy deposition of lignin ?

Answer:

  1. A- T.S sclerenchyma fibres.
    B- T.S parenchyma cells.
    C- T. S collenchyma.
  2. Mechanical strength and flexibility: Collenchyma.
  3. Hemp and Jute: Sclerenchyma fibres.
  4. Aerenchyma: Modification of parenchyma.
  5. Deposition of Lignin: Sclerenchyma.

Question 3.

  1. Identify the figure. What is its function ?
    HOTS Questions for Class 9 Science Chapter 6 Tissues image - 2
  2. Label X, Y and Z .
  3. Which ones of them develop from the same mother cell ?
  4. X loses its nucleus in mature state. Still it remains alive. How ?

Answer:

  1. Identification: Phloem tissue.
    Function: Conduction of organic food.
  2. X – Sieve tube cell.
    Y – Sieve plate
    Z – Companion cell.
  3. From Same Mother Cell: Sieve tube cell and companion cell.
  4. Living of Sieve Tube Cells: It is controlled by nucleated companion cells with which they are connected by plasmodesmata.

Question 4.

  1. Identify figures A and B.
    HOTS Questions for Class 9 Science Chapter 6 Tissues image - 3
  2. Which is called tesselated and pavement epithelium ?
  3. Which one lines gastro-intestinal tract and epiglottis ?
  4. Which one allows diffusion of substances ?

Answer:

  1. Identification:
    A- Squamous epithelium
    B- Ciliated columnar epithelium.
  2. Pavement Epithelium: Squamous epithelium.
    Tesselated Epithelium: Squamous epithelium.
  3. Gastro-intesinal Tract: Simple columnar epithelium.
    Epiglottis: Stratified columnar epithelium.
  4. Diffusion of Substances: Simple epithelium, especially squamous one.

Question 5.

  1. Identify the figure.
    HOTS Questions for Class 9 Science Chapter 6 Tissues image - 4
  2. Label X, Y and Z.
  3. What is the chemical composition of material of the figure ?
  4. What is the function of Y

Answer:

  1. Identification: Section of bone.
  2. X- Periosteum.
    Y- Haversian canal
    Z- Canaliculus (plural canaliculi)
  3. Composition,
    1. Ossein (30-40%)
    2. Mineral matter
      (60-70%) of calcium and magnesium carbonate and phosphate.
  4. Function of Haversian Canal: To carry nutrients to the interior of the bone.

Question 6.
What will happen if

  1. Bone is dipped in HCl
  2. Bone is dried ?

Answer:

  1. Bone Dipped in HCl. Mineral matter dissolves. Only organic matter is left.
  2. Dried Bone. Organic matter destroyed. Only mineral matter is left.

Question 7.
What will happen if

  1. Ligament gets overstretched ?
  2. Heparin is absent in blood ?
  3. Striated muscles contract rapidly for longer duration ?

Answer:

  1. Oversretching of Ligament: Sprain.
  2. Absence of Heparin in Blood: Blood coagulation occurs inside the blood vessels.
  3. Striated Muscle Contraction: Fatigue due to accumulation of lactic acid.

Question 8.

  1. Identify figures A and B .
    HOTS Questions for Class 9 Science Chapter 6 Tissues image - 5
  2. Label X, Y and Z.
  3. Which one acts as impulse booster ?
  4. Which one is under control of our will ?

Answer:

  1. Identification:
    A – Cardiac muscle fibres.
    B – Striated muscle fibre.
  2. X- Intercalated disc.
    Y- Dark band.
    Z- Light band.
  3. Impulse Booster: Intercalated disc.
  4. Control of Will: Striated or skeletal muscle.

Hope given HOTS Questions for Class 9 Science Chapter 6 Tissues are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS

by

RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS

Other Exercises

Question 1.
If the ratio of mode and median of a certain data is 6 : 5, then find the ratio of its mean and median.
Solution:
We know that
Mode = 3 median – 2 mean…(i)
and \(\frac { mode }{ median } \) = \(\frac { 6 }{ 5 } \)
Mode = \(\frac { 6 }{ 5 } \)median
∴From (i), \(\frac { 6 }{ 5 } \) median = 3 median – 2 mean
=> 2 mean = 3 median – \(\frac { 6 }{ 5 } \)median
2 mean = \(\frac { 15-6 }{ 5 } \)median = \(\frac { 9 }{ 5 } \)median
\(\frac { mean }{ median } \) = \(\frac { 9 }{ 5X2 } \) = \(\frac { 9 }{ 10 } \)
∴Ratio = 9:10

Question 2.
If the mean of x + 2, 2x + 3, 3x + 4, 4x + 5 is x + 2, find x.
Solution:
Mean of x + 2, 2x + 3, 3x + 4, 4x + 5 = x + 2
=> \(\frac { x + 2+2x + 3+3x + 4+4x + 5 }{ 4 } \) = x + 2
=> 10x + 14 = 4x + 8
=> 10x – 4x = 8 – 14
=> 6x= – 6
∴ x = – 1

Question 3.
If the median of scores ,\(\frac { x }{ 2 } \), \(\frac { x }{ 3 } \), \(\frac { x }{ 4 } \), \(\frac { x }{ 5} \) and \(\frac { x }{ 6 } \) (where x > 0) is 6, then find the value \(\frac { x }{ 6 } \)
Solution:
\(\frac { x }{ 2 } \), \(\frac { x }{ 3 } \), \(\frac { x }{ 4 } \), \(\frac { x }{ 5} \), \(\frac { x }{ 6 } \)
Here n = 5
Median = \(\frac { n+1 }{ 2 } \) th term = \(\frac { 5+1 }{ 2 } \) th
\(\frac { 6 }{ 2 } \) = 3rd term = \(\frac { x }{ 4 } \)
\(\frac { x }{ 4 } \) = 6 => x = 24
\(\frac { x }{ 6 } \) = \(\frac { 24 }{ 6 } \) = 4
∴Hence = \(\frac { x }{ 6 } \) = 4

Question 4.
If the mean of 2, 4, 6, 8, x, y is 5, then find the value of x + y.
Solution:
Mean of 2, 4, 6, 8, x, y is 5
\(\frac { 2+4+6+8+x+y }{ 6 } \) = 5
\(\frac { 20+x+y }{ 6 } \) = 5
=> 20 + (x +y) = 30
=> x + y = 30 – 20 = 10
∴x + y = 10

Question 5.
If the mode of scores 3, 4, 3, 5, 4, 6, 6, x is 4, find the value of x.
Solution:
Mode of 3, 4, 3, 5, 4, 6, 6, x is 4
∴ 4 comes in maximum times
But here ,
3 2
4 2
5 1
6 2
3, 4 and 6 are equal in number
∴ x must be 4 so that it becomes in maximum times

Question 6.
If the median of 33, 28, 20. 25, 34, x is 29. find the maximum possible value of x.
Solution:
Median of 33, 28, 20, 25, 34, x is 29
Now arranging in ascending order 20, 25, 28, x, 33, 34
Here n = 6
Median = \(\frac { 1 }{ 2 } \left[ \frac { 6 }{ 2 } th\quad term+\left( \frac { 6 }{ 2 } +1 \right) th\quad term \right] \)
29 = \(\frac { 1 }{ 2 } \) [3rd term + 4th term]
29 = \(\frac { 1 }{ 2 } \) [28+x]
58 = 28 + x
=> x = 58 – 28 = 30
∴Possible value of x = 30

Question 7.
If the median of the scores 1, 2, x, 4, 5 (where 1 <2 <x <4 <5) is 3, then find the mean of the scores.
Solution:
Scores are 1, 2, x, 4, 5 and median 3
Here n = 5 which is odd
Median = \(\frac { n+1 }{ 2 } \) th term = \(\frac { 5+1 }{ 2 } \) = \(\frac { 6 }{ 2 } \) th
=> 3 = 3rd term = x
=> 3 = x
∴ x = 3
Mean of the score = \(\frac { 1+2+3+4+5 }{ 5 } \) = 3

Question 8.
If the ratio of mean and median of a certain data is 2 : 3, then find the ratio of its mode and mean.
Solution:
We know that mode = 3 median – 2 mean
RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS 8.1
\(\frac { mode }{ mean } \) = \(\frac { 5 }{ 2 } \)
Ratio in mode and mean = 5 : 2

Question 9.
The arithmetic mean and mode of a data are 24 and 12 respectively, then find the median of the data.
Solution:
Mean = 24
Mode = 12
We know that mode = 3 median – 2 mean
12 = 3 median – 2 x 24
12 = 3 median – 48
3 median 12 + 48 = 60
Median = \(\frac { 60 }{ 3 } \) = 20

Question 10.
If the difference of mode and median of a data is 24, then find the difference of median and mean.
Solution:
Mode – Median = 24
Mode = 24 + median
But mode = 3 median – 2 mean
3 median – 2 mean = 24 + median
3 median – median – 2 mean = 24
=> 2 median – 2 mean = 24
=> Median – Mean = 12 (Dividing by 2)

Hope given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.4

by

RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.4

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.4

Other Exercises

Question 1.
Find out the mode of the following marks obtained by 15 students in a class:
Marks : 4, 6, 5, 7, 9, 8, 10, 4, 7, 6, 5, 9, 8, 7, 7.
Solution:
Marks obtained are in ascending order,
4, 4, 5, 5, 6, 6, 7, 7, 7, 7, 8, 8, 9, 9, 10
Here we see that 7 is the number which is maximum times i.e. 4 times
Mode = 7

Question 2.
Find the mode for the following data:
125, 175, 225, 125, 225, 175, 325, 125, 375, 225, 125
Solution:
Arranging in ascending order,
125, 125, 125, 125, 175, 175, 225, 225, 225, 325, 375
We see that, 125 is the number which is in maximum times
Mode = 125

Question 3.
Find the mode for the following series:
7.5, 7.3, 7.2, 7.2, 7.4, 7.7, 7.7, 7.5, 7.3, 7.2, 7.6, 7.2
Solution:
Arranging in ascending order,
7.2, 7.2, 7.2, 7.2, 7.3, 7.3, 7.4, 7.5, 7.5, 7.6, 7.7, 7.7
We see that 7.2 comes in maximum times
Mode = 7.2

Question 4.
Find the mode of the following data in each case:
(i) 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18
(ii) 7, 9, 12, 13, 7, 12, 15, 7, 12, 7, 25, 18, 7
Solution:
(i) 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18
Arranging in ascending order,
14, 14, 14,. 14, 17, 18, 18, 18, 22, 23, 25, 28
Here we see that 14 comes in maximum times
Mode = 14
(ii) 7, 9, 12, 13, 7, 12, 15, 7, 12, 7, 25, 18, 7
Arranging in order,
7, 7, 7, 7, 7, 9, 12, 12, 12, 13, 15, 18, 25
Here we see that 7 comes in maximum times
Mode = 7

Question 5.
The demand of different shirt sizes, as obtained by a survey, is given below:
RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.4 5.1
Find the modal shirt sizes, as observed from the survey.
Solution:
From the given data
RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.4 5.2
From above, we see that
Modal size is 39 as it has maximum times persons

Hope given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

NCERT Solutions for Class 9 Science Chapter 6 Tissues

by

NCERT Solutions for Class 9 Science Chapter 6 Tissues

These Solutions are part of NCERT Solutions for Class 9 Science. Here we have given NCERT Solutions for Class 9 Science Chapter 6 Tissues. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 9 Science (Biology) Chapter 6 – Tissues solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 6 – Tissues Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

More Resources

NCERT TEXT BOOK QUESTIONS

IN TEXT QUESTIONS

Question 1.
What is a tissue ?
Answer:
Tissue is a group of related cells that have a common origin and perform a common function.

Question 2.
What is the utility of tissues in multicellular organisms ?
Answer:

  1. Division of Labour: Tissues bring about division of labour in multicellular organisms. It increases efficiency.
  2. Higher Organisation: Tissues become organised to form organs and organ systems.
  3. Individual Cells: Work load of individual cells has decreased.
  4. Higher Survival: Because of division of labour, higher efficiency and organisation, the multicellular organisms have high survival.

Question 3.
Name types of simple tissues.
Answer:
Three – parenchyma, collenchyma and sclerenchyma. (Meristematic tissue is also a simple tissue).

Question 4.
Where is apical meristem found ?
Answer:
Apical meristem occurs at root and stem tips.

Question 5.
Which tissue makes up the husk of coconut ?
Answer:
Sclerenchyma.

Question 6.
What are the constituents of phloem ?
Answer:
Sieve tubes, companion cells, phloem parenchyma and phloem fibres.

Question 7.
Name the tissue responsible for movement of our body.
Answer:
Muscular tissue.

Question 8.
What does a neuron look-like ?
Answer:
A miniature tree with thin hair like parts arising from its ends.

Question 9.
Give three features of cardiac muscles.
Answer:

  1. Cells/Fibres: They are small, cylindrical, uninucleate striated with short lateral branches.
  2. Intercalated Discs: In the area of union becween the two adjacent cardiac muscle fibres, zig-zag junctions called intercalated discs develop. The intercalated discs function as impulse boosters.
  3. Rhythmic Contractions: The muscles are involuntary and nonfatigued which continue to contract and relax tirelessly throughout life.

Question 10.
What are the functions of areolar tissue ?
Answer:

  1. Packing: Areolar tissue provides packing material in various organs.
  2. Binding: It binds various structures with one another in such a way as to prevent their dislocation while allowing Macrophage limited movement.
  3. Covering: It provides covering over nerves, muscles and blood vessels.
  4. Repair: The tissue provides materials for repair of injury.

NCERT CHAPTER END EXERCISES

Question 1.
Define the term “tissue”.
Answer:
Tissue is a group of related cells that have a common origin and perform a common function.

Question 2.
How many types of elements together make up the xylem tissue ? Name them.
Answer:
Xylem tissue is formed of four types of elements. They are tracheids, vessels, xylem parenchyma and xylem fibres.

Question 3.
How are simple tissues different from complex tissues in plants ? (CCE 2014)
Answer:
Differences between Simple and Complex Tissues

Simple TissuesComplex Tissues
1. Cells: A simple tissue is formed of only one type of cells.A complex tissue is made of more than one type of cells.
2. Activity: All the cells perform the same function.The different cells perform different fractions of a function.
3. Types: There are three types of simple plant tissues— . parenchyma, collenchyma and sclerenchyma.There are two types of complex plant tissues— xylem and phloem.
4. Function: They form primary structure of the plant.They form transport system of the plant.

Question 4.
Differentiate amongst parenchyma, collenchyma and sclerenchyma on the basis of the cell wall.
Answer:

ParenchymaCollenchymaSclerenchyma

1. Thickness: The cell wall is thin.

2. Smoothness: It is smooth.

3. Nature: Wall is formed of cellulose.

It is thickened.

It is unevenly thickened.

The thickening is pectocellulosic.

It is thickened.

The wall is uniformly thickened.

The thickening is generally of lignin.

Question 5.
What are the functions of stomata ?
Answer:
Functions of Stomata:

  1. Gaseous Exchange: Stomata are sites where exchange of gases (carbon dioxide and oxygen) occurs between the plant interior and external environment.
  2. Transpiration: Major part of transpiration occurs through stomata. Transpiration removes excess water and keeps plant surfaces cool even in bright sun.
  3. Regulation: They regulate both gaseous exchange and transpiration.

Question 6.
Diagramatically show the difference amongst three types of muscle fibres. (CCE 2014)
Answer:
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 1
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 2
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 3

Question 7.
What is the specific function of cardiac muscle ?
Answer:
Rhythmic contraction and relaxation continuously throughout life without getting fatigued.

Question 8.
Differentiate amongst striated, unstriated and cardiac muscles on the basis of their structure and site/ location in the body.
Answer:

Striated Muscle Fibres

Smooth Muscle Fibres

Cardiac Muscle Fibres

1. Cells. They are long cylindrical cells.The fibres are elongated and spindle­shaped.The cells are small and cylindrical.
2. Ends. The fibres have blunt ends.The fibres have pointed ends.The fibres have broad ends.
3. Striations. They possess striationsStriations or light and dark bandsStriations are present but they are
or alternate light and dark bands.are absent.fainter than those of striated muscle fibres.
4. Intercalated Discs and CrossIntercalated discs and cross-Intercalated discs and cross-
Connections. They are absent.connections are absent.connections are present.
5. Nucleus. The muscle fibre isSmooth muscle fibre is uninucleate.The cells are uninucleate. Nucleus
multinucleate. Nuclei are oval inNucleus is centrally placed, oval orin oval-rounded. It is centrally
outline. They occur peripherally below the sarcolemma.elongated.placed.
6. Arrangement. They occur in bundles.They generally form sheets.They form a network.

Question 9.
Draw a labelled diagram of a neuron. (CCE 2013)
Answer:
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 4

Question 10.
Name the following :
(a) Tissue that forms the inner lining of our mouth,
(b) Tissue that connects muscle to bone in humans.
(c) Tissue that transports food in plants.
(d) Tissue that stores fat in our body.
(e) Connective tissue with a fluid matrix.
(f) Tissue present in the brain.
Answer:
(a) Epithelial tissue
(b) Tendon
(c) Phloem
(d) Adipose
(e) Blood
(f) Nervous tissue.

Question 11.
Identify the types of tissue in the following : skin, bark of tree, bone, lining of kidney tubule, vascular bundle.
Answer:
(a) Skin— Epithelial tissue
(b) Bark of Tree— Cork (Protective tissue)
(c) Bone— Connective tissue with solid matrix
(d) Lining of Kidney Tubule— Epithelial tissue
(e) Vascular Bundle— Complex or vascular tissues, xylem and phloem.

Question 12.
Name the regions in which parenchyma tissue is present.
Answer:
It occurs in almost all nonwoody parts of the plants— cortex, pith, medullary rays of stem, cortex and pith of root, chlorenchyma of leaves, flowers, pith of fruits, etc. Epidermis is special type of parenchyma.

Question 13.
What is the role of epidermis in plants ?
Answer:

  1. Protection,
  2. Regulation of transpiration,
  3. Formation of insulating stationary air layer with the help of hair,
  4. Exchange of gases.

Question 14.
Flow does the cork act as a protective tissue ?
Answer:
Cork acts as a protective layer because its cells are dead, filled with tannins, resin and air, impermeable due to deposition of suberin over the cell walls and absence of intercellular spaces. It is insulating (heat proof), fire proof, shock proof, water proof and repellent to microbes and animals.

Question 15.
Complete the table :
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 5
Answer:
(i) Parenchyma
(ii) Sclerenchyma
(iii) Phloem.

SELECTION TYPE QUESTIONS

Alternate Response Type Questions
(True/False, Right/Wrong, Yes/No)

Question 1.
Vacuoles are absent in meristematic plant cells.
Question 2.
Sderenchyma has irregularly thickened cells.
Question 3.
Absorptive surface areas of roots are increased by the presence of root hair.
Question 4.
Cells of connective tissue are compactly packed with no intercellular spaces.
Question 5.
Cardiac muscles undergo rhythmic contraction and relaxation throughout life.
Question 6.
Areolar connective tissue binds muscles with bones.
Question 7.
Cells of cork are dead, suberised and compacdy arranged.
Question 8.
Voluntary muscles control the movement of iris of eye.

Matching Type Questions

Question 9.
Match the contents of the columns A and B (single matching)
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 6

Question 10.
Match the contents of columns I, II and III (Double matching)
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 7

Question 11.
Which one of the following tissues are involved in growth (G), absorption (A), transportation (T) (Key or check list items)
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 8

Question 12.
Match the Stimulus with Appropriate Response.
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 9

Fill in the Blanks

Question 13. Animals move around in search of ………….. mate and shelter.
Question 14. A thick waterproof coating of ………… occurs over the epidermis in desert plants.
Question 15. ……………. consists of tracheids, vessels, parenchyma and fibres.
Question 16. ………….. epithelium occurs in the lining of renal tubules and ducts of salivary glands.
Question 17. Tendons connect …………….. with bones.

Answers:
NCERT Solutions for Class 9 Science Chapter 6 Tissues image - 10

SOME ACTIVITY BASED QUESTIONS

Question 1.
Why is blood considered to be connective tissue ?
Answer:

  1. Like other connective tissues, blood consists of living cells scattered in an abundant matrix. The matrix is liquid or plasma in blood.
  2. Blood circulates throughout the body, receiving and providing materials to all tissues and organs of the body. It thus connects all parts of the body.

Question 2.
What is aerenchyma ? Give its functions.
Answer:
Definition: It is a specialised parenchyma found in aquatic plants which consists of network of small cells that enclose large air cavities.
Functions:

  1. Storage of Gases: It stores metabolic gases, O2 and CO2, for use inside the plants in respiration and photosynthesis.
  2. Buoyancy: It makes the aquatic plant buoyant both inside and over the surface of water.

Question 3.
What is skeletal connective tissue ? Give its functions.
Answer:
Definition: Skeletal connective tissue is that connective tissue in which the matrix is solid and the living cells occur inside fluid filled spaces called lacunae. It is of two types, cartilage and bone.
Functions:

  1. Endoskeleton: It forms the internal supporting framework of the animal body.
  2. Protection: The tissue protects the vital organs like brain, spinal cord, heart, lungs, etc.
  3. Joints: The tissue forms joints which allow for growth and movement of body parts.
  4. Muscles: It provides surface for attachement to muscles.
  5. Blood Cells: They develop inside red marrow of bones.
  6. Minerals: Bony skeleton stores minerals, some of which are withdrawn by the body in case of emergency.

Question 4.
What is synapse ? Explain.
Answer:
It is a junction between two neurons without developing an organic union. The terminal knobbed branch end of an axon of one neuron comes in near contact with a dendrite terminal of the next neuron. A narrow fluid filled space occurs between the two. An activated axon end passes out a neurotransmitter like acetylcholine which provides sensation to dendrite terminal. This helps in transfer of impulse from one neuron to the next.

NCERT Solutions for Class 9 Science Chapter 6 Tissues

Hope given NCERT Solutions for Class 9 Science Chapter 6 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1

by

RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1

Other Exercises

Question 1.
In a ∆ABC, if ∠A = 55°, ∠B = 40°, find ∠C.
Solution:
∵ Sum of three angles of a triangle is 180°
∴ In ∆ABC, ∠A = 55°, ∠B = 40°
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1 Q1.1
⇒ 55° + 40° + ∠C = 180°
⇒ 95° + ∠C = 180°
∴ ∠C= 180° -95° = 85°

Question 2.
If the angles of a triangle are in the ratio 1:2:3, determine three angles.
Solution:
Ratio in three angles of a triangle =1:2:3
Let first angle = x
Then second angle = 2x
and third angle = 3x
∴ x + 2x + 3x = 180° (Sum of angles of a triangle)
⇒6x = 180°
⇒x = \(\frac { { 180 }^{ \circ } }{ 6 }\)  = 30°
∴ First angle = x = 30°
Second angle = 2x = 2 x 30° = 60°
and third angle = 3x = 3 x 30° = 90°
∴ Angles are 30°, 60°, 90°

Question 3.
The angles of a triangle are (x – 40)°, (x – 20)° and (\(\frac { 1 }{ 2 }\) x – 10)°. Find the value of x.
Solution:
∵ Sum of three angles of a triangle = 180°
∴ (x – 40)° + (x – 20)° + (\(\frac { 1 }{ 2 }\)x-10)0 = 180°
⇒ x – 40° + x – 20° + \(\frac { 1 }{ 2 }\)x – 10° = 180°
⇒ x + x+ \(\frac { 1 }{ 2 }\)x – 70° = 180°
⇒ \(\frac { 5 }{ 2 }\)x = 180° + 70° = 250°
⇒ x = \(\frac { { 250 }^{ \circ }x 2 }{ 5 }\)  = 100°
∴ x = 100°

Question 4.
Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°. Determine all the angles of the triangle.
Solution:
Let each of the two equal angles = x
Then third angle = x + 30°
But sum of the three angles of a triangle is 180°
∴ x + x + x + 30° = 180°
⇒ 3x + 30° = 180°
⇒3x = 150° ⇒x = \(\frac { { 150 }^{ \circ } }{ 3 }\) = 50°
∴ Each equal angle = 50°
and third angle = 50° + 30° = 80°
∴ Angles are 50°, 50° and 80°

Question 5.
If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.
Solution:
In the triangle ABC,
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1 Q5.1
∠B = ∠A + ∠C
But ∠A + ∠B + ∠C = 180°
⇒∠B + ∠A + ∠C = 180°
⇒∠B + ∠B = 180°
⇒2∠B = 180°
∴ ∠B = \(\frac { { 180 }^{ \circ } }{ 2 }\) = 90°
∵ One angle of the triangle is 90°
∴ ∆ABC is a right triangle.

Question 6.
Can a triangle have:
(i) Two right angles?
(ii) Two obtuse angles?
(iii) Two acute angles?
(iv) All angles more than 60°?
(v) All angles less than 60°?
(vi) All angles equal to 60°?
Justify your answer in each case.
Solution:
(i) In a triangle, two right-angles cannot be possible. We know that sum of three angles is 180° and if there are two right-angles, then the third angle will be zero which is not possible.
(ii) In a triangle, two obtuse angle cannot be possible. We know that the sum of the three angles of a triangle is 180° and if there are
two obtuse angle, then the third angle will be negative which is not possible.
(iii) In a triangle, two acute angles are possible as sum of three angles of a trianlge is 180°.
(iv) All angles more than 60°, they are also not possible as the sum will be more than 180°.
(v) All angles less than 60°. They are also not possible as the sum will be less than 180°.
(vi) All angles equal to 60°. This is possible as the sum will be 60° x 3 = 180°.

Question 7.
The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angle is 10°, find the three angles.
Solution:
Let three angles of a triangle be x°, (x + 10)°, (x + 20)°
But sum of three angles of a triangle is 180°
∴ x + (x+ 10)° + (x + 20) = 180°
⇒ x + x+10°+ x + 20 = 180°
⇒ 3x + 30° = 180°
⇒ 3x = 180° – 30° = 150°
∴ x = \(\frac { { 180 }^{ \circ } }{ 2 }\) = 50°
∴ Angle are 50°, 50 + 10, 50 + 20
i.e. 50°, 60°, 70°

Question 8.
ABC is a triangle is which ∠A = 72°, the internal bisectors of angles B and C meet in O. Find the magnitude of ∠BOC.
Solution:
In ∆ABC, ∠A = 12° and bisectors of ∠B and ∠C meet at O
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1 Q8.1
Now ∠B + ∠C = 180° – 12° = 108°
∵ OB and OC are the bisectors of ∠B and ∠C respectively
∴ ∠OBC + ∠OCB = \(\frac { 1 }{ 2 }\) (B + C)
= \(\frac { 1 }{ 2 }\) x 108° = 54°
But in ∆OBC,
∴ ∠OBC + ∠OCB + ∠BOC = 180°
⇒ 54° + ∠BOC = 180°
∠BOC = 180°-54°= 126°
OR
According to corollary,
∠BOC = 90°+ \(\frac { 1 }{ 2 }\) ∠A
= 90+ \(\frac { 1 }{ 2 }\) x 72° = 90° + 36° = 126°

Question 9.
The bisectors of base angles of a triangle cannot enclose a right angle in any case.
Solution:
In right ∆ABC, ∠A is the vertex angle and OB and OC are the bisectors of ∠B and ∠C respectively
To prove : ∠BOC cannot be a right angle
Proof: ∵ OB and OC are the bisectors of ∠B and ∠C respectively
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1 Q9.1
∴ ∠BOC = 90° x \(\frac { 1 }{ 2 }\) ∠A
Let ∠BOC = 90°, then
\(\frac { 1 }{ 2 }\) ∠A = O
⇒∠A = O
Which is not possible because the points A, B and C will be on the same line Hence, ∠BOC cannot be a right angle.

Question 10.
If the bisectors of the base angles of a triangle enclose an angle of 135°. Prove that the triangle is a right triangle.
Solution:
Given : In ∆ABC, OB and OC are the bisectors of ∠B and ∠C and ∠BOC = 135°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1 Q10.1
To prove : ∆ABC is a right angled triangle
Proof: ∵ Bisectors of base angles ∠B and ∠C of the ∆ABC meet at O
∴ ∠BOC = 90°+ \(\frac { 1 }{ 2 }\)∠A
But ∠BOC =135°
∴ 90°+ \(\frac { 1 }{ 2 }\) ∠A = 135°
⇒ \(\frac { 1 }{ 2 }\)∠A= 135° -90° = 45°
∴ ∠A = 45° x 2 = 90°
∴ ∆ABC is a right angled triangle

Question 11.
In a ∆ABC, ∠ABC = ∠ACB and the bisectors of ∠ABC and ∠ACB intersect at O such that ∠BOC = 120°. Show that ∠A = ∠B = ∠C = 60°.
Solution:
Given : In ∠ABC, BO and CO are the bisectors of ∠B and ∠C respectively and ∠BOC = 120° and ∠ABC = ∠ACB
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1 Q11.1
To prove : ∠A = ∠B = ∠C = 60°
Proof : ∵ BO and CO are the bisectors of ∠B and ∠C
∴ ∠BOC = 90° + \(\frac { 1 }{ 2 }\)∠A
But ∠BOC = 120°
∴ 90°+ \(\frac { 1 }{ 2 }\) ∠A = 120°
∴ \(\frac { 1 }{ 2 }\) ∠A = 120° – 90° = 30°
∴ ∠A = 60°
∵ ∠A + ∠B + ∠C = 180° (Angles of a triangle)
∠B + ∠C = 180° – 60° = 120° and ∠B = ∠C
∵ ∠B = ∠C = \(\frac { { 120 }^{ \circ } }{ 2 }\) = 60°
Hence ∠A = ∠B = ∠C = 60°

Question 12.
If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.
Solution:
In a ∆ABC,
Let ∠A < ∠B + ∠C
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1 Q12.1
⇒∠A + ∠A < ∠A + ∠B + ∠C
⇒ 2∠A < 180°
⇒ ∠A < 90° (∵ Sum of angles of a triangle is 180°)
Similarly, we can prove that
∠B < 90° and ∠C < 90°
∴ Each angle of the triangle are acute angle.

Hope given RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.