RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS

Other Exercises

Question 1.
Solution:
In two congruent triangles ABC and DEF, if AB = DE and BC = EF. Name the pairs of equal angles.

In AABC and ADEF,
∆ABC ≅ ∆DEF
and AB = DE, BC = EF
∴ ∠A = ∠D, ∠B = ∠E and ∠C = ∠F
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS Q1.1

Question 2.
Solution:
In two triangles ABC and DEF, it is given that ∠A = ∠D, ∠B = ∠E and ∠C = ∠F. Are the two triangles necessarily congruent?

No, as the triangles are equiangular, so similar.

Question 3.
Solution:
If ABC and DEF are two triangles such that AC = 2.5 cm, BC = 5 cm, ∠C = 75°, DE = 2.5 cm, DF = 5 cm and ∠D = 75°. Are two triangles congruent?
Yes, triangles are congruent (SAS axiom)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS Q3.1

Question 4.
Solution:
In two triangles ABC and ADC, if AB = AD and BC = CD. Are they congruent?
Yes, these are congruent
In two triangles ABC are ADC,
AB = AD (Given)
BC = CD (Given)
and AC = AC (Common)
∴ ∆sABC ≅ AADC (SSS axiom)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS Q4.1

Question 5.
Solution:
In triangles ABC and CDE, if AC = CE, BC = CD, ∠A = 60°, ∠C – 30° and ∠D = 90°. Are two triangles congruent?
Yes, triangles are congruent because,
In ∆ABC, and ∆CDE,
AC = CE
BC = CD ∠C = 30°
∴ ∆ABC ≅ ∆CDE (SAS axiom)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS Q5.1

Question 6.
Solution:
ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.

Given : In ∆ABC, AB = AC
BE and CF are two medians
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS Q6.1
To prove : BE = CF
Proof: In ∆ABE and ∆ACF.
AB = AC (Given)
∠A = ∠A (Common)
AE = AF (Half of equal sides)
∴ ∆ABE ≅ ∆ACF (SAS axiom)
∴ BE = CF (c.p.c.t.)

Question 7.
Solution:
Find the measure of each angle of an equilateral triangle.

In ∆ABC,
AB = AC = BC
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS Q7.1
∵ AB = AC
∴ ∠C = ∠B …(i)
(Angles opposite to equal sides)
Similarly,
AC = BC
∴ ∠B = ∠A …(ii)
From (i) and (ii),
∠A = ∠B = ∠C
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
∴ ∠A + ∠B + ∠C = \(\frac { { 180 }^{ \circ } }{ 3 }\)  = 60°

Question 8.
Solution:
CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that ∆ADE ≅ ∆BCE.

Given : An equilateral ACDE is formed on the side of square ABCD. AE and BE are joined
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS Q8.1
To prove : ∆ADE ≅ ∆BCE
Proof : In ∆ADE and ∆BCE,
AD = BC (Sides of a square)
DE = CE (Sides of equilateral triangle)
∠ADE = ∠BCE(Each = 90° + 60° = 150°)
∴ AADE ≅ ABCE (SAS axiom)

Question 9.
Solution:
Prove that the sum of three altitude of a triangle is less than the sum of its sides.

Given : In ∆ABC, AD, BE and CF are the altitude of ∆ABC
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS Q9.1
To prove : AD + BE + CF < AB + BC + CA
Proof : In right ∆ABD, ∠D = 90°
Then other two angles are acute
∵ ∠B < ∠D
∴ AD < AB …(i)
Similarly, in ∆BEC and ∆ABE we can prove thatBE and CF < CA …(iii)
Adding (i), (ii), (iii)
AD + BE -t CF < AB + BC + CA

Question 10.
Solution:
In the figure, if AB = AC and ∠B = ∠C. Prove that BQ = CP.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS Q10.1

Given : In the figure, AB = AC, ∠B = ∠C
To prove : BQ = CP
Proof : In ∆ABQ and ∆ACP
AB = AC (Given)
∠A = ∠A (Common)
∠B = ∠C (Given)
∴ ∆ABQ ≅ ∆ACP (ASA axiom)
∴ BQ = CP (c.p.c.t.)

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6

Other Exercises

Question 1.
In ∆ABC, if ∠A = 40° and ∠B = 60°. Determine the longest and shortest sides of the triangle.
Solution:
In ∆ABC, ∠A = 40°, ∠B = 60°
But ∠A + ∠B + ∠C = 180°
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 Q1.1
⇒ 40° + 60° + ∠C = 180°
⇒ ∠C = 180° = (40° + 60°)
= 180° – 100° = 80°
∵ ∠C = 80°, which is the greatest angle and
∠A = 40° is the smallest angle
∴ Side AB which is opposite to the greatest angle is the longest and side BC which is opposite to the smallest angle is the shortest.

Question 2.
In a ∆ABC, if ∠B = ∠C = 45°. which is the longest side?
Solution:
In ∆ABC, ∠B = ∠C = 45°
But ∠A + ∠B + ∠C = 180°
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 Q2.1
⇒ ∠A + 45° + 45° = 180°
⇒ ∠A + 90° = 180°
∴ ∠A = 180°-90° = 90°
∴∠A is the greatest
∴ Side BC opposite to it is the longest

Question 3.
In ∆ABC, side AB is produced to D so that BD = BC. If ∠B = 60° and ∠A = 70°, prove that :
(i) AD > CD
(ii) AD > AC
Solution:
Given : In AABC, side BC is produced to D such that BD = BC
∠A = 70° and ∠B = 60°
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 Q3.1
To prove :
(i) AD > CD (ii) AD > AC
Proof: In ∆ABC,
∠A = 70°, ∠B = 60°
But Ext. ∠CBD + ∠CBA = 180° (Linear pair)
∠CBD + 60° = 180° 3
⇒ ∠CBD = 180° – 60° = 120°
But in ∆BCD,
BD = BC
∴ ∠D = ∠BCD
But ∠D + ∠BCD = 180° – 120° = 60°
∴∠D = ∠BCD = \(\frac { { 60 }^{ \circ } }{ 2 }\)  = 30°
and in ∆ABC,
∠A + ∠B + ∠C = 180°
⇒ 70° + 60° + ∠C = 180°
⇒ 130° + ∠C = 180°
∴ ∠C =180°- 130° = 50°
Now ∠ACD = ∠ACB + ∠BCD = 50° + 30° = 80°
(i) Now in ∆ACB,
∠ACD = 80° and ∠A = 70°
∴ Side AD > CD
(Greater angle has greatest side opposite to it)
(ii) ∵ ∠ACD = 80° and ∠D = 30°
∴ AD > AC

Question 4.
Is it possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?
Solution:
We know that in a triangle, sum of any two sides is greater than the third side and 2 cm + 3 cm = 5 cm and 5 cm < 7 cm
∴ This triangle is not possible to draw

Question 5.
In ∆ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.
Solution:
In ∆ABC, ∠B = 35°, ∠C = 65° and AP is the bisector of ∠BAC which meets BC in P.
Arrange PA, PB and PC in descending order In ∆ABC,
∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 Q5.1
⇒ ∠A + 35° + 65° = 180°
∠A + 100°= 180°
∴ ∠A =180°- 100° = 80°
∵ PA is a bisector of ∠BAC
∴ ∠1 = ∠2 = \(\frac { { 80 }^{ \circ } }{ 2 }\)  = 40°
Now in ∆ACP, ∠ACP > ∠CAP
⇒ ∠C > ∠2
∴ AP > CP …(i)
Similarly, in ∆ABP,
∠BAP > ∠ABP ⇒ ∠1 > ∠B
∴ BP > AP …(ii)
From (i) and (ii)
BP > AP > CP

Question 6.
Prove that the perimeter of a triangle is greater than the sum of its altitudes
Solution:
Given : In ∆ABC,
AD, BE and CF are altitudes
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 Q6.1
To prove : AB + BC + CA > AD + BC + CF
Proof : We know that side opposite to greater angle is greater.
In ∆ABD, ∠D = 90°
∴ ∠D > ∠B
∴ AB >AD …(i)
Similarly, we can prove that
BC > BE and
CA > CF
Adding we get,
AB + BC + CA > AD + BE + CF

Question 7.
In the figure, prove that:
(i) CD + DA + AB + BC > 2AC
(ii) CD + DA + AB > BC
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 Q7.1
Solution:
Given : In the figure, ABCD is a quadrilateral and AC is joined
To prove :
(i) CD + DA + AB + BC > 2AC
(ii) CD + DA + AB > BC
Proof:
(i) In ∆ABC,
AB + BC > AC …(i)
(Sum of two sides of a triangle is greater than its third side)
Similarly in ∆ADC,
CD + DA > AC …(ii)
Adding (i) and (ii)
CD + DA + AB + BC > AC + AC
⇒ CD + DA + AB + BC > 2AC
(ii) In ∆ACD,
CD + DA > CA
(Sum of two sides of a triangle is greater than its third side)
Adding AB to both sides,
CD + DA + AB > CA + AB
But CA + AB > BC (in ∆ABC)
∴ CD + DA + AD > BC

Question 8.
Which of the following statements are true (T) and which are false (F)?
(i) Sum of the three sides of a triangle is less than the sum of its three altitudes.
(ii) Sum of any two sides of a triangle is greater than twice the median drawn to the third side.
(iii) Sum of any two sides of a triangle is greater than the third side.
(iv) Difference of any two sides of a triangle is equal to the third side.
(v) If two angles of a triangle are unequal, then the greater angle has the larger side opposite to it.
(vi) Of all the line segments that can be drawn from a point to a line not containing it, the perpendicular line segment is the shortest one.
Solution:
(i) False. Sum of three sides of a triangle is greater than the sum of its altitudes.
(ii) True.
(iii) True.
(iv) False. Difference of any two sides is less than the third side.
(v) True.
(vi) True.

Question 9.
Fill in the blanks to make the following statements true.
(i) In a right triangle, the hypotenuse is the ……. side.
(ii) The sum of three altitudes of a triangle is ……. than its perimeter.
(iii) The sum of any two sides of a triangle is …….. than the third side.
(iv) If two angles of a triangle are unequal, then the smaller angle has the ….. side opposite to it.
(v) Difference of any two sides of a triangle is……. than the third side.
(vi) If two sides of a triangle are unequal, then the larger side has ……… angle opposite to it.
Solution:
(i) In a right triangle, the hypotenuse is the longest side.
(ii) The sum of three altitudes of a triangle is less than its perimeter.
(iii) The sum of any two sides of a triangle is greater than the third side.
(iv) If two angles of a triangle are unequal, then the smaller angle has the smaller side opposite to it.
(v) Difference of any two sides of a triangle is less than the third side.
(vi) If two sides of a triangle are unequal, then the larger side has greater angle opposite to it.

Question 10.
O is any point in the interior of ∆ABC. Prove that
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + OB + OC
(iii) OA + OB + OC > \(\frac { 1 }{ 2 }\) (AB + BC + CA)
Solution:
Given : In ∆ABC, O is any point in the interior of the ∆ABC, OA, OB and OC are joined
To prove :
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + OB + OC
(iii) OA + OB + OC > \(\frac { 1 }{ 2 }\) (AB + BC + CA)
Construction : Produce BO to meet AC in D.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 Q10.1
Proof: In ∆ABD,
(i) AB + AD > BD (Sum of any two sides of a triangle is greater than third)
⇒ AB + AD > BO + OD …(i)
Similarly, in ∆ODC,
OD + DC > OC …(ii)
Adding (i) and (ii)
AB + AD + OD + DC > OB + OD + OC
⇒ AB + AD + DC > OB + OC
⇒ AB + AC > OB + OC
(ii) Similarly, we can prove that
BC + AB > OA + OC
and CA + BC > OA + OB
(iii) In ∆OAB, AOBC and ∆OCA,
OA + OB > AB
OB + OC > BC
and OC + OA > CA
Adding, we get
2(OA + OB + OC) > AB + BC + CA
∴ OA + OB + OO > \(\frac { 1 }{ 2 }\) (AB + BC + CA)

Question 11.
Prove that in a quadrilateral the sum of all the sides is greater than the sum of its diagonals.
Solution:
Given : In quadrilateral ABCD, AC and BD are its diagonals,
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 Q11.1
To prove : AB + BC + CD + DA > AC + BD
Proof: In ∆ABC,
AB + BC > AC …(i)
(Sum of any two sides of a triangle is greater than its third side)
Similarly, in ∆ADC,
DA + CD > AC …(ii)
In ∆ABD,
AB + DA > BD …(iii)
In ∆BCD,
BC + CD > BD …(iv)
Adding (i), (ii), (iii) and (iv)
2(AB + BC + CD + DA) > 2AC + 2BD
⇒ 2(AB + BC + CD + DA) > 2(AC + BD)
∴ AB + BC + CD + DA > AC + BD

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 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 12 Heron’s Formula Ex 12.4

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4

Other Exercises

Question 1.
In the figure, it is given that AB = CD and AD = BC. Prove that ∆ADC ≅ ∆CBA.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4 Q1.1
Solution:
Given : In the figure, AB = CD, AD = BC
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4 Q1.2
To prove : ∆ADC = ∆CBA
Proof : In ∆ADC and ∆CBA
CD = AB (Given)
AD = BC (Given)
CA = CA (Common)
∴ ∆ADC ≅ ∆CBA (SSS axiom)

Question 2.
In a APQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.
Solution:
Given : In ∆PQR, PQ = QR
L, M and N are the mid-points of sides PQ, QR and RP respectively. Join LM, MN and LN
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4 Q2.1
To prove : ∠PNM = ∠PLM
Proof : In ∆PQR,
∵ M and N are the mid points of sides PR and QR respectively
∴ MN || PQ and MN = \(\frac { 1 }{ 2 }\) PQ …(i)
∴ MN = PL
Similarly, we can prove that
LM = PN
Now in ∆NML and ∆LPN
MN = PL (Proved)
LM = PN (Proved)
LN = LN (Common)
∴ ∆NML = ∆LPN (SSS axiom)
∴ ∠MNL = ∠PLN (c.p.c.t.)
and ∠MLN = ∠LNP (c.p.c.t.)
⇒ ∠MNL = ∠LNP = ∠PLM = ∠MLN
⇒ ∠PNM = ∠PLM

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4 are helpful to complete your math homework.

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HOTS Questions for Class 9 Science Chapter 7 Diversity in Living Organisms

HOTS Questions for Class 9 Science Chapter 7 Diversity in Living Organisms

These Solutions are part of HOTS Questions for Class 9 Science. Here we have given HOTS Questions for Class 9 Science Chapter 7 Diversity in Living Organisms

Question 1.
(a) Identify figures A to F.
HOTS Questions for Class 9 Science Chapter 7 Diversity in Living Organisms image - 1
(b) Which one of them is unicellular and eukaryotic organism ?
(c) Which one of them shows

  1. Heterotrophic nutrition
  2. Mixotrophic nutrition ?

(d) Which one of them is non-vascular embryophyte ?
(e) In which of them, xylem lacks vessels while phloem is devoid of companion cells ?
(f) In which of them endosperm is halpoid ?
(g) Which of them is commonly called

  1. Bread Mould
  2. Male Shield Fern ?

Answer:
(a) Identification.
A-Fern (Male Shield Fern, Dryopteris)
B-Moss (Funaria).
C—Pinus.
D—Spirogyra.
E—Euglena.
F-Rhizopus.
(b) Unicellular Eukaryotic Organism. Euglena.
(c)

  1. Heterotrophic Nutrition. Rhizopus.
  2. Mixotrophic Nutrition. Euglena.

(d) Non-vascular Embryophyte. Moss (Funaria)
(e) Xylem without Vessels and Phloem without Companion Cells. Pinus.
(f) Haploid Endosperm. Pinus
(g)

  1. Bread Mould._Rhizopus.
  2. Male Shield Fern. Dryopteris.

More Resources

Question 2.
(a) Identify figure A to D.
HOTS Questions for Class 9 Science Chapter 7 Diversity in Living Organisms image - 2
(b) Which one belongs to

  1. Platyhelminthes
  2. Arthropoda
  3. Annelida ?

(c) Which one of them has

  1. Tissue level organisation
  2. Primitive organ level organisation ?

(d) Which one of them has

  1. Flame cells as exretory organs
  2. Nephridia as excretory organs ?

(e) Which one of them has poison claw ?
(f) Which one of them is

  1. Diploblastic
  2. Triploblastic ?

Answer:
(a) Identification.
A-Centipede.
B-Hydra.
C-Liverfluke {Fasciola).
D- Earthworm {Pheretima posthuma).
(b)

  1. Plathyhelminthes—Liverfluke (Fasciola)
  2. Arthropoda-Centipede (Scolopendra).
  3. Annelida-Earthworm {Pheretima posthuma).

(c)

  1. Tissue Level Organisation. Hydra.
  2. Primitive Organ Level Organisation. Liverfluke (Fasciola)

(d)

  1. Flame Cells. Liverfluke (Fasciola)
  2. Nephridia. Earthworm {Pheretima posthuma)

(e) Poison Claw. Centipede {Scolopendra)
(f)

  1. Diploblastic. Hydra.
  2. Triploblastic. Liverfluke {Fasciola), Centipede (Scolopendra), Earthworm {Pheretima posthuma).

Question 3.
(a) Identify the figure and write down the phylum to which it belongs.
(b) Which type of
HOTS Questions for Class 9 Science Chapter 7 Diversity in Living Organisms image - 3

  1. Locomotory organ is present in it.

Circulatory system is present in it.
(c) Which type of symmetry is present ?
Answer:
(a) Identification. Pila (Apple Snail). Phylum. Mollusca.
(b)

  1. Locomotory Organ. Foot,
  2. Circulatory System. Open (Haemocoel).

(c) Symmetry. Asymmetry due to torsion (spirally coiled).

Question 4.
(a) Label W, X, Y and Z.
HOTS Questions for Class 9 Science Chapter 7 Diversity in Living Organisms image - 4
(b) Identify the above diagram.
(c) Name the phylum in which notochord is present.
(d) Name the subphylum in which notochord is present throughout life.
Answer:
(a) W-Anus.
X-Gill slits.
Y-Notochord.
Z-Nerve cord.
(b) Identification. Basic characteristics of chordates.
(c) Phylum with Notochord. Chordata.
(d) Subphylum with Notochord throughout Life. Cephalochordata.

Hope given HOTS Questions for Class 9 Science Chapter 7 Diversity in Living Organisms are helpful to complete your science homework.

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RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS

RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS

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

Other Exercises

Mark the correct alternative in each of the following:
Question 1.
If all the three angles of a triangle are equal, then each one of them is equal to
(a) 90°
(b) 45°
(c) 60°
(d) 30°
Solution:
∵ Sum of three angles of a triangle = 180°
∴ Each angle = \(\frac { { 180 }^{ \circ } }{ 3 }\)  = 60° (c)

Question 2.
If two acute angles of a right triangle are equal, then each acute is equal to
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
In a right triangle, one angle = 90°
∴ Sum of other two acute angles = 180° – 90° = 90°
∵ Both angles are equal
∴ Each angle will be = \(\frac { { 90 }^{ \circ } }{ 2 }\)  = 45° (b)

Question 3.
An exterior angle of a triangle is equal to 100° and two interior opposite angles are equal. Each of these angles is equal to
(a) 75°
(b) 80°
(c) 40°
(d) 50°
Solution:
In a triangle, exterior angles is equal to the sum of its interior opposite angles
∴ Sum of interior opposite angles = 100°
∵ Both angles are equal
∴ Each angle will be = \(\frac { { 100 }^{ \circ } }{ 2 }\)  = 50° (d)

Question 4.
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
(a) an isosceles triangle
(b) an obtuse triangle
(c) an equilateral triangle
(d) a right triangle
Solution:
Let ∠A, ∠B, ∠C be the angles of a ∆ABC and let ∠A = ∠B + ∠C
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q4.1
But ∠A + ∠B + ∠C = 180°
( Sum of angles of a triangle)
∴ ∠A + ∠A = 180° ⇒ 2∠A = 180°
⇒ ∠A = \(\frac { { 180 }^{ \circ } }{ 2 }\)  = 90°
∴ ∆ is a right triangle (d)

Question 5.
Side BC of a triangle ABC has been produced to a point D such that ∠ACD = 120°. If ∠B = \(\frac { 1 }{ 2 }\)∠A, then ∠A is equal to
(a) 80°
(b) 75°
(c) 60°
(d) 90°
Solution:
Side BC of ∆ABC is produced to D, then
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q5.1
Ext. ∠ACB = ∠A + ∠B
(Exterior angle of a triangle is equal to the sum of its interior opposite angles)
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q5.2

Question 6.
In ∆ABC, ∠B = ∠C and ray AX bisects the exterior angle ∠DAC. If ∠DAX = 70°, then ∠ACB =
(a) 35°
(b) 90°
(c) 70°
(d) 55°
Solution:
In ∆ABC, ∠B = ∠C
AX is the bisector of ext. ∠CAD
∠DAX = 70°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q6.1
∴ ∠DAC = 70° x 2 = 140°
But Ext. ∠DAC = ∠B + ∠C
= ∠C + ∠C (∵ ∠B = ∠C)
= 2∠C
∴ 2∠C = 140° ⇒ ∠C = \(\frac { { 140 }^{ \circ } }{ 2 }\) = 70°
∴ ∠ACB = 70° (c)

Question 7.
In a triangle, an exterior angle at a vertex is 95° and its one of the interior opposite angle is 55°, then the measure of the other interior angle is
(a) 55°
(b) 85°
(c) 40°
(d) 9.0°
Solution:
In ∆ABC, BA is produced to D such that ∠CAD = 95°
and let ∠C = 55° and ∠B = x°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q7.1
∵ Exterior angle of a triangle is equal to the sum of its opposite interior angle
∴ ∠CAD = ∠B + ∠C ⇒ 95° = x + 55°
⇒ x = 95° – 55° = 40°
∴ Other interior angle = 40° (c)

Question 8.
If the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is
(a) 90°
(b) 180°
(c) 270°
(d) 360°
Solution:
In ∆ABC, sides AB, BC and CA are produced in order, then exterior ∠FAB, ∠DBC and ∠ACE are formed
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q8.1
We know an exterior angles of a triangle is equal to the sum of its interior opposite angles
∴ ∠FAB = ∠B + ∠C
∠DBC = ∠C + ∠A and
∠ACE = ∠A + ∠B Adding we get,
∠FAB + ∠DBC + ∠ACE = ∠B + ∠C + ∠C + ∠A + ∠A + ∠B
= 2(∠A + ∠B + ∠C)
= 2 x 180° (Sum of angles of a triangle)
= 360° (d)

Question 9.
In ∆ABC, if ∠A = 100°, AD bisects ∠A and AD⊥ BC. Then, ∠B =
(a) 50°
(b) 90°
(c) 40°
(d) 100°
Solution:
In ∆ABC, ∠A = 100°
AD is bisector of ∠A and AD ⊥ BC
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q9.1
Now, ∠BAD = \(\frac { { 100 }^{ \circ } }{ 2 }\) = 50°
In ∆ABD,
∠BAD + ∠B + ∠D= 180°
(Sum of angles of a triangle)
⇒ ∠50° + ∠B + 90° = 180°
∠B + 140° = 180°
⇒ ∠B = 180° – 140° ∠B = 40° (c)

Question 10.
An exterior angle of a triangle is 108° and its interior opposite angles are in the ratio 4:5. The angles of the triangle are
(a) 48°, 60°, 72°
(b) 50°, 60°, 70°
(c) 52°, 56°, 72°
(d) 42°, 60°, 76°
Solution:
In ∆ABC, BC is produced to D and ∠ACD = 108°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q10.1
Ratio in ∠A : ∠B = 4:5
∵ Exterior angle of a triangle is equal to the sum of its opposite interior angles
∴ ∠ACD = ∠A + ∠B = 108°
Ratio in ∠A : ∠B = 4:5
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q10.2

Question 11.
In a ∆ABC, if ∠A = 60°, ∠B = 80° and the bisectors of ∠B and ∠C meet at O, then ∠BOC =
(a) 60°
(b) 120°
(c) 150°
(d) 30°
Solution:
In ∆ABC, ∠A = 60°, ∠B = 80°
∴ ∠C = 180° – (∠A + ∠B)
= 180° – (60° + 80°)
= 180° – 140° = 40°
Bisectors of ∠B and ∠C meet at O

RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q11.1

Question 12.
Line segments AB and CD intersect at O such that AC || DB. If ∠CAB = 45° and ∠CDB = 55°, then ∠BOD =
(a) 100°
(b) 80°
(c) 90°
(d) 135°
Solution:
In the figure,
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q12.1
AB and CD intersect at O
and AC || DB, ∠CAB = 45°
and ∠CDB = 55°
∵ AC || DB
∴ ∠CAB = ∠ABD (Alternate angles)
In ∆OBD,
∠BOD = 180° – (∠CDB + ∠ABD)
= 180° – (55° + 45°)
= 180° – 100° = 80° (b)

Question 13.
In the figure, if EC || AB, ∠ECD = 70° and ∠BDO = 20°, then ∠OBD is
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q13.1
(a) 20°
(b) 50°
(c) 60°
(d) 70°
Solution:
In the figure, EC || AB
∠ECD = 70°, ∠BDO = 20°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q13.2
∵ EC || AB
∠AOD = ∠ECD (Corresponding angles)
⇒ ∠AOD = 70°
In ∆OBD,
Ext. ∠AOD = ∠OBD + ∠BDO
70° = ∠OBD + 20°
⇒ ∠OBD = 70° – 20° = 50° (b)

Question 14.
In the figure, x + y =
(a) 270
(b) 230
(c) 210
(d) 190°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q14.1
Solution:
In the figure
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q14.2
Ext. ∠OAE = ∠AOC + ∠ACO
⇒ x = 40° + 80° = 120°
Similarly,
Ext. ∠DBF = ∠ODB + ∠DOB
y = 70° + ∠DOB
[(∵ ∠AOC = ∠DOB) (vertically opp. angles)]
= 70° + 40° = 110°
∴ x+y= 120°+ 110° = 230° (b)

Question 15.
If the measures of angles of a triangle are in the ratio of 3 : 4 : 5, what is the measure of the smallest angle of the triangle?
(a) 25°
(b) 30°
(c) 45°
(d) 60°
Solution:
Ratio in the measures of the triangle =3:4:5
Sum of angles of a triangle = 180°
Let angles be 3x, 4x, 5x
Sum of angles = 3x + 4x + 5x = 12x
∴ Smallest angle = \(\frac { 180 x 3x }{ 12x }\) = 45° (c)

Question 16.
In the figure, if AB ⊥ BC, then x =
(a) 18
(b) 22
(c) 25
(d) 32
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q16.1
Solution:
In the figure, AB ⊥ BC
∠AGF = 32°
∴ ∠CGB = ∠AGF (Vertically opposite angles)
= 32°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q16.2
In ∆GCB, ∠B = 90°
∴ ∠CGB + ∠GCB = 90°
⇒ 32° + ∠GCB = 90°
⇒ ∠GCB = 90° – 32° = 58°
Now in ∆GDC,
Ext. ∠GCB = ∠CDG + ∠DGC
⇒ 58° = x + 14° + x
⇒ 2x + 14° = 58°
⇒ 2x = 58 – 14° = 44
⇒ x = \(\frac { 44 }{ 2 }\) = 22°
∴ x = 22° (b)

Question 17.
In the figure, what is ∠ in terms of x and y?
(a) x + y + 180
(b) x + y – 180
(c) 180° -(x+y)
(d) x+y + 360°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q17.1
Solution:
In the figure, BC is produced both sides CA and BA are also produced
In ∆ABC,
∠B = 180° -y
and ∠C 180° – x
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q17.2
∴ z = ∠A = 180° – (B + C)
= 180° – (180 – y + 180 -x)
= 180° – (360° – x – y)
= 180° – 360° + x + y = x + y – 180° (b)

Question 18.
In the figure, for which value of x is l1 || l2?
(a) 37
(b) 43
(c) 45
(d) 47
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q18.1
Solution:
In the figure, l1 || l2
∴ ∠EBA = ∠BAH (Alternate angles)
∴ ∠BAH = 78°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q18.2
⇒ ∠BAC + ∠CAH = 78°
⇒ ∠BAC + 35° = 78°
⇒ ∠BAC = 78° – 35° = 43°
In ∆ABC, ∠C = 90°
∴ ∠ABC + ∠BAC = 90°
⇒ x + 43° = 90° ⇒ x = 90° – 43°
∴ x = 47° (d)

Question 19.
In the figure, what is y in terms of x?
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q19.1
Solution:
In ∆ABC,
∠ACB = 180° – (x + 2x)
= 180° – 3x …(i)
and in ∆BDG,
∠BED = 180° – (2x + y) …(ii)
∠EGC = ∠AGD (Vertically opposite angles)
= 3y
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q19.2
In quad. BCGE,
∠B + ∠ACB + ∠CGE + ∠BED = 360° (Sum of angles of a quadrilateral)
⇒ 2x+ 180° – 3x + 3y + 180°- 2x-y = 360°
⇒ -3x + 2y = 0
⇒ 3x = 2y ⇒ y = \(\frac { 3 }{ 2 }\)x (a)

Question 20.
In the figure, what is the value of x?
(a) 35
(b) 45
(c) 50
(d) 60
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q20.1
Solution:
In the figure, side AB is produced to D
∴ ∠CBA + ∠CBD = 180° (Linear pair)
⇒ 7y + 5y = 180°
⇒ 12y = 180°
⇒ y = \(\frac { 180 }{ 12 }\) = 15
and Ext. ∠CBD = ∠A + ∠C
⇒ 7y = 3y + x
⇒ 7y -3y = x
⇒ 4y = x
∴ x = 4 x 15 = 60 (d)

Question 21.
In the figure, the value of x is
(a) 65°
(b) 80°
(c) 95°
(d) 120°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q21.1
Solution:
In the figure, ∠A = 55°, ∠D = 25° and ∠C = 40°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q21.2
Now in ∆ABD,
Ext. ∠DBC = ∠A + ∠D
= 55° + 25° = 80°
Similarly, in ∆BCE,
Ext. ∠DEC = ∠EBC + ∠ECB
= 80° + 40° = 120° (d)

Question 22.
In the figure, if BP || CQ and AC = BC, then the measure of x is
(a) 20°
(b) 25°
(c) 30°
(d) 35°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q22.1
Solution:
In the figure, AC = BC, BP || CQ
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q22.2
∵ BP || CQ
∴ ∠PBC – ∠QCD
⇒ 20° + ∠ABC = 70°
⇒ ∠ABC = 70° – 20° = 50°
∵ BC = AC
∴ ∠ACB = ∠ABC (Angles opposite to equal sides)
= 50°
Now in ∆ABC,
Ext. ∠ACD = ∠B + ∠A
⇒ x + 70° = 50° + 50°
⇒ x + 70° = 100°
∴ x = 100° – 70° = 30° (c)

Question 23.
In the figure, AB and CD are parallel lines and transversal EF intersects them at P and Q respectively. If ∠APR = 25°, ∠RQC = 30° and ∠CQF = 65°, then
(a) x = 55°, y = 40°
(b) x = 50°, y = 45°
(c) x = 60°, y = 35°
(d) x = 35°, y = 60°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q23.1
Solution:
In the figure,
∵ AB || CD, EF intersects them at P and Q respectively,
∠APR = 25°, ∠RQC = 30°, ∠CQF = 65°
∵ AB || CD
∴ ∠APQ = ∠CQF (Corresponding anlges)
⇒ y + 25° = 65°
⇒ y = 65° – 25° = 40°
and APQ + PQC = 180° (Co-interior angles)
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q23.2
y + 25° + ∠1 +30°= 180°
40° + 25° + ∠1 + 30° = 180°
⇒ ∠1 + 95° = 180°
∴ ∠1 = 180° – 95° = 85°
Now, ∆PQR,
∠RPQ + ∠PQR + ∠PRQ = 180° (Sum of angles of a triangle)
⇒ 40° + x + 85° = 180°
⇒ 125° + x = 180°
⇒ x = 180° – 125° = 55°
∴ x = 55°, y = 40° (a)

Question 24.
The base BC of triangle ABC is produced both ways and the measure of exterior angles formed are 94° and 126°. Then, ∠BAC = ?
(a) 94°
(b) 54°
(c) 40°
(d) 44°
Solution:
In ∆ABC, base BC is produced both ways and ∠ACD = 94°, ∠ABE = 126°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q24.1
Ext. ∠ACD = ∠BAC + ∠ABC
⇒ 94° = ∠BAC + ∠ABC
Similarly, ∠ABE = ∠BAC + ∠ACB
⇒ 126° = ∠BAC + ∠ACB
Adding,
94° + 126° = ∠BAC + ∠ABC + ∠ACB + ∠BAC
220° = 180° + ∠BAC
∴ ∠BAC = 220° -180° = 40° (c)

Question 25.
If the bisectors of the acute angles of a right triangle meet at O, then the angle at O between the two bisectors is
(a) 45°
(b) 95°
(c) 135°
(d) 90°
Solution:
In right ∆ABC, ∠A = 90°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q25.1
Bisectors of ∠B and ∠C meet at O, then 1
∠BOC = 90° + \(\frac { 1 }{ 2 }\) ∠A
= 90°+ \(\frac { 1 }{ 2 }\) x 90° = 90° + 45°= 135° (c)

Question 26.
The bisects of exterior angles at B and C of ∆ABC, meet at O. If ∠A = .x°, then ∠BOC=
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q26.1
Solution:
In ∆ABC, ∠A = x°
and bisectors of ∠B and ∠C meet at O.
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q26.2

Question 27.
In a ∆ABC, ∠A = 50° and BC is produced to a point D. If the bisectors of ∠ABC and ∠ACD meet at E, then ∠E =
(a) 25°
(b) 50°
(c) 100°
(d) 75°
Solution:
In ∆ABC, ∠A = 50°
BC is produced
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q27.1
Bisectors of ∠ABC and ∠ACD meet at ∠E
∴ ∠E = \(\frac { 1 }{ 2 }\) ∠A = \(\frac { 1 }{ 2 }\) x 50° = 25° (a)

Question 28.
The side BC of AABC is produced to a point D. The bisector of ∠A meets side BC in L. If ∠ABC = 30° and ∠ACD =115°,then ∠ALC =
(a) 85°
(b) 72\(\frac { 1 }{ 2 }\) °
(c) 145°
(d) none of these
Solution:
In ∆ABC, BC is produced to D
∠B = 30°, ∠ACD = 115°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q28.1
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q28.2

Question 29.
In the figure , if l1 || l2, the value of x is
(a) 22 \(\frac { 1 }{ 2 }\)
(b) 30
(c) 45
(d) 60
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q29.1
Solution:
In the figure, l1 || l2
EC, EB are the bisectors of ∠DCB and ∠CBA respectively EF is the bisector of ∠GEB
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q29.2
∵ EC and EB are the bisectors of ∠DCB and ∠CBA respectively
∴ ∠CEB = 90°
∴ a + b = 90° ,
and ∠GEB = 90° (∵ ∠CEB = 90°)
2x = 90° ⇒ x = \(\frac { 90 }{ 2 }\) = 45 (c)

Question 30.
In ∆RST (in the figure), what is the value of x?
(a) 40°
(b) 90°
(c) 80°
(d) 100°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q30.1
Solution:
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q30.2

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

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RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1

Other Exercises

Question 1.
In the figure, the sides BA and CA have been produced such that BA = AD and CA = AE. Prove the segment DE || BC.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q1.1
Solution:
Given : Sides BA and CA of ∆ABC are produced such that BA = AD are CA = AE. ED is joined.
To prove : DE || BC
Proof: In ∆ABC and ∆DAE AB=AD (Given)
AC = AE (Given)
∠BAC = ∠DAE (Vertically opposite angles)
∴ ∆ABC ≅ ∆DAE (SAS axiom)
∴ ∠ABC = ∠ADE (c.p.c.t.)
But there are alternate angles
∴ DE || BC

Question 2.
In a ∆PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.
Solution:
Given : In ∆PQR, PQ = QR
L, M and N are the mid points of the sides PQ, QR and PR respectively
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q2.1
To prove : LM = MN
Proof : In ∆LPN and ∆MRH
PN = RN (∵ M is mid point of PR)
LP = MR (Half of equal sides)
∠P = ∠R (Angles opposite to equal sides)
∴ ALPN ≅ AMRH (SAS axiom)
∴ LN = MN (c.p.c.t.)

Question 3.
Prove that the medians of an equilateral triangle are equal.
Solution:
Given : In ∆ABC, AD, BE and CF are the medians of triangle and AB = BC = CA
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q3.1
To prove : AD = BE = CF
Proof : In ∆BCE and ∆BCF,
BC = BC (Common side)
CE = BF (Half of equal sides)
∠C = ∠B (Angles opposite to equal sides)
∴ ABCE ≅ ABCF (SAS axiom)
∴ BE = CF (c.p.c.t.) …(i)
Similarly, we can prove that
∴ ∆CAD ≅ ∆CAF
∴ AD = CF …(ii)
From (i) and (ii)
BE = CF = AD
⇒ AD = BE = CF

Question 4.
In a ∆ABC, if ∠A = 120° and AB = AC. Find ∠B and ∠C.
Solution:
In ∆ABC, ∠A = 120° and AB = AC
∴ ∠B = ∠C (Angles opposite to equal sides)
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q4.1
⇒ 120° + ∠B + ∠B = 180°
⇒ 2∠B = 180° – 120° = 60°
∴ ∠B = \(\frac { { 60 }^{ \circ } }{ 2 }\) = 30°
and ∠C = ∠B = 30°
Hence ∠B = 30° and ∠C = 30°

Question 5.
In a ∆ABC, if AB = AC and ∠B = 70°, find ∠A.
Solution:
In ∆ABC, ∠B = 70°
AB =AC
∴ ∠B = ∠C (Angles opposite to equal sides)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q5.1
But ∠B = 70°
∴ ∠C = 70°
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ ∠A + 70° + 70° = 180°
⇒ ∠A + 140°= 180°
∴∠A = 180°- 140° = 40°

Question 6.
The vertical angle of an isosceles triangle is 100°. Find its base angles.
Solution:
In ∆ABC, AB = AC and ∠A = 100°
But AB = AC (In isosceles triangle)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q6.1
∴ ∠C = ∠B (Angles opposite to equal sides)
∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ 100° + ∠B + ∠B = 180° (∵ ∠C = ∠B)
⇒ 2∠B = 180° – 100° = 80°
∴ ∠C = ∠B = 40°
Hence ∠B = 40°, ∠C = 40°

Question 7.
In the figure, AB = AC and ∠ACD = 105°, find ∠BAC.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q7.1 Solution:
In ∆ABC, AB = AC
∴ ∠B = ∠C (Angles opposite to equal sides)
But ∠ACB + ∠ACD = 180° (Linear pair)
⇒ ∠ACB + 105°= 180°
⇒ ∠ACB = 180°-105° = 75°
∴ ∠ABC = ∠ACB = 75°
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ ∠A + 75° + 75° = 180°
⇒ ∠A + 150°= 180°
⇒ ∠A= 180°- 150° = 30°
∴ ∠BAC = 30°

Question 8.
Find the measure of each exterior angle of an equilateral triangle.
Solution:
In an equilateral triangle, each interior angle is 60°
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q8.1
But interior angle + exterior angle at each vertex = 180°
∴ Each exterior angle = 180° – 60° = 120°

Question 9.
If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.
Solution:
Given : In an isosceles ∆ABC, AB = AC
and base BC is produced both ways
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q9.1
To prove : ∠ACD = ∠ABE
Proof: In ∆ABC,
∵ AB = AC
∴∠C = ∠B (Angles opposite to equal sides)
⇒ ∠ACB = ∠ABC
But ∠ACD + ∠ACB = 180° (Linear pair)
and ∠ABE + ∠ABC = 180°
∴ ∠ACD + ∠ACB = ∠ABE + ∠ABC
But ∠ACB = ∠ABC (Proved)
∴ ∠ACD = ∠ABE
Hence proved.

Question 10.
In the figure, AB = AC and DB = DC, find the ratio ∠ABD : ∠ACD.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q10.1
Solution:
In the given figure,
In ∆ABC,
AB = AC and DB = DC
In ∆ABC,
∵ AB = AC
∴ ∠ACD = ∠ABE …(i) (Angles opposite to equal sides)
Similarly, in ∆DBC,
DB = DC
∴ ∠DCB = ∠DBC .. (ii)
Subtracting (ii) from (i)
∠ACB – ∠DCB = ∠ABC – ∠DBC
⇒ ∠ACD = ∠ABD
∴ Ratio ∠ABD : ∠ACD = 1 : 1

Question 11.
Determine the measure of each of the equal angles of a rightangled isosceles triangle.
OR
ABC is a rightangled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Solution:
Given : In a right angled isosceles ∆ABC, ∠A = 90° and AB = AC
To determine, each equal angle of the triangle

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q11.1
∵ ∠A = 90°
∴ ∠B + ∠C = 90°
But ∠B = ∠C
∴ ∠B + ∠B = 90°
⇒ 2∠B = 90°
90°
⇒ ∠B = \(\frac { { 90 }^{ \circ } }{ 2 }\)  = 45°
and ∠C = ∠B = 45°
Hence ∠B = ∠C = 45°

Question 12.
In the figure, PQRS is a square and SRT is an equilateral triangle. Prove that
(i) PT = QT
(ii) ∠TQR = 15°
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q12.1
Solution:
Given : PQRS is a square and SRT is an equilateral triangle. PT and QT are joined.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q12.2
To prove : (i) PT = QT; (ii) ∠TQR = 15°
Proof : In ∆TSP and ∆TQR
ST = RT (Sides of equilateral triangle)
SP = PQ (Sides of square)
and ∠TSP = ∠TRQ (Each = 60° + 90°)
∴ ∆TSP ≅ ∆TQR (SAS axiom)
∴ PT = QT (c.p.c.t.)
In ∆TQR,
∵ RT = RQ (Square sides)
∠RTQ = ∠RQT
But ∠TRQ = 60° + 90° = 150°
∴ ∠RTQ + ∠RQT = 180° – 150° = 30°
∵ ∠PTQ = ∠RQT (Proved)
∠RQT = \(\frac { { 30 }^{ \circ } }{ 2 }\)  = 15°
⇒ ∠TQR = 15°

Question 13.
AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the ponits A and B (see figure). Show that the line PQ is perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q13.1
Solution:
Given : AB is a line segment.
P and Q are points such that they are equidistant from A and B
i.e. PA = PB and QA = QB AP, PB, QA, QB, PQ are joined
To prove : PQ is perpendicular bisector of AB
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q13.2
Proof : In ∆PAQ and ∆PBQ,
PA = PB (Given)
QA = QB (Given)
PQ = PQ (Common)
∴ ∆PAQ ≅ ∆PBQ (SSS axiom)
∴ ∠APQ = ∠BPQ (c.p.c.t.)
Now in ∆APC = ∆BPC
PA = PB (Given)
∆APC ≅ ∆BPC (Proved)
PC = PC (Common)
∴ ∆APC = ∆BPC (SAS axiom)
∴ AC = BC (c.p.c.t.)
and ∠PCA = ∠PCB (c.p.c.t.)
But ∠PCA + ∠PCB = 180° (Linear pair)
∴ ∠PCA = ∠PCB = 90°
∴ PC or PQ is perpendicular bisector of AB

 

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 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.

HOTS Questions for Class 9 Science Chapter 15 Improvement in Food Resources

HOTS Questions for Class 9 Science Chapter 15 Improvement in Food Resources

These Solutions are part of HOTS Questions for Class 9 Science. Here we have given HOTS Questions for Class 9 Science Chapter 15 Improvement in Food Resources

Question 1.
Identify the plants and animals shown in the figures.
HOTS Questions for Class 9 Science Chapter 15 Improvement in Food Resources image - 2
To which category each one belongs ?
Answer:
A. White Leghorn …………….. exotic bread of poultry, acclimatised to India, very good layer and moderately good broiler.
B. jersey ……………… exotic cattle, very high milk yielding
C. Gundhi Bug …………….. Insect pest of Paddy.
D. Wild Oat ………………. Rabi season narrow leaved monocot weed.

More Resources

Question 2.
Name (a) A milch animal with maximum fat content in its milk.
HOTS Questions for Class 9 Science Chapter 15 Improvement in Food Resources image - 1
(b) Kharif season narrow leaved and broad leaved weeds.
(c) Cattle disease transferable to human beings.
(d) Bird used in cock fighting.
Answer:
(a) Surti breed of buffalo with fat content of 8-12% in its milk.
(b)

  1. Narrow Leaved Monocot Weed ……………….. Nutgrass (Cyperus/ Cyperinus rotundus).
  2. Broad Leaved Dicot Weed ………………… Amaranthus.

(c) Anthrax, amoebiasis, tuberculosis.
(d) Aseel ……………… local (indigenous) breed of poultry.

Question 3.
What is royal jelly ? How does it differ from bee bread ?
Answer:
Royal jelly is a protein rich food secreted by hypophyseal glands of young workers for feeding the queen.
Bee bread is prepared by adult workers for all other members of the colony. Bee bread is a mixture of honey and pollen.

Hope given HOTS Questions for Class 9 Science Chapter 15 Improvement in Food Resources are helpful to complete your science homework.

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RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1

RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1

Other Exercises

Question 1.
Three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angle.
Solution:
Sum of four angles of a quadrilateral = 360°
Three angles are = 110°, 50° and 40°
∴ Fourth angle = 360° – (110° + 50° + 40°)
= 360° – 200° = 160°

Question 2.
In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1 : 2 : 4 : 5. Find the measures of each angle of the quadrilateral.
Solution:
Sum of angles of a quadrilateral ABCD = 360°
Ratio in angles = 1 : 2 : 4 : 5
Let first angle = x
Second angle = 2x
Third angle = 4x
and fourth angle = 5x
∴ x + 2x + 4x + 5x = 360°
⇒ 12x = 360° ⇒ x = \(\frac { { 360 }^{ \circ } }{ 12 }\)  = 30°
∴ First angle = 30°
Second angle = 30° x 2 = 60°
Third angle = 30° x 4 = 120°
Fourth angle = 30° x 5 = 150°

Question 3.
The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. [NCERT]
Solution:
Sum of four angles of a quadrilateral = 360°
Ratio in the angles = 3 : 5 : 9 : 13
Let first angle = 3x
Then second angle = 5x
Third angle = 9x
and fourth angle = 13x
∴ 3x + 5x + 9x+ 13x = 360°
⇒ 30x = 360°
⇒ x = \(\frac { { 360 }^{ \circ } }{ 30 }\) = 12°
∴ First angle = 3x = 3 x 12° = 36°
Second angle = 5x = 5 x 12° = 60°
Third angle = 9x = 9 x 12° = 108°
Fourth angle = 13 x 12° = 156°

Question 4.
In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively.
Prove that ∠COD = \(\frac { 1 }{ 2 }\) (∠A + ∠B).
Solution:
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1 Q4.1

RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1 Q4.2

 

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NCERT Exemplar Solutions for Class 9 Science Chapter 15 Improvement in Food Resources

NCERT Exemplar Solutions for Class 9 Science Chapter 15 Improvement in Food Resources

These Solutions are part of NCERT Exemplar Solutions for Class 9 Science . Here we have given NCERT Exemplar Solutions for Class 9 Science Chapter 15 Improvement in Food Resources

Question 1.
Match the items of column A with those of column B.
NCERT Exemplar Solutions for Class 9 Science Chapter 15 Improvement in Food Resources image - 1
Answer:
a — ii,
b — iii,
c — i,
d — iv.

More Resources

Question 2.
Fill in the blanks
(a) Pigeon Pea is a good source of ………… .
(b) Berseem is an important …………. crop.
(c) The crops which are grown in rainy season are called ………….. crops.
(d) ………… are rich in vitamins.
(e) ………… crop grows in winter seasons.
Answer:
(a) protein
(b) fodder
(c) kharif
(d) Vegetables
(e) Rabi.

Question 3.
What is GM crop ? Name any one crop which is grown in India.
Answer:
GM or genetically modified crop is the one which has been developed through introduction of some specific genes from other sources, e.g, insect resistant B Cotton (being grown in India), vitamin A rich Golden Rice.

Question 4.
List out some useful traits in improved crop.
Answer:

  1. Higher yield
  2. Improved quality
  3. Resistance to biotic and abiotic stresses
  4. Change in maturity
  5. Wider adaptability
  6. Desirable agronomic traits.

Question 5.
Why is organic matter important for crop production ?
Answer:
Organic matter forms humus. It is important for crop production because it makes the soil fertile.

  1. Organic matter improves soil structure by forming soil crumbs.
  2. It increases water holding capacity of sandy soils.
  3. It improves aeration of clayey soils.
  4. During its decomposition, it liberates minerals which enrich the soil.
  5. Biochemicals present in decaying organic matter improve growth of crop plants.

Question 6.
Why is use of excess fertilizer detrimental for environment ?
Answer:
Use of excess fertilisers will cause :

  1. Mineral loading of underground water.
  2. Excess minerals in the crop plants.
  3. Salination of soil.
  4. Run off from fertilizers rich soil, will cause eutrophication of water bodies.

Question 7.
Give one word for the following :
(a) Farming without the use of chemicals as fertilizers, herbicides and pesticides is known as ……….. .
(b) Growing of Wheat and Groundnut on the same field is called ………… .
(c) Planting of Soyabean and Maize in alternate rows in the same field is called as ………… .
(d) Growing different crops on a piece of land in pre-planned succession is known as ………….. .
(e) Xanthium and Parthenium are commonly known as …………. .
(f) Causal organism of any disease is called as …………. .
Answer:
(a) Organic farming
(b) mixed cropping
(c) Inter-cropping
(d) crop rotation
(e) weeds
(f) pathogen.

Question 8.
Match the columns A and B.
NCERT Exemplar Solutions for Class 9 Science Chapter 15 Improvement in Food Resources image - 2
Answer:
a — iii,
b — v,
c — iv,
d — i,
e — ii.

Question 9.
If there is low rainfall in a village throughout the year, what measures will you suggest to the farmers for better cropping.
Answer:

  1. Reduced tilling
  2. Enrich soil with humus which increases its water holding capacity
  3. Use of drought resistant and early maturing varieties of crops.

Question 10.
Group the following and tabulate them as energy yielding, protein yielding, oil yielding and fodder crop :
Wheat, Rice, Berseem, Maize, Gram, Oat, Pigeon Gram, Sudan Grass, Lentil, Soyabean, Groundnut, Castor and Mustard.
Answer:

  1. Wheat, Rice, Maize, Oat Energy yielding.
  2. Gram, Pigeon Gram, Lentil, Soyabean Protein yielding.
  3. Ground nut, Castor, Mustard, Soyabean Oil yeilding.
  4. Berseem, Oat, Sudan Grass Fodder crops.

Question 11.
Define the terms hybridisation and photoperiod.
Answer:
Hybridisation. It is crossing of two (or more) types of individuals with different useful traits in order to bring them together in the progeny.
Photoperiod. It is duration of day light that influences plants and other organisms in their growth, reproduction and maturation.

Question 12.
Fill in the blanks :
(a) Photoperiod affect the …………. .
(b) Kharif crops are cultivated from ……….. to ……….. .
(c) Rabi crops are cultivated from ………… to ……….. .
(d) Paddy, Maize, Green Gram and Black Gram are ……….. crops.
(e) Wheat, Gram, Pea and Mustard are …………. crops.
Answer:
(a) flowering (of plants)
(b) June, October
(c) November, April
(d) kharif
(e) rabi.

Question 13.
Cultivation practices and crop yield are related to environmental conditions. Explain.
Answer:
All crops do not grow under similar conditions. Some require high temperature, some low temperature, longer duration of sunlight, shorter duration of sunlight, more humidity, low humidity, moderate humidity, loam soil, sandy soil, etc. Apple cannot be grown in plains because it requires several days of low temperature. In plains, there are two major seasons of crop
plants kharif (rainy season) and rabi (winter season).

Question 14.
Fill in the blanks :
(a) A total of …………. nutrients are essential to plants.
(b) ……… and ………. are supplied by air to plants.
(c) ………. is supplied by water to plants.
(d) Soil supplies ………. nutrients to plants.
(e) ……….. nutrients are required in large quantity and called as
(f) ……….. nutrients are needed in small quantity for plants and are called ………… .
Answer:
(a) 16
(b) Carbon, oxygen
(c) Hydrogen
(d) 13
(e) six, macronutrients
(f) seven, micronutrients.

Question 15.
Differentiate between compost and vermicompost.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 15 Improvement in Food Resources image - 3

Question 16.
Arrange these statements in correct sequence of preparation of green manure.
(a) Green plants are decomposed in soil.
(b) Green plants are cultivated for preparing manure or crop plant parts are used.
(c) Plants are ploughed and mixed in soil.
(d) After decomposition, it becomes green manure.
Answer:
(b) > (c) > (a) > (d).

Question 17.
An Italian bee variety A melliftra has been introduced in India for honey production. Write about its merits over other varieties.
Answer:
There are three common species of honey bees in our country used for commercial production of honey. These are

  1. Apis cerana indica — Indian Bee
  2. Apis dorsata — Rock Bee
  3. Apis florae {A. florea) — Little Bee

To produce honey at commercial level, the Italian variety of honey bee {Apis mellifera) is commonly used. This variety is domesticated in our country. It is preferred because

  1. It is gentle in nature
  2. It has good honey collection capacity
  3. It has the ability to protect itself from enemies and
  4. Less swarming.

Question 18.
In agricultural practices, higher input gives higher yield. Discuss how ?
Answer:
In agriculture, higher yield can be obtained only by employing higher yielding varieties, improved farming practices, modern technology, latest agricultural machines and implements, nutrient supply, etc. All these require high cost and knowledge of new techniques and improvements. Therefore, a farmer’s purchasing capacity for inputs determines the cropping system and production practices.

Question 19.
Discuss the role of hybridisation in crop improvement.
Answer:

  1. Choice of Parents: Two older varieities of crop having different desirable characteristics, are selected. For example, if we want to obtain a variety having higher yield as well as disease resistance, we should select two existing varieties of crops, one having higher yield and the other having more resistance to diseases.
  2. Cross-breeding: The Two Parents. Pollen grains of plants of one variety are dusted over the stigmas of plants of the other variety and vice-versa. It produces a new variety which has good chacateristics of both the parents. The process of crossing plants of two varieties having different traits to produce a hybrid having good traits of both is called The crossing may be intervarietal (between different varieties), intergeneric (between different genera) or interspecific (between different species of the same genus). The most common type of breeding is intervarietal.

Question 20.
Define

  1. Vermicompost
  2. Green Manure
  3. Biofertilizers.

Answer:

  1. Vermicompost: It is a manure rich in pulverised organic matter and worm castings. Vermicompost is formed by the activity of earth worms on organic remains.
  2. Green Manure: It is manure formed inside soil from young green crop plants ploughed hack into soil. The green manure crops are generally quick growing legume crops which are mulched by ploughing them back into field in tender stage (about 6 — 8 weeks) only, generally at the time of flowering. These crops are completely decomposed in about 1—2 months when the next crop can be sown.
  3. Biofertilizers: Living organisms like nitrogen fixing bacteria, nitrogen fixing blue-green algae and mineral solubilising bacteria that supply nutrients to crop plants are called biofertilizers.

Question 21.
Discuss various methods of weed control.
Answer:
Mechanical Methods :

  1. Removal of weeds by pulling them out with hand.
  2. Removal of weeds by using trowel (khurpa).
  3. Removal of weeds by ploughing, burning and flooding before sowing.
  4. Removal of weeds by interculture.

Cultural Methods :

  1. Timely sowing of crops.
  2. Proper seed bed preparation
  3. Intercropping
  4. Crop rotation

Chemical Methods :
Destroying the weeds by spraying special chemicals called weedicides like, 2, 4—D (2, 4-dichlorophenoxy acetic acid), Butachlor, Atrazine, isoproturon, etc.
Biological Methods :
In the biological control of weeds, some appropriate insects or some other organisms are deliberately put into the crop field having weeds. The insects or other organisms selectively destroy the weed plants but do not harm the crop plants. For example, Cochineal insects are used to eradicate the weed called Opuntia. Mexican Beetle and Cassia can check the growth of Parthenium. Aquatic weeds are controlled by the fish called grass carp.

Question 22.
Differentiate between the following :
(i) Capture fishery and culture fishery (CCE 2011, 2012)
(ii) Mixed cropping and inter cropping
(iii) Bee keeping and poultry farming.
Answer:
(i) Differences between Capture Fishery and Culture Fishery
NCERT Exemplar Solutions for Class 9 Science Chapter 15 Improvement in Food Resources image - 4
(ii) Differences between Mixed Cropping and Intercropping. 

Mixed cropping Intercropping
Seeds: Seeds of different crops are mixed before sowing. Seeds of different crops are not mixed. They are sown separately.
Pattern: There is no pattern of sowing. The different crops are sown in separate rows or strips.
Target: It minimises the risk of crop failure. It increases crop productivity per unit area.
Fertilizers: Only a common type of fertilizer can be added. Specific fertilizers can be provided to each crop.
Pesticides: Crop sepecific pesticides cannot be sprayed. Crop specific pesticides can be sprayed without difficulty.
Harvesting: Harvesting of early maturing crop provides a lot of difficulty. There is little difficulty in harvesting individual crops.
Produce: There is some mixing of the produce of different crops. There is no mixing of produce of different crops.
Inputs: Lesser inputs of irrigation and nutrients are required. Requirement of inputs is comparatively more.
Ex. Barley/wheat and gram/mustard. Ex. Soya bean and Maize, Finger Millet and Cow Pea.

(iii) Differences between Bee Keeping and Poultry Farming
NCERT Exemplar Solutions for Class 9 Science Chapter 15 Improvement in Food Resources image - 5

Question 23.
Give merits and demerits of fish culture.
Answer:
Merits:

  1. Desired Fish: Economically important desired fishes are made available.
  2. Smaller Area: A large number of fishes are raised in a small area.
  3. Fish Seeding: Fishes are made to breed in different seasons.
  4. Mortality: There is little mortality in the younger stages of the fishes.
  5. Improvement: Through selective hybridisation, yield and quality of fishes are improved.

Demerits:

  1. Genetic Loss: Only some selected high yielding and economically important breeds are reared. Other breeds and varieties are lost.
  2. Threat to Biodiversity: Other forms of aquatic organisms are being ignored. Even natural waters are being seeded with economically important fishes so that natural biodiversity has come under threat.

Question 24.
What do you understand by composite fish culture ?
Answer:
Composite culture is highly advantageous:

  1. These fishes do not compete with one another for food because they have different types of food habits.
  2. The food available in all the zones (parts) of the pond is utilized due to their food habits.
  3. The carps consume only natural foods.
  4. Six different types of fish are reared in the same pond.
  5. The yield is very high.

Question 25.
Why bee keeping should be done in good pasturage ?
Answer:
Pasturage/Crop/Flora: Quality, taste and quantity of honey depend upon the flora or pasturage available for nectar and pollen collection. Pasturage is different in different geographical regions and locations, plains and mountains. The best honey is almond honey of Kashmir.

Question 26.
Write the modes by which insects affect the crop yield.
Answer:
The important types of crops are :

  1. Cereal Crops (Grain Crops): Wheat, rice, maize, (major cereals), barley, sorghum, minor millets (minor cereals). They are rich source of carbohydrates that meet the energy requirement of body. Other useful ingredients are dietary fibres, proteins, vitamins and minerals.
  2. Pulses (Legume Seeds): Gram (chana), pea (matar), black gram (urad), green gram (moong), pigeon pea (arhar), lentil (masoor). Pulses are rich in proteins.
  3. Oil Seed Crops: Groundnut (peanut), mustard, linseed, sunflower, sesame, soyabean, niger, castor[1]. They are a source of oils, fats and fatty acids.                                                                                                 .
  4. Fruit and Vegetable Crops: Orange, apple, mango, banana, pineapple, cabbage, cauliflower, brinjal, okra, carrot, beans, potato. They are a source of minerals, vitamins, roughage and small quantities of protein, carbohydrates and oil.
  5. Chilly, black pepper, ginger, turmeric. They are used for enhancing palatability of foods.
  6. Other Crops: Crop plants also yield fibres (e.g., cotton), sugar (e.g., sugarcane, sugar beet), tobacco, tea, etc.
  7. Fodder Crops: Berseem, oat, Sudan grass, sorghum. They provide green fodder to cattle.
  8. Castor oil is not edible oil. It is mainly used as a lubricant, purgative, phenyls, hair fixers, etc.

Question 27.
Discuss why pesticides are used in very accurate concentration and in very appropriate manner ?
Answer:
Pesticides should not be used indiscriminately. Only very accurate dose should be applied at appropriate time. Even slight excess of the pesticide is extremely harmful.

  1. Pesticides are harmful to soil biota. Therefore, they reduce soil fertility.
  2. They pass into ground water and make it toxic.
  3. They enter the crop plants and make them toxic.
  4. Pesticides pass into surface water through run off from sprayed fields and harm the aquatic biota.

Question 28.
Name two types of animal feed and write their functions.
Answer:

  1. The sick animal looks tired. It becomes inactive and tries to remain isolated.
  2. There is little interest in taking food.
  3. The animal moves slowly, limping well behind others in the herd.
  4. The milk yield, egg laying capacity or working capacity of the animal is reduced drastically.
  5. The animal passes loose or watery dung.
  6. Urine may become coloured.
  7. Eyes may turn red.
  8. Ears may droop.
  9. There may be wheezy breathing, coughing or sneezing.
  10. Excessive salivation.
  11. Watery nose.
  12. The animal may run temperature and shiver.

Question 29.
What would happen if poultry birds are larger in size and have no summer adaptation capacity ? In order to get small sized poultry birds having summer adaptability, what methods will be employed ?
Answer:
Larger sized birds require more feed. Summer adaptation is connected with egg laying. Little summer adaptation reduces egg laying. In order to obtain small sized poultry birds having summer adaptability, it is desirable to

  1. Either introduce the required exotic birds from outside and
  2. cross breed the local birds with exotic birds from outside. Small sized birds are preferred for
    1. Lower requirement of feed
    2. Higher egg laying capacity
    3. Lower requirement for space.

Question 30.
Suggest some preventive measures for the diseases of poultry birds.
Answer:

  1. The poultry birds should be kept in good spacious, airy and ventilated shelter.
  2. The shelter should be cleaned properly and regularly. Quick and hygienic disposal of excreta should be ensured.
  3. External parasites should be controlled, preferably by applying dilute insecticide solution. Disinfectant should be sprayed to kill mosquitoes and other external parasites.
  4. Every animal should be vaccinated to immunise it against common infections and diseases.

Question 31.
The figure shows two crop fields (plots A and B) have been treated by manures and chemical fertilizers respectively, keeping other environmental factors same.
NCERT Exemplar Solutions for Class 9 Science Chapter 15 Improvement in Food Resources image - 6
Observe the graph and answer the following questions :

  1. Why does plot B show sudden increase and then gradual decrease in yield ?
  2. Why is the highest peak in plot A graph slightly delayed ?
  3. What is the reason for the different pattern of the two graphs.

Answer:

  1. Sudden Increase: Chemical fertilizer supplies the minerals immediately in good quantity. Gradual Decrease. It is due to depletion of nutrients caused by absorption by plants, leaching to lower layers of the soil and killing of decomposer microbes.
  2. Manures decay slowly so that release of minerals is also delayed. Manures take time to mix up with the soil and form crumbs, that increase water holding and aeration of the soil.
  3. The difference in the two graphs indicates that manuring the soil is more beneficial than the use of chemical fertilizers. Rather, use of chemical fertilizers is harmful in the longer run.

Hope given NCERT Exemplar Solutions for Class 9 Science Chapter 15 Improvement in Food Resources are helpful to complete your science homework.

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RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2

Other Exercises

Question 1.
BD and CE are bisectors of ∠B and ∠C of an isosceles ∠ABC with AB = AC. Prove that BD = CE.
Solution:
Given : In ∆ABC, AB = AC
BD and CE are the bisectors of ∠B and ∠C respectively
To prove : BD = CE
Proof: In ∆ABC, AB = AC
∴ ∠B = ∠C (Angles opposite to equal sides)
∴ \(\frac { 1 }{ 2 }\) ∠B = \(\frac { 1 }{ 2 }\) ∠C
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2 Q1.1
∠DBC = ∠ECB
Now, in ∆DBC and ∆EBC,
BC = BC (Common)
∠C = ∠B (Equal angles)
∠DBC = ∠ECB (Proved)
∴ ∆DBC ≅ ∆EBC (ASA axiom)
∴ BD = CE

Question 2.
In the figure, it is given that RT = TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that: ∆RBT = ∆SAT.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2 Q2.1
Solution:
Given : In the figure, RT = TS
∠1 = 2∠2 and ∠4 = 2∠3
To prove : ∆RBT ≅ ∆SAT
Proof : ∵ ∠1 = ∠4 (Vertically opposite angles)
But ∠1 = 2∠2 and 4 = 2∠3
∴ 2∠2 = 2∠3 ⇒ ∠2 = ∠3
∵ RT = ST (Given)
∴∠R = ∠S (Angles opposite to equal sides)
∴ ∠R – ∠2 = ∠S – ∠3
⇒ ∠TRB = ∠AST
Now in ∆RBT and ∆SAT
∠TRB = ∠SAT (prove)
RT = ST (Given)
∠T = ∠T (Common)
∴ ∆RBT ≅ ∆SAT (SAS axiom)

Question 3.
Two lines AB and CD intersect at O such that BC is equal and parallel to AD. Prove that the lines AB and CD bisect at O.
Solution:
Given : Two lines AB and CD intersect each other at O such that AD = BC and AD \(\parallel\)
BC
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2 Q3.1
To prove : AB and CD bisect each other
i. e. AO = OB and CO = OD
Proof: In ∆AOD and ∆BOC,
AD = BC (Given)
∠A = ∠B (Alternate angles)
∠D = ∠C (Alternate angles)
∴ ∆AOD ≅ ∆BOC (ASA axiom)
AO = OB and AO = OC (c.p.c.t.)
Hence AB and CD bisect each other.

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2 are helpful to complete your math homework.

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NCERT Solutions for Class 9 Science Chapter 15 Improvement in Food Resources

NCERT Solutions for Class 9 Science Chapter 15 Improvement in Food Resources

These Solutions are part of NCERT Solutions for Class 9 Science. Here we have given NCERT Solutions for Class 9 Science Chapter 15 Improvement in Food Resources. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 9 Science (Biology) Chapter 15 – Improvement in Food Resources solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 15 – Improvement in Food Resources 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 do we get from cereals, pulses, fruits and vegetables ?
Answer:
Cereals: Mostly carbohydrates
Pulses: Mostly proteins
Fruits: Vitamins, minerals, organic acids.
Vegetables: Vitamins, minerals, small quantities of proteins, carbohydrates and oil.

Question 2.
How do biotic and abiotic factors adversely affect crop production ?
Answer:
Biotic factors are living organisms that reduce crop productivity due to either directly feeding on them
(e.g., insects, rodents) or causing diseases (e.g., nematodes, mycoplasmas, bacteria, viruses, fungi).
Abiotic factors are non-living components of environment that affect growth of crop plants like excess of water (water-logging), scarcity of water (drought), salinity, heat, cold or frost. Water logging reduces aeration of soil which is harmful to growth and functioning of roots. In drought, water is not available to meet the requirement of the plants for transpiration, growth and photosynthesis. Frost, cold and heat reduce or inhibit metabolic activities and are, therefore, harmful.

Question 3.
What are the desirable agronomic characteristics for crop improvement ? ( CCE 2015)
Answer:
They are different for different crops.
Cereals should be dwarf but with large ears. Dwarfness makes their stem stronger. They can withstand lodging effect of strong winds. Nutrient requirement is also less. Large ears produce more grains.
Legumes should have more pods which generally develop in relation to stem branching. Therefore, more branching and good foliage increase their productivity.
Fodder crops meant for feeding catde must have profuse branching, good foliage, juicy stems and large size.

Question 4.
What are macronutrients and why are they called macronutrients ?
Answer:
Macronutrients are essential elements required for growth and reproduction of plants. They are called macronutrients because they are required in larger quantities forming more than 1 mg per 1 gm of dry matter.

Question 5.
How do plants get nutrients ?
Answer:
Plants obtain nutrients from air, water and soil. Air is the source of carbon and oxygen. Hydrogen is obtained from water. The remaining thirteen elements are got directly from soil through root absorption.

Question 6.
Compare the use of manure and fertilizers in maintaining soil fertility.
Answer:

Manure Fertilizer
Nature. Manure is semidecomposed organic matter. Fertilizer is a chemical formulation.
Preparation. It is prepared from natural materials like plant residues and animal residues. It is synthetic being formed from chemical salts.
Mineral Content. It contains only a small quantity of mineral salts. Fertilizers contain pure mineral salts or their precursors.
Specificity. It is not nutrient specific. It is nutrient specific.
Organic Matter. It adds organic matter to the soil. There is no addition oforganic matter.
Quantity. It is required in large quantity. It is required in small quantity.
Nutrient Availability. Nutrient availability is moderate. Nutrients are released slowly. Fertilizer possesses readily available plant nutrients.
Transport. Manure is bulky. It is very difficult to transport it to longer distances. It has smaller bulk. Fertilizers can, therefore, be transported easily to long distances.
Storage. Manure cannot be stored for long. Fertilizers can be stored for long duration.
Soil. It helps in maintaining soil texture, its hydration and aeration. It can harm soil texture and other soil characters.
Excess. Excess manure is not much harmful. Excess fertilizer is harmful to plants. It also causes pollution.

Question 7.
Which of the following conditions will give the most benefits ? Why ?
(a) Farmers use high quality seed but do not adopt irrigation or use fertilizers.
(b) Farmers use ordinary seeds, adopt irrigation and use fertilizers.
(c) Farmers use quality seeds, adopt irrigation, use fertilizers and crop protection measures.
Answer:
Farmers are benefitted when they use quality seeds, irrigation, fertilizers and crop protection measures (choice c). Ordinary seeds cannot yield very high due to poor quality while quality seeds without necessary inputs yield low.

Question 8.
Why should preventive measures and biological control methods are preferred for protecting crops ?
Answer:
Preventive measures and biological control methods do not allow any measurable loss in quality and quantity of crops. They also do not cause any degradation of the environment. Cost is also very small. Preventive measures protect the crops from pests. Biological control methods eliminate the pests without harming crops and other human interests.

Question 9.
What factors may be responsible for huge losses during storage ?
Answer:

  1. Abiotic Factors: Excess moisture in grains, dampness and high temperature in storage place.
  2. Biotic Factors: Insects, mites, rodents, birds, fungi and bacteria.

Question 10.
Which method is commonly used for improving the cattle breeds and why ? (CCE 2011)
Answer:
Cross breeding indigenous breeds with exotic breeds. Foreign or exotic breeds have higher milk yield and longer lactation period as compared to indigenous breeds. Therefore, indigenous breeds should be cross¬bred with exotic breeds. The local breeds are hardy and resistant to several diseases. There are two methods of cross breeding — natural and artificial insemination. Artificial insemination is preferred as frozen semen can be transported, required in small quantity and protects the cows from contagious diseases.

Question 11.
Discuss the implications of the following statement “It is interesting to note that poultry is India s most efficient converter of low fibre food stuff (which is unfit for humant consumption) into highly nutritious animal protein food. ”
Answer:
India is basically agriculture society where a lot of wastes are produced during food processing, e.g., fish meal, meat meal, rice bran, etc. They are profitably used in forming poultry feed. In return poultry provides us with egg and meat rich in animal protein.

Question 12.
What management practices are common in dairy and poultry farming ? (CCE 2011, 2012, 2013, 2015)
Answer:

  1. Proper shelter, its hygiene, aeration and lighting.
  2. Proper feed and feed additives.
  3. Proper drinking water.
  4. Health care including vaccination.

Question 13.
What is the difference between broilers and layers and in their management ? (CCE 2011, 2012)
Answer:
Broilers are fast growing young chicken of 6-10 weeks age which are known for the good quality and taste of their meat.
Layers are sexually mature hens which are raised for egg laying.
Broilers are given diet rich in protein, with adequate fat, vitamins A and K. They are provided with best of space, hygiene and temperature. Layers are given inferior quality feed. Light is required for good egg laying. Temperature variations may occur to some degree.

Question 14.
How are fishes obtained ?
Answer:
There are two methods of obtaining fish, capture fishery (capturing fish) from natural waters and culture fishery in impounded waters. In both cases the fish are caught with the help of nets.

Question 15.
What are the advantages of composite fish culture ?
Answer:

  1. There is no competition for food or space amongst different types of fish.
  2. Food available in different parts of the pond are utilised due to their different food habits and different habitats.
  3. The fish yield is high as some six types of fish are growing simultaneously.

Question 16.
What are the desirable characters of bee varieties suitable for bee keeping ? (CCE 2012, 2015)
Answer:

  1. Gentleness in nature,
  2. Good honey collection.
  3. Prolific queen,
  4. Less swarming,
  5. Ability to protect itself from enemies, e.g. Italian Honey Bee (Apis mellifera).

Question 17.
What is pasturage and how it is related to honey production ? (CCE 2013)
Answer:
Pasturage is flora or crop available to honey bees for collection of nectar and pollen. Pollen is food for honey bees. Nectar is transformed into honey. The amount and quality of honey depend upon type and extent of pasturage.

NCERT CHAPTER END EXERCISES

Question 1.
Explain any one method of crop production which ensures high yield.
Answer:
Rotation of crops is a method of crop production which ensures high yield. Crop rotation is growing of different crops on the same piece of land in a preplanned succession. Crop rotation is done for one year, two year and three year cycle. Crops chosen are such that they withdraw nutrients from different layers of the soil. Crop rotation involving a leguminous crop ensures that the soil gets naturally enriched with nitrogen.
Crop rotation is useful in

  1. Weed control,
  2. Reduction in pest infestation,
  3. Elimination of soil borne diseases,
  4. Saving of nitrogen fertilizer
  5. Improving soil structure and fertility,
  6. Raising of 2-4 crops in a year from the same land giving higher returns to the farmer,
  7. Yield of individual crops is also higher due to improved soil structure, soil fertility, fewer weeds, fewer insects and diseases.

One Year Rotation    :  Rice—Wheat, Maize—Mustard
Two Year Rotation    :  Maize-Potato-Sugarcane-Pea
Three Year Rotation  :  Rice-Wheat-Mung-Mustard-Sugarcane-Burseem

Question 2.
Why are manures and fertilizers used in fields ? (CCE 2011)
Answer:
Manures and fertilizers are added to fields mainly to replenish minerals which get depleted due to withdrawal by crop plants and leaching down to lower strata of soil.

  1. Manures add small quantity of all minerals to the soil. They improve soil hydration, soil aeration and activity of soil micro-organisms, some of which are required for solubilisation of heavy minerals.
  2. Fertilizers are nutrient specific which contain one or more minerals in concentrated form. They meet the immediate and complete mineral requirement of high yielding varieties. However, they harm soil structure and cause pollution of crops, soil, ground water and nearby surface waters. A combination of both manure and fertilizer is highly useful.

Question 3.
What are advantages of intercropping and crop rotation ?
Answer:

Advantages of Intercropping :

  1. Intercropping increases productivity per unit area.
  2. It saves time and labour of the farmer and makes better use of resources.                            ‘
  3. No Mixing. There is no mixing of produce of the different crops.
  4. Soil erosion is checked as the field is not left uncovered for any long period.
  5. Specific fertilizers required for each crop can be added.
  6. Pesticide and weedicide required for each crop can be applied.
  7. Seeds of different crops can be sown separately.
  8. Harvesting. Each crop can be harvested, threshed and marketed separately.

Advantages of Crop Rotation:

  1. Attack by insects and fungi is minimized because different pests are associated with different crops. By varying the crops, the insects and fungi associated with a particular crop usually disappear.
  2. Rotation of crops helps in weed control. This is because weeds are associated with specific crops. When the crop is changed, the weeds associated with the previous crop usually disappear.
  3. Rotation of crops improves the fertility of the soil and hence brings about an increase in the production of food grains.
  4. It saves a lot of nitrogenous fertilizer, because growing leguminous crop during the rotation fixes atmospheric nitrogen with the help of their nitrogen fixing bacteria and there is no need to add nitrogenous fertilizer to the soil.
  5. The chemical nature of the soil is not altered as different crops require different types of fertilizers.
  6. There is optimum utilization of nutrients as different crops obtain nutrients from different layers of the
  7. The land is not kept free of cultivation. Two to four crops are raised which increases the income of farmers.

Question 4.
What is genetic manipulation ? How is it useful in agricultural practices (CCE 2013)
Answer:
Genetic manipulation is incorporation of new genes for various traits from other genotypes into a crop variety so as to bring about desired change. It is carried out by means of hybridisation, mutation breeding, polyploidy and DNA recombination technology.
Uses:
In agriculture, higher yield can be obtained only by employing higher yielding varieties, improved farming practices, modern technology, latest agricultural machines and implements, nutrient supply, etc. All these require high cost and knowledge of new techniques and improvements. Therefore, a farmer’s purchasing capacity for inputs determines the cropping system and production practices.

Question 5.
How do storage grain losses occur ?
Answer:
Both abiotic and biotic factors damage stored grains.
Abiotic Factors

  1. Moisture Content of Grains: Moisture content of grains is generally higher than optimum 14%. Higher moisture content of grains increases their respiration, which heats the grains and reduces their keeping quality. Microorganisms, fungi and insects attack such grains.
  2. Dampness and Humidity of Air: Dampness of godowns and humidity of air cause growth ol moulds over and inside the grains.
  3. Temperature: Temperature of 30°C and above is harmful to stored grains due to activity of microorganisms, insects, pests and activation of enzymes of the grains.

Biotic Factors

  1. Rodents: Six rats consume food equivalent to an average human being. They damage 5-6 times more grains by cutting and contamination (by urine, hair and excreta).
  2. Birds: Birds are often seen in large number around godowns. They are able to puncture bags and eat the stored grains. The birds also contaminate the stored grains with their excreta and feathers. Bird excreta often contains Salmonella, the bacterium causing food poisoning.
  3. Insects and Worms: Insects and their larvae feed on stored grains either internally (internal feeders like Pulse Beetle, Rice Weevil, Lesser Grain Borer) or externally (external feeders, e.g., Khapra Beetle, Rust Red Flour Beetle). They damage the grains, decrease their quality and cause contamination with webs, cocoons, excreta, dead remains and toxins.
  4. Microorganisms: Bacteria, yeasts and moulds attack stored grains and cause their rotting. Rotting brings about discolouration, loss of weight, bad odour and aflatoxin contamination of grains which also lose their ability to germinate.

Question 6.
How do good animal husbandry practices benefit farmers ? (CCE 2011)
Answer:
Good animal husbandry practices keep the animals healthy and more productive. There is higher yield of animal products— milk, eggs, meat.

  1. Shelter: Diseases spread if the animal shelters are dingy and crowded. If they are spacious, hygienic and well lighted, the animals remain healthy.
  2. Vaccination: Vaccination of young animals prevents the occurrence of common diseases.
  3. Segregation of Sick Animals: When sick animals are noticed, they are immediately segregated. Cleanliness drive is undertaken and the remaining animals are given prophylactic doses of medicines to prevent the spread of disease. The livestock remains healthy and productive.
  4. Proper Diet: A proper optimum diet with feed additives enhances growth and yield of animals.
  5. Breeding: Breeding for more milk, longer lactation period, more egg laying, better convertibility of food and other good characteristics have allowed various branches of animal husbandry to give better yield.

Question 7.
What are the benefits of cattle farming ? (CCE 2011)
Answer:

  1. Milk: It provides milk. Better breeds and better diet have increased milk availability in India. The phenomenal rise in milk availability is called white revolution. Dr. Verghese Kurien is father of white revolution in India. Today, India produces more milk (over 133 million tonnes) than any other country though per capita milk availability is still low at 295 gms/day.
  2. Utilisation of Spare Time: Farmers and others can utilize their spare time in the morning and evening for cattle farming.
  3. Extra Income: Cattle farming provides a good amount of extra income.
  4. Fodder: Growing fodder is helpful to farmers as the land is not left vacant: Selling of fodder gives good return.
  5. Organic Wastes: Wheat bran, Rice bran, gram chaff and oil cakes are organic wastes which form a good part of cattle feed.
  6. Soil Fertility: Cattle dung is useful ingredient for biogas generation and manure formation.

Question 8.
For increasing production, what is common in poultry, fisheries and bee-keeping ?
Answer:

  1. Good quality breeds,
  2. Proper feeds.
  3. Proper accomodation and care.

Question 9.
How do you differentiate amongst capture fishing, mariculture and aquaculture.
Answer:
NCERT Solutions for Class 9 Science Chapter 15 Improvement in Food Resources image - 1

SELECTION TYPE QUESTIONS

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

Question 1.
Most fish production in inland water body is capture fishing.
Question 2.
Little Bee is Apis florae.
Question 3.
Light soil has good water retention capacity.
Question 4.
A common method of introduction of desired trait in a crop plant is hybridisation.
Question 5.
Food security depends upon availability and access to food.
Question 6.
Fish breed throughout the year.
Question 7.
Mullets are marine fish farmed in sea water.
Question 8.
A common cereal is gram.

Matching Type Questions :

Question 9.
Match the articles given in column I and column II (Single Matching)
NCERT Solutions for Class 9 Science Chapter 15 Improvement in Food Resources image - 2

Question 10.
Match the contents of columns I, II and III (Double Matching)
NCERT Solutions for Class 9 Science Chapter 15 Improvement in Food Resources image - 3

Question 11.
In composite fish culture mark the fish as top feeder (T), midzone feeder (M) and bottom feeder (B) (Check List or Key Items)
NCERT Solutions for Class 9 Science Chapter 15 Improvement in Food Resources image - 4

Question 12.
Match the Stimulus with Appropriate Response.
NCERT Solutions for Class 9 Science Chapter 15 Improvement in Food Resources image - 5

Fill in the Blanks

Question 13. Most of our food comes from …………….. and animals husbandry. ,
Question 14. Vegetables, spices and ……………. provide vitamins, minerals and small amount of nutrients.
Question 15. Out of 16 nutrients, seven are ………………. .
Question 16. Milk production can be increased by increasing ……………… period.
Question 17. Shell fish include prawns and ……………….. .

Answers:
NCERT Solutions for Class 9 Science Chapter 15 Improvement in Food Resources image - 6

SOME ACTIVITY BASED QUESTIONS

Question 1.
During heavy rains in a village, the rain water carried away excess of nitrogenous compounds present in the soil to a pond. How will it affect the growth of algae and phytoplankton in the pond ?
Answer:
Algae and phytoplankton will grow faster causing bloom.

Question 2.
Name two good fumigants.
Answer:
Ethylene dibromide and aluminium phosphide.

Question 3.
How do moisture and temperature affect the life of food materials ?
Answer:
High moisture content, and high temperature at the time of storage of any food material will decrease the life of food material due to :
(a) growth of micro-organisms,
(b) increased enzymatic spoilage of food.

Question 4.
State one indicator each for infestation by insects and rodents in stored food grains. Describe one method each for controlling the population of insects and rodents.
Answer:
Presence of cocoons, webs and weevilled grain indicates the infestation by insects.
Presence of excreta and holes in the bags indicates the infestation by rats.
Method of Controlling Insects in Storage by Fumigation. Fumigation is the most effective method for controlling the insect population in storage. Fumigants consist of volatile chemicals which quickly vapourise and kill the insects without affecting the grains, e.g. ethylene dibromide.
Depending on the quantity of grains to be fumigated, a certain number of EDB ampules wrapped in cloth are inserted a little below the surface of the grain and broken gently. The structure is left air-tight and undisturbed for a week. Toxic fumes diffuse and kill the insects.
Method of Controlling Rodents. Using single or multiple doses of rodenticide is the effective method of killing rodents. The baits can be prepared by mixing flour of cereals with jaggery or sugar with edible oil and rodenticide in right proportion, e.g., zinc phosphide.

Question 5.
Which one of the following is not a part of the biotic environment—
Man, air, trees, insects ?
Answer:
Air.

Question 6.
Name any two storage pests of foodgrains found commonly in India.
Answer:
Rice weevil and khapra beetle.

Question 7.
Why is chemical method of controlling pests not considered good ?
Answer:
The chemicals used are sprayed on the crop to prevent diseases which leads to environmental pollution.
Part of the chemicals penetrate the grains. They are harmful to animals and human beings.

Question 8.
“If we excessively use pesticides to safeguard the crop from blight, it may cause long term damage to the mankind. ” Justify the statement. (CCE 2011 )
Answer:
Pesticides are toxic, poisonous and often non-biodegradable chemical compounds. They have the following ill-effects on our body.

  1. They cause irritation to the skin and respiratory system.
  2. They can enter our system either directly or through the crop produce and harm us. So, due to these ill-effects, pesticides should be handled with care.

Question 9.
Why are kharif crops more susceptible to infestation than the rabi crops ?
Answer:
Kharif crops are more susceptible to infestation by pests because the humid and warm conditions at that time are favourable for infestation.

Question 10.
Give one example each of where fumigation and spraying is more suitable.
Answer:
Spraying is suitable for disinfecting the storage godown or structure before the food grains are stored into it. Fumigation is more useful in protecting the stored grains from pests.

Question 11.
Name two types of leguminous fodder which are relished by the cattle.
Answer:
Barseem and lucerne.

Question 12.
What does each of the following yield ?
(a) Jersey
(b) Murrah
(c) White leghorn
(d) 1R-8.
Answer:
(a) Milk
(b) Milk
(c) Egg
(d) Paddy.

Question 13.
Mention the percent protein content in milk, egg and fish.
Answer:
The percent protein content in
(a) Milk is 4%
(b) Egg is 13%
(c) Fish is 19%.

Question 14.
Name two cows that can yield 5000-6000 litres of milk during lactation period.
Answer:
Frieswal and Holstein friesian.

Question 15.
Name two varieties of Indian fishes.
Answer:
Fresh water fishes — Catla and Rohu
Marine fishes – Hilsa, Sardine

Question 16.
Name the two external factors that have favourable effect on egg laying of hens.
Answer:
Light intensity and duration of light affect the egg laying of hens.

Question 17.
Write two infectious diseases of each of cow and poultry.
Answer:
Diseases of cow – Foot and mouth disease and Cow Pox.
Diseases of poultry – Chick Pox and Aspergillosis

Question 18.
Name the main food component found in egg white.
Answer:
Proteins.

Question 19.
How many cows can be approximately impregnated by the semen from one bull ?
Answer:
3000 cows.

Question 20.
These are the varieties of poultry birds : ‘A ’and ‘B’ are Aseel and Busra, ‘C’ and ‘D’ are White Leghorn and Rhode Island Red. Which are indigenous and which are exotic ? What would you obtain if indigenous species is bred with exotic ? What would be the advantages of this process ?
Answer:
Aseel and Busra are indigenous. White Leghorn and Rhode Island Red are exotic.
Advantages of the Process. It produces hybrids that possess better characteristics of both the indigenous and exotic varieties like high yield and resistance to diseases.

Question 21.
Explain the term : varietal improvement.
Answer:
Varietal improvement is the development of new variety through scientific breeding which has better yield, better quality and higher resistance to diseases.

Question 22.
What is meant by mixed farming ? What are its advantages ?
Answer:
Mixed farming is the raising of different crops and animals on the same farm. It provides good income to the farmer.

Question 23.
Which one of the following crops would require a minimum quantity of NPK or urea for its proper growth : Paddy, Peas, Wheat, Sugarcane ?
Answer:
Peas.

Question 24.
What does the number, 1 : 2 mean in intercropping ?
Answer:
1 : 2 means, the row pattern with one row of the one crop and two rows of the intercrop.

Question 25.
A farmer was advised not to use a particular type of fertilizer for his crop after harvesting a crop of peas. Give one reason for such an advice. Name any two crops other than pea crop which might have shown the same effect.
Answer:
The farmer is following the practice of crop rotation. By this practice, the nitrogen lost due to wheat crop is replenished by peas, a leguminous crop. As a result, he does not have to use any nitrogenous fertilizer. Gram, Bean.

NCERT Solutions for Class 9 Science Chapter 15 Improvement in Food Resources

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