RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6

RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6

Other Exercises

Factorize each of the following algebraic expressions :
Question 1.
4x2 + 12xy + 9y2
Solution:
4x2 + 12xy + 9y2 = (2x)2 + 2 x 2x x 3y + (3y)2 {∵ a2 + 2ab + b2 = (a +b)2}
= (2x + 3y)2    

Question 2.
9a2 – 24ab + 16b2
Solution:
9a2 – 24ab + 16b2
= (3a)2 – 2 x 3a x 4b + (4b)2     {∵ a2 – 2ab + b2 = (a – b)2}
= (3a – 4b)2

Question 3.
36a2 – 6pqr + 9r2
Solution:
p2q2 – 6pqr + 9r2
= (pq)2 – 2 x pq x3r + (3r)2 {∵ a2 – 2ab + b2 = (a -b)2}
= (pq-3r)2

Question 4.
36a2 + 36a + 9
Solution:
36a2 + 36a + 9
= (6a)2 + 2 x 6a x 3 + (3)2   {∵ a2 + 2ab + b2 = (a + b)2
= (6a + 3)2

Question 5.
a2 + 2ab + b2 – 16
Solution:
a2 + 2ab + b2 – 16
= (a + b)2 – (4)2     {∵ a2 + 2ab + b2 = (a + b)2 and a2 – b2 = (a + b) (a – b)}
= (a + b + 4) (a + b – 4)

Question 6.
9z2 – x2 + 4xy – 4y2
Solution:
9z2 – x2 + 4xy – 4y2    {∵ a2 – b2 = (a + b) (a – b) and a2 – 2ab + b2 (a – b)2}
= 9z2 – (x2 – 4xy + 4y2)
= (3z)2 – [(x)2 – 2 x x x 2y + (2y)2]
= (3z)2-(x-2y)2
= (3z + x – 2y) (3z – x + 2y)

Question 7.
9a4 – 24a2b2 + 16b4 – 256
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6 1

Question 8.
16 – a6 + 4a3b3 – 4b6
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6 2

Question 9.
a2 – 2ab + b2 – c2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6 3

Question 10.
x2 + 2x + 1 – 9y2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6 4

Question 11.
a2 + 4ab + 3b2
Solution:
a2 + 4ab + 3b2
= a2 + 4ab+ 4b2 – b2
= (a)2 + 2 x a x 2b + (2b)2 – b2 (∵ 3b2 = 4b2 – b2)
= (a + 2b)2 – (b)2   {∵ a2 – b2 = (a +b) (a – b)}
= (a + 2b + b) (a + 2b- b)

Question 12.
96 – 4x-x2
Solution:
96 – 4x – x2 = 96 – (4x + x2)
= 96 – [(x)2 + 2 x x x 2 + (2)2] + (2)2   (on completing the square)
= 96 + 4 – (x + 2)2 = 100 – (x + 2)2
= (10)2 – (x + 2)2
= (10 + x + 2) (10 – x- 2)
= (x + 12) (-x + 8)

Question 13.
a4 + 3a2 + 4
Solution:
a4 + 3a2 + 4
= (a2)2 + (2)2 + 2 x a2 x 2 – a2   (on completing the square)
= (a2 + 2)2 – (a)2
= (a2 + 2 + a) (a2 + 2 – a)
= (a1 + a + 2) (a2 – a + 2)

Question 14.
4a4 + 1
Solution:
4x4 + 1 = (2a2)2 + (1)2 + 2 x 2x2 x 1 – 2 x 2x2 x 1  (completing the square)
= (2x2 + 1)2 – 4a2
= (2x2 + 1)2 – (2a)2   {a2 – b2 = (a + b) (a – b)}
= (2x2 + 1 + 2a) (2a2 + 1 – 2a)
= (2a2 + 2a + 1) (2a2 – 2a + 1)

Question 15.
4x4+y4
Solution:
4a4 + y4 = (2x2)2 + (y2)2 + 2 x 2x2y2 – 2 x 2x2y2
= (2x2 + y2)2 – 4x2y2
= (2x2 + y2)2 – (2xy)2
= (2x2 + y2 + 2xy) (2x2 + y2 – 2xy)
= (2x2 + 2xy + y2) (2x2 – 2xy + y2)

Question 16.
(x+ 2)2 – 6 (a + 2) + 9
Solution:
(x + 2)2 – 6 (x + 2) + 9
=  (x + 2)2 – 2 x (x + 2) x 3 + (3)2
= (x + 2 – 3)2
= (x-1)2 = (x-1)(x-1)

Question 17.
25 – p2 – q2 – 2pq
Solution:

RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6 5

Question 18.
a2 + 9y2 – 6xy – 25a2
Solution:
a2 + 9y2 – 6xy – 25a2
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6 6

Question 19.
49 – a2 + 8ab – 16b2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6 7

Question 20.
a2 – 8ab + 16b2 – 25c2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6 8

Question 21.
x2 -y2+ 6y- 9
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6 9

Question 22.
25x2 – 10x + 1 – 36y2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6 10
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6 11

Question 23.
a2-b2 + 2bc – c2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6 12

Question 24.
a2 + 2ab + b2 -c2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6 13

Question 25.
49 -x2 – y2 + 2xy
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6 14
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6 15

Question 26.
a2 + 4b2 – 4ab – 4c2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6 16

Question 27.
x2 -y2 – 4xz + 4z2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6 17

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RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.2

RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.2

Other Exercises

Divide :

Question 1.
6x3y2z2 by 3x2yz
Solution:
RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.2 1

Question 2.
15m2n3 by 5m2n2
Solution:
RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.2 2
Question 3.
24a3b3 by -8ab
Solution:
RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.2 3
Question 4.
-21abc2 by 7abc
Solution:
RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.2 4
Question 5.
72xyz2 by – 9xz
Solution:
RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.2 5
Question 6.
-72a4b5c8 by – 9a2b2c3
Solution:
RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.2 6

Simplify :

Question 7.
RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.2 7
Solution:
RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.2 8
Question 8.
RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.2 9
Solution:
RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.2 10

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RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.1

RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.1

Other Exercises

Question 1.
Write the degree of each of the following polynomials :
(i) 2x3+5x2-7
(ii) 5x2 – 3x +7 ’
(iii) 2x + x2 –-8
(iv) \(\frac { 1 }{ 2 }\) y7 -12y5 + 48y6 – 10
(v) 3x3 + 1
(vi) 5
(vii) 20x3 + 12x2y2– 10y2 + 20
Solution:
(i) 2x3 + 5x2-7: The degree of this polynomial is 3.
(ii) 5x2 – 3x + 2 : The degree of this polynomial is 2.
(iii) 2x + x2 – 8 : The degree of this polynomial is 2.
(iv) \(\frac { 1 }{ 2 }\) y7 – 12y6 + 48y5 – 10 : The degree of this polynomial is 7.
(v) 3x3 + 1 : The degree of this polynomial is 3.
(vi) 5 : The degree of this polynomial is 0 as it is only constant term
(vii) 20x3 + 12x2y2 – 10y2 + 20: The degree of this polynomial is 2 + 2 = 4.

Question 2.
Which of the following expressions are not polynomials :
(i) x2 + 2x2                   
(ii) √a x + x2-x3
(iii) 3y3 – √5y + 9      
(iv) ax1/2 + ax + 9x2 + 4
(v) 3x2 + 2x-1 + 4x + 5
Solution:
(i) x2 + 2x-2 = x2 + 2x \(\frac { 1 }{ { x }^{ 2 } }\) =x2 + \(\frac { 1 }{ { x }^{ 2 } }\)
: It is not xx polynomial as it has negative integral power.
(ii) √ax + x2 – x3: It is polynomial.
(iii) 3y3  √5y + 9 : It is a polynomial.
(iv) ax1/2+ ax + 9x2 + 4: It is not a polynomial as  the degree of \(\frac { 1 }{ { x }^{ 2 } }\) is an integer.
(v) 3x2 + 2x-1 + 4x + 5 : It is not a polynomial as the degree of x–2, x-1 are negative.

Question 3.
Write each of the following polynomials in the standard form. Also write their degree.
(i) x2 + 3 + 6x + 5x4
(ii) a1 + 4 + 5a6
(iii) (x3 – 1) (x3 – 4)
(iv) (y3 – 2) (y3 + 11)
(v) \(\left( { a }^{ 3 }-\frac { 3 }{ 8 } \right)\) \(\left( { a }^{ 3 }-\frac { 16 }{ 17 } \right)\)
(vi) \(\left( { a }+\frac { 3 }{ 4 } \right)\) \(\left( { a }+\frac { 3 }{ 4 } \right)\)
Solution:
Polynomial in standard form is the polynomial in ascending order or descending order.
RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.1 1

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RS Aggarwal Class 6 Solutions Chapter 11 Line Segment, Ray and Line Ex 11A

RS Aggarwal Class 6 Solutions Chapter 11 Line Segment, Ray and Line Ex 11A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 11 Line Segment, Ray and Line Ex 11A.

Other Exercises

Question 1.
Solution:
Line segment
(i) In figure (i) are \(\overline { XY } \) and \(\overline { YZ } \)
(ii) In figure (ii) \(\overline { AD } ,\overline { AB } ,\overline { AC } ,\overline { AE } ,\overline { BD } ,\overline { BC } ,\overline { CE } \)
(iii) In figure (iii) \(\overline { PQ } ,\overline { PR } ,\overline { PS } ,\overline { QR } ,\overline { QS } ,\overline { RS } \)

Question 2.
Solution:
(i) In fig. (i), line segments is \(\overline { AB } \), and
rays are \(\overrightarrow { AC }\) and \(\overrightarrow { BD }\).
(ii) In fig. (ii), line segments are
\(\overline { GE } ,\overline { GP } ,\overline { EP } \) and rays are
\(\overrightarrow { EF } ,\overrightarrow { GH } ,\overrightarrow { PQ }\)
(iii) In fig. (iii), line segments are \(\overline { OL } ,\overline { OP } \)
and rays are \(\overrightarrow { LM } ,\overrightarrow { PQ } \).

Question 3.
Solution:
(i) Four line segments are \(\overline { PR } ,\overline { QR } ,\overline { PQ } ,\overline { RS } \).
(ii) Four ray can be \(\overrightarrow { PA },\overrightarrow { QC }, \overrightarrow { RB } ,\overrightarrow { SD } \)
(iii) \(\overline { PR } ,\overline { QS } \) are two non-intersecting lines.

Question 4.
Solution:
(i) Three or more points in a plane are said to be collinear if they all lie on the same line.
RS Aggarwal Class 6 Solutions Chapter 11 Line Segment, Ray and Line Ex 11A Q4.1
(ii) In the figure given above, points A, B, C are collinear points.
We can draw exactly one line passing through three collinear points

Question 5.
Solution:
(i) Four pairs of intersecting lines are : (AB, PQ) ; (AB, RS) ; (CD, PQ) ; (CD, RS)
(ii) Four collinear points are : A, Q, S, B
(iii) Three non-collinear points are : A, Q, P
(iv) Three concurrent lines are : AB, PS and RS.
(v) Three lines whose point of intersection is P are : CD, PQ and PS.

Question 6.
Solution:
The lines drawn through given points A, B, C are as shown below. The names of these lines are AB, BC and AC.
RS Aggarwal Class 6 Solutions Chapter 11 Line Segment, Ray and Line Ex 11A Q6.1
Also it is clear that three different lines can be drawn.

Question 7.
Solution:
(i) In the the given figure, there are six line segments, namely AB, AC, AD, BD, BC, DC.
RS Aggarwal Class 6 Solutions Chapter 11 Line Segment, Ray and Line Ex 11A Q7.1
(ii) In the given figure, there are ten line segments, namely, AD, AB, AC, AO, OC, BC, BD, BO, OD, CD.
RS Aggarwal Class 6 Solutions Chapter 11 Line Segment, Ray and Line Ex 11A Q7.2
(iii) In the given figure, there are six line segments, namely AB, AF, BF, CD, DE, CE.
RS Aggarwal Class 6 Solutions Chapter 11 Line Segment, Ray and Line Ex 11A Q7.3
(iv) In the given figure, there are twelve line segments, namely, AB, BC, CD, AD, BF, CG, DH, AE, EF, FG, GH, EH.
RS Aggarwal Class 6 Solutions Chapter 11 Line Segment, Ray and Line Ex 11A Q7.4

Question 8.
Solution:
\(\overleftrightarrow { PQ } \) is a line
RS Aggarwal Class 6 Solutions Chapter 11 Line Segment, Ray and Line Ex 11A Q7.5
(i) False, as M does not lie on \(\overrightarrow { NQ } \)
(ii) True
(iii) True
(iv) True
(v) True

Question 9.
Solution:
(i) False
Point has no dimensions.
(ii) False
A line segment has a length.
(iii) False
A ray has no finite length.
(iv) False
If \(\overrightarrow { AB } \) and ray \(\overrightarrow { BA } \) have opposite directions.
(v) True
Length of \(\overline { AB } \) and \(\overline { BA } \) is same.
(vi) True
Line \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { BA } \) are same.
(vii) True
Distance between A and B or B and A is same, so they determine a unique line segment.
(viii) True
Two lines intersect each other at one point.
(ix) False
Two intersecting planes intersect at one line not at one point.
(x) False
If A, B, C are collinear and points D, E are collinear then it is not necessary that there points A, B, C, D and E are collinear.
(xi) False
Infinite number of rays can be drawn with a given end point.
(xii) True
We can draw one and only one line passing through two given points.-
(xiii) True
We can draw infinite number of lines pass through a given point.

Question 10.
Solution:
(i) definite
(ii) one
(iii) no
(iv) definite
(v) cannot. Ans.

 

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RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7

RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7

Other Exercises

Factorize each of the following algebraic expressions :
Question 1.
x2 + 12x – 45
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7 1
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7 2

Question 2.
40 + 3x – x2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7 3
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7 3.1

Question 3.
a2 + 3a-88
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7 4

Question 4.
a2 – 14a – 51
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7 5

Question 5.
x2 + 14x + 45
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7 6

Question 6.
x2 – 22x + 120
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7 7

Question 7.
x2– 11x – 42
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7 8
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7 9

Question 8.
a2 + 2a – 3
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7 10

Question 9.
a2 + 14a + 48
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7 11

Question 10.
x2 – 4x – 21
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7 12

Question 11.
y2 – 5y-36
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7 13

Question 12.
(a2-5a)2-36
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7 14
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7 15

Question 13.
(a + 7) (a – 10) + 16
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7 16
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7 17

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NCERT Exemplar Solutions for Class 10 Science Chapter 9 Heredity and Evolution

NCERT Exemplar Solutions for Class 10 Science Chapter 9 Heredity and Evolution

These Solutions are part of NCERT Exemplar Solutions for Class 10 Science. Here we have given NCERT Exemplar Solutions for Class 10 Science Chapter 9 Heredity and Evolution

NCERT Exemplar Solutions for Class 10 Science Chapter 9 Short Answer Questions

Question 1.
How is a sex of new born determined in humans ?
Answer:
Natural selection is preferential survival and differential reproduction of individuals of a population which possess variations that provide better adaptability to the existing environment. Peppered Moth of England has two forms, light grey and dark grey. Prior to industrial revolution, tree trunks in the forests around.
Manchester were light greyish green due to presence of lichens. Most of the Peppered Moth found in the area were light coloured with dark spots which could not be spotted easily by predator birds. During 1845 to 1890, air pollution killed the lichen flora. The deposition of soot changed the colour of the tree trunks to blackish one. Peppered Moth of the area also exhibited switch over to melanic or blackish form.
It provided better survival value against dark background. The light grey form largely disappeared as it could be easily spotted by predator birds and devoured.

More Resources

Question 2.
Does genetic combination of mothers play a significant role in determining the sex of a new born ?
Answer:
No. Mothers have no role in determining the sex of the new born. Mothers are homogametic, that is, they produce only one type of ova (22 + X). Fathers are heterogametic, that is, they produce two types of sperms, gynosperms (22 + X ) and androsperms (22 + Y). If gynosperm (22 + X) fertilises the ovum (22 + X), the sex of new born will be female (44 + 2X). If androsperm (22 + Y) fuses with ovum (22 + X), the new born will be boy (44 + XY).

Question 3.
Mention three important features of fossils which help in the study of evolution.
Answer:
Fossils are remains or impressions of past organisms that are found in the rocks. Fossils of lower strata belong to early periods while those of upper strata are of later periods. Arranging the fossils stratumwise will indicate the occurrence of different forms of life at different times. It is found that the early fossils generally belong to simple organisms. Complexity and elaboration increased gradually with evolution. Evolution has never been linear or straight. A number of variants or branches appeared, some of which were more complex while others were less complex.

  1. Fossils indicate the path of evolution of different groups.
  2. They can indicate the phylogeny of some organisms, e.g, Horse, Elephant.
  3. Some fossils have characteristics intermediate between two groups,
    e.g., toothed bird Archaeopteryx. They indicate how one group has evolved from another.

Question 4.
Why do all the gametes formed in human females have an X-chromosome ?
Answer:
Human females are homogametic. Their generic constitution is 44 + XX The two sex chromosomes are similar. Their ova which are produced after meiosis carry a gametic constitution of 22 + X No other combination is possible.

Question 5.
In human beings the satistical possibility of getting either a male or female child is 50 : 50. Give suitable explanation.
Answer:
Human females (44 + XX) are homogametic, that is, they produce only one type of ova (22 + X). Human males are heterogametic. They produce two types of sperms (22 + X and 22 + Y) in equal proportion, that is, 50 : 50 ratio. The chance of male or female child is also 50 : 50, as there is equal chance of androsperm (22 + Y) or gynosperm (22 + X) fertilizing an ovum.

Question 6.
A very small population of a species faces a greater threat of extinction than a larger population. Provide a suitable genetic explanation.
Answer:
A small population is always at a risk of degeneration and extinction due to

  1. Excessive inbreeding that brings about inbreeding depression or degeneration,
  2. Fewer recombinations and variations which are otherwise essential for maintaining vitality and vigour of the species.
  3. Lesser adaptability to changes in the environment,
  4. Increased threat to survival due to poaching, habitat destruction and environmental change.

Question 7.
What are homologous structures ? Give an example. Is it necessary that homologous structures always have a common ancestor ?
Answer:
Homologous structures or organs are those structures which have similar origin, similar development, similar internal structure and similar basic plan but show different external form and function, e.g., forelimbs of amphibians, reptiles, mammals and birds. Homologous structures are always have a common ancestry because otherwise there cannot be any similarity in basic plan, internal structure, development or origin. Modifications have occurred in them due to varied adaptations.

Question 8.
Does the occurrence of diversity of animals on earth suggest their diverse ancestry also ? Discuss this point in the light of evolution.
Answer:
Diversity of animals does not mean that they have diverse ancestry. Animals can be grouped into distinct lineages (e.g., mammals, birds, reptiles, annelids). Many of the lineages further show some similarities in basic traits indicating a common ancestry, e.g.,
vertebrates. Therefore, animals having a common ancestor in the remote past have successively developed newer and newer traits forming various groups of animals.

Question 9.
Give the pair of contrasting traits of the following characters in Pea plant and mention which is dominant and recessive

  1. Yellow seed
  2. Round seed.

Answer:

  1. Seed Colour: Yellow (dominant), green (recessive).
  2. Seed Shape: Round (dominant), wrinkled (recessive).

Question 10.
Why did Mendel choose Pea plant for his experiments ?
Answer:
Mendel chose Garden Pea or Edible Pea (Pisum sativum) as his experimental material because of

  1. Easy availability of detectable contrasting traits of several characters,
  2. Flower structure normally allows self fertilization but cross fertilization can also be carried out easily.
  3. Annual nature of the plant,
  4. Formation of a large number of seeds by each plant,
  5. Requirement of little care except at the time of cross breeding.

Question 11.
A woman has only daughters. Analyse the situation genetically and provide a suitable, explanation.
Answer:
A woman produces only one type of ova (22 + X) while her husband produces two types of sperms, gynosperms (22 + X) and androsperms (22 + Y) in equal proportion. It is a chance that each time the woman conceived, only the gynosperm fertilised the egg so that only daughters were born.

NCERT Exemplar Solutions for Class 10 Science Chapter 9 Long Answer Questions

Question 12.
Does geographical isolation of individuals of a species lead to formation of a new species ? Provide a suitable explanation.
Answer:
Yes. Geographical isolation of a population will lead to genetic drift as there will be no gene flow between it and the parent species. Inbreeding in small population will result in fixation of certain alleles and elimination of others. There will be change in gene frequency. Mutations will produce new alleles and hence a new gene pool. Accumulation of new alleles and hence new variations over several generations will ultimately lead to the formation of new species.

Question 13.
Bacteria have a simpler body plan when compared with human beings. Does it mean that human beings are more evolved than bacteria. Provide a suitable explanation.
Answer:
Both bacteria and human beings perform all the activities of life and live comfortably in their environments. They, therefore, seem
to be equally evolved. However, human beings have a far more complex organisation and differentiation which are absent in bacteria. Since complex organisation and differentiation develop only through evolution, humans are far more evolved than bacteria.

Question 14.
All the human races like Africans, Asians, Europeans, Americans and others might have evolved from a common ancestor. Provide a few evidences in support of this view.
Answer:
All the human races have evolved from a common ancestor because they possess

  1. Common body plan
  2. Common structure
  3. Similar physiology
  4. Similar metabolism
  5. Similar chromosome number
  6. Common genes or genetic blue print
  7. Free interbreeding.

Question 15.
Differentiate between inherited and acquired characters. Give one example of each type.
Answer:

Acquired Traits

 Inherited Traits
1.     Development. The traits develop during life time of an individual.

2.     Nature. They are somatic variations.

3.     Cause. Acquired traits develop due to direct effect of environment, use and disuse and conscious efforts.

4.     Fate. They die with the death of the individual.

Ex. Muscular body of a wrestler.

 The traits are obtained from the parents.

They are genetic variations.

The traits develop due to mutations and reshuffling of genetic material.

They are passed on to the next generation.

Ex. Fused and free ear lobes.

Question 16.
Give reasons why acquired characters are not inherited.
Answer:
Many of the variations have no immediate benefit to the species. They function as preadaptations which can be beneficial under certain environmental conditions like heat tolerance variation if the temperature of the area rises.

Question 17.
Evolution has exhibited greater stability of molecular structures when compared with morphological structures. Comment on the statement and justify your opinion.
Answer:
Life is a highly organised system of biochemicals and their reactions. It is expressed inside the cells. They are the same right from bacteria to human beings. However, the method of procurement of the raw materials for forming bio-chemicals is not the same in various organisms. It has helped in avoiding cut throat competition amongst the living beings. They developed different morphological features and modes of obtaining nutrients. As a result a huge diversity has appeared in the living world.

Question 18.
In the following crosses, write the characteristics of the progency
NCERT Exemplar Solutions for Class 10 Science Chapter 9 Heredity and Evolution image - 1
Answer:
(a) All Round Yellow
(b) Round Yellow, Round Green, Wrinkled Yellow and Wrinkled Green in the ratio of 9 : 3 : 3 : 1.
(c) All Wrinkled Green (et) All Round Yellow (Rr Yy).

Question 19.
Study the following cross showing self pollination in F1. Fill in the blank and answer the question that follows :
NCERT Exemplar Solutions for Class 10 Science Chapter 9 Heredity and Evolution image - 2
Answer:
Rr Yy (Round Yellow ).

Question 20.
In the above question, what are combination of characters in F2 progeny ? What are their ratios ?
Answer:
Round Yellow – 9,
Round Green – 3,
Wrinkled Yellow – 3,
Wrinkled Green -1
i.e., 9 : 3 : 3 : 1.

Question 21.
Give the basic features of the mechanism of inheritance.
Answer:
Basic Features of Inheritance

  1. Unit Characters: An organism is made of a large number of characters, each of which behaves as a unit.
  2. Genes: Characters are controlled by genes.
  3. One Gene-One Character: A single gene generally controls one character.
  4. Location of Genes: Genes are located on chromosomes.
  5. Alleles: A gene may have two or more forms called alleles. They represent different traits of a character.
  6. Paired Alleles: An individual possesses two alleles of every gene. The two may be similar or dissimilar.
  7. Dominance: Where there are two different forms or alleles of the gene, generally one expresses its effect. It is called dominant allele. The other which does not express its effect in presence of dominant allele, is called recessive allele.
  8. Segregation: The two alleles separate at the time of gamete formation. A gamete has only one allele or form of the gene.
  9. Independent Assortment: The alleles of different genes located on separate chromosomes behave independent of one another.
  10. Pairing: Fusion of gametes during fertilization, brings together the two forms of a gene in the zygote.

Question 22.
Give reasons for the appearance of new combination of characters in the F2 progeny.
Answer:
Independent Assortment: The two forms of a gene separate and pair independent of the two forms of other genes during gametogenesis and fertilisation. It causes new combination of characters, e.g.;
NCERT Exemplar Solutions for Class 10 Science Chapter 9 Heredity and Evolution image - 3

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RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8

RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8

Other Exercises

Resolve each of the following quadratic trinomials into factors :
Question 1.
2x2 + 5x + 3
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 1

Question 2.
2x2– 3x – 2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 2

Question 3.
3x2 + 10x + 3
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 3
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 4

Question 4.
7x – 6 – 2x2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 5

Question 5.
7x2 – 19x – 6
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 6

Question 6.
28-31x -5x2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 7
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 8

Question 7.
3 + 23y – 8y2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 9

Question 8.
11x2 – 54x + 63
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 10

Question 9.
7x-6x2 + 20
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 11

Question 10.
3x2 + 22x + 35
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 12
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 13

Question 11.
12x2 – 17xy + 6y2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 14

Question 12.
6x2 – 5xy – 6y2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 15

Question 13.
6x2 + 13xy + 2y2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 16

Question 14.
14x2 + 11xy – 15y2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 17
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 18

Question 15.
6a2 + 17ab – 3b2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 19

Question 16.
36a2 + 12abc – 15b2c2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 20

Question 17.
15x2 – 16xyz – 15y2z2
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 21

Question 18.
(x – 2y)2 -5 (x- 2y) + 6
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 22

Question 19.
(2a – b)2 + 2 (2a – b) – 8
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8 23

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RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.3

RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.3

Other Exercises

Factorize each of the following algebraic expressions.

Question 1.
6x (2x – y) + 7y (2x – y)
Solution:
6x (2x – y) + 7y (2x – y)
= (2x – y) (6x + 7y)
[∵ (2x – y) is common]

Question 2.
2r (y – x) + s (x – y)
Solution:
2r (y – x) + s (x – y)
-2r (x – y) +s (x – y)
= (x – y) (-2r + s)                   [(x – y) is common]
= (x-y) (s-2r)

Question 3.
7a (2x – 3) + 2b (2x – 3)
Solution:
7a (2x – 3) + 3b (2x – 3)
= (2x – 3) (7a + 3b)               [(2x – 3) is common]

Question 4.
9a (6a – 5b) – 12a2 (6a – 5b)
Solution:
9a (6a – 5b) – 12a2 (6a – 5b)
HCF of 9 and 12 = 3
∴ 3a (6a – 5b) (3 – 4a)
{(6a – 5b) is common}

Question 5.
5 (x – 2y)2 + 3 (x – 2y)
Solution:
5 (x – 2y)2 + 3 (x – 2y)
= 5 (x – 2y) (x – 2y) + 3 (x – 2y)
= (x – 2y) {5 (x – 2y) + 3}
{(x – 2y) is common}
= (x – 2y) (5x – 10y + 3)

Question 6.
16 (2l – 3m)2 – 12 (3m – 2l)
Solution:
16 (2l – 3m)2 – 12 (3m-2l)
= 16 (2l – 3m) (2l – 3m) + 12 (2l – 3m)
HCF of 16, 12 = 4 4 (2l-3m) {4 (2l- 3m) + 3}
{(2l – 3m) is common}
= 4 (2l -3m) (8l- 12m+ 3)

Question 7.
3a (x – 2y) – b (x – 2y)
Solution:
3a (x – 2y) – b (x – 2y)
= (x – 2y) (3a – b)
{(x – 2y) is common}

Question 8.
a2 (x + y) + b2 (x + y) + c2 (x + y)
Solution:
a2 (x + y) + b2 (x + y) + c2 (x + 3’)
= (x + y) (a2 + b2 + c2)
{(x + y) is common}

Question 9.
(x-y)2 + (x -y)
Solution:
(x – y)2 + (x- y) = (x – y) (x – y) + (x – y)
= (x – y) (x – y + 1)                          {(a – y) is common}

Question 10.
6 (a + 2b) – 4 (a + 2b)2
Solution:
6 (a + 2b) – 4 (a + 2b)2
= 6 (a + 2b) – 4 (a + 2b) (a + 2b)
HCF of 6, 4 = 2
= 2 {a + 2b) {3 – 2 {a + 2b)
{2 (a + b) is common}
= 2 (a + 2b) (3-2 a- 4b)

Question 11.
a (x -y) + 2b (y – x) + c (x -y)2
Solution:
a (x -y) + 2b (y – x) + c (x -y)2
= a (x – y) – 2b (x – y) + c (x – y) {x – y)
= (x – y) {x – 2b + c (x – y)}
{(a – y) is common}
= (a – y) (a – 2b + cx – cy)

Question 12.
– 4 (a – 2y)2 + 8 (a – 2y)
Solution:
– 4 (x – 2y)2 + 8 (x – 2y)
= – 4 (x – 2y) (x – 2y) + 8 (x – 2y)
{- 4 (x – 2y) is common}
= – 4 (x – 2y) (x – 2y – 2)
= 4 (x – 2y) (2 – x + 2y)

Question 13.
x3 (a – 2b) + a2 (a – 2b)
Solution:
x3 (a – 2b) + x2 (a – 2b)
HCF of x3, x2 = x2
∴ x2 (a – 2b) (x + 1)
{x2 (x – 2b) is common}
= x2 (x – 2b) (x + 1)

Question 14.
(2x – 3y) (a + b) + (3x – 2y) (a + b)
Solution:
(2x – 3y) (a + b) + (3x – 2y) (a + b)
= (a + b) {2x – 3y + 3x – 2y}
{(x + b) is common}
= (a + b) (5x – 5y)
= 5 (a + b) (x – y)

Question 15.
4 (x + y) (3a – b) + 6 (a + y) (2b – 3a)
Solution:
4 (x + y) (3a – b) + 6 (a + y) (2b – 3a)
= 4 (x + y) (3a – b) – 6 (x + y) (3a – 2b)
HCF of 4, 6 = 2
= 2 (x + y) {2 (3a – b) – 3 (3a – 2b)}
= 2 (x + 3) {6a – 2b – 9a + 6b}
= 2 (x +y) {-3a + 4b}
= 2 (x + y) (4b – 3a)

Hope given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.3 are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10D

RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10D

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10D.

Other Exercises

OBJECTIVE QUESTIONS
Mark against the correct answer in each of the following :

Question 1
Solution:
(d) \(\\ \frac { 92 }{ 115 } \)
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10D Q1.1

Question 2.
Solution:
(a) \(\\ \frac { 57 }{ x } \)
= \(\\ \frac { 51 }{ 85 } \)
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10D Q2.1

Question 3.
Solution:
(a) \(\\ \frac { 25 }{ 35 } \)
= \(\\ \frac { 45 }{ x } \)
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10D Q3.1

Question 4.
Solution:
(c) \(\\ \frac { 4 }{ 5 } \)
= \(\\ \frac { x }{ 35 } \)
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10D Q4.1

Question 5.
Solution:
(b) \(\\ \frac { a }{ b } \)
= \(\\ \frac { c }{ d } \)
=>ad = bc

Question 6.
Solution:
(b) a : b :: b : c
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10D Q6.1

Question 7.
Solution:
(b) \(\\ \frac { 5 }{ 8 } \) < \(\\ \frac { 3 }{ 4 } \) => 4 x 5 < 3 x 8 => 20 < 24

Question 8.
Solution:
Total amount = Rs 760
Ratio A : B = 8 : 11
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10D Q8.1

Question 9.
Solution:
(d) largest = \(\\ \frac { 252\times 7 }{ 5+7 } \)
= \(\\ \frac { 252\times 7 }{ 12 } \)
= 21 x 7
= 147

Question 10.
Solution:
(b) largest = \(\\ \frac { 90\times 5 }{ 1+3+5 } \)
= \(\\ \frac { 90\times 5 }{ 9 } \)
= 50 cm

Question 11.
Solution:
(c) total strength of school
largest = \(\\ \frac { 840 }{ 5 } \) x (12 + 5)
= \(\\ \frac { 840\times 17 }{ 5 } \)
= 168 x 17
= 2856

Question 12.
Solution:
(b) Cost of 12 pens = Rs 138
Cost of 1 pen = Rs \(\\ \frac { 138\times 14 }{ 12 } \)
and cost of 14 pens = Rs 161

Question 13.
Solution:
(b) \(\\ \frac { 24\times 15 }{ 8 } \)
= 45 days

Question 14.
Solution:
(a) \(\\ \frac { 26\times 40 }{ 20 } \)
= 52 men

Question 15.
Solution:
(b) In 6 L of petrol, a car covers = 111 km
In 1 L, it will cover = \(\\ \frac { 111 }{ 6 } \) km
and in 10 L it will cover = \(\\ \frac { 111\times 10 }{ 6 } \) km
= 185 km

Question 16.
Solution:
(a) \(\\ \frac { 28\times 550 }{ 700 } \)
= 22 days

Question 17.
Solution:
Ratio in the angles of triangle
= 3 : 1 : 2
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10D Q17.1

Question 18.
Solution:
(b) Ratio in length and breadth of a rectangle = 5 : 4
Width = 36 m
Length = \(\\ \frac { 5 }{ 4 } \) x 36 m
= 45m

Question 19.
Solution:
Bus covers in 3 hrs = 195 km
It will cover in 1 hr = \(\\ \frac { 195 }{ 3 } \) = 65 km
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10D Q19.1

Question 20.
Solution:
1 dozen = 12 bars
Cost of 5 bars of soap = Rs.82.50
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10D Q20.1

Question 21.
Solution:
Total pencils in 30 packets of 8 pencils
= 30 x 8= 240
and total pencils of 25 packets of 12
pencils = 25 x 12 = 300
Now cost of 240 pencils = Rs. 600
Then cost of 1 percent = Rs. \(\\ \frac { 600 }{ 24 } \)
and cost of 300 pencils = Rs. \(\\ \frac { 600\times 300 }{ 240 } \)
= Rs 750 (b)

Question 22.
Solution:
Journey of 75 km costs = Rs 215
Cost of 1 km = Rs \(\\ \frac { 215 }{ 75 } \)
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10D Q22.1

Question 23.
Solution:
1st term = 12
2nd term = 21
fourth term = 14
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10D Q23.1

Question 24.
Solution:
10 boys dig a patch in = 12 hrs
1 boy will dig it in = 12 x 10 hours
RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10D Q24.1

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RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9

RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9

Other Exercises

Factorize each of the following quadratic polynomials by using the method of completing the square.
Question 1.
p2 + 6p + 8
Solution:
p2 + 6p + 8
= p2 + 2 x p x 3 + 32 – 32 + 8   (completing the square)
= (p2 + 6p + 32) – 1
= (p + 3)2 – 12
= (P + 3)2 – (1)2       { ∵ a2 + b2  = (a+b) (a-b)}
= (p +3+1) (p + 3 -1)
= (p+4) (p+ 2)

Question 2.
q2 – 10q + 21
Solution:
q2 – 10q + 21
= (q)2 – 2 x q x 5 + (5)2 – (5)2 + 21   (completing the square)
= (q)2 – 2 x q x 5 + (5)2 -25+21
= (q)2-2 x q x 5 + (5)2 – 25 +21
= (q)2-2 x q x 5 + (5)2 – 4
= (q – 5)2 – (2)     {∵ a2 – b2 = (a + b) (a – b)}
= (q- 5 + 2) (q-5-2)
=(q- 3) (q-7)

Question 3.
4y2 + 12y + 5
Solution:
4y2 +12y + 5
= (2y)2 + 2 x 2y x 3 + (3)2 – (3)2 + 5    (completing the square)
= (2y + 3)2 – 9 + 5
= (2y + 3)2 – 4
= (2y + 3)2-(2)2   {∵ a2 – b2 = (a + b) (a – b)}
= (2y + 3 + 2) (2y + 3 – 2)
= (2y + 5) (2y+ 1)

Question 4.
p2 + 6p- 16
Solution:
p2 + 6p – 16
= (p)2 + 2 x  p x 3 + (3)2 – (3)2 – 16    (completing the square)
= (p)2 + 2 x p x 3 + (3)2 – 9 – 16
= (p + 3)2 – 25
= (p + 3)2 – (5)2     {∵ a2 -b2 = {a + b) (a – b)}
= (p + 3 + 5)(p + 3-5)
= (p + 8) (p – 2)

Question 5.
x2 + 12x + 20
Solution:
x2 + 12x + 20
= (x)2 + 2 x x x 6 + (6)2 – (6)2 + 20   (completing the square)
= (x)2 + 2 x x x6 + (6)2 -36 + 20
= (x + 6)2 -16
= (x + 6)2 – (4)2   {∵ a2 – b2 = (a + b) (a – b)}
= (x + 6 + 4) (x + 6 – 4)
= (x + 10) (x + 2)

Question 6.
a2 – 14a – 51
Solution:
a2 – 14a-51
= (a)2 – 2 x x 7 + (7)2 – (7)2 – 51       (completing the square)
= (a)2 – 2 x a x 7 + (7)2 – 49 – 51
= (a – 7)2 – 100
= (a – 7)2 – (10)2    {∵  a2 – b2 = (a + b) (a – b)}
= (a – 7 + 10) (a – 7 – 10)
= (a + 3) (a – 17)

Question 7.
a2 + 2a – 3
Solution:
a2 + 2a – 3
= (a)2 + 2 x a x 1 + (1)2 – (1)2 – 3   (completing the square)
= (a)2 + 2 x a x 1 + (1)2 – 1 – 3
= (a + 1)2 – 4
= (a + 1)2 – (2)2 {∵ a2 – b2 = (a + b) (a – b)}
= (a + 1 + 2) (a + 1 – 2)
= (a + 3) (a – 1)

Question 8.
4x2 – 12x + 5
Solution:
4x2 – 12x + 5
= (2x)2 – 2 x 2x x 3 + (3)2 – (3)2 + 5  (completing the square)
= (2x)2 – 2 x 2x x 3 + (3)2 -9 + 5
= (2x – 3)2 – 4
= (2x – 3)2 – (2)2      {∵ a2 – b2 = (a + b) (a – b)}
= (2x – 3 + 2) (2x – 3 – 2)
= (2x – 1) (2x – 5)

Question 9.
y2 – 7y + 12
Solution:
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9 1
RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9 2

Question 10.
z2-4z-12
Solution:
z2 – 4z – 12
= (z)2 – 2 x z x 2 + (2)2 – (2)2 – 12  (completing the square)
= (z)2 – 2 x z x 2 + (2)2 – 4 – 12
= (z-2)2-16
= (z-2)2-(4)2   {∵ a2 – b2 = (a + b) (a – b)}
= (z – 2 + 4) (z – 2 – 4)
= (z + 2)(z-6)

Hope given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9 are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 17 Quadrilaterals Ex 17A

RS Aggarwal Class 6 Solutions Chapter 17 Quadrilaterals Ex 17A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 17 Quadrilaterals Ex 17A

Other Exercises

Question 1.
Solution:
In the figure, a quadrilateral
(i) Its diagonals are AC and BD
(ii) Two pairs of opposite sides are AB, CD and AD, BC
(iii) Two pairs of opposite angles are ∠A, ∠C and ∠B, ∠D
(iv) Two pairs of adjacent sides are AB, BC and CD and DA
(v) Two pairs of adjacent angles are ∠A, ∠B and ∠B, ∠C

Question 2.
Solution:
Steps of construction :
(i) Draw a line segment AB = 6.5 cm.
RS Aggarwal Class 6 Solutions Chapter 17 Quadrilaterals Ex 17A Q2.1
(ii) At A, draw a ray AE making an angle of 70° with the help of the protractor and cut off AD = 4.8 cm.
(iii) With centre B and radius 4.8 cm and with centre D and radius 6.5 cm, draw two arcs intersecting each other at C.
(iv) Join BC and DC. Then ABCD is the required parallelogram.
(v) Join AC and BD which measures 9.3 cm and 6.6 cm respectively.

Question 3.
Solution:
Perimeter of the parallelogram = 56 cm
Ratio in sides = 4 : 3
Let first side = 4x
Then second side = 3x
Perimeter = 2 x sum of two sides
=> 56 = 2 x (4x + 3x)
=> 7x × 2 = 56
=> 14x = 56
=> x = \(\\ \frac { 56 }{ 14 } \)
= 4
First side = 4x = 4 × 4 = 16 cm and second side = 3x = 3 × 4 = 12 cm. Ans.

Question 4.
Solution:
(a) A parallelograms whose diagonals are equal and adjacent sides are unequal, is a rectangle.
(b) A parallelogram whose diagonal are equal and also side are equal, is a. square.
(c) A parallelogram whose diagonal are unequal but adjacent sides are equal is a rhombus.

Question 5.
Solution:
A quadrilateral whose one pair of opposite sides are equal but other pair non parallel, is called a trapezium
When the non-parallel sides of a trapezium are equal, then it is called an isosceles trapezium.
ABCD is an isosceles trapezium in which
AD = BC
Then ∠DAB = ∠CBA
On measuring, AD = BC = 3 cm
and ∠DAB = ∠CBA = 60°
RS Aggarwal Class 6 Solutions Chapter 17 Quadrilaterals Ex 17A Q5.1

Question 6.
Solution:
(a) False , Diagonals of a parallelogram are not equal.
(b) False , Diagonals of a rectangle do not bisect each other at right angles.
(c) False , Diagonals of a rhombus are not equal.

Question 7.
Solution:
(a) Because if each side of a rectangle are equal it is called a square.
(b) Square is a special rhombus if its each angle is equal i.e., of 90°.
(c) If in a parallelogram, if each angle is of 90°, it is called a rectangle.
(d) A square is a parallelogram whose each side and each angle are equal.

Question 8.
Solution:
A regular quadrilateral is a quadrilateral if its each side and angles are equal square is a regular quadrilateral.

 

Hope given RS Aggarwal Solutions Class 6 Chapter 17 Quadrilaterals Ex 17A are helpful to complete your math homework.

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