Class 6 Maths RS Aggarwal Solutions

RS Aggarwal Class 6 Solutions Chapter 12 Parallel Lines Ex 12

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 12 Parallel Lines Ex 12.

Question 1.
Solution:
In the given figure, the pairs of parallel edges are :
AB || CD and BC || AD

Question 2.
Solution:
In the given figure, pairs of all possible parallel edges are:
AB | | HE | | DC || GF ; BE | | CF || AH
| | DG ; AD | | GH | | BC | | EF

Question 3.
Solution:
(i) In the given figure,
DE || BC

(ii) In the given figure,
AB || DC ; AD || BC

(iii) In the given figure,
AB || DC ; AD || BC

(iv) In the given figure,
LM || RQ, MP || RS, PQ || SL.

(v) In the given figure,
AB || CD, CD || EF, AC || BD, CE || DF, AB || EF.

Question 4.
Solution:
(i) Place the rular so that one of its measuring edges lies along the line. Hold it firmly with one hand. Now place a set square with one arm of the right angle coinciding with the edge of the rular. Read off the distance between i and m on the set square which is 1.7cm.

(ii) Place the rular so that one of the measuring edges of the rular lies along the line /. Hold it firmly with one hand and place a set square with one arm of the right angle coinciding with the edge of the rular. Read off the distance between the lines / and m on the set square which is 1.2 cm.

Question 5.
Solution:
It is given that l || m.
Also AB ⊥ l i.e. AB is the perpendicular distance between two parallel lines l and m.
Again CD ⊥ l i.e. CD is the perpendicular distance between two parallel lines l and m.
But the perpendicular distance between two parallel lines is always same everywhere.
CD = AB = 2.3 cm.

Question 6.
Solution:
In the given figure, we see that the line segments AB and CD do not intersect. But, the corresponding lines determined by them will clearly intersect. So, the segment AB and CD are not parallel.

Question 7.
Solution:
(i) Place the rular so that one of its measuring edges lies along of its measuring edges lies along the line l. Hold it firmly and place a set square with one arm of the right angle coinciding with the edge of the rular. Draw the line segment AB along the edge of the set square as shown in figure.

Slide the set square along the rular and draw some more segments CD and EF. We observe that AB = CD = EF.
l || m.
(ii) Place the rular so that one of its measuring edges lies along the line l. Hold it firmly and place a set square with one arm of the right angle conciding with the edge of the rular. Draw the line segment AB along the edge of the set square.

Slide the set square along the rular and draw some more segments CD and EF as shown in the figure.
We observe that AB ≠ CD ≠ EF
Hence l is not parallel to m.

Question 8.
Solution:
(i) True
(ii) True
(iii) False
(iv) False.

 

Hope given RS Aggarwal Solutions Class 6 Chapter 12 Parallel Lines Ex 12 are helpful to complete your math homework.

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

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 11B.

Other Exercises

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

Mark against the correct answer in each of following.

Question 1.
Solution:
(c) a line has no end points

Question 2.
Solution:
(b) a ray has no end points

Question 3.
Solution:
(a) a line segment has two end points

Question 4.
Solution:
(b) a line segment has definite length

Question 5.
Solution:
(b) a line segment can be drawn on a piece of paper

Question 6.
Solution:
(d) unlimited number can be drawn passing through a given point

Question 7.
Solution:
(a) one only can be drawn passing through two given points

Question 8.
Solution:
Two planes intersect in a line. (c)

Question 9.
Solution:
Two lines intersect at a point. (a)

Question 10.
Solution:
Two points in a plane determine exactly one line segment. (a)

Question 11.
Solution:
The minimum number of points of intersection of three lines in a plane is 0.(d)

Question 12.
Solution:
The maximum number of a points of intersection of three lines in a plane is 3 (d)

Question 13.
Solution:
Every line segment has a definite length. (c)

Question 14.
Solution:
Ray \(\overrightarrow { AB } \) not same as ray \(\overrightarrow { BA } \) Both are different. Hence \(\overrightarrow { AB } \) same as \(\overrightarrow { BA } \) is false.(b)

Question 15.
Solution:
An unlimited number of rays can be drawn with a given point as the initial point. (c)

Hope given RS Aggarwal Solutions Class 6 Chapter 11 Line Segment, Ray and Line Ex 11B are helpful to complete your math homework.

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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

• RS Aggarwal Solutions Class 6 Chapter 17 Quadrilaterals Ex 17A
• RS Aggarwal Solutions Class 6 Chapter 17 Quadrilaterals Ex 17B

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.

(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°

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.

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

RS Aggarwal Class 6 Solutions Chapter 16 Triangles Ex 16B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 16 Triangles Ex 16B

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 16 Triangles Ex 16A
• RS Aggarwal Solutions Class 6 Chapter 16 Triangles Ex 16B

Objective questions
Mark against the correct answer in each of following.

Question 1.
Solution:
(c) ∵ It has three sides and three angles i.e. six.

Question 2.
Solution:
(b) ∵ Sum of three angles of a triangle is 180°.

Question 3.
Solution:
(b) ∵ Largest angle
\(=\frac { { 180 }^{ O }\times 4 }{ 2+3+4 } =\frac { { 180 }^{ O }\times 4 }{ 9 } \)
= 80°

Question 4.
Solution:
(d) ∵ A triangle has 180° and if two angles are complementary i.e. sum of two angles is 90°, then third angle will be 180° – 90° = 90°.

Question 5.
Solution:
(c) ∵ Sum of three angles is 180° and sum of two equal angles = 70° + 70° = 140°, then third angle will be 180°- 140° = 40°.

Question 6.
Solution:
(c) ∵ A scalene triangle has different sides.

Question 7.
Solution:
In an isosceles ∆ABC, ∠B = ∠C and bisector of ∠B and ∠C meet at O and ∠A = 40°

Question 8.
Solution:
Side of a triangle are in the ratio 3:2:5 and perimeter = 30 m
Length of longest side = \(\frac { 30\times 5 }{ 3+2+5 } \)
= \(\frac { 30\times 5 }{ 10 } \) cm
= 15 cm (b)

Question 9.
Solution:
Two angles of a triangle are 30° and 25° But sum of three angles of a triangle – 180°
Third angle = 180° – (30 + 25°)
= 180° – 55° = 125° (d)

Question 10.
Solution:
Each angles of an equilateral triangle = 60°
as each angle of an equilateral triangle are equal
Each angle = \(\\ \frac { 180 }{ 3 } \) = 60° (c)

Question 11.
Solution:
In the figure, P lies on AB
Its lies on the ∆ABC (c)

Hope given RS Aggarwal Solutions Class 6 Chapter 16 Triangles Ex 16B 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.

RS Aggarwal Class 6 Solutions Chapter 16 Triangles Ex 16A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 16 Triangles Ex 16A

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 16 Triangles Ex 16A
• RS Aggarwal Solutions Class 6 Chapter 16 Triangles Ex 16B

Question 1.
Solution:
A, B and C are three non-collinear points in a plane. AB, BC and CA are joined.

(i) The side opposite ∠C is AB
(ii) The angle opposite to the side BC is ∠A
(iii) The vertex opposite to the side CA is B
(iv) The side opposite to the vertex B is CA

Question 2.
Solution:
The measures of two angles of a triangle are 72° and 58°
But measure of three angles of a triangle is 180°
Third angle will be = 180 – (72° + 58°)
= 180° – 130°
= 50°

Question 3.
Solution:
Sum of three angles of a triangle = 180°
Ratio of three angles = 1 : 3 : 5

Hence, three angles are 20°, 60° and 100°
Ans.

Question 4.
Solution:
Sum of three angles of a right triangle = 180°
Sum of two acute angles = 180°- 90°
= 90°
Measure of one angle = 50°
Second acute angle = 90° – 50° = 40°
Ans.

Question 5.
Solution:
Let the measure of each of the equal angles be x°. Then,
x° + x°+ 110°= 180°
(Angle sum property of a triangle)
=> 2x°+110°= 180°
=> 2x° = 180° – 110° = 70°
=> \({ x }^{ O }={ \left( \frac { 70 }{ 2 } \right) }^{ O }={ 35 }^{ O }\)
The measure of each of the equal angles is 35°.

Question 6.
Solution:
Let the three angles of a triangle be ∠A, ∠B, ∠C. Then, ∠A = ∠B + ∠C
Adding ∠A to both sides, we get ∠A + ∠A = ∠A + ∠B + ∠C
=> 2 ∠A = 180°
(Angle sum property of a triangle)
=> ∠A = \({ \left( \frac { 180 }{ 2 } \right) }^{ O }={ 90 }^{ O }\)
One of the angles of the triangle is a right angle.
Hence, the triangle is a right triangle

Question 7.
Solution:
In a ∆ABC,
3∠A = 4∠B = 6∠C = 1 (say)

Question 8.
Solution:
(i) It is obtuse triangle.
(ii) It is acute triangle.
(iii) It is right triangle.
(iv) It is obtuse triangle.

Question 9.
Solution:
(i) It is an isosceles triangle as it has two equal sides.
(ii) It is an isosceles triangle as it has two equal sides.
(iii) It is a scalene triangle as its sides are different in length.
(iv) It is an equilateral triangle as its all sides are equal.
(v) It is an equilateral triangles as its angles are equal, so its sides will also be equal.
(vi) It is an isosceles triangle as its two base angles are equal, so its two sides are equal.
(vii) It is a scalene triangle as its angles are different, so its sides will also be different or unequal.

Question 10.
Solution:
In ∆ABC, D is a point on BC and AD is joined
Now we get triangles ∆ABC, ∆ABD and ∆ADC

Question 11.
Solution:
(i) No
(ii) No
(iii) Yes
(iv) No
(v) No
(vi) Yes.

Question 12.
Solution:
(i) three, three, three.
(ii) 180°
(iii) different
(iv) 60°
(v) equal
(vi) perimeter.

 

Hope given RS Aggarwal Solutions Class 6 Chapter 16 Triangles Ex 16A 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.