CBSE Class 7

RS Aggarwal Class 7 Solutions Chapter 19 Three-Dimensional Shapes Ex 19

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 19 Three-Dimensional Shapes Ex 19.

Question 1.
Solution:
(i) A cuboid has six rectangular faces 12 edges and 8 vertices.
(ii) A cylinder has one curved face and two flat faces.
(iii) A cone has one curved face and one flat face..
(iv) A sphere has a curved surface.

Question 2.
Solution:
(i) True : A cylinder has no vertex
(ii) True : A cube has 6 faces, 12 edges and 8 vertices.
(iii) True : A cone has one vertex.
(iv) False : A sphere has no edge.
(v) True : A sphere has one curved surface

Question 3.
Solution:
(i) Examples of cone : Ice cream cone, conical tent, conical cap, conical vessal
(ii) Examples of a sphere : A ball, a football, a volleyball, a basket ball, a hand ball.
(iii) Examples of a cuboid : A tin, a cardboard box, a book, a room, a matchbox.
(iv) Examples of a cylinder : Circular pipe, a pencil, road roller, a gas cylinder, a jar.

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RS Aggarwal Class 7 Solutions Chapter 18 Reflection and Rotational Symmetry Ex 18B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 18 Reflection and Rotational Symmetry Ex 18B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 18 Reflection and Rotational Symmetry Ex 18A
• RS Aggarwal Solutions Class 7 Chapter 18 Reflection and Rotational Symmetry Ex 18B

Question 1.
Solution:
(a) An equilateral triangle has three lines of symmetry which are the angle bisectors.
(b) It has three order of rotational symmetry.

Question 2.
Solution:
The rectangle should be rotated through 180° and 360° to be in symmetrical position with the original position as given below :

Question 3.
Solution:
A square has four orders of rotational symmetry and angles through which the rotational symmetry are 90°, 180°, 270° and 360° as given below:

Question 4.
Solution:
(i) A rhombus has two lines of symmetry which are its diagonal.
(ii) Order of rotational symmetry of a rhombus is not possible. Therefore it has no rotational symmetry.

Question 5.
Solution:
Three letters of the English Alphabet which have two lines of symmetry and rotational symmetry of order 2 are H, I and N.

Question 6.
Solution:
The figure which has only on line of symmetry but no rotational symmetry order is an isosceles triangle.

Question 7.
Solution:
No, only isosceles trapezium has a line of symmetry but not every trapezium.

Question 8.
Solution:
The line of symmetry of a semicircle is the perpendicular bisector of the diameter No, it has not any rotational symmetry.

Question 9.
Solution:
A scalene triangle has neither any line of symmetry nor a rotational symmetry.

Question 10.
Solution:
In the given figure, the line of symmetry has been drawn which is one. There is no rotational symmetry of this figure.

Question 11.
Solution:
(i) The given figure has two lines of symmetry which has been drawn.

(ii) It has three orders of the rotational symmetry which are 90°, 270° and 360°.

Question 12.
Solution:
There is one letter of the English Alphabet Z which has no line of symmetry but it has order two of rotational symmetry of 180° and 360°.

Hope given RS Aggarwal Solutions Class 7 Chapter 18 Reflection and Rotational Symmetry Ex 18B are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 18 Reflection and Rotational Symmetry Ex 18A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 18 Reflection and Rotational Symmetry Ex 18A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 18 Reflection and Rotational Symmetry Ex 18A
• RS Aggarwal Solutions Class 7 Chapter 18 Reflection and Rotational Symmetry Ex 18B

Tick (✓) the correct answer in each of Q.1 to Q.9
Question 1.
Solution:
(a) A scalene triangle has no line of symmetry

Question 2.
Solution:
(c) A rectangle has two lines of symmetry which are the lines joining the mid points of opposite sides.

Question 3.
Solution:
(d) A square has four lines of symmetry which are two diagonal and two the lines joining the mid points of opposite sides.

Question 4.
Solution:
(b) A rhombus has two lines of symmetry which are the diagonals.

Question 5.
Solution:
(d) A circle has an unlimited number of lines of symmetry as its lines of symmetry is its diameter which are infinite in number.

Question 6.
Solution:
(a) ∆ABC in which AB = AC, is an isosceles triangle and AD ⊥ BC.

AD is the line of symmetry

Question 7.
Solution:
(a) ABCD is a kite in which AB = AD and BC = DC

Its line of symmetry will be one diagonal AC.

Question 8.
Solution:
(c) The letter O of the English Alphabet has two lines of symmetry as shown here in the figure.

Question 9.
Solution:
(a) The letter Z of the English Alphabet has no line of symmetry.

Question 10.
Solution:
The line/lines of symmetry have been drawn as given below :

Question 11.
Solution:
(i) True
(ii) True
(iii) True : The bisectors of angles are its lines of symmetry.
(iv) False : A rhombus has two lines of symmetry which are its diagonals.
(v) True : The two diagonals and two perpendicular bisector of its opposite sides are the lines of symmetry.
(vi) True : The perpendicular bisectors of opposite sides are the two lines of symmetry of the rectangle.
(vii) True : Each of the English Alphabet H, I, O and X has two lines of symmetry.

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RS Aggarwal Class 7 Solutions Chapter 17 Constructions CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 17 Constructions CCE Test Paper.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17A
• RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17B
• RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17C
• RS Aggarwal Solutions Class 7 Chapter 17 Constructions CCE Test Paper

Question 1.
Solution:
Given, ∠ABO = 60°
∠CDO = 40°
⇒ ∠ABO = ∠BOC = 60° [alternate angles]

Question 2.
Solution:
Here, AB || EC
∠BAC = ∠ACE = 70° (alternate angles)
⇒ ∠BCA = 180° – ∠BAC
⇒ ∠BCA = 180°- 120°
⇒ ∠BCA = 60°

Question 3.
Solution:
(i) ∠AOC = ∠BOD = 50° [vertically opposite angles]
(ii) ∠BOC = 180° – 50° (linear pair)
= 130°

Question 4.
Solution:
Here, 3x + 20 + 2x – 10 = 180
⇒ 5x + 10 = 180
⇒ 5x = 170
⇒ x = 34
∠AOC = (3 x 34 + 20)° = (102 + 20)° = 122°
∠BOC = (2 x 34 – 10)° = (68 – 10)° = 58°

Question 5.
Solution:
In ∆ABC, ∠A + ∠B + ∠C = 180°
⇒ 65° + 45° + ∠C= 180°
⇒ ∠C = 180° – 110° = 90°

Question 6.
Solution:
Let x = 2k and y = 3k
2k + 3k = 120° [Exterior angle property]
⇒ 5k = 120°
⇒ k = 24°
x = 2 x 24° = 48° and y = 3 x 24° = 72°
In ∆ABC :
∠A + ∠B + ∠C = 180°
⇒ 48° + 72° + ∠C = 180°
⇒ ∠C = 180°- 120°
⇒ ∠C = 60°
z = 60°

Question 7.
Solution:
Since it is a right triangle, by using the Pythagoras theorem:
Length of the hypotenuse = √(8² + 15²) = √(64 + 225) = √289 = ± 17 cm
The length of the side can not be negative.

Question 8.
Solution:
Given:
∠BAD = ∠DAC …..(i)
To show that ∆ABC is isosceles, we should show that ∠B = ∠C
AD ⊥ BC, ∠ADB = ∠ADC = 90°
∠ADC = ∠ADB
∠BAD + ∠ABD = ∠DAC + ∠ACD (exterior angle property)
∠DAC + ∠ABD = ∠DAC + ∠ ACD [from equation (i)]
∠ABD = ∠ACD
This is because opposite angles of a triangle ∆ABC are equal.
Hence, ∆ABC is an isosceles triangle.

Mark (✓) against the correct answer in each of the following :
Question 9.
Solution:
(c) 145°
The supplement of 35° = 180° – 35° = 145°

Question 10.
Solution:
(d) 124
x° + 56° = 180° (linear pair)
⇒ x = 180° – 56°
⇒ x = 124
x = 124°

Question 11.
Solution:
(c) 65°
∠ACD = 125°
∠ACD = ∠CAB + ∠ABC (the exterior angles are equal to the sum of its interior opposite angles)
∠ABC = 125° – 60° = 65°

Question 12.
Solution:
(c) 105°
∠A + ∠B + ∠C = 180°
⇒ ∠A = 180° – (40° + 35°)
⇒ ∠A = 105°

Question 13.
Solution:
(c) 60°
Given:
2∠A = 3∠B

Question 14.
Solution:
(b) 55°
In ∆ABC :
A + B + C = 180° …(i)
Given, A – B = 33°
A = 33° + B …(ii)
B – C = 18°
C = B + 18° …(iii)
Putting the values of A and B in equation (i):
⇒ B + 33° + B + B – 18° = 180°
⇒ 3B = 180°
⇒ B = 55°

Question 15.
Solution:
(b) 3√2 cm
Here, AB = AC
In right angled isosceles triangle:
BC² = AB² + AC²
⇒ BC² = AB² + AB²
⇒ BC² = 2AB²
⇒ 36 = 2AB²
⇒ AB² = 18
⇒ AB = √18
⇒ AB = 3√2

Question 16.
Solution:
(i) The sum of the angles of a triangle is 180°.
(ii) The sum of any two sides of a triangle is always greater than the third side.
(iii) In ∆ABC, if ∠A = 90°, then BC² = (AB²) + (BC²)
(iv) In ∆ABC :
AB = AC
AD ⊥ BC
Then, BD = DC
This is because in an isosceles triangle, the perpendicular dropped from the vertex joining the equal sides, bisects the base.
(v) In the given figure, side BC of ∆ABC has produced to D and CE || BA.
If ∠ABC = 50°, then ∠ACE = 50°
AB || CE
∠BAC = ∠ACE = 50° (alternate angles)

Question 17.
Solution:
(i) True
(ii) True
(iii) False. Each acute angle of an isosceles right triangle measures 45°.
(iv) True.

Hope given RS Aggarwal Solutions Class 7 Chapter 17 Constructions CCE Test Paper are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17C.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17A
• RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17B
• RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17C
• RS Aggarwal Solutions Class 7 Chapter 17 Constructions CCE Test Paper

Mark (✓) against the correct answer in each of the following:
Question 1.
Solution:
(c) Supplement of 45° is 135°
135°+ 45° = 180°

Question 2.
Solution:
(b) Complement of 80° is 10°
10° + 80° = 90°

Question 3.
Solution:
(b) The angle is its own complement.
The measure of the angles will be 45° (45° + 45° = 90°)

Question 4.
Solution:
(a) The angle is one-fifth of its supplement
Let angle be x, then
x + 5x = 180°
⇒ 6x = 180°
⇒ x = 30°
Angle is 30°

Question 5.
Solution:
(b) Let angle is x
Then its complement angle=x-24° But x + x- 24° = 90°
⇒ 2x = 90° + 24° = 114°
⇒ x = 57°
The required angle is 57°

Question 6.
Solution:
(b) Let required angle = x
Then its supplement angle = x + 32
But x + x + 32° = 180°
⇒ 2x = 180° – 32 = 148°
⇒ x = 74°
Required angle = 74°

Question 7.
Solution:
(c) Two supplementary angle are in the ratio = 3 : 2
Let first angle = 3x
Second angle = 2x
But 3x + 2x = 180°
⇒ 5x = 180°
⇒ x = 36°
Smaller angle = 2x = 2 x 36° = 72°

Question 8.
Solution:
(b) In the figure ∠BOC = 132°
But ∠AOC + ∠BOC =180° (Linear pair)
⇒ ∠AOC + 132° = 180°
⇒ ∠AOC = 180° – 132° = 48°

Question 9.
Solution:
(c) In the figure, ∠AOC = 68°
But ∠AOC + ∠BOC = 180° (Linear pair)
⇒ 68° + x = 180°
⇒ x = 180° – 68° = 112°

Question 10.
Solution:
(b) In the figure,
AOB is a straight line
∠AOC + ∠BOC = 180° (Linear pair)
⇒ 2x – 10° + 3x + 15° = 180°
⇒ 5x = 180° + 10° – 15° = 175°
⇒ x = 35°
x = 35

Question 11.
Solution:
(d) In the figure,
AOB is a straight line
∠AOC + ∠COD + ∠DOB = 180°
⇒ 55° + x + 45° = 180°
⇒ x + 100° = 180°
⇒ x = 180° – 100° = 80°

Question 12.
Solution:
(a) AOB is a straight line
x + y = 180°
But 4x = 5y

Question 13.
Solution:
(b) AB and CD intersect each other at O and ∠AOC = 50°
∠BOD = ∠AOC = 50° (Vertically opposite angles)

Question 14.
Solution:
(a) AOB is a straight line
∠AOC + ∠COD + ∠DOB = 180°
⇒ 3x – 8° + 50° + x + 10° = 180°
⇒ 4x = 180° + 8° – 50° – 10°
⇒ 4x = 128°
⇒ x = 32°

Question 15.
Solution:
(b) In ∆ABC, side BC is produced to D
∠ACD = 132° and ∠A = 54°
Ext. ∠ACD = ∠A + ∠B
⇒ 132° = 54° + ∠B
⇒ ∠B = 132° – 54° = 78°

Question 16.
Solution:
(c) In ∆ABC,
Side BC is produced to D
∠A = 45°, ∠B = 55°
Ext. ∠ACD = ∠A + ∠B = 45° + 55° = 100°

Question 17.
Solution:
(b) In ∆ABC, side BC is produced to D
∠ABC = 70° and ∠ACD = 120°
Ext. ∠ACD = ∠BAC + ∠ABC
⇒ 120° = ∠BAC + 70°
⇒ ∠BAC = 120° – 70° = 50°

Question 18.
Solution:
(c) In the figure,
∠AOB = 50°, ∠BOC = 90°
∠COD = 70°, ∠AOD = x.
But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360° (Angles at a point)
⇒ 50° + 90° + 70° + x = 360°
⇒ 210 + x = 360°
⇒ x = 360° – 210°
⇒ x = 150°

Question 19.
Solution:
(c) In the figure,
Side BC of ∆ABC is produced to D
CE || BA is drawn
∠A = 50° and ∠ECD = 60°
AB || CE
∠ABC = ∠ECD (corresponding angle) = 60°
But in ∆ABC,
∠A + ∠B + ∠ACB = 180° (Angles of a triangles)
⇒ 50° + 60° + ∠ACB = 180°
⇒ ∠ACB = 180° – 50° – 60° = 70°

Question 20.
Solution:
(b) In ∆ABC,
∠A = 65°, ∠C = 85°
But ∠A + ∠B + ∠C = 180° (Angles of a triangle)
⇒ 65° +∠B+ 85° = 180°
⇒ 150° + ∠B = 180°
⇒ ∠B = 180° – 150° = 30°

Question 21.
Solution:
(d) Sum of angles of a triangle = 180°

Question 22.
Solution:
(c) Sum of angles of a quadrilateral = 360°

Question 23.
Solution:
(b) In the figure, AB || CD
∠OAB = 150°, ∠OCD = 120°

From O, draw OE || AB or CD
AB || DE
∠OAB + ∠AOE =180°
⇒ 150° + ∠AOE = 180°
⇒ ∠AOE = 180° – 150° = 30°
Similarly DE || CD
∠EOC + ∠OCD = 180°
⇒ ∠EOC + 120° = 180°
⇒ ∠EOC = 180° – 120° = 60°
Now ∠AOC = ∠AOE + ∠EOC = 30° + 60° = 90°

Question 24.
Solution:
(a) In the given figure,
PQ || RS,
∠PAB = 60° and ∠ACS = 100°
PQ || RS
∠ABC = ∠PAB (alternate angles) = 60°
But Ext. ∠ACS = ∠BAC + ∠ABC
⇒ 100° = ∠BAC + 60°
⇒ ∠BAC = 100° – 60° = 40°

Question 25.
Solution:
(c) In the figure, AB || CD || EF
∠ABG =110° and ∠GCD = 100°
∠BGC = x°
AB || EF
∠ABG + ∠BGE = 180°
⇒ 110° + ∠BGE = 180°
⇒ ∠BGE = 180° – 110° = 70°
Similarly CD || EF
∠GCD + ∠CGF = 180°
⇒ 100° + ∠CGF = 180°
⇒ ∠CGF = 180° – 100° = 80°
But ∠BGE + ∠BGC + ∠CGF = 180°
⇒ 70° + x + 80° = 180°
⇒ 150° + x = 180°
⇒ x = 180° – 150° = 30°

Question 26.
Solution:
(d) Sum of any two sides of a triangle is always greater than the third side

Question 27.
Solution:
(d) The diagonals of a rhombus always bisect each other at right angles.

Question 28.
Solution:
(c) In ∆ABC, ∠B = 90°
AB = 5 cm and AC = 13 cm
But AC² = AB² + BC² (By Pythagoras Theorem)
⇒ (13)² = (5)² + BC²
⇒ 169 = 25 + BC2
⇒ BC² = 169 – 25 = 144 = (12)²
BC = 12 cm

Question 29.
Solution:
(c) In ∆ABC, ∠B = 37°, ∠G = 29°
But ∠A + ∠B + ∠C = 180° (angles of a triangle)
⇒ ∠A + 37° + 29° = 180°
⇒ ∠A + 66° = 180°
⇒ ∠A = 180° – 66° = 114°

Question 30.
Solution:
(c) The ratio of angles of a triangle is 2 : 3 : 7
But sum of angles of a triangle = 180°

Question 31.
Solution:
In ∆ABC,
Let 2∠A = 3∠B = 6∠C = x

Question 32.
Solution:
(a) In ∆ABC,
∠A + ∠B = 65°, ∠B + ∠C = 140°
∠A = 65°
∠C = 140° – ∠B
But ∠A + ∠B + ∠C = 180° (Angles of a triangle)
⇒ 65° – ∠B + 140° – ∠B + ∠B = 180°
⇒ 205° – ∠B = 180°
⇒ ∠B = 205° – 180° = 25°

Question 33.
Solution:
(b) In ∆ABC, ∠A – ∠B = 33°
and ∠B – ∠C = 18°
∠A = 33° + ∠B and ∠C = ∠B – 18°
But ∠A + ∠B + ∠C = 180°
⇒ 33° + ∠B + ∠B + ∠B – 18° = 180°
⇒ 3∠B = 180° – 33° + 18° = 165°
⇒ ∠B = 55°

Question 34.
Solution:
(c) In ∆ABC
∠A + ∠B + ∠C= 180° (Sum of angles of a triangle)
But angles are (3x)°, (2x – 7)° and (4x – 11)°
3x + (2x – 7) + (4x – 11)° = 180°
⇒ 3x + 2x – 7 + 4x – 11° = 180°
⇒ 9x – 18° = 180°
⇒ 9x = 180° + 18° = 198°
⇒ x = 22°

Question 35.
Solution:
(c) ∆ABC is a right angled, ∠A = 90°
AB = 24 cm, AC = 7 cm

but BC² = AB² + AC²
⇒ BC² = (24)² + (7)² = 576 + 49 = 625 = (25)²
BC = 25 cm

Question 36.
Solution:
(b) Let AB is a ladder and A is the window
BC = 15 m, AC = 20 m

Now in right ∆ABC
AB² = BC² + AC² = (15)² + (20)² = 225 + 400 = 625 = (25)²
AB = 25 m
Length of ladder = 25 m

Question 37.
Solution:
(a) Let AB and CD are two poles such that
AB = 6 m, CD = 11 m
and distance between two poles BD = 12m
From A, draw AE || BD
AE = BD = 12m
CE = CD – ED = 11 – 6 = 5 m

Now in right ∆AEC
AC² = AE² + CE² = (12)² + (5)² = 144 + 25 = 169 = (13)²
AC = 13 m
Distance between tops of poles = 13 m

Question 38.
Solution:
(d) ∆ABC is an isosceles triangle
∠C = 90°,
AC = 5 cm
BC = AC = 5 cm

In right ∆ABC
AB² = AC² + BC² = (5)² + (5)² = 25 + 25 = 50 = 2 x 25
AB = √(2 x 25) = 5√2 cm

Hope given RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17C are helpful to complete your math homework.

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