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Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.7 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.7

Question 1.
Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm
Answer:
For the given cone, radius r = 6 cm and height h = 7 cm.
Volume of a cone
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 6 × 6 × 7 cm³
= 264 cm³

(ii) radius 3.5 cm, height 12 cm
Answer:
For the given cone, radius
r = 3.5 cm = \(\frac{7}{2}\) cm and height h = 12 cm.
Volume of a cone
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 12 cm³
= 154 cm³

Question 2.
Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm
Answer:
For the given conical vessel, radius r = 7 cm and slant height l = 25 cm.
h = \(\sqrt{l^{2}-r^{2}}\)
= \(\sqrt{25^{2}-7^{2}}\)
= \(\sqrt{625-49}\)
= √576
∴ h = 24 cm
Capacity of conical vessel
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 7 × 7 × 24 cm³
= 1232 cm³
= \(\frac{1232}{1000}\) liters
= 1.232 litres

(ii) height 12 cm, slant height 13 cm
Answer:
For the given conical vessel, height
h = 12 cm and slant height l = 13 cm.
r = \(\sqrt{l^{2}-h^{2}}\)
= \(\sqrt{13^{2}-12^{2}}\)
= \(\sqrt{169-144}\)
= \(\sqrt{25}\)
∴ r = 5 cm
Capacity of conical vessel
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 5 × 5 × 12 cm³
= \(\frac{6600}{21}\)
= \(\frac{6600}{21 \times 1000}\) liters
= \(\frac{11}{35}\) liters

Question 3.
The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14.)
Answer:
For the given cone, height h = 15 cm and
volume = 1570 cm³.
Volume of a cone = \(\frac{1}{3}\) πr²h
∴ 1570 cm³ = \(\frac{1}{3}\) × 3.14 × r² × 15 cm³
∴ 1570 cm³ = 15.7 × r² cm³
∴ r² = \(\frac{1570}{15.7}\) cm²
∴ r² = 100 cm²
∴ r = 10 cm

Question 4.
If the volume of a right circular cone of height 9 cm is 48 πcm³, find the diameter of its base.
Answer:
For the given right circular cone, height h = 9 cm and volume = 48 πcm³
Volume of a cone = \(\frac{1}{3}\) πr²h
∴ 48π cm³ = \(\frac{1}{3}\) × π × r² × 9 cm³
∴ r² = \(\frac{48 \pi \times 3}{\pi \times 9}\) cm²
∴ r² = 16 cm²
∴ r² = 16 cm²
∴ r = 4 cm
Now, diameter = 2r = 2 × 4 cm = 8 cm
Thus, the diameter of the right circular cone is 8 cm.

Question 5.
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kiloliters ?
Answer:
For the given conical pit,
radius r = \(\frac{\text { diameter }}{\mathbf{2}}\) = \(\frac{3.5}{2}\) m = \(\frac{35}{2}\) m
and height (depth) h = 12 m
Capacity of the conical pit
= Volume of the cone
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{35}{20}\) × \(\frac{35}{20}\) × 12 m³
= 38.5 m³
= 38.5 kilioliters

Question 6.
The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find
(i) height of the cone,
(ii) slant height of the cone and
(iii) curved surface area of the cone.
Answer:
For the given right circular cone,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{28}{2}\) cm = 14 cm and
volume = 9856 cm³.

(i) Volume of a cone = \(\frac{1}{3}\) πr²h
∴ 9856 cm³ = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 14 × 14 × h
∴ h = \(\frac{9856 \times 3 \times 7}{22 \times 14 \times 14}\) cm
∴ h = 48 cm

(ii) l = \(\sqrt{h^{2}+r^{2}}\)
= \(\sqrt{48^{2}+14^{2}}\)
= \(\sqrt{2304+196}\)
= \(\sqrt{2500}\)
∴ l = 50 cm

(iii) Curved surface area of a cone
= πrl
= \(\frac{22}{7}\) × 14 × 50 cm²
= 2200 cm²

Question 7.
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm.. Find the volume of the solid so obtained.
Answer:
A right circular cone is received when ∆ ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm.
For the cone so obtained, radius r = 5 cm, height h = 12 cm and slant height l = 13 cm.
Volume of the cone obtained
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × π × 5 × 5 × 12 cm³
= 100π cm³

Question 8.
If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two ‘solids obtained in Questions 7 and 8.
Answer:
Now, if ∆ ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm, again a right circular cone is received.
For the cone so obtained, radiqs r = 12cm; height h = 5 cm and slant height l = 13 cm.
Volume of the cone obtained
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × π × 12 × 12 × 5 cm³
= 240π cm³
Ratio of the volumes of two cones obtained in question 7 and 8 = \(\frac{100 \pi}{240 \pi}\) = \(\frac{5}{12}\) = 5 : 12.

Question 9.
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Answer:
For the conical heap of wheat,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{10.5}{2}\) m = \(\frac{105}{2}\) m and
height h = 3 m
Volume of the conical heap of wheat
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{105}{20}\) × \(\frac{105}{20}\) × 3 m³
= 86.625 m³
To cover the heap with canvas, the area of the canvas required will be equal to the curved surface area of the heap.
Now, l = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{\left(\frac{105}{20}\right)^{2}+(3)^{2}}\)
= \(\sqrt{(5.25)^{2}+9}\)
= \(\sqrt{27.5625+9}\)
= \(\sqrt{36.5625}\)
= 6.05 m (approx.)
Curved surface area of the conical heap
= πrl
= \(\frac{22}{7}\) × \(\frac{105}{20}\) × 6.05 m²
= 99.825 m²
Thus, 99.825 m² canvas is required to cover the conical heap of wheat to protect it from rain.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6

Question 1.
The circumference of the base of a 7 cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm³ = 1l)
Answer:
For the given cylindrical vessel, height h = 25 cm and circumference of the base = 132 cm.
Circumference of base = 2πr
∴ 132 cm = 2 × \(\frac{22}{7}\) × r cm
∴ r = \(\frac{132 \times 7}{2 \times 22}\) cm
∴ r = 21 cm
Hence, for the cylindrical vessel, radius r = 21 cm
Capacity of cylindrical vessel
= Volume of cylinder
= πr²h
= \(\frac{22}{7}\) × 21 × 21 × 25 cm³
= 34650 cm³
= \(\frac{34650}{1000}\) liters
= 34.65 litres
Thus, the cylindrical vessel can hold 34.65 litres of water.

Question 2.
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ of wood has a mass of 0.6 g.
Answer:
For the cylindrical wooden pipe,
outer radius R = \(\frac{\text { diameter }}{2}\) = \(\frac{28}{2}\) cm = 14 cm,
inner radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{24}{2}\) cm = 12 cm and
height (length) h = 35 cm.
Volume of the cylindrical wooden pipe
= Volume of outer cylinder – Volume of inner cylinder
= πR²h – πr²h
= πh (R² – r²)
= πh (R + r) (R – r)
= \(\frac{22}{7}\) × 35 × (14 + 12) (14 – 12) cm³
= 110 × 26 × 2 cm³
= 5720 cm³
Now, mass of 1 cm³ of wood = 0.6 g
∴ Mass of 5720 cm³ of wood = 5720 × 0.6 g
= 3432 g
= 3.432 kg
Thus, the mass of the given pipe is 3.432 kg.

Question 3.
A soft drink is available in two packs
(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm.
Which container has greater capacity and by how much?
Answer:
(i) For the cuboidal container with rectangular base, length l = 5 cm; breadth b = 4 cm and height h = 15 cm.
Capacity of the cuboidal container
= Volume of cuboid
= l × b × h
= 5 × 4 × 15 cm³
= 300 cm³

(ii) For the cylindrical container,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{7}{2}\) cm and
height h = 10 cm.
Capacity of the cylindrical container = Volume of cylinder
= πr²h
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) × 10 cm³
= 385 cm³
Hence, the capacity of the cylindrical container is more than the cuboidal container by 385 – 300 = 85 cm³.

Question 4.
If the lateral surface area of a cylinder is 94.2 cm2 and its height is 5 cm, then find
(i) radius of its base and
(ii) its volume. (Use π = 3.14)
Answer:
(i) For the given cylinder, height h = 5 cm and lateral (curved) surface area = 94.2 cm².
Curved surface area of a cylinder = 2 πrh
∴ 94.2 cm² = 2 × 3.14 × r × 5 cm²
∴ r = \(\frac{94.2}{2 \times 3.14 \times 5}\) cm
∴ r = 3 cm
Thus, the radius of the cylinder is 3 cm.

(ii) Volume of a cylinder
= πr²h
= 3.14 × 3 × 3 × 5 cm³
= 141.3 cm³
Thus, the volume of the cylinder is 141.3 cm³.

Question 5.
It costs ₹ 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m², find
(i) inner curved surface area of the vessel,
(ii) radius of the base and
(iii) capacity of the vessel.
Answer:
(i) Area of the region painted at the cost of ₹ 20 = 1 m²
∴ Area of the region painted at the cost of ₹ 2200 = \(\frac{2200}{20}\) m² = 110m²
Thus, the inner curved surface area of the vessel is 110 m².

(ii) For the cylindrical vessel, height (depth) h- 10 m and curved surface area = 110 m².
Curved surface area of cylindrical vessel = 2πrh
∴ 110 m² = 2 × \(\frac{22}{7}\) × r × 10 m²
∴ r = \(\frac{110 \times 7}{2 \times 22 \times 10}\) m
∴ r = \(\frac{7}{4}\) m
∴ r = 1.75 m
Thus, the radius of the cylindrical vessel is 1.75 m.

(iii) Capacity of the cylindrical vessel
= Volume of a cylinder
= πr²h
= \(\frac{22}{7}\) × 1.75 × 1.75 × 10 m³
= 96.25 m³
= 96.25 kilolitres
Thus, the capacity of the cylindrical vessel is 96.25 kilolitres.

Question 6.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it ?
Answer:
For the closed cylindrical vessel, height h = 1 m and capacity = 15.4 litres
∴ Volume of the vessel = 15.4 litres
= \(\frac{15.4}{1000}\) m³
= 0.0154 m³
Volume of cylindrical vessel = πr²h
∴ 0.0154 m³ = \(\frac{22}{7}\) × r² × 1 m³
∴ r² = \(\frac{154}{10000} \times \frac{7}{22}\) m²
∴ r² = \(\frac{49}{10000}\) m²
∴ r² = \(\frac{7}{100}\) m
∴ r = 0.07 m
Thus, the radius of the cylindrical vessel is 0. 07 m.
Area of the metal sheet required to make closed cylindrical vessel
= Total surface area of a cylinder
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × 0.07 (0.07 + 1) m²
= 0.44 × 1.07 m²
= 0.4708 m²
Thus, 0.4708 m² of metal sheet would be needed to make the closed cylindrical vessel.

Question 7.
A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
Answer:
For the solid cylinder of graphite,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{1}{2}\) mm = \(\frac{1}{20}\) cm and
height h = 14 cm.
Volume of cylinder of graphite
= πr²h
= \(\frac{22}{7}\) × \(\frac{1}{20}\) × \(\frac{1}{20}\) × 14 cm³ = 011 cm³
For the hollow cylinder of wood,
outer radius R = \(\frac{\text { diameter }}{2}\) = \(\frac{7}{2}\) mm = \(\frac{7}{20}\) cm,
inner radius r = \(\frac{1}{20}\) cm and height h = 14 cm.
Volume of hollow cylinder of wood
= πR²h – πr²h
= πh (R² – r²)
= πh (R + r) (R – r)
= \(\frac{22}{7}\) × 14 \(\left(\frac{7}{20}+\frac{1}{20}\right)\left(\frac{7}{20}-\frac{1}{20}\right)\) cm³
= 44 × \(\frac{8}{20}\) × \(\frac{8}{20}\) cm³
= \(\frac{528}{100}\) cm³
= 5.28 cm³
Thus, in the given pencil, the volume of wood is 5.28 cm³ and the volume of graphite is 0.11cm³.

Question 8.
A patient in a hospital is given soup daily in a cylindrical howl of diameter 7 cm. If the bowl is Oiled with soup to a height of 4 cm, how much soup the hospital has ‘ to prepare daily to serve 250 patients ?
Answer:
For the soup served in cylindrical bowl, radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{7}{2}\) cm and height h = 4 cm.
Volume of soup served to one patient = Volume of a cylinder
= πr²h
= \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 4 cm³
= 154 cm³
Thus, the volume of soup served to 1 patient =154 cm³
∴ The volume of soup served to 250 patients
= 154 × 250 cm³
= 38500 cm³
= \(\frac{38500}{1000}\) litres
= 38.5 litres
Thus, the hospital has to prepare 38500 cm³, i.e., 38.5 litres of soup daily.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 15 Visualising Solid Shapes Ex 15.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.3

1. Count the number of cubes in each of the following figures :


Solution:
(i) 6
(ii) 21
(iii) 32
(iv) 13

2. If three cubes of dimensions 2 cm × 2 cm × 2 cm are placed side by side, what would be the dimensions of resulting cuboid ?

Solution:
Length 6 cm, breadth 2 cm and height 2 cm.

3. If we throw light an the following solids from the top name the shape of shadow obtained in each case and also give a rough sketch of the shadow.
(i) DVD player

Solution:

(ii) Sandwich

Solution:

(iii) Straw

Solution:

4. What cross-sections do you get when you given a.
(i) Vertical cut
(ii) Horizontal cut.
to the following solids ?
(a) A die
(b) A square pyramid
(c) A round melon
(d) A circular pipe
(e) A brick
(f) An ice cream cone.
Solution:
(a) Square, Square,
(b) Triangle, Square,
(c) Circle, circle
(d) Circle, Rectangle
(e) Rectangle, Rectangle
(f) Triangle, Circle.

5. If we throw light on following solids, from left name the shape of shadow in each and also give a rough sketch of the shadow.
(i)

Solution:

(ii)

Solution:

6. Here are the shadows of some 3-D objects, when seen under the lamp of an overhead projector. Identify the solids that match each shadow (There may be multiple answers for these)

Solution:
(i) Dice, chalk box etc
(ii) Book, Mobile, DVD Player etc
(iii) Cricket ball, Disc etc.
(iv) Birthday cap, Icecream cone etc.

7. Sketch the front, side and top view of the following figures :
(i)

Solution:

(ii)

Solution:

(iii)

Solution:

(iv)

Solution:

8. Multiple choice questions :

Question (i).
The number of cubes in the given structure is ?

(a) 12
(b) 10
(c) 9
(d) 8
Answer:
(a) 12

Question (ii).
The number of unit cubes to be added in above students to make a cuboid of dimensions 4 unit × 2 unit × 3 unit is ?
(a) 11
(b) 12
(c) 13
(d) 14
Answer:
(b) 12

Question (iii).
What cross-section is made by vertical cut in a cuboid
(a) Square
(b) Rectangle
(c) Circle
(d) Triangle
Answer:
(b) Rectangle

Question (iv).
What cross section is made by horizontal cut in a cone
(a) Triangle
(b) Circle
(c) Square
(d) Rectangle
Answer:
(b) Circle

Question (v).
Which solid cost a shadow of triangle under the effect of light
(a) Sphere
(b) Cylinder
(c) Cone
(d) Cube
Answer:
(c) Cone

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 15 Visualising Solid Shapes Ex 15.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.2

1. Use isometric dot paper and make an isometric sketch of the following figures :
Question (i).

Solution:

Question (ii).

Solution:

Question (iii).

Solution:

Question (iv).

Solution:

2. Draw (i) an oblique sketch (ii) Isometric sketch for :
(a) A cube with a edge of 4 cm long
(b) A cuboid of length 6 cm, breadth 4 cm and height 3 cm
Solution:
(a) (i) Oblique sketch of cube with a edge 4 cm long.

(ii) Iso metric sketch for a cube with a edge of 4 cm long.

(b) (i) an oblique sketch for a cuboid of length 6 cm, breadth 4 cm and height 3 cm.

(ii) Isometric sketch for a cuboid of length 6 cm, breadth 4 cm and height 3 cm.

3. Two cubes each with edge 3 cm are placed side by side to form a cuboid, sketch oblique and isometric sketch of this cuboid.
Solution:

4. Draw an Isometric sketch of triangular pyramid with base as equilateral triangle of 6 cm and height 4 cm.
Solution:

5. Draw an Isometric sketch of square pyramid.
Solution:

6. Make an oblique sketch for each of the given Isometric shapes.

Question (i).

Solution:

Question (ii).

Solution:

7. Using an isometric dot paper draw the solid shape formed by the given net.

Solution:

8. Multiple choice questions:

Question (i).
An oblique sheet is made up of:
(a) Rectangles
(b) Squares
(c) Right angled triangles
(d) Equilateral triangles.
Answer:
(b) Squares

Question (ii).
An isometric sheet is made up of dots forming :
(a) Squares
(b) Rectangles
(c) Equilateral triangles
(d) Right angled triangle
Answer:
(c) Equilateral triangles

Question (iii).
An oblique sketch has :
(a) Proportional lengths
(b) Parallel lengths
(c) Non-Proportional lengths
(d) Perpendicular lengths.
Answer:
(c) Non-Proportional lengths

Question (iv).
An Isometric sketch has :
(a) Non proportional lengths
(b) Parallel lengths
(c) Perpendicular lengths
(d) Proportional lengths.
Answer:
(d) Proportional lengths.

Question (v).
Isometric sketches shows objects of:
(a) Two dimensions
(b) Shadows
(c) Three dimensions
(d) One dimension.
Answer:
(c) Three dimensions

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 10 Practical Geometry Ex 10.1

1. What is the use of instrument ruler?
Solution:
We use instrument ruler to draw line segment and to measure their lengths.

2. What is the use of protractor?
Solution:
We use a protractor to draw and measure angle.

3. What is the use of compasses?
Solution:
We use compasses to mark equal lengths, draw arcs and circles.

4. Construct the following angles using set squares.

Question (i)
(i) 30°
(ii) 45°
(iii) 60°
(iv) 75°
(v) 90°.
Solution:
(i) Steps of Construction:

1. Draw a ray OA.

2. To construct an angle of 30° we use 30° set square. Place the set square in such a way that one of its edges containing the 30° angle coincides with the ray OA as shown in the figure.

3. Draw a ray OB starting from the vertex O along the 30° edge of the set square as shown in figure.
4. Remove the set square.
Thus, the required \(\angle \mathrm{AOB}\) = 30°.

(ii) Steps of Construction:

1. Draw a ray OA.

2. To construct an angle of 45° we use 45° set square.
Place a 45° set square along ray OA as shown in the figure.

3. Draw a ray OB starting from the vertex O along 45° the, edge of set square.
4. Remove the 45° set square. Thus, the required \(\angle \mathrm{AOB}\) = 45°

(iii) Steps of Construction.

1. Draw a ray OA.

2. To construct an angle of 60°. We use 30° set square.

Place a 30° set square with 60° edge along ray OA as shown in the figure.
3. Draw a ray OB starting from the vertex O along 60° edge of the set square.
4. Remove the 30° set square. Thus the required \(\angle \mathrm{AOB}\) = 60°.

(iv) Steps of Construction

1. Draw a ray of OA.

2. To draw angle of 75° we use both set squares in combination as 45° + 30° = 75°.
Place 45° set square with 45° edge along OA

3. Place 30° set square adjacent to 45° set square as show. Draw a ray starting from the vertex O along the edge of 30° set square.
4. Remove both the set squares. Thus required \(\angle \mathrm{AOB}\) = 75°.

(v) Steps of Construction

1. Draw a ray OA.

2. To construct angle of 90° place any set square with 90° corner at O along OA.

3. Draw a ray OB starting from the vertex O along 90° edge of the set square.
4. Remove the set square. Thus, the required \(\angle \mathrm{AOB}\) = 90°.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 9 Understanding Elementary Shapes MCQ Questions

Multiple Choice Questions

Question 1.
In the figure, which of the following is true?
(a) PR = PQ
(b) PR > QR
(c) PS > PR
(d) PR < PQ.
Answer:
(b) PR > QR

Question 2.
Which angle is represented in the given figure?
(a) Reflex
(b) Acute
(c) Obtuse
(d) Right angle.
Answer:
(a) Reflex

Question 3.
Which angle is represented in the given figure?
(a) Acute
(b) Right angle
(c) Obtuse
(d) Reflex
Answer:
(b) Right angle

Question 4.
Which of the following is the example of perpendicular lines?
(a) Railway lines
(b) Line segment forming letter ‘X’
(c) Adjacent edges of a table
(d) Line segment forming line ‘M’.
Answer:
(c) Adjacent edges of a table

Question 5.
Which of the following forms triangles?
(a) 60°, 72°, 48°
(b) 73°, 54°, 59°
(c) 60°, 51°, 70°
(d) 100°, 42°, 39°.
Answer:
(a) 60°, 72°, 48°

Question 6.
Which of the following are sides of a triangle?
(a) 1, 2, 3
(b) 2, 2,1
(c) 3, 4, 2
(d) 5, 6, 12.
Answer:
(c) 3, 4, 2

Question 7.
A parallelogram having adjacent sides equal is called a …………… .
(a) Trapezium
(b) Rhombus
(c) Rectangle
(d) Square.
Answer:

Question 8.
Which of the following is not true for rectangle?
(a) Diagonals are equal
(b) Diagonals bisect each other
(c) Each angle is 90°
(d) All sides are equal.
Answer:
(d) All sides are equal

Question 9.
Which of the following is not true?
(a) Every rhombus is a parallelogram
(b) Each square is rhombus.
(c) Each rectangle is a square.
(d) Each square is parallelogram.
Answer:
(c) Each rectangle is a square

Question 10.
A cuboid has ………….. edges.
(a) 10
(b) 6
(c) 12
(d) 8.
Answer:
(c) 12

Question 11.
The following angle is an:

(a) acute angle
(b) obtuse angle
(c) right angle
(d) straight angle.
Answer:
(a) acute angle

Question (ii)
What is the angle name for half a revolution?
(a) acute angle
(b) obtuse angle
(c) straight angle
(d) right angle.
Answer:
(c) straight angle

Question (iii)
What is the angle name for one-fourth revolution?
(a) right angle
(b) straight angle
(c) complete angle
(d) acute angle.
Answer:
(a) right angle

Question (iv)
An angle whose measure is less than that of a right angle is …………….. .
(a) complete angle
(b) acute angle
(c) obtuse angle
(d) straight angle.
Answer:
(b) acute angle

Question (v)
An angle whose measure is greater than that of a right angle:
(a) acute angle
(b) complete angle
(c) obtuse angle
(d) straight angle.
Answer:
(c) obtuse angle

Fill in the blanks:

Question (i)
An angle whose measure is equal to 90°, is called ……………….. .
Answer:
right angle

Question (ii)
An angle whose measure is the sum of the measure of two right angle is ………………. .
Answer:
straight angle

Question (iii)
Your instrument box looks like a ………………. .
Answer:
cuboid

Question (iv)
A road-roller looks like a ……………… .
Answer:
cylinder

Question (v)
A cuboid has ……………. faces.
Answer:
six

Write True/False:

Question (i)
Every rhombus is a parallelogram. (True/False)
Answer:
True

Question (ii)
Every square is a rhombus. (True/False)
Answer:
True

Question (iii)
All sides of a rectangle are equal. (True/False)
Answer:
False

Question (iv)
Each rectangle is a square. (True/False)
Answer:
False

Question (v)
Each angle of a rectangle is of 90°. (True/False)
Answer:
True.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes Ex 9.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 9 Understanding Elementary Shapes Ex 9.6

1. Give two examples of each of the following shapes from your surroundings:

Question (i)
Cube
Solution:
Cube. Examples:
(i) Dice,
(ii) Sugar cubes.

Question (ii)
Cuboid
Solution:
Cuboid. Examples:
(i) Matchbox,
(ii) Geometry box.

Question (iii)
Cone
Solution:
Cone: Examples:
(i) Ice cream cone,
(ii) Joker cap.

Question (iv)
Cylinder
Solution:
Cylinder. Examples:
(i) Drum,
(ii) Circular pipe.

Question (v)
Shpere.
Solution:
Shpere. Examples:
(i) Globe,
(ii) Ball.

2. Classify the following as plane figures and solid figures:

Question (i)
(i) Rectangle
(ii) Sphere
(iii) Cylinder
(iv) Circle
(v) Cube
(vi) Cuboid
(vii) Triangle
(viiii) Cone
(ix) Square
(x) Prism.
Solution:
Plane figures:
(i) Rectangle
(iv) Circle
(vii) Triangle
(ix) Square.

Solid figures:
(ii) Sphere
(iii) Cylinder
(v) Cube
(vi) Cuboid
(viii) Cone
(x) Prism.

3. Write the name of shapes in the base of the following solids:

Question (i)
Cube
Solution:
Square

Question (ii)
Cylinder
Solution:
Circle

Question (iii)
Tetrahedron
Solution:
Equilateral triangle

Question (iv)
Cuboid
Solution:
Rectangle

Question (v)
Square Pyramid.
Solution:
Square.

4. Fill in the table:

Shape	Number of Flat Faces	Number of Curved Faces	Number of Vertices	Number of Edges
(i) Cuboid
(ii) Cube
(iii) Cylinder
(iv) Cone
(v) Sphere
(vi) Triangular Prism
(vii) Square Pyramid
(viii) Tetrahedron

Solution:

Shape	Number of	Number of	Number of	Number of
	Flat Faces	Curved Facet!	Vertices	Edges
(i) Cuboid	6	Nil	8	12
(ii) Cube	6	Nil	8	12
(iii) Cylinder	2	1	Nil	2
(iv) Cone	1	1	1	1
(v) Sphere	Nil	1	Nil	Nil
(vi) Triangular Prism	5	Nil	6	9
(vii) Square Pyramid	5	Nil	5	8
(viii) Tetrahedron	4	Nil	4	6

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes Ex 9.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 9 Understanding Elementary Shapes Ex 9.5

1. Which of the following are polygons and there is no polygon. Give the reason:

Solution:
(i) It is not a closed figure. Therefore it is not a polygon.
(ii) It is made up of lines segment. Therefore it is polygon.
(iii) It is not a polygon, because it is not made of line segments.
(iv) It is not closed by line segment. Therefore, it is not a polygon.
(v) It is not polygon because line segments are intersecting each other.
(vi) It is made up of line segments, therefore it is a polygon.

2. Classify the following as concave or convex polygons:

Solution:
(i) Concave Polygon
(ii) Convex Polygon
(iii) Concave Polygon
(iv) Concave Polygon
(v) Convex Polygon
(vi) Convex Polygon.

3. Tick in the boxes, if the property holds true for a particular quadrilateral otherwise eroes out ‘x’:

Quadrilateral Properties	Rectangle	Parallelogram	Rhombus	Trapezium	Square
All sides are equal
Only opposite sides are equal
Diagonals are equal
Diagonals bisect each other
Diagonals are perpendicular to each other
Each angle is 90°

Solution:

Quadrilateral Properties	Rectangle	Parallelogram	Rhombus	Trapezium	Square
All sides are equal	×	×	√	×	√
Only opposite sides are equal	√	√	×	×	×
Diagonals are equal	√	×	×	×	√
Diagonals bisect each other	√	√	√	×	√
Diagonals are perpendicular to each other	×	×	√	×	√
Each angle is 90°	√	×	X	×	√

4. Fill in the blanks:

Question (i)
…………… is a quadrilateral with only one pair of opposite sides parallel.
Solution:
Trapezium

Question (ii)
…………….. is a quadrilateral with all sides equal and diagonals of equal length.
Solution:
Square

Question (iii)
A polygon with atleast one angle is reflex is called ……………….. .
Solution:
Concave polygon

Question (iv)
………….. is a regular quadrilateral.
Solution:
Square

Question (v)
…………… is a quadrilateral with opposite sides equal and diagonals of unequal length.
Solution:
Parallelogram.

5. State True or False:

Question (i)
A rectangle is always a rhombus.
Solution:
False

Question (ii)
The diagonals of a rectangle are perpendicular to each other.
Solution:
False

Question (iii)
A square is a parallelogram.
Solution:
True

Question (iv)
A trapezium is a parallelogram.
Solution:
False

Question (v)
Opposite sides of a parallelogram are parallel.
Solution:
True.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 15 Visualising Solid Shapes Ex 15.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.1

1. Match the two dimensional figure with the names.

Answer:
(i) (e)
(ii) (d)
(iii) (a)
(iv) (b)
(v) (c)

2. Match the three dimension shapes with the names.

Answer:
(i) (d)
(ii) (e)
(iii) (a)
(iv) (c)
(v) (b)

3. Identify the nets which can be used to make cubes (cut out copies of the nets and try it).


Answer:
(i), (iv)

4. Draw the net for a square pyramid with base as square of sides 5 cm and slant edges 7 cm.
Answer:

5. Draw a net for the following cylinder.

Answer:

6. Draw the net of the solid given in figure.

Answer:

7. Dice are cubes with dots on each face opposite faces of a die always have a total of seven dots on them following are two nets to make dice (cuber) the number inserted in each square indicate the number of dots in that box insert suitable number in the blank squares.


Solution:

8. Which solid will be obtained by folding the following net ?

Solution:

9. Complete the following table.


Solution:
(i) Faces : 6.
(ii) Edges : 2. vertices : NIL,
(iii) Faces : 7. Edges : 15.
(iv) faces : 5, vertices : 5.

10. Multiple Choice questions :

Question (i).
Out of following which is 3-D figure ?
(a) Square
(b) Triangle
(c) Sphere
(d) Circle
Answer:
(c) Sphere

Question (ii).
Total number of faces a cylinder has :
(a) 0
(b) 2
(c) 1
(d) 3
Answer:
(d) 3

Question (iii).
How many edges are there in a square pyramid ?
(a) 5
(b) 8
(c) 1
(d) 4
Answer:
(b) 8

Question (iv).
Sum of number on the opposite faces of a die is :
(a) 8
(b) 7
(c) 9
(d) 6
Answer:
(b) 7

Question (v).
Which is not a solid figure ?
(a) Cuboid
(b) Sphere
(c) Quadrilateral
(d) Pyramid
Answer:
(c) Quadrilateral

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes Ex 9.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 9 Understanding Elementary Shapes Ex 9.4

1. Classify each of the following triangles as scalene, isosceles or equilateral:

Solution:
(i) Here, two sides of triangle are equal in length.
∴ It is an isosceles triangle.
(ii) Here, all the three sides of the triangle are equal in length.
∴ It is an equilateral triangle.
(iii) Here, no two sides are equal in length.
∴ It is scalene triangle.
(iv) Here, two sides of triangle are equal in length.
∴ It is an isosceles triangle.
(v) Here, no two sides are equal in length.
∴ It is scalene triangle.
(vi) Here, all the three sides of the triangle are equal in length.
∴ It is equilateral triangle.

2. Classify each of the following triangles as acute, obtuse or right triangle:

Solution:
(i) Here, one angle is 120°, which is obtuse angle.
∴ It is an obtuse-angled triangle.
(ii) Here, one angle is 90°, which is right angle.
∴ It is an right-angled triangle.
(iii) Here, each angle is acute angle.
∴ It is an acute-angled triangle.
(iv) Here, one angle is 90°, which is right angle.
∴ It is an right-angled triangle.
(v) Here, one angle is 120°, which is obtuse angle.
∴ It is an obtuse-angled triangle.
(vi) Here, each angle is 60°, which is actute angle.
∴ It is an actute angled triangle.

3. Which of the following triangles are possible with the given angles?

Question (i)
60°, 60°, 60°
Solution:
In a triangle sum of the three angles of a triangle is equal to 180°.
Here, sum of the three angles of triangle is:
60° + 60° + 60° = 180°
∴ This triangle is possible.

Question (ii)
110°, 50°, 30°
Solution:
Here, sum of the three angles of triangle is:
110° + 50° + 30°= 190° ≠ 180°
∴ This triangle is not possible.

Question (iii)
65°, 55°, 60°
Solution:
Here, sum of the three angles of triangle is:
65°+ 55°+ 60°= 180°
∴ This triangle is possible.

Question (iv)
90°, 40°, 50°
Solution:
Here, sum of the three angles of triangle is:
90°+ 40°+ 50°= 180°
∴ This triangle is possible.

Question (v)
48°, 62°, 50°
Solution:
Here, sum of the three angles of triangle is:
48°+ 62°+ 50°= 160° ≠ 180°
∴ This triangle is not possible.

Question (vi)
90°, 95°, 30°.
Solution:
Here, sum of the three angles of triangle is:
90°+ 95°+ 30° =215° ≠ 180°
∴ This triangle is not possible.

4. Classify each of the following triangles as scalene, isosceles or equilateral triangle:

Question (i)
4 cm, 5 cm, 6 cm
Solution:
The sides of triangle are 4 cm, 5 cm, 6 cm
No, two sides of this triangle are equal.
∴ This is a scalene triangle.

Question (ii)
5 cm, 7 cm, 5 cm
Solution:
The sides of triangle are 5 cm, 7 cm, 5 cm
Here, two sides are equal each of 5 cm in length.
∴ This is an isosceles triangle.

Question (iii)
4.2 m, S3 m, 6.1 m
Solution:
The sides of triangle are 4.2 m, 5.3 m, 6.1 m
Here, all sides are of different length.
∴ This is a scalene triangle.

Question (iv)
3.5 cm, 3.5 cm, 33 cm
Solution:
The sides of triangle are 3.5 cm, 3.5 cm, 3.5 cm
All the sides of triangle are of equal length.
∴ This is an equilateal triangle.

Question (v)
8 cm, 4.2 cm, 4.2 cm
Solution:
The sides of triangle are 8 cm, 4.2 cm, 4.2 cm
Here, two sides of the triangle are of equal length.
∴ This is an isosceles triangle.

Question (vi)
2 cm, 3 cm, 4 cm.
Solution:
The sides of triangle are 2 cm, 3 cm, 4 cm
All the sides of the triangle are of different lengths
∴ This is a scalene triangle.

5. Name the following triangles in both ways: (Based on sides and angles)

Solution:
(i) Based on sides: In this triangle, no two sides of the triangle are equal.
∴ This is a scalene triangle.
Based on angles: All the three angles of the triangle are acute.
∴ This is an acute-angled triangle.

(ii) Based on sides: In this triangle, two sides are of equal length each is 4 cm.
∴ This is an isosceles triangle.
Based on angles: In this triangle, one angle is of 90° which is a right angle.
∴ This is a right-angled triangle.

(iii) Based on sides: In this triangle, two sides are of equal length.
∴ This is an isosceles triangle.
Based on angles: In this triangle one angle is of 110°, which is obtuse angle.
∴ This is an obtuse-angled triangle.

(iv) Based on sides: In this triangle, all the sides are of equal length i.e. each = 4 cm.
∴ This is an equilateral triangle.
Based on angles: In this triangle, all the angles are acute angles.
∴ This is an acute-angled triangle.

(v) Based on sides: In this triangle, all the three sides are of different lengths.
∴ This is a scalene triangle.
Based on angles: In this triangle, one angle is 105°, which is obtuse angle.
∴ This is an obtuse-angled triangle.

6. Fill in the blanks:

Question (i)
A triangle has …………. sides.
Solution:
3

Question (ii)
A triangle has …………. vertices.
Solution:
3

Question (iii)
A triangle has …………. angles.
Solution:
3

Question (iv)
A triangle has …………. parts.
Solution:
6

Question (v)
A triangle whose all sides are different is known as ………………. .
Solution:
Scalene triangle

Question (vi)
A triangle whose all angles are acute is known as ……………….. .
Solution:
Acute angled triangle

Question (vii)
A triangle whose two sides are equal is known as ……………….. .
Solution:
Isosceles triangle

Question (viii)
A triangle whose one angle is obtuse is known as ……………….. .
Solution:
obtuse-angled triangle

Question (ix)
A triangle whose all sides are equal is known as ……………….. .
Solution:
Equilateral triangle

Question (x)
A triangle whose one angle is right angle is known as ……………….. .
Solution:
Right-angled triangle

7. State True or False:

Question (i)
Each equilateral triangle is an isosceles triangle.
Solution:
True

Question (ii)
Each acute-angled triangle is a scalene triangle.
Solution:
False

Question (iii)
Each isosceles triangle is an equilateral triangle.
Solution:
False

Question (iv)
There are two obtuse angles in an obtuse triangle.
Solution:
False

Question (v)
In right triangle, there is only one right angle.
Solution:
True

Question (vi)
Right triangle can never be isosceles.
Solution:
False.