Bhagya

Punjab State Board PSEB 8th Class English Book Solutions English Vocabulary Prefixes and Suffixes Exercise Questions and Answers, Notes.

PSEB 8th Class English Vocabulary Prefixes and Suffixes

A- Prefixes

Word / Example:
in + active = inactive
in + accurate = inaccurate
in + capable = incapable
in + complete = incomplete
in + correct = incorrect
in + credible = incredible
in + curable = incurable
in + definite = indefinite
in + dependent = independent
in + direct = indirect

in + efficient = inefficient
in + finite = infinite
in + firm = infirm
in + flexible = inflexible
in.+ formal = informal
in + gratitude = ingratitude
in + hospitable = inhospitable
in + human = inhuman
in + secure = insecure
in + significance = insignificance
in + sincerity = insincerity
in + sufficient = insufficient
in + tangible = intangible
in + temperate = intemperate
in + tolerable = intolerable
in + visible = invisible
im + mature = immature
im + mobile = immobile
im + mobilize = inmodilize
im + moderate = immoderate
im + polite = impolite
im + possible = impossible
im + proper = improper
dis + advantage = inadvantage
dis + agree = disagree
dis + appear = disappear
dis + arm = disarm
dis + honour = dishonour
dis + inherit = disinherit
dis + loyal = irloyal
dis + obedient = ofirobedient
dis + orderly = orderly
dis + place = displace
dis + please = displease.
un + able = unable
un + accustomed = unaccustomed
un + affected = unaffected
un + armed = unarmed
un + avoidable = unavoidable
un + certain = uncertain
un + chartered = unchartered
un + common = uncommon
ir + reparable = irreparable
ir + resolute = irresolute
ir + responsible = irresponsible
ir + resistible = irresistible
ir + reverent = – irreverent
ir + revocable = irrevocable
non + aggression = nonaggression
non + aligned = nonaligned
non + cooperate = noncooperate
non + conformist = nonconformist
non + contentious = noncontentious
non + payment = nonpayment
non + poisonous = nonpoisonous
non + political = nonpolitical
non + resident = nonresident
non + religious = nowreligious
non + sense = nonsense
non + smoker = nonsmoker
non + taxable = nontaxable
non + vegetarian = nonvegetarian
non + violence = nonviolence
de + centralize = decentralize
de + increase = decrease
de + inflate = deflate
de + ascend = descend
il + legal = illegal
il + legible = illegible
il + literate = illiterate
il + literacy = illiteracy

B- suffixes

Word:
-able
Examples:
favour + able = favourable
objection + able = objectionable
reason + able = reasonable
remark + able = remarkable
value + able = valueable
prefer + able = preferable
knowledge + able = knowledgeable
notice + able = noticeable
like + able = likeable.

Word:
-ance
Examples:
accept + ance = acceptance
allow + ance = allowance
defy + ance = defiance
guide + ance = guidance
import + ance = importance
resist + ance = resistance
appear + ance = appearance
accept + ance = acceptance
perform + ance = performance.

Word:
-ment
Examples:
fulfill + ment = fulfillment
agree + ment = agreement
attach + ment = attachment
fulfil + ment = fulfilment
pay + ment = payment
refresh + ment = refreshment
settle + ment = settlement
advertise + ment = advertisement
entertain + ment = entertainment
misplace + ment = misplacement
replace + ment = replacement
refresh + ment = refreshment
pay + ment = payment

Word:
-fill
Examples:
law + ful = lawful
beauty + fill = beautiful
care + ful = careful
grace + fill = graceful
hope + fid = hopeful
use + ful = useful
success + fill = successful
power + fill = powerful
doubt + ful = doubtful
pain + fill = painful.

Word:
-er
Examples:
begin + er = beginner
build + er = builder
invade + er = invadeer
make + er = maker
command + er = commander
fast + er = faster
air + er = airer
act + er = acter
bike + er = biker
able + er = abler
old + er = older.

Word:
-less
Examples:
art + less = artless
hope + less = hopeless
noise + less = noiseless
taste + less = tasteless
weight + less = weightless
help + less = helpless
wire + less = wireless
need + less = needless
home + less = homeless
number + less = numberless
mother + less = motherless
driver + less = driverless
system + less = systemless
power + less = powerless
thank + less = thankless
pain + less = painless
fear + less = fearless.

Word:
-y
Examples:
cloud + y = cloudy
fault + y = faulty
sand + y = sandy
sleep + y = sleepy
rain + y = rainy
cream + y = creamy
noise + y = noisy.
bush + y = bushy
juice + y = juicy
snow + y = snowy
rock + y = rocky.

Word:
-en
Examples:
gold + en = golden
silk + en = silken
light + en = lighten
soft + en = soften
bright + en = brighten
straight + en = straighten
earth + en = earthen
wood + en = wooden
wool + en = woolen
dark + en = darken
sharp + en = sharpen

Word:
-ous
Examples:
grace + ous = grancious
nerve + ous = nervous
courage + ous = courageous
mystery + ous = mysterious
danger + ous = dangerous
teacher + ous = teacherous
continue + ous = continuous
luxury + ous = luxurious
glory + ous = glorious

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 10 Practical Geometry Ex 10.2

1. With the help of a ruler, construct line segments of given lengths:

Question (i)
5 cm
Solution:
Steps of Construction.

1. Place the ruler on a paper and hold it firmly.
2. Mark a point A with the pencil against 0 of the ruler and the point B against 5 cm mark of the ruler.

3. Join the two points A and B by moving the pencil along the ruler.

Thus, AB = 5 cm is the required line segment.

Question (ii)
6.5 cm
Solution:
Steps of Construction.

1. Place the ruler on a paper and hold it firmly.
2. Mark a point A with the pencil against 0 of the ruler and another point B against 6.5 cm mark of the ruler.

3. Join the two points A and B by moving the pencil along the ruler. Thus, AB = 6.5 cm is the required line segment.)

Question (iii)
5.2 cm
Solution:
Steps of Construction.

1. Place the ruler on a paper and hold it firmly.
2. Mark a point P with the pencil against 0 of the ruler and another point Q against 5.2 cm mark of the ruler.

3. Join the two points P and Q by moving the pencil along the ruler. Thus, PQ = 5.2 cm is the required line segment.

Question (iv)
6.8 cm
Solution:
Steps of Construction.

1. Place the ruler on a paper and hold it firmly.
2. Mark a point C with the pencil against o of the ruler and another point D against 6.8 cm mark of the ruler.
3. Join the two points C and D by moving the pencil along the ruler.

Thus CD = 6.5 cm is the required line segment.

Question (v)
9.7 cm
Solution:
Steps of Construction.

1. Place the ruler on a paper and hold it firmly.
2. Mark a point L with the pencil against zero mark of the ruler and another point M against 9.7 cm mark of the ruler.

3. Join the two points L and M by moving the pencil along the ruler.

4. Thus, LM = 9.7 cm is the required line segment.

Question (vi)
8.4 cm.
Solution:
Steps of Construction.

1. Place the ruler on a paper and hold it firmly.
2. Mark a point X with the pencil against zero of the ruler and another point Y against 8.4 cm mark of the ruler.

3. Join the two points X and Y by moving the pencil along the ruler.

4. Thus XY = 8.4 cm is the required line segment.

2. Draw line segments given in Question by using a ruler and compasses.
Solution:
(i) Steps of Construction.

1. Draw a line l and mark a point A on it.

2. Place the compasses pointer on the zero mark of the ruler. Open it to place the pencil point upto the 5 cm mark.
3. Now without changing the opening of compasses, place the pointer on A and draw an arc to cut the line l at point B.

4. Thus AB = 5 cm is the required line segment.

(ii) Steps of Construction.

1. Draw a line l and mark a point A on it.

2. Place the compasses pointer on the zero mark of the ruler. Open it to place the pencil point upto the 6.5 cm mark.
3. Now without changing the opening of compasses, place the pointer on A and draw an arc to cut the line l at point B.

4. Thus AB = 6.5 cm is the required line segment.

(iii) Steps of Construction.

1. Draw a line l and mark a point A on it.

2. Place the compasses pointer on the zero mark of the ruler.
Open it to place the pencil point up to the 5.2 cm mark.
3. Now without changing the opening of compasses, place the pointer on A and draw an arc to cut the line l at point B.

4. Thus AB = 5.2 cm is the required line segment.

(iv) Steps of Construction.

1. Draw a line l and mark a point P on it.

2. Place the compasses pointer on the zero mark of the ruler. Open it to place the pencil point upto the 6.8 cm mark.
3. Now without changing the opening of compasses, place the pointer on P and draw an arc to cut the line l at point Q.

4. Thus PQ = 6.8 cm is the required line segment.

(v) Steps of Construction.

1. Draw a line l and mark a point L on it.

2. Place the compasses pointer on the zero mark of the ruler. Open it to place the pencil point upto the 9.7 cm mark.
3. Now without changing the opening of compasses, place the pointer on L and draw an arc to cut the line l at point M.

4. Thus LM = 9.7 cm is the required line segment.

(vi) Steps of Construction.

1. Draw a line l and mark a point X on it.

2. Place the compasses pointer on the zero mark of the ruler. Open it to place the pencil point upto the 8.4 cm mark.
3. Now without changing the opening of compasses, place the pointer on X and draw an arc to cut the line l at point Y.

4. Thus XY = 8.4 cm is the required line segment.

3. Construct AB of length 8.4 cm. From it cut off AC of length 5.3 cm. Measure BC.
Solution:
Steps of Construction.
1. Draw a line l and mark a point A on it.

2. Place the compasses pointer on the zero mark of the ruler. Open it to place the pencil point upto the 8.4 cm mark.
3. Now without changing the opening of compasses, place the pointer on A and draw an arc to cut the line l at point B.
4. Thus AB = 8.4 cm.
5. Now open the compasses equal to AC = 5.3 cm.
6. Place the metal point of compasses on A. Then point with pencil point draw an arc, interecting the line l at C.
7. Now AC = 5.3 cm.
8. By measurement BC = 3.1 cm.

4. Draw two line segments AB and CD of lengths 8.4 cm and 4.5 cm respectively. Construct the line segments of the following lengths:

Question (i)
AB + CD
Solution:
AB + CD = 8.4 cm + 4.5 = 12.9 cm. We can construct a line segment AD of length 12.9 cm. Using a ruler and compasses.

Steps of Construction:

1. Draw line segment AB = 8.4 cm and segment CD = 4.5 cm line.

2. Draw a line l longer than combined length of AB and CD i. e. 12.9 cm. Mark a point A on it.

3. Take the compasses and measure AB. Without changing the opening of the compasses place its needle at A and draw an arc intersecting line l at B.

4. Again adjust the 8.4 cm compass and measure the line segment CD.
5. Without changing the opening of the compasses place the pointer at B on the line l and draw an arc cutting the line l at P.

6. Then AP is the required line segment whose length is equal to the sum of lengths of line segments AB and CD.
7. On measurement AP = 12.9 cm.

Verification:
AP= AB + BP
= AB + CD
= 8.4 cm + 4.5 cm
= 12.9 cm.

Question (ii)
AB – CD
Solution:
Steps of Construction:

1. Draw a line l and mark point A on it.
2. Take the compasses and measures AB. Without changing the opening of the compasses place its needle at A and draw an arc intersecting l at B.

3. Again adjust the compasses and measure the line segment CD.
4. Without changing the opening of the compasses place the pointer at B and draw an arc intersecting AB at Q.
5. AQ is the required line segment whose length is equal to difference of lengths of line segments AB and CD.

6. On measurement AQ = 3.9 cm.

Verification:
AQ = AB – OB
= AB – CD
= 8.4 cm – 4.5 cm
= 3.9 cm.

Question (iii)
2 CD.
Solution:
Steps of Construction.

1. Draw a line l and mark point P on it.

2. Open out the compasses and adjust measure CD without changing the opening of the compasses place the needle at point P and draw an arc intersecting line l at point X such that PX = CD.

3. Now again without changing the opening of compasses place the needle at point X and draw an arc cutting the line l at Q. Such that XQ = CD.

Thus, PQ is the required line segment which is equal to 2 CD.
4. Measure PQ, PQ = 9 cm.
Verification.
Now, PQ = PX + XQ
= CD + CD = 2CD
= 2 × 4.5 cm = 9 cm
Hence, PQ = 2 CD.

5. Draw two line segments PQ and RS of lengths 6.4 cm and 3.6 cm respectively. Construct the line segments of the following lengths:

Question (i)
PQ + RS
Solution:
Steps of Construction.

1. Draw line segment PQ = 6.4 cm and line segment RS = 3.6 cm.

2. Draw a line l longer than combined length of PQ and RS i. e. 10 cm. Take a point P on it.

3. Take the compasses and measure PQ. Without changing the opening of the compasses place its needle at P and draw an arc cutting line l at Q.

4. Again adjust the compasses and measure the line segment RS.
5. Without changing the opening of the compasses place the pointer at Q on the line l and draw an arc cutting the line l at R.

Thus PT is the required line segment whose length is equal to the sum of line segments PQ and RS.
7. Measure PT = 10 cm.

Verification:

PT = PQ + QT
= PQ + RS
= 6.4 cm + 3.6 cm
= 10 cm

Question (ii)
PQ – RS
Solution:
Steps of Construction:

1. Draw a line l and mark a point P on it.

2. Take the compasses and measure PQ. Without changing the opening of the compasses place its needle at P and draw an arc intersecting l at Q.

3. Again adjust the compasses and measure the line segment RS.
4. Without changing the opening of the compasses place the pointer at Q and draw an arc intersecting PQ at T.

5. PT is the required line segment whose length is equal to difference of lengths of line segments PQ and RS.
6. Measure PT, PT = 2.8 cm
Verification:
PT = PQ – QT
= PQ – RS
= 6.4 cm – 3.6 cm
= 2.8 cm.

Question (iii)
2 PQ
Solution:
Steps of Construction:

1. Draw a line l and mark a point A on it.

2. Take the compasses and measure PQ. Without changing the opening of the compasses place its needle at point A and draw an arc intersecting line l at point B such that AB = PQ.

3. Now again without changing the compasses, place the needle at point B and draw an arc cutting the line l at C such that BC = PQ.

4. Then AC is the required line segment whose length is equal to 2 PQ.
5. Measure AC, AC = 12.8 cm

Verification:
AC = AB + BC
= PQ + PQ
= 2PQ = 2 × 6.4 cm
= 12.8 cm.

Question (iv)
2 RS
Solution:
Steps of Construction

1. Draw a line l and mark a point A on it.

2. Take the compasses and measure RS. Without changing the compasses, place the needle at point A and draw an arc intersecting the line l at point B such that AB = RS.

3. Again without changing the opening of compasses, place the needle at point B and draw an arc cutting line l at C such that BC = RS.
4. Then AC is the required line segment whose length is equal to 2RS.

5. Measure AC, AC = 7.2 cm

Verification:
AC = AB + BC
= RS + RS
= 2 RS = 2 × 3.6 cm
= 7.2 cm.

Question (v)
3 RS.
Solution:
Steps of Construction.

1. Draw a line l and mark point A on it.

2. Take the compasses and measure RS. Without changing the compasses, place the needle at point A and draw an arc cutting the line l at point B such that AB = RS.

3. Again, without changing the opening of the compasses, place the needle at point B and draw an arc cutting line l at C such that BC = RS.

4. Once again, without changing the opening of the compasses, place the needle at point C and draw an arc cutting line l at D such that CD = RS. Then AD is the required line segment whose length is equal to 3 RS.

5. Measure AD, AD = 10.8 cm
Verification:
AD = AB + BC + CD
= RS + RS + RS
= 3RS = 3 × 3.6 cm
= 10.8 cm

6. Draw a line segment PQ of any length. Now without measuring it draw a copy of PQ.
Solution:
Steps of Construction.

1. Draw given line segment PQ

2. Draw a line l and mark a point A on it.

3. Take the compass and measure PQ. Without disturbing the compasses, place the needle of the compasses at point A on l and draw an arc, which intersect the line l at point B.

4. Then AB is the required line segment which is equal to the length of PQ. Thus AB = PQ

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.8 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

Question 1.
Find the volume of a sphere whose radius is
(i) 7 cm
Answer:
For the given sphere, radius r = 7 cm.
Volume of a sphere
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 7 × 7 × 7 cm³
= \(\frac{4312}{3}\) cm³
= 1437\(\frac{1}{3}\) cm³

(ii) 0.63 m.
Answer:
For the given sphere, radius r = 0.63 m.
Volume of a sphere
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.63 × 0.63 × 0.63 m³
= 1.05 m³ (approx.)

Question 2.
Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm
(ii) 0.21 m
Answer:
Amount of water displaced by a solid spherical ball = Volume of spherical ball

(i) For the given spherical ball, diameter 28
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{28}{2}\) cm = 14 cm
Volume of spherical ball
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 14 × 14 × 14 cm³
= \(\frac{34496}{3}\) cm³
= 11498\(\frac{2}{3}\) cm³
Thus, the amount of water displaced by the given solid spherical ball is = 11498\(\frac{2}{3}\) cm³

(ii) For the given spherical ball, diameter 28
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{0.21}{2}\) m
Volume of spherical ball
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × \(\frac{0.21}{2}\) × \(\frac{0.21}{2}\) × \(\frac{0.21}{2}\) m³
= 11 × 0.01 × 0.21 × 0.21 m³
= 0.004851 m³
Thus, the amount of water displaced by the given solid spherical ball is 0.004851 m³.

Question 3.
The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9g per cm³?
Answer:
For the given spherical ball,
radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{4.2}{2}\) cm
= 2.1 cm
= \(\frac{21}{10}\) cm
Volume of a sphere
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × \(\frac{21}{10}\) × \(\frac{21}{10}\) × \(\frac{21}{10}\) cm³
= 38.808 cm³
Now, the density of the metal of the ball is 8.9 g per cm³.
∴ Mass of the ball = Volume × Density
= 38.808 cm³ × 8.9 g/cm³
= 345.39 g (approx.)
Thus, the mass of the metallic ball is 345.39 g (approx.).

Question 4.
The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon ?
Answer:
As the diameter of the moon is one-fourth of the diameter of the earth, the radius of the moon is also one-fourth of the radius of the earth. In other words, the radius of the earth is four times the radius of the moon. Let, the radius of the moon be r and the radius of the earth be R.
Then, R = 4r
Now, \(=\frac{\text { volume of the moon }}{\text { volume of the earth }}\) = \(\frac{\frac{4}{3} \pi r^{3}}{\frac{4}{3} \pi R^{3}}\)
= \(\left(\frac{r}{R}\right)^{3}\)
= \(\left(\frac{r}{4 r}\right)^{3}\)
= \(\left(\frac{1}{4}\right)^{3}\)
= \(\frac{1}{64}\)
∴ Volume of the moon
= \(\frac{1}{64}\) × Volume of the earth
Thus, the volume of the moon is \(\frac{1}{64}\) times the volume of the earth.

Question 5.
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Answer:
For the hemispherical bowl,
radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{10.5}{2}\) cm
= 5.25 cm
= \(\frac{21}{4}\) cm
Capacity of the hemispherical bowl
= Volume of a hemisphere
= \(\frac{2}{3}\) πr³
= \(\frac{2}{3} \times \frac{22}{7} \times \frac{21}{4} \times \frac{21}{4} \times \frac{21}{4}\) cm³
= 303.19 cm³ (approx.)
= \(\frac{303.19}{1000}\) liters (approx.)
= 0.303 liters (approx)
Thus, the given hemispherical bowl can hold 0. 303 litres (approx.) of milk.

Question 6.
A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
Answer:
For the hemispherical tank, inner radius r = 1 m and the thickness of the iron sheet = 1 cm = 0.01 m.
∴ For the hemispherical tank, outer radius
R = 1 + 0.01 m = 1.01 m.
Volume of the iron used in the tank
= Volume of outer hemisphere – Volume of Inner hemisphere
= \(\frac{2}{3}\) πR³ – \(\frac{2}{3}\) πr³
= \(\frac{2}{3}\) π (R³ – r³)
= \(\frac{2}{3}\) × \(\frac{22}{7}\) (1.01³ – 1³) m³
= \(\frac{44}{21}\) (1.030301 – 1) m³
Thus, the volume of the iron used to make the tank is 0.06349 m³ (approx.).

Question 7.
Find the volume of a sphere whose surface area is 154 cm².
Answer:
For the given sphere, surface area = 154 cm².
Surface area of a sphere = 4πr²
∴ 154 cm² = 4 × \(\frac{22}{7}\) × r²cm²
∴ r² = \(\frac{154 \times 7}{4 \times 22}\) cm²
∴ r² = \(\frac{49}{4}\) cm²
∴ r = \(\frac{7}{2}\)
Thus, the radius of the given sphere is \(\frac{7}{2}\) cm.
Volume of a sphere
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}\) cm³
= \(\frac{539}{3}\) cm³
= 179\(\frac{2}{3}\) cm³
Thus, the volume of the given sphere is 179\(\frac{2}{3}\) cm³.

Question 8.
A dome of a building is in the form of a hemisphere. From inside, it was white washed at the cost of ₹ 4989.60. If the s cost of whitewashing is ₹ 20 per square metre, find the
(i) inside surface area of s the dome.
Answer:
(i) Area of the region whitewashed at the
cost of ₹ 20 = 1 m²
∴ Area of the region whitewashed at the cost of ₹ 4989.60 = \(\frac{4989.60}{20}\) m² = 249.48 m²
Hence, the inner surface area of the dome is 249.48 m²

(ii) volume of the air inside the dome.
Answer:
Curved surface area of hemispherical dome = 2πr²
∴ 249.48 m² = 2 × \(\frac{22}{7}\) × r² m²
∴ r² = \(\frac{249.48 \times 7}{2 \times 22}\) m²
∴ r² = 39.69 m²
∴ r² = \(\sqrt{39.69}\) m
∴ r = 6.3 m
Thus. the radius of the hemispherical dome is 6.3m.
Volume of air inside the hemispherical dome = Volume of a hemisphere
= \(\frac{2}{3}\) πr³
= \(\frac{2}{3}\) × \(\frac{22}{7}\) × 6.3 × 6.3 × 6.3 m³
= 523.908 m³
= 5239 m³ (approx.)
Thus, the volume of the air inside the dome is 523.9 m³ (approx.).

Question 9.
Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’, Find the
(i) radius r’ of the new sphere,
Answer:
(i) 27 solid iron spheres of radius r are melted to form 1 iron sphere of radius r’.
∴ Volume of 1 sphere of radius r’
= Volume of 27 spheres of radius r
∴ \(\frac{4}{3}\) πr’³ = 27 × \(\frac{4}{3}\) πr’³
∴ r’³ = 27r³
∴ r’³ = (3r)³
∴ r’ = 3r

(ii) ratio of S and S’.
Answer:
The surface area of the sphere with radius r is S and the surface area of the sphere with radius r’ is S’.
Then,
\(\frac{\mathrm{s}}{\mathrm{S}^{\prime}}=\frac{4 \pi r^{2}}{4 \pi r^{\prime 2}}=\frac{r^{2}}{r^{\prime 2}}=\frac{r^{2}}{(3 r)^{2}}=\frac{r^{2}}{9 r^{2}}=\frac{1}{9}\) = 1 : 9
Thus, the ratio of S and S’ is 1 : 9.

Question 10.
A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule ?
Answer:
For the spherical capsule, radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{3.5}{2}\) mm
= 1.75 mm
Capacity of the spherical capsule
= Volume of a sphere
= \(\frac{4}{3}\) πr’³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 1.75 × 1.75 × 1.75 mm³
= 22.46 mm³ (approx.)
Thus, 22.46 mm³ (approx.) medicine is needed to fill the given capsule.

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 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 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.