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Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

Note: Assume π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.
Find the surface area of a sphere of radius:
(i) 10.5 cm
Answer:
For the given sphere,
radius r = 10.5 cm = \(\frac{21}{2}\) cm.
Surface area of a sphere
= 4πr²
= 4 × \(\frac{22}{7}\) × \(\frac{21}{2}\) × \(\frac{21}{2}\) cm² = 1386 cm²

(ii) 5.6 cm
Answer:
For the given sphere, radius r = 5.6cm.
Surface area of a sphere
= 4πr²
= 4 × \(\frac{22}{7}\) × 5.6 × 5.6 cm²
= 394.24 cm²

(iii) 14 cm
Answer:
For the given sphere, radius r= 14 cm.
Surface area of a sphere
= 4πr²
= 4 × \(\frac{22}{7}\) × 14 × 14 cm²
= 2464 cm²

Question 2.
Find the surface area of a sphere of diameter:
(i) 14cm
Answer:
For the given sphere, diameter d = 14 cm.
Then, radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{14}{2}\) cm = 7 cm
Surface area of a sphere
= 4πr²
= 4 × \(\frac{22}{7}\) × 7 × 7 cm²
= 616 cm²

(ii) 21cm
Answer:
For the given sphere, diameter d = 21 cm.
Then, radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{21}{2}\) cm
Surface area of a sphere
= 4πr²
= 4 × \(\frac{22}{7}\) × \(\frac{21}{2}\) × \(\frac{21}{2}\) cm²
= 1386 cm²

(iii) 3.5 m
Answer:
For the given sphere, diameter d = 3.5 cm.
Then, radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{3.5}{2}\) m
= \(\frac{35}{20}\) m
Surface area of a sphere
= 4πr²
= 4 × \(\frac{22}{7}\) × \(\frac{35}{20}\) × \(\frac{35}{20}\) m²
= 38.5 m²

Note: We can also use the formula “Surface area of a sphere = πd²” as
4πr² = π × 4r² = π × (2r)² = πd², where r and d are radius and diameter of the sphere respectively.

Question 3.
Find the total surface area of a hemisphere of radius 10 cm. (Use π =3.14)
Answer:
For the given hemisphere, radius r = 10 cm.
Total surface area of a hemisphere
= 3πr²
= 3 × 3.14 × 10 × 10 cm²
= 942 cm²

Question 4.
The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Answer:
For the first case, radius r₁ of the spherical balloon is 7 cm.
Surface area of the spherical balloon in the
first case = 4πr₁²
= 4 × \(\frac{22}{7}\) × 7 × 7cm²
For the second case, radius r₂ of the spherical balloon is 14 cm.
Surface area of the spherical balloon in the second case = 4πr₂²
Then, the required ratio of surface areas in two cases
= \(\frac{4 \times \frac{22}{7} \times 7 \times 7}{4 \times \frac{22}{7} \times 14 \times 14}\)
= \(\frac{1}{4}\) = 1 : 4
Thus, the required ratio is 1 : 4.
Note: Here, the ratio of radii = 7 : 14 = 1 : 2
Hence, the ratio of surface areas = \(\left(\frac{1}{2}\right)^{2}\) = \(\frac{1}{4}\) = 1 : 4, because in the formula of surface area of sphere, the degree of r is 2.

Question 5.
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹ 16 per 100 cm².
For the given hemispherical bowl, diameter = 10.5 cm.
Then, the radius r of the bowl = \(\frac{\text { diameter }}{2}\)
= \(\frac{10.5}{2}\) cm
= \(\frac{\frac{21}{2}}{2}\) cm
= \(\frac{21}{4}\) cm
Inner curved surface area of the hemispherical bowl
= 2 πr²
= 2 × \(\frac{22}{7}\) × \(\frac{21}{4}\) × \(\frac{21}{4}\) cm²
= \(\) cm²
= 173.25 cm²
Cost of tin-plating 100 cm² region = ₹ 16
∴ Cost of tin-plating 173.25 cm² region
= ₹ \(\left(\frac{16 \times 173.25}{100}\right)\)
= ₹ 27.72
Thus, the cost of tin-plating on the inner surface of the bowl is ₹ 27.72.

Question 6.
Find the radius of a sphere whose surface area is 154 cm².
Answer:
For the given sphere, surface area = 154 cm².
Surface area of a sphere = 4 πr²
∴ 154 cm² = 4 × \(\frac{22}{7}\) × r² cm²
∴ r² = \(\frac{154 \times 7}{4 \times 22}\) cm²
∴ r² = \(\frac{49}{4}\) cm²
∴ r = \(\frac{7}{2}\) cm
∴ r = 3.5 cm
Thus, the radius of the given sphere is 3.5 cm.

Question 7.
The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Answer:
Suppose, the diameter of the moon = d₁
The diameter of the earth = 4 × d₁ = 4d₁
Then, the radius of the moon r₁ = \(\frac{d_{1}}{2}\) and
the radius of the earth r₂ = \(\frac{4 d_{1}}{2}\) = 2d₁.
Now, \(\frac{\text { The surface area of the moon }}{\text { The surface area of the earth }}\) = \(\frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}\)
= \(\frac{r_{1}^{2}}{r_{2}^{2}}\)
= \(\frac{\left(\frac{d_{1}}{2}\right)^{2}}{\left(2 d_{1}\right)^{2}}\)
= \(\frac{d_{1}^{2}}{4} \times \frac{1}{4 d_{1}^{2}}\)
= \(\frac{1}{16}\)
= 1 : 16
Thus, the ratio of the surface area of the moon and the surface area of the earth is 1 : 16.

Question 8.
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Answer:
For the given hemispherical bowl, the inner radius is 5 cm and the thickness of steel is 0.25 cm.
∴ Outer radius r of the given hemispherical bowl = 5 + 0.25 cm = 5.25 cm.
Curved surface area of a hemisphere
= 2πr²
= 2 × \(\frac{22}{7}\) × 5.25 × 5.25 cm²
= 2 × \(\frac{22}{7}\) × \(\frac{525}{100}\) × \(\frac{525}{100}\) cm²
= \(\frac{693}{4}\) cm²
= 173.25 cm²
Thus, the outer curved surface area of the given hemispherical bowl is 173.25 cm².

Question 9.
A right circular cylinder just encloses a sphere of radius r (see the given figure). Find :
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).

Answer:
Here,
radius of the cylinder = radius of the sphere = r and height of the cylinder h
= 2 × radius of the sphere = 2r
(i) Surface area of the sphere = 4πr²

(ii) Curved surface area of the cylinder
= 2 πrh
= 2 × 1 × r × 2r
= 4 πr²

(iii) Ratio of areas obtained in ( i ) and (ii)
= \(\frac{4 \pi r^{2}}{4 \pi r^{2}}\)
= \(\frac{1}{1}\)
= 1 : 1

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes Ex 9.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 9 Understanding Elementary Shapes Ex 9.3

1. Identify the shape having perpendicular lines:

Solution:
(ii), (iii), (v).

2. Identify the examples having perpendicular lines:

Question (i)
(i) Lines of railway track.
(ii) Adjacent edges of a table.
(iii) Line segment forming letter ‘L’.
Solution:
(ii) Adjacent edges of a table.
(iii) Line segment forming letter ‘L’.

3. Let \(\overrightarrow{\mathbf{AB}}\) be perpendicular to \(\overrightarrow{\mathbf{PQ}}\) and they intersect at O. What is the measure of \(\angle \mathbf{AOP}\)?
Solution:

\(\angle \mathbf{AOP}\) = 90°, because AB ⊥ PQ.

4. Line m is perpendicular to line l in the given figure. Each point on the line l is marked at equal intervals. Study the diagram and state true or false.

Question (i)
Line m is ⊥ bisector of line segment AI.
Solution:
True

Question (ii)
CE = EG
Solution:
True

Question (iii)
DF = 2DE.
Solution:
True.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes Ex 9.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 9 Understanding Elementary Shapes Ex 9.2

1. Classify the angles as acute, obtuse, right, straight or reflex angles:

Solution:
(i) Acute angle
(ii) Obtuse angle
(iii) Reflex angle
(iv) Straight angle
(v) Acute angle
(vi) Right angle
(vii) Obtuse angle
(viii) Right angle
(ix) Reflex angle
(x) Acute angle.

2. Classify the angles:

Question (i)
80°
Solution:
80° is between 0° and 90°.
∴ It is an acute angle.

Question (ii)
172°
Solution:
172° is between 90° and 180°
∴ It is an obtuse angle.

Question (iii)
90°
Solution:
90° is a right angle.

Question (iv)
0°
Solution:
0° is a zero angle.

Question (v)
179°
Solution:
179° is between 90° and 180°.
∴ It is an obtuse angle.

Question (vi)
215°
Solution:
215° is between 180° and 360°.
∴ It is an reflex angle.

Question (vii)
360°
Solution:
360° is a complete angle.

Question (viii)
350°
Solution:
350° is between 180° and 360°.
∴ It is a reflex angle.

Question (ix)
15°
Solution:
15° is between 0° and 90°.
∴ It is an acute angle.

Question (x)
180°
Solution:
180° is a straight angle.

3. Measure the following angles with protractor and write their measurement:

Solution:
(i) 60°
(ii) 125°
(iii) 110°
(iv) 80°
(v) 120°
(vi) 105°
(vii) 80°
(viii) 135°
(ix) 88°
(x) 90°.

4. How many degrees are there in

Question (i)
Two right angles
Solution:
1 right angle = 90°
∴ Two right angles = 2 × 90°
= 180°

Question (ii)
\(\frac {2}{3}\) right angles
Solution:
1 right angle = 90°
∴ \(\frac {2}{3}\) right angles = \(\frac {2}{3}\) × 90°
= 2 × 30°
= 60°

Question (iii)
Four right angles?
Solution:
1 right angle = 90°
∴ Four right angles = 4 × 90°
= 360°

5. What fraction of a clockwise revolution does the hour hand of a clock turn through when it goes from:

Question (i)
3 to 9
Solution:
3 to 9 : Half or \(\frac {1}{2}\)

Question (ii)
5 to 8
Solution:
5 to 8 : Quarter or \(\frac {1}{4}\)

Question (iii)
10 to 4
Solution:
10 to 4 : Half or \(\frac {1}{2}\)

Question (iv)
2 to 11
Solution:
2 to 11 : 3 Quarters or \(\frac {3}{4}\)

Question (v)
6 to 3
Solution:
6 to 3 : 3 Quarters or \(\frac {3}{4}\)

Question (vi)
2 to 7.
Solution:
2 to 7 : \(\frac {5}{12}\)

6. Find the number of right angles turned through by the hour hand of a dock when it goes from

Question (i)
5 to 8
Solution:
5 to 8 : 1 right angle

Question (ii)
1 to 7
Solution:
1 to 7 : 2 right angles

Question (iii)
4 to 10
Solution:
4 to 10 : 2 right angles

Question (iv)
9 to 12
Solution:
9 to 12 : 1 right angles

Question (v)
11 to 2
Solution:
11 to 2 : 1 right angles

Question (vi)
9 to 6
Solution:
9 to 6 : 3 right angles

Question (vii)
2 to 11
Solution:
2 to 11 : 3 right angles

Question (viii)
10 to 1
Solution:
10 to 1 : 1 right angles

Question (ix)
12 to 6
Solution:
12 to 6 : 2 right angles

Question (x)
5 to 2.
Solution:
5 to 2 : 3 right angles.

7. Where will be the hand of a clock stop if it starts at:

Question (i)
12 and make \(\frac {1}{4}\) revolution clock-wise.
Solution:
For 1 revolution, the hour hand takes 12 hours.
For \(\frac {1}{4}\) revolution, the hour hand takes \(\frac {1}{4}\) × 12 hours = 3 hours.
If hour hand starts at 12 and make \(\frac {1}{4}\) revolution clockwise it will stop at 3.

Question (ii)
2 and make \(\frac {1}{2}\) revolution clock-wise.
Solution:
For 1 revolution, the hour hand takes 12 hours.
For \(\frac {1}{2}\) revolution, the hour hand takes \(\frac {1}{2}\) × 12 hours = 6 hours.
If hour hand starts at 2 and make \(\frac {1}{2}\) revolution clockwise it will stop at 8.

Question (iii)
5 and make \(\frac {1}{4}\) revolution clock-wise.
Solution:
For 1 revolution, the hour hand takes 12 hours.
For \(\frac {1}{4}\) revolution, the hour hand takes \(\frac {1}{4}\) × 12 hours = 3 hours
If hour hand starts at 5 and make \(\frac {1}{4}\) revolution clockwise it will stop at 8.

Question (iv)
5 and make \(\frac {3}{4}\) revolution clock-wise.
Solution:
For 1 revolution, the hour hand takes 12 hours .
For \(\frac {3}{4}\) revolution, the hour hand takes \(\frac {3}{4}\) × 12 hours = 9 hours.
If hour hand starts at 5 and make \(\frac {3}{4}\) revolution clockwise it will stop at 2.

8. What part of revolution have you turned through if you stand facing:

Question (i)
East and turn clockwise to North
Solution:
I turned through \(\frac {3}{4}\) part of a revolution.

Question (ii)
South and turn clockwise to North
Solution:
I turned through \(\frac {1}{2}\) part of a revolution.

Question (iii)
South and turn clockwise to East
Solution:
I turned through \(\frac {3}{4}\) part of a revolution.

Question (iv)
West and turn clockwise to East
Solution:
I turned through \(\frac {1}{2}\) part of at revolution.

9. Find the angle measure between the hands of the clock in each figure:

Solution:
(i) Angle measure between the hands of the clock at 3.00 a.m.
= \(\frac {3}{12}\) × 360° = 90°
(ii) Angle measure between the hands of the clock at 6.00 a.m.
= \(\frac {6}{12}\) × 360° = 180°
(iii) Angle measure between the hands of the clock at 2.00 a.m.
= \(\frac {2}{12}\) × 360° = 60°

10. Draw the following angles by protractor:

Question (i)
(i) 40°
(ii) 75°
(iii) 105°
(iv) 90°
(v) 130°
Solution:

11. State true or false:

Question (i)
The sum of two right angles is always a straight angle.
Solution:
True

Question (ii)
The sum of two acute angles is always a reflex angle.
Solution:
False

Question (iiii)
The obtuse angle has measurement between 90° to 180°.
Solution:
True

Question (iv)
A complete revolution has four right angles.
Solution:
True

12. Fill in the blanks:

Question (i)
The angle which is greater than 0° and less than 90° is called ………….. .
Solution:
acute angle

Question (ii)
The angle whose measurement equal to two right angle is …………….. .
Solution:
straight angle

Question (iii)
The angle between 90° and 180° is ……………. .
Solution:
obtuse angle.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 6 Triangles Ex 6.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6

Question 1.
In figure, PS is bisector of ∠QPR of ∆PQR. Prove that = \(\frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{PQ}}{\mathrm{PR}}\).

Solution:
Given: ∆PQR. PS is bisector of ∠QPR
i.e., ∠1 = ∠2
To prove: \(\frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{PQ}}{\mathrm{PR}}\)
Construction : Through R draw a line parallel to PS to meet QP produced at T.
Proof: In ∆QRT, PS || TR

∠1 = ∠4 (Corresponding angle)
but ∠1 = ∠2 (given)
∴ ∠3 = ∠4
In ∆PRT,
∠3 = ∠4 (Proved)
PT = PR
[Equal side have equal angle opposite to it]
In ∆QRT,
PS || TR
∴ \(\frac{\mathrm{QP}}{\mathrm{PT}}=\frac{\mathrm{QS}}{\mathrm{SR}}\)
[By Basic Proportionality Theorem]
\(\frac{\mathrm{QP}}{\mathrm{PR}}=\frac{\mathrm{QS}}{\mathrm{SR}}\) (PT = PR)
\(\frac{\mathrm{PQ}}{\mathrm{PR}}=\frac{\mathrm{QS}}{\mathrm{SR}}\)
Which is the required result.

Question 2.
In the given fig., D is a point on hypotenuse AC of ∆ABC, DM ⊥ BC, DN ⊥ AB, prove that:
(i) DM² = DN.MC
(ii) DN² = OMAN.

Solution:
Given: ∆ABC, DM ⊥ BC, DN ⊥ AB
To prove: DM² = DN . AC
DN² = DM . AN.
Proof: BD ⊥ AC (Given)
⇒ ∠BDC = 90°
⇒ ∠BDM + ∠MDC = 90°
In ∠DMC, ∠DMC = 90°
[∵ DM ⊥ BC (Given)]
⇒ ∠C + ∠MDC = 90°
From (1) and (2),
∠BDM + ∠MDC = ∠C + ∠MDC
∠BDM =∠C
[Cancelling ∠MDC from both sides]
Now in ∆BMD and ∆MDC,
∠BDM = ∠C [Proved)
∠BMD = ∠DMC [Each 90°]
∆BMD ~ ∆MDC [By AA criterion of similarity]
⇒ \(\frac{\mathrm{DM}}{\mathrm{BM}}=\frac{\mathrm{MC}}{\mathrm{DM}}\)
[∵ Corresponding sides of similar triangles are proportional]
⇒ DM² = BM × MC
⇒ DM² = DN × MC [∵ BM = DN]
Similarly, ∆NDA ~ ∆NBD
⇒ \(\frac{\mathrm{DN}}{\mathrm{BN}}=\frac{\mathrm{AN}}{\mathrm{DN}}\)
[∵ Corresponding sides of similar triangles are próportional]
⇒ DN² = BN × AN
⇒ DN² = DM × AN .
Hence proved.

Question 3.
In fig., ABC is triangle in which ∠ABC > 90° and AD ⊥ BC produced, prove that AC² = AB² + BC² + 2BC.BD.

Solution:
Given: ∠ABC, AD ⊥ BC when produced, ∠ABC > 90°.
To prove : AC² = AB² + BC² + 2BC. BD.
Proof: Let BC = a,
CA = b,
AB = c,
AD = h
and BD = x.
In right-angled ∆ADB,
Using Pythagoras Theorem.
AB² = BD² + AD²
i.e., c² = x² + h²
Again, in right-angled AADC,
AC² = CD² + AD²
i.e.. b² = (a + x)² + h²
= a² + 2ax + x² + h²
= a² + 2ax + c²; [Using (1)]
b² = a² + b² + 2w.
Hence, AB² = BC² + AC² + 2BC × CD.

Question 4.
In fig., ABC is a triangle in which ∠ABC < 90°, AD ⊥ BC, prove that AC² = AB² + BC² – 2BC.BD.

Solution:
Given: ∆ ABC, ∠ABC < 90°, AD ⊥ BC.
To prove : AC² = AB² + BC² – 2BC BD.
Proof: ADC is right-angled z at D
AC² = CD² + DA² (Pythagora’s Theorem) ……………..(1)
Also, ADB is right angled ∆ at D
AB² = AD² + DB² ……………….(2)
From (1), we get:
AC² = AD² + (CB – BD)²
= AD² + CB² + BD² – 2CB × BD
or AC² = (BD² + AD²) + CB² – 2CB × BD
AC² = AB² + BC² – 2BC × BD. [Using (2)]

Question 5.
In fig., AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
(i) AC² = AD² + BC. DM + \(\left(\frac{B C}{2}\right)^{2}\)
(ii) AB² = AD² ± BC.DM + \(\left(\frac{B C}{2}\right)^{2}\)
(iii) AC² + AB² = 2 AD² + \(\frac{1}{2}\) BC²

Solution:
Given: ∆ABC, AM ⊥ BC,
AD is median of ¿ABC.
To prove:
(i) AC² = AD² + BC. DM + \(\left(\frac{B C}{2}\right)^{2}\)
(ii) AB² = AD² ± BC.DM + \(\left(\frac{B C}{2}\right)^{2}\)
(iii) AC² + AB² = 2 AD² + \(\frac{1}{2}\) BC²
Proof: In ∆AMC.
AC² = AM² + MC²
= AM² + (MD + DC)²
AC² = AM² + MD² + DC² + 2MD × DC
AC² = (AM² + MD²) + \(\left(\frac{\mathrm{BC}}{2}\right)^{2}\) + 2 . MD \(\left(\frac{\mathrm{BC}}{2}\right)\)
AC² = AD² + BC . MD + \(\frac{\mathrm{BC}^{2}}{4}\) …………(1)

(ii) In right angled triangle AME,
AB² = AM² + BM²
= AM² + (BD – MD)²
=AM² + BD² + MD² – 2BD × MD
= (AM² + MD²) + BD² – 2(\(\frac{1}{2}\) BC) MD
= AD² + (\(\frac{1}{2}\) BC)² – BC . MD
[∵ In ∆ AMD; AD² = MA² + MD²]
AB² + AD² (\(\left(\frac{\mathrm{BC}}{2}\right)^{2}\)) – BC . MD ………….(2)

(iii) Adding (1) and (2),
AB² + AC² = AD² + BC.MD + (\(\left(\frac{\mathrm{BC}}{2}\right)^{2}\)) + AD² + (\(\left(\frac{\mathrm{BC}}{2}\right)^{2}\)) – BC . MD
= 2 AD² + \(\frac{\mathrm{BC}^{2}}{4}+\frac{\mathrm{BC}^{2}}{4}\)
= 2AD² + 2 \(\frac{\mathrm{BC}^{2}}{4}\)
AB² + AC² = 2AD² + \(\frac{\mathrm{BC}^{2}}{2}\)
Which is the required result.

Question 6.
Prove that sum of squares of the diagonals of a parallelogram is equal to sum of squares of its sides.
Solution:

Given:
Let ABCD be a parallelogram in which diagonalš AC and BD intersect at point M.
To prove: AB² + BC² + CD² + DA² = AC² + BD²
Solution:
Proof: Diagonals of a parallelogram bisect each other.
∴ In || gm ABCD,
Diagonal BD and AC bisect each other.
Or MB and MD are medians of ∆ABC and ∆ADC respectively.
We know that, if AD is a medians of ¿ABC,
then AB² + AC² = 2AD² + BC²
Using this result, we get:
AB² + BC² = 2 BM² + \(\frac{1}{2}\) AC² ………..(1)
and AD² + CD² = 2 DM² + \(\frac{1}{2}\) AC² ………….(2)
Adding (1) and (2), we get:
AB² + BC² + AD² + CD² = 2 (BM² + DM²) + (AC² + AC²)
AB² + BC² + AD² + CD² = 2 (\(\frac{1}{2}\) BD² + \(\frac{1}{4}\) BD²) + AC²
AB² + BC² + AD² + CD² = BD² + AC²
Hence, sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Question 7.
In fig., two chords AB and CD intersect each other at the point P prove that:
(i) ∆APC ~ ∆DPB
(ii) AP.PB = CP.DP.

Solution:
Given: Circle, AB and CD are two chords intersects each other at P.

To prove:
(i) ∆APC ~ ∆DPB
(ii) AP.PB = CP.DP.
Proof:
(i) In ∆APC and ∆DPB,
∠1 = ∠2 (Vertically opposite angle)
∠3 = ∠4 (angle on same segment)
∴ ∆APC ~ ∆DPB [AA similarity criterion]

(ii) ∆APC ~ ∆DPB (Proved above)
\(\frac{\mathrm{AP}}{\mathrm{DP}}=\frac{\mathrm{PC}}{\mathrm{PB}}\)
(If two triangles are sitnilar corresponding sides are proportional)
AP.PB = PC.DP
Which is the required result.

Question 8.
In fig., two chords AB and CD of a circle intersect each other at point P (when produced) outside the circle prove:
(i) ∆PAC ~ ∆PDB
(ii) PA.PB = PC.PD.

Solution:
Given: AB and CD are two chord of circle intersects each other at P (when produced)
To prove:
(i) ∆PAC ~ ∆PDB
(ii) PA.PB = PC.PD.
Proof:
(i) In ∆PAC and ∆PDB,
∠P = ∠P (Common)
∠PAC = ∠PDB.
(Exterior angle of cyclic quadrilqteral is equal to interior opposite angle)
∴ ∆PAC ~ ∆PDB [AA similarity criterion]

(ii) ∆PAB ~ ∆WDB
∴ \(\frac{\mathrm{PA}}{\mathrm{PD}}=\frac{\mathrm{PC}}{\mathrm{PB}}\)
[If two triangles are similar corresponding sides are proportional]
PA × PB = PC × PD.

Question 9.
In fig., D is a point on side BC of BD AB ∆ABC such that \(\frac{\mathbf{B D}}{\mathbf{D C}}=\frac{\mathbf{A B}}{\mathbf{A C}}\). Prove that: AD is bisector of ∠BAC.

Solution:
Given: A ∆ABC, D is a point on BC such that \(\frac{\mathbf{B D}}{\mathbf{D C}}=\frac{\mathbf{A B}}{\mathbf{A C}}\)
To prove: AD bisects ∠BAC
i.e., ∠1 = ∠2
Construction: Through C draw CE || DA meeting BA produced at E.

Proof:
In ∆BCE, we have:
AD || CE ………(const.)
So, by Basic Proportionality Theorem,
But \(\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AE}}\)
\(\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}\)

⇒ \(\frac{\mathrm{AB}}{\mathrm{AE}}=\frac{\mathrm{AB}}{\mathrm{AC}}\)
⇒ AE = AC

In ∆ACE, we have:
AE = AC
⇒ ∠3 = ∠4 ………. (∠s opp. to equal sides)
Since CE || DA and AC cuts them, then:
∠2 = ∠4 ……….(alt ∠s)
Also CE || DA and BAE cuts them, then:
∠1 = ∠3 …………(Corr. ∠s)
Thus we have:
∠3 = ∠4
⇒ ∠3 = ∠1
But ∠4 = ∠2
⇒ ∠1 = ∠2.
HenCe AD bisects ∠BAC.

Question 10.
Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out? If she pulls in the string at the rate of 5 cm per second, what will the horizontal distance of the fly from her after 12 seconds?

Solution:
A right angled triangle, ABC, in which,
AB = 1.8 cm,
BC = 2.4 cm.
∠B = 90°
By Pythagoras Theorem,
AC² = AB² + BC²
AC² = (1.8)² + (2.4)²
AC² = 3.24 + 5.76 =9
AC² = (3)²
AC = 3 cm
Now, when Nazima pulls in the string at the rate of 5 cm/sec ; then the length of the string decrease = 5 × 12 = 60 cm
= 0.6 m in 12 seconds.
Let after 12 seconds, position of the fly will be at D.
∴ AD = AC – distance covered in 12 seconds
AD = (3 – 0.6) m
AD = 2.4 m
Also, in right angled ∆ABD,
Using Pythagoras Theorem,
AD² = AB² + BD²
(2.4)² = (1.8)² + BD²
BD² = 5.76 – 3.24
BD² = 2.52 m
BD = 1.587 m.
∴ Horizontal distance of the fly from Nazima = BD + 1.2 m
= (1.587 + 1.2) m
= 2.787 m
= 2.79 m
Hence, original length of string and horizontal distance of the fly from Nazima is 3 m and 2.79 m.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

Note: Assume π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.
Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Answer:
For the given cone, diameter d = 10.5 cm.
Then, radius r = \(\frac{10.5}{2}\) cm and slant height
l = 10 cm.
Curved surface area of a cone
= πrl
= \(\frac{22}{7}\) × \(\frac{10.5}{2}\) × 10 cm²
= \(\frac{22}{7}\) × \(\frac{105}{2}\) × cm²
= 11 × 15 cm²
= 165 cm²
Thus, the curved surface area of the given cone is 165 cm².

Question 2.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Answer:
For the given cone, diameter d = 24 m.
Then, radius r = \(\frac{24}{2}\) = 12 m and slant height l = 21 m.
Total surface area of a cone
= πr (l + r)
= \(\frac{22}{7}\) × 12(21 + 12) m²
= \(\frac{22 \times 12 \times 33}{7}\) m²
= \(\frac{8712}{7}\) m²
= 1244.57 m²
Thus, the total surface area of the given cone is 1244.57 m².

Question 3.
Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find,
(i) radius of the base and
(ii) total surface area of the cone.
Answer:
For the given cone, slant height l = 14 cm and curved surface area = 308 cm²
(i) Curved surface area of a cone = πrl
∴ 308 cm² = \(\frac{22}{7}\) × r × 14 cm
∴ \(\frac{308 \times 7}{22 \times 14}\) cm = r
∴ r = 7 cm
Thus, the radius of the base of the cone is 7 cm.

(ii) Total surface area of a cone
= πrl + πr²
= 308 + \(\frac{22}{7}\) × 7 × 7 cm²
= 308 + 154 cm²
= 462 cm²
Thus, the total surface area of the cone is 462 cm².

Question 4.
A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is ₹ 70.
Answer:
For the conical tent,
radius r = 24 m and height h = 10 m.

(i) l = \(\sqrt{h^{2}+r^{2}}\)
= \(\sqrt{10^{2}+24^{2}}\)
= \(\sqrt{100+576}\)
= \(\sqrt{676}\)
∴ l = 26 m
Thus, the slant height of the tent is 26 m.

(ii) Area of the canvas used to make tent
= Curved surface area of the conical tent
= πrl
= \(\frac{22}{7}\) × 24 × 26 m²
= \(\frac{13728}{7}\) m²
Cost of 1 m² canvas = ₹ 70
∴ Cost of \(\frac{13728}{7}\) m² canvas
= ₹ \(\left(70 \times \frac{13728}{7}\right)\)
= ₹ 1,37,280
Thus, the cost of canvas required is ₹ 1,37,280.

Question 5.
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm.
(Use π = 3.14)
Answer:
For the conical tent to be made, radius r = 6 m and height h = 8 m.
l² = h² + r² = 8² + 6² = 64 + 36 = 100
∴ l = √100 = 10 m
Area of the tarpaulin used in making tent
= Curved surface area of conical tent
= πrl
= 3.14 × 6 × 10 m²
= 188.4 m²
Now, the width of the tarpaulin is 3 m.
∴ Length of tarpaulin required = \(\frac{188.4}{3}\) m
= 62.8 m
But, 20 cm, i.e., 0.2 m of tarpaulin is required more for margins and wastage.
∴ Total length of the tarpaulin required = 62.8 + 0.2 m = 63 m
Thus, total length of tarpaulin required is 63 m.

Question 6.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of ₹ 210 per 100 m².
Answer:
For the given conical tomb,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{14}{2}\) = 7 m
and slant height l = 25 m. .
Area of the region to be whitewashed
= Curved surface area of the conical tomb
= πrl
= \(\frac{22}{7}\) × 7 × 25 m²
= 550 m²
Cost of whitewashing 100 m² region = ₹ 210
∴ Cost of whitewashing 550 m² region
= ₹ \(\left(\frac{210 \times 550}{100}\right)\)
= ₹ 1155
Thus, the cost of whitewashing the curved surface of the tomb is ₹ 1155.

Question 7.
A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Answer:
For the conical cap, radius r = 7 cm and height h = 24 cm.
l = \(\sqrt{h^{2}+r^{2}}\)
= \(\sqrt{24^{2}+7^{2}}\)
= \(\sqrt{576+49}\)
= \(\sqrt{625}\) = 25 cm
Area of the sheet required to make 1 conical cap
= Curved surface area of the conical cap
= πrl
= \(\frac{22}{7}\) × 7 × 25 cm²
= 550 cm²
Area of sheet required to make 1 cap = 550 cm²
∴ Area of sheet required to make 10 caps
= 550 × 10 cm²
= 5500 cm²
Thus, the area of the sheet required to make 10 caps is 5500 cm²

Question 8.
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled carboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹ 12 per m², what will be the cost of painting all these cones? (Use π = 3.14 and take \(\sqrt{1.04}\) = 1.02)
Answer:
For the given cone,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{40}{2}\) = 20 cm = 0.2 m and
height h = 1 m.
l = \(\sqrt{h^{2}+r^{2}}\)
= \(\sqrt{1^{2}+0.2^{2}}\)
= \(\sqrt{1.04}\)
= 1.02 m
Curved surface area of a cone
= πrl
= 3.14 × 0.2 × 1.02 m²
∴ Curved surface area of 50 cones
= 50 × 3.14 × 0.2 × 1.02 m²
= 32.028 m²
Cost of painting 1 m² region = ₹ 12
∴ Cost of painting 32.028 m² region
= ₹ (12 × 32.028)
= ₹ 384.34 (approx.)
Thus, the cost of painting all the 50 cones is ₹ 384.34 (approx.)

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

Note: Assume π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.
The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder.
Answer:
Height of cylinder h = 14 cm.
Curved surface area of a cylinder = 2 πrh
∴ 88 cm² = 2 × r × 14cm
∴ \(\frac{88 \times 7}{2 \times 22 \times 14}\) cm = r
∴ r = 1 cm
Now, diameter of the cylinder = 2r = 2 × 1 cm
= 2 cm
Thus, the diameter of the base of the cylinder is 2 cm.

Question 2.
It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same ?
Answer:
Height of cylindrical tank h = 1 m
Diameter of the cylinder =140 cm
∴ Radius of the cylinder r = \(\frac{\text { diameter }}{2}\)
= \(\frac{140}{2}\) cm
= 70 cm
= 0.7 m
Total surface area of the closed cylindrical tank
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × 0.7 (0.7 + 1) m²
= 4.4 × 1.7 m²
= 7.48 m²
Thus, 7.48 m² sheet is required to make the closed cylindrical tank.

Question 3.
A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see the given figure). Find its

(i) inner curved surface area,
Answer:
For inner cylinder, diameter = 4 cm
∴ For inner cylinder,
radius r = \(\frac{\text { diameter }}{2}\) = 2 cm
and height (length) h = 77 cm.
Inner curved surface area of the pipe
= 2πrh
= 2 × \(\frac{22}{7}\) × 2 × 77 cm²
= 968 cm²
Thus, the inner curved surface area is 968 cm².

(ii) outer curved surface, area,
Answer:
For outer cylinder, diameter = 4.4 cm
∴ For outer cylinder,
radius R = \(\frac{\text { diameter }}{2}\) = \(\frac{4.4}{2}\) = 2.2
and height h = 77 cm.
Outer curved surface area of the pipe
= 2πRh
= 2 × \(\frac{22}{7}\) × 2 × 77 cm²
= 1064.8 cm²
Thus, the outer curved surface area is
1064.8 cm².

(iii) total surface area.
Answer:
Total surface area includes the area of two circular rings at the ends together with the inner and outer curved surface areas.
For each circular ring, outer radius R = 2.2 cm and inner radius r = 2 cm
Area of one circular ring
= π(R² – r²)
= \(\frac{22}{7}\)(2.2² – 2²)cm²
= \(\frac{22}{7}\) (4.84 – 4) cm²
= \(\frac{22}{7}\) × 0.84 cm²
= 2.64 cm²
∴ Area of two circular rings.
= 2 × 2.64 cm²
= 5.28 cm²
Now, total surface area of the pipe = Inner curved surface area + outer curved surface area + area of two circular rings
= 968 + 1064.8 + 5.28 cm²
= 2038.08 cm²

Question 4.
The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².
Answer:
For the cylindrical roller, diameter d = 84 cm and height (length) h = 120 cm.
Curved surface area of the cylindrical roller
= πdh
= \(\frac{22}{7}\) × 84 × 120 cm²
= 31680 cm²
= \(\frac{31680}{10000}\) m²
= 3.168 m²
Thus, the area of playground levelled in 1 complete revolution of the roller = 3.168 m²
∴ The area of playground levelled in 500 complete revolutions of the roller
= 3.168 × 500 m² = 1584 m²
Thus, the area of the playground is 1584 m².

Question 5.
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹ 12.50 per m².
Answer:
For the cylindrical pillar, diameter d = 50 cm = 0.5 m and height h = 3.5 m.
Curved surface area of the cylindrical pillar
= πdh
= \(\frac{22}{7}\) × 0.5 × 3.5 m²
= 5.5 m²
Cost of painting 1 m² area = ₹ 12.50
∴ Cost of painting 5.5 m2 area = ₹ (12.50 x 5.5)
= ₹ 68.75
Thus, the cost of painting the curved surface of the pillar is ₹ 68.75.

Question 6.
Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.
Answer:
For the given cylinder, radius r = 0.7 m and
curved surface area = 4.4 m².
Curved surface area of a cylinder = 2πrh
∴ 4.4 m² = 2 × \(\frac{22}{7}\) × 0.7m × h
∴ h = \(\frac{4.4 \times 7}{2 \times 22 \times 0.7}\)m
∴ h = 1 m
Thus, the height of the cylinder is 1 m.

Question 7.
The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of ₹ 40 per m².
Answer:
A circular well means a cylindrical well. For the cylindrical well, diameter d = 3.5 m and height (depth) h = 10 m.
(i) Curved surface area of the well
= πdh
= \(\frac{22}{7}\) × 3.5 × 10 m²
= 110 m²

(ii) Cost of plastering 1 m² region = ₹ 40
∴ Cost of plastering 110 m2 region
= ₹ (40 × 110)
= ₹ 4400

Question 8.
In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Answer:
For the cylindrical pipe, diameter d = 5 cm = 0.05 m and height (length) h = 28 m.
The radiation surface in the system is the •curved surface of the pipe.
Hence, we find the curved surface area of the cylindrical pipe.
Curved surface area of the cylindrical pipe
= πdh
= \(\frac{22}{7}\) × 0.05 × 28 m²
= 4.4 m²
Thus, the total radiating surface in the system is 4.4 m².

Question 9.
Find: (i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if \(\frac{1}{12}\) of the steel actually used was wasted in making the tank.
Answer:
For the closed cylindrical tank, diameter d = 4.2 m, hence radius
r = \(\frac{4.2}{2}\) = 2.1 m and height h = 4.5 m.

(i) Curved surface area of the cylindrical tank
= 2 πrh
= 2 × \(\frac{22}{7}\) × 2.1 × 4.5 m²
= 59.4 m²

(ii) Total surface area of the closed cylindrical tank
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × 2.1 (2.1 + 4.5) m²
= 13.2 × 6.6 m²
= 87.12 m²
Suppose, x m² steel was used for making the tank. But during production, \(\frac{1}{12}\) of the steel was wasted.
∴ Actual quantity of steel used = \(\frac{11}{12}\)x m².
Hence, \(\frac{11}{12}\)x = 87.12
∴ x = \(\frac{8712}{100} \times \frac{12}{11}\)
∴ x = 95.04 m²
Thus, the quantity of steel actually used during the preparation of the tank is 95.04 m².

Question 10.
In the given figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Answer:
The shape of the decorative cloth will be cylindrical.
For the cylinder of cloth, diameter d = 20 cm and height h = 30 cm + 2.5 cm + 2.5 cm = 35 cm.
Curved surface area of the cylinder of cloth
= πdh
= \(\frac{22}{7}\) × 20 × 35 cm²
Thus, 2200 cm² cloth is required for covering the lampshade.

Question 11.
The students of Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboards. If there were 35 competitors, how much cardboard was required to be bought for the competition ?
Answer:
The cylindrical penholders to be made have base but open at the top. Thus, to prepare a penholder, the area of the cardboard required will be given by the curved surface area of the cylinder and the area of base.

For cylindrical penholder, radius r = 3 cm and height h = 10.5 cm.
Area of cardboard required for 1 penholder
= Curved surface area of cylinder + Area of base
= 2πrh + πr²
= πr (2h + r)
= \(\frac{22}{7}\) × 3(2 × 10.5 + 3) cm2
= \(\frac{66 \times 24}{7}\) cm²
∴ Area of the cardboard required for 35 penholders
= 35 × \(\frac{66 \times 24}{7}\) cm²
= 7920 cm²
Thus, 7920 cm² cardboard was required to be bought for the competition.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

Question 1.
A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is open at the top. Ignoring the thickness of the plastic sheet, determine: (i) The area of the sheet required for making the box. (ii) The cost of sheet for it, if a sheet measuring 1 m² costs ₹ 20.
Answer:
The plastic box to be made is open at the top. Hence, the plastic sheet is required for the lateral surfaces and the base.
Here, for the box to be made,
length l = 1.5 m;
breadth b = 1.25 m and
height h = 65 cm = 0.65 m.
Area of the plastic sheet required for open box = Lateral surface area + Area of base
= 2 h(l + b) + l × b
= 2 × 0.65 (1.5 + 1.25) + 1.5 × 1.25 m²
= 1.3 × 2.75 + 1.875 m²
= 3.575 + 1.875 m²
= 5.45 m²
Cost of 1 m² sheet = ₹ 20
∴ Cost of 5.45 m² sheet = ₹ (5.45 × ₹ 20)
= ₹ 109

Question 2.
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹ 7.50 per m².
Answer:
For the given room, length 1 = 5 m; breadth b = 4 m and height h = 3 m.
Area of the region to be white washed
= Area of four walls + Area of ceiling
= 2 h(l + b) + l × b
= 2 × 3 (5 + 4) + 5 × 4 m²
= 54 + 20 m²
= 74 m²
Cost of white washing 1 m² region = ₹ 7.5
∴ Cost of white washing 74 m² region
= ₹ (74 × 7.5)
= ₹ 555

Question 3.
The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ₹ 10 per m² is ₹ 15,000, find the height of the hall. [Hint: Area of the four walls = Lateral surface area.]
Answer:
Area painted at the cost of ₹ 10 = 1 m²
∴ Area painted at the cost of ₹ 15,000
= \(\frac{15000}{10}\)
= 1500 m²
∴ Area of the four walls = 1500m²
∴ Lateral surface area = 1500 m²
∴ Perimeter Of the floor × Height = 1500 m²
∴ 250 m × Height = 1500 m²
∴ Height = \(\frac{15000}{250}\)
∴ Height = 6 m

Question 4.
The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimensions 22.5 cm × 10 cm× 7.5 cm can be painted out of this container?
Answer:
For each brick, length l = 22.5 cm; breadth b = 10 cm and height h = 7.5 cm.
Total surface area of one brick
= 2 (lb + bh + hl)
= 2 (22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm²
= 2 (225 + 75 + 168.75) cm²
= 2 (468.75) cm²
= 937.5 cm²
= \(\frac{937.5}{10000}\) m² = 0.09375 m²
No. of bricks that can be painted with paint sufficient to paint 0.09375 m² area = 1
∴ No. of bricks that can be painted with paint sufficient to paint 9.375 m² area 9.375
= \(\frac{9.375}{0.09375}\) = 100

Question 5.
A cubical box has each edge 10 cm and |> another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much ?
Answer:
For the cubical box, edge a = 10 cm and for the cuboidal box, length l = 12.5 cm; breadth b = 10 cm and height h = 8 cm
(i) Lateral surface area of cubical box
= 4a²
= 4 (10)² cm²
= 400 cm
Lateral surface area of cuboidal box
= 2h(l + b)
= 2 × 8(12.5 + 10) cm²
= 16 × 22.5 cm²
= 360 cm²
Thus, the lateral surface area of cubical box is greater by 40 cm² (400 – 360).

(ii) Total surface area of cubical box = 6a2
= 6 (10)² cm²
= 600 cm²
Total surface area of cuboidal box
= 2 (lb + bh + hl)
= 2(12.5 × 10 + 10 × 8 + 8 × 12.5) cm²
= 2 (125 + 80 + 100) cm²
= 2 (305) cm²
= 610 cm²
Thus, the total surface area of cubical box is smaller by 10 cm² (610 – 600).

Question 6.
A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass ?
(ii) How much of tape is needed for all the 12 edges ?
Answer:
(i) For the cuboidal greenhouse, length l = 30 cm; breadth fa = 25 cm and height h = 25 cm.
Area of glass used
= Total surface area of cuboid
= 2 (lb + bh + hl)
= 2 (30 × 25 + 25 × 25 + 25 × 30) cm²
= 2 (750 + 625 + 750) cm²
= 2 (2125) cm²
= 4250 cm²

(ii) 12 edges of the cuboidal greenhouse is made-up of 4 lengths, 4 breadths and 4 heights.
∴ Length of tape needed for 12 edges
= 4l + 4b + 4h
= 4 (l + b + h)
= 4 (30 + 25 + 25) cm
= 4 (80) cm
= 320 cm

Question 7.
Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5 % of the total surface area is required extra. If the cost of the cardboard is ₹ 4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.
Answer:
For bigger cuboidal boxes, length l = 25 cm;
breadth b = 20 cm and height h = 5 cm.
Total surface area of a bigger box
= 2 (lb + bh + hl)
= 2 (25 × 20 + 20 × 5 + 5 × 25) cm²
= 2 (500 + 100 + 125) cm²
= 1450 cm²
Area of cardboard required for overlap
= 5 % of 1450 cm²
= 72.5 cm²
Thus, the total area of cardboard required for 1 bigger box = 1450 + 72.5 cm²
= 1522.5 cm²
∴ The total area of cardboard required for 250 bigger boxes = (1522.5 × 250) cm²
For smaller cuboidal boxes, length l = 15 cm; breadth b = 12 cm and height h = 5 cm.
Total surface area of a smaller box
= 2 (lb + bh + hl)
= 2 (15 × 12 + 12 × 5 + 5 × 15) cm²
= 2(180 +60 + 75) cm²
= 2 (315) cm²
= 630 cm²
Area of cardboard required for overlap
= 5% of 630 cm²
= 31.5 cm²
Thus, the total area of cardboard required for 1 smaller box = 630 + 31.5 cm² = 661.5 cm²
∴ The total area of cardboard required for 250 smaller boxes = (661.5 × 250) cm²
Now, the total area of cardboard required for all the boxes
= (1522.5 × 250) + (661.5 × 250) cm²
= 250(1522.5 + 661.5) cm²
= 250 × 2184 cm²
Cost of 1000 cm² cardboard = ₹ 4
∴ Cost of 250 × 2184 cm² cardboard
= ₹ \(\left(\frac{4 \times 250 \times 2184}{1000}\right)\)
= ₹ 2184

Question 8.
Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m ?
Answer:
For the box-like structure without base, length
l = 4m; breadth b = 3m and height h = 2.5m.
Area of tarpaulin required
= Area of lateral surfaces + Area of top
= 2 h(l + b) + l × b
= 2 × 2.5 (4 + 3) + 4 × 3 m²
= 35 + 12 m²
= 47 m²

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.