Class 10

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter Statistics Ex 14.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.2

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution:
For mode:
In the given data, Maximum frequency is 23 and it corresponds to the class interval 35 – 45
Modal class = 35 – 45
So, l = 35; f₁ = 23; f₀ = 21; f₂ = 14 and h = 10
Using fonnula, Mode l = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
Mode = 35 + \(\left[\frac{23-21}{2(23)-21-14}\right]\) × 10
= 35 + \(\frac{2}{46-35}\) × 10
= 35 + \(\frac{20}{11}\) = 35 + 1.8 = 36.8.

For Mean:

From above data,
Assumed mean (a) = 30
Width of class (h) = 10
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=\frac{43}{80}\) = 0.5375
Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 30 + 10 (0.5375)
= 30 + 5.375 = 35.375 = 35.37
Hence, mode of given data is 36.8 years and mean of the given data is 35.37 years. Also, it is clear from above discussion that average age of a patient admitted in the hospital is 35.37 years and maximum number of patients admitted in the hospital are of age 36.8 years.

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the ¡nodal lifetimes of the components.
Solution:
In the given data.
Maximum frequency is 61 and it corresponds to the class interval 60 – 80.
∴ Model class = 60 – 80
So, l = 60; f₁ = 61 ; f₀ = 52; f₂ = 38 and h = 20
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

Mode = 60 + \(\left(\frac{61-52}{2(61)-52-38}\right)\) × 20

= 60 + \(\frac{9}{122-52-38}\) × 20

= 60 + \(\frac{9}{32}\) × 20

= 60 + \(\frac{180}{32}\)

= 60 + 5.625 = 65.625
Hence, modal Lifetimes of the components is 65.625 hours.

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution:
For Mode: In the given data.
Maximum frequency is 40, and it corresponds to the class interval 1500 – 2000.
∴ Model class = 1500 – 2000
So, l = 1500; f₁ = 40; f₀ = 24; f₂ = 33 and h = 500
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

= 1500 + \(\left\{\frac{40-24}{2(40)-24-33}\right\}\) × 500

= 1500 + \(\left\{\frac{16}{80-24-33}\right\}\) × 500

= 1500 + \(\frac{16 \times 500}{23}\)

= 1500 + \(\frac{8000}{23}\) = 1500 + 347.83 = 1847.83

For Mean:

From above data,

Assumed Mean (a) = 2750
Length of width (h) = 500
\(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=-\frac{35}{200}\) = – 0.175
Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\overline{\mathbf{X}}\) = 2750 + 500 (- 0.175)
= 2750 – 87.50 = 2662.50
Hence, the modal monthly expenditure of family is 1847.83 and the mean monthly expenditure is 2662.50.

Question 4.
The following distribution gives the slate-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Solution:
For Mode:
In the given data,
Maximum frequency is 10 and it corresponds to the class interval is 30 – 35.
∴ Modal class = 30 – 35.
So, l = 30; f₁ = 10; f₀ = 9; f₂ = 3 and h = 5
using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
Mode = 30 + \(\left(\frac{10-9}{2(10)-9-3}\right)\) × 5
= 30 + \(\frac{1}{20-12}\) × 5
= 30 + \(\frac{5}{8}\) = 30 + 0.625 = 30.625 = 30.63 (approx.)

For mean:

From above data, Assumed Mean (a) = 32.5
Width of class (h) = 5
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=-\frac{23}{35}\) = – 0.65

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\overline{\mathbf{X}}\) = 32.5 + 5 (- 0.65)
= 32..5 – 3.25 = 29.25 (approx.)
Hence, mode and mean of given data is 30.63 and 29.25. Also, from above discussion, it clear that states/U.T. have student per teacher is 30.63 and on average, this ratio is 29.25.

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches

Find the mode of the data.
Solution:
In the given data,
Maximum frequency is 18 and it corresponds to the class interval 4000 – 5000.
∴ Modal class = 4000 – 5000
So, l = 4000; f₁ = 18; f₀ = 4; f₂ = 9 and h = 1000
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

Mode = 4000 + \(\left(\frac{18-4}{2(18)-4-9}\right)\) × 1000

= 4000 + \(\frac{14}{36-13}\) × 1000

= 4000 + \(\frac{14000}{23}\) = 4000 + 608.6956

= 4000 + 608.7 = 4608.7 (approx.)
Hence, mode of the given data is 4608.7.

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised It in the table given below. Find the mode of the data:

Solution:
In the given data,
Maximum frequency is 20 and it corresponds to the class interval 40 – 50
∴ Modal Class = 40 – 50
So, l = 40; f₁ = 20; f₀ = 12; f₂= 11 and h = 10
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

Mode = 40 + \(\left(\frac{20-12}{2(20)-12-11}\right)\) × 10

= 40 + \(\frac{8}{40-23}\) × 10

= 40 + \(\frac{80}{17}\) = 40 + 4.70588

= 40 + 4.7 = 44.7 (approx.)
Hence, mode of the given data is 44.7 cars.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter Statistics Ex 14.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1

Question 1.
A survey was conducted by a group of students as a part of their enviroment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?
Solution:
Since, number of plants and houses are small in their values; so we must use direct method

Mean X = \(\overline{\mathrm{X}}=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}\)

\(\frac{162}{20}\) = 8.1

Hence, mean number of plants per house is 8.1.

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:

From given data,
Assumed Mean (a) = 150
and width of the class (h) = 20
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\)
= \(\frac{-12}{50}\) = – 0.24

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
= 150 + (20) (- 0.24)
= 150 – 4.8 = 145.2
Hence,mean daily wages of the workers of factory is ₹ 145.20.

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is 18. Find the missing frequency f.

Solution;

From the above data,
Assumed mean (a) = 18
Using formula, Mean \((\overline{\mathrm{X}})=a+\frac{\sum f_{i} d_{i}}{\Sigma f_{i}}\)

\(\bar{X}=18+\frac{2 f-40}{44+f}\)
But. Mean of data \((\bar{x})\) = 18 …(Given)
∴ 18 = 18 + \(\frac{2 f-40}{44+f}\)
or \(\frac{2 f-40}{44+f}\) = 18 – 18 = 0
or 2f – 40 = 0
or 2f = 40
or f = \(\frac{40}{2}\) = 20
Hence, missing frequency f is 20.

Question 4.
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Solution:

From above data,
Assumed Mean (a) = 75.5
Width of Class (h) = 3
∴ \(\bar{u}=\frac{\sum f_{i} u_{i}}{\sum f_{i}}\)
= \(\frac{4}{30}\) (approx.)
Using formula,
Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
= 75.5 + 3 (0.13) = 75.5 + 0.39
\(\overline{\mathrm{X}}\) = 75.89
Hence, mean heart beats per minute for women is 75.89.

Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution:
Since, values of number of mamgoes and number of boxes are large numerically. So, we must use step-deviation method.

From above data,
Assumed Mean (a) = 57
Width of class (h) = 3
∴ \(\bar{u}=\frac{\sum f_{i} u_{i}}{\sum f_{i}}\)

\(\bar{u}=\frac{25}{400}\) = 0.0625

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 57 + 3(0.0625)
= 57 + 0.1875 = 57.1875 = 57.19.
Hence, mean number of mangoes kept in a packing box is 57.19.

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.

Solution:

From above data,
Assumed mean (a) = 225
Width of class (h) = 50
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\)

\(\bar{u}=-\frac{7}{25}\) = 0.28
Using formula,
Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 225 + 50 (- 0.28)
\(\bar{X}\) = 225 – 14
\(\bar{X}\) = 211
Hence, mean daily expenditure on food is ₹ 211.

Question 7.
To find out the concentration of SO₂ in the air (in parts per million, e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO₂ in the air.
Solution:

From above data,
Assumed mean (a) = 0.10
Width of class (h) = 0.04
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\)

\(\bar{u}=-\frac{1}{30}\) = – 0.33(approx.)
Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 0.10 + 0.04 (- 0.33)
= 0.10 – 0.0013 = 0.0987 (approx.)
Hence, mean concentration of SO₂ in the air is 0.0987 ppm.

Question 8.
A class teacher has the folloing absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:

From above data,
Assumed Mean (a) = 17
Using formula, Mean(\(\bar{X}\)) = a + \(\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}\)
\(\bar{X}\) = 17 + \(-\frac{181}{40}\)
= 17 – 4.52 = 12.48.
Hence, mean 12.48 number of days a student was absent.

Question 9.
The following table gives the literacy rate (in percentage) of 35 citIes. Find the mean literacy rate.

Solution:

From the above data,
Assumed Mean (a) = 70
Width of class (h) = 10
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=\frac{-2}{35}\) = – 0.057

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 70 + 10 (- 0.057)
= 70 – 0.57 = 69.43
Hence, mean literacy rate is 69.43%.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5

Question 1.
A copper wire 3 mm in diameter is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to 8.88 g per cm³.
Solution:
Diametcr of wire (d) = 3 mm
∴ Radius of wire (r) = \(\frac{3}{2}\) mm = \(\frac{3}{20}\) cm

Diameter of cylinder = 10 cm
Radius of cylinder (R) = 5 cm
Height of cylinder (H) = 12 cm
Circumference of cylinder = length of wire used in one turn
2πR = length of wire used in one turn
2 × \(\frac{22}{7}\) × 5 = length of wire used in one turn
\(\frac{220}{7}\) = length of wire used in one turn
Number of turn used = \(\frac{\text { Height of cylinder }}{\text { Diameter of wire }}\)
= \(\frac{12 \mathrm{~cm}}{3 \mathrm{~mm}}=\frac{12}{3} \times 10\) [1 mm = \(\frac{1}{10}\) cm]
= \(\frac{120}{3}\) = 40

∴ length of wire used = Number of turns × length of wire used in one turn
H = 40 × \(\frac{220}{7}\) = 1257.14 cm
Volume of wire used = πr²H
= \(\frac{22}{7} \times \frac{3}{20} \times \frac{3}{20} \times 1257.14\) = 88.89 cm³
Mass of 1 cm³ = 8.88 gm
Mass of 88.89 cm³ = 8.88 × 88.89 = 789.41 gm
Hence, length of wire is 1257.14 cm
and mass of wire is 789.41 gm.

Question 2.
A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area the double cone so formed. (Choose value of t as found
appropriate.)
Solution:
Let ∆ABC be the right triangle right angled at A whose sides AB and AC measure 3 cm and 4 cm respectively.
The length of the side BC (hypotenuse) = \(\sqrt{3^{2}+4^{2}}=\sqrt{9+16}\) = 5 cm
Here, AO (or A’O) is the radius of the common base of the double cone formed revolving the right triangle about BC.
Height of the cone BAA’ is BO and slant height is 3 cm.
Height of the cone CAA’ is CO and slant height is 4 cm.
Now, ∆AOB ~ ∆CAB (AA similarity)

∴ \(\frac{\mathrm{AO}}{4}=\frac{3}{5}\)

⇒ AO = \(\frac{4 \times 3}{5}=\frac{12}{5}\) cm

Also \(\frac{\mathrm{BO}}{3}=\frac{3}{5}\)

⇒ BO = \(\frac{\mathrm{BO}}{3}=\frac{3}{5}\) cm
Thus CO = BC – OB
= 5 – \(\frac{9}{5}\) = \(\frac{16}{5}\) cm
Now, Volume of double cone = [Volume of Cone ABA’] + [Volume of cone ACA’]

∴ Volume of double cone = 30 cm³.
Also, surface area of double cone = [Surface area of Cone ABA’] + [surface area of Cone ACA’]
= π .AO.AB + π. AO.A’C
= \(\left(\frac{22}{7} \times \frac{12}{5} \times 3\right)+\left(\frac{22}{7} \times \frac{12}{5} \times 4\right)\)

= \(\left(\frac{792}{35} \times \frac{1056}{35}\right)\)
= \(\frac{1848}{35}\) = 52.75 cm²

Question 3.
A cistern, Internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without the water overflowing, each brick being 22.5 cm × 7.5 cm × 6.5 cm?
Solution:
Volume of one brick = 22.5 × 7.5 × 6.5 cm³ = 1096.87 cm³
Volume of cistern = 150 × 120 × 110 cm³ = 1980000 cm³
Let number of bricks used = n
Total volume of n bricks = n (volume of one brick) = n [1096.871] cm³
Volume of water = 129600 cm³
Volume of water available for bricks = 1980000 – 129600 = 1850400 cm³

Each bricks absorb \(\frac{1}{17}\)th of its volume in water
Volume of water available for bricks = \(\frac{17}{16}\) × volume of water available for bricks
= \(\frac{17}{10}\) × 1850400
Volume of water available for bricks = 1966050 cm³
Total volume of n bricks = Volume of water available for bricks
n[1096.87] cm³ = 1966050 cm³

n = \(\frac{1966050}{1096.87}\)
n = 1792.42
Hence, Number of bricks used = 1792.

Question 4.
In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km², show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.
Solution:
Area of valley = 97280 km²
Rainfall in valley = 10 cm
∴ Volume of total rainfall = 97280 × \(\frac{10}{100}\) × \(\frac{1}{1000}\) km³
= 9.728 km³

Question 5.
An oil funnel made of tin sheet consists of a cylindrical portion 10 cm long attached to a frust.un of a cone. if the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.

Solution:
Diameter of top of funnel = 18 cm
∴ Radius of top of funnel (R) = \(\frac{18}{2}\) cm = 9 cm
Diameter of bottom of funnel = 8 cm
Radius of bottom of funnel (r) = 4 cm
Height of cylindrical portion (h) = 10 cm
Height of frustum (H) = (22 – 10) = 12 cm
Slant height of frustum (l)
= \sqrt{\mathrm{H}^{2}+(\mathrm{R}-r)^{2}}\(\)

= \(\sqrt{(12)^{2}+(9-4)^{2}}\)

= \(\sqrt{144+(5)^{2}}\)

= \(\sqrt{144+25}\) = \(\sqrt{169}\)
Slant height of frustum (l) = 13 cm
Area of metal sheet required curved
surface area of cylindrical base + curved surface of frustum = 2πrh + πL [R + r]
= 2 × \(\frac{22}{7}\) × 4 × 10 + \(\frac{212}{7}\) × 13 [9 + 4] cm²
= 251.42 + 531.14 = 782.56 cm²

Question 6.
Derive the formula for the curved surface area and total surface area of a frustum of a cone, given to you in Section 13.5, using the symbols as explamed.
Solution:
A frustum of a right circular cone has two unequal flat circular bases and a curved surface. Let ACDB be the frustum of the cone obtained by removing the portion VCÐ. The line-segment OP joining the centres of two bases is called the height of the frustum. Each of the line-segments AC and BD of frustum ACDB is called its slant height.

Let R and r be the radii of the circular end faces (R > r) of the frustum ACDB of the cone (V, AB). We complete the conical part VCD. Let h and l be the vertical height and slant height respectively. Then OP = h and AC = BD = l.

The frustum of the right circular cone can be viewed as the difference of the two right circular VAB and VCD.
Let the height of the cone VAB be h₁ and its slant height l₁ i.e. VP = h₁ and VA = VB = l₁.

Now from right ∆DEB, we have
DB² = DE² + BE²

⇒ l² = h² + (R – r)²
⇒ l = \(\sqrt{h^{2}+(\mathrm{R}-r)^{2}}\)
Again ∆VOD ~ ∆VPB

⇒ \(\frac{V D}{V B}=\frac{O D}{P B}\)

⇒ \(\frac{l_{1}-l}{l_{1}}=\frac{r}{R}\)

⇒ 1 – \(\frac{l}{l_{1}}=\frac{r}{\mathrm{R}}\)

⇒ \(\frac{l}{l_{1}}=1-\frac{r}{\mathrm{R}}=\frac{\mathrm{R}-r}{\mathrm{R}}\)

⇒ l₁ = \(\frac{l \mathrm{R}}{\mathrm{R}-r}\)

⇒ Now, l₁ – l₁ = \(\frac{l \mathrm{R}}{\mathrm{R}-r}-l=\frac{l r}{\mathrm{R}-r}\)

Curved surface area of the frustum of cone = πRl₁ – πr(l₁ – l)
[Curved surface area of a cone = π × r × 1]

= πR. \(\frac{l R}{R-r}\) – πr. \(\frac{l r}{\mathrm{R}-r}\)

= πl (\(\frac{\mathrm{R}^{2}-r^{2}}{\mathrm{R}-r}\)) = \(\frac{\pi l(\mathrm{R}-r)(\mathrm{R}+r)}{(\mathrm{R}-r)}\)
= πl (R + r) sq. units.
∴ Curved snafaue area of the frustum of right circular cone = πl (R + r) sq. units,
where l = \(\sqrt{h^{2}+(\mathrm{R}-r)^{2}}\)
and total surface area of frustum of right circular cone = Curved surface area + Area of base + Area of top
= πl (R + r) + πR² + πr²
∴ π [R² + r² + l (R + r)] sq. units.

Question 7.
Derive the formula for the volume of the frustum of a cone, given to you in section 13.5, using the symbols as explained.
Solution:
A frustum of a right circular cone has two unequal flat circular bases and a curved surface. Let ACDB be the frustum of the cone obtained by removing the portion VCD. The line-segment OP joining the centres of two bases is called the height of the frustum. Each of the line-segments AC and BD of frustum ACDB is called its slant height.

Ler R and r be the radii of the circular end faces (R > r) of the fnistum ACDB of the cone (V, AB). We complete the conical part VCD. Let h and l be the vertical height and slant height respectively. Then OP = h and AC = BD = l. The frustum of the right circular cone can be viewed as the difference of the two right circular VAB and VCD.
Let theheight of the cone VAB be h₁ and its slant height l₁ i.e.VP = h₁ and VA = VB = l₁.
∴ The height of the cone VCD = VP – OP = h₁ – h
Since right ∆s VOD and VPB are similar
⇒ \(\frac{\mathrm{VO}}{\mathrm{VP}}=\frac{\mathrm{OD}}{\mathrm{PB}}=\frac{h_{1}-h}{h_{1}}=\frac{r}{\mathrm{R}}\)

⇒ 1 – \(\frac{h}{h_{1}}=\frac{r}{\mathrm{R}}\)

⇒ \(\frac{h}{h_{1}}=1-\frac{r}{\mathrm{R}}=\frac{\mathrm{R}-r}{\mathrm{R}}\)

⇒ h₁ = \(\frac{h \mathrm{R}}{\mathrm{R}-r}\)

⇒ Height of the cone VCD = \(\frac{h \mathrm{R}}{\mathrm{R}-r}-h=\frac{h \mathrm{R}-h \mathrm{R}+h r}{\mathrm{R}-r}=\frac{h r}{\mathrm{R}-r}\)

Volume of the frustrum ACDB of the cone (V,AB) = Volume of the cone (V, AB) – Volume of the cone (V, CD)
= \(\)

= \(\frac{\pi}{3}\left(\mathrm{R}^{2} \cdot \frac{h \mathrm{R}}{\mathrm{R}-r}-r^{2} \cdot \frac{h r}{\mathrm{R}-r}\right)\)

= \(\frac{\pi h}{3}\left(\frac{\mathrm{R}^{3}-r^{3}}{\mathrm{R}-r}\right)\)

= \(\frac{1}{3} \pi h \times \frac{(\mathrm{R}-r)\left(\mathrm{R}^{2}+\mathrm{R} r+r^{2}\right)}{(\mathrm{R}-r)}\)

= \(\frac{1}{3}\) πh (R² + Rr + r²)
Hence, volume of the frustum of cone is \(\frac{1}{3}\) πh (R² + Rr + r²).

Again if A₁ and A₂ (A₁ > A₂) are the surface areas of the two circular bases, then A₁ = πR² andA₂ = πr²

Now volume of the frustum of cone,

= \(\frac{1}{3}\) πh (R² + r² + Rr)

= \(\frac{h}{3}\) (πR² + πr² + \(\sqrt{\pi \mathrm{R}^{2}} \sqrt{\pi r^{2}}\))

= \(\frac{h}{3}\) (A₁ + A₂ + \(\sqrt{A_{1} A_{2}}\))

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

Question 1.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Solution:

Radius of upper end (R) = 2 cm
Radius of lower end (r) = 1 cm
Height of glass (H) = 14 cm
Glass is in the shape of frustum
Volume of frustum = \(\frac{1}{3}\) π [R² + r² +Rr]H

= \(\frac{1}{3}\) × \(\frac{22}{7}\) [(2)² + (1)² + 2 × 1] 14

= \(\frac{1}{3}\) × \(\frac{22}{7}\) [4 + 1 + 2] 14

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 7 × 14

= \(\frac{22 \times 14}{3}\)
Hence, Volume of glass = 102.67 cm³.

Question 2.
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution:
Slant height of frustum = 4 cm

Let radius of upper end and lower ends are R and r
Circumference of upper end = 18 cm
2πR = 18
R = \(\frac{18}{2 \pi}=\frac{9}{\pi}\) cm
Circumference of lower end = 6 cm
2πr = 6 cm
r = \(\frac{6}{2 \pi}=\frac{3}{\pi}\) cm
Curved surface area of frustum = π [R + r] l
= π \(\left[\frac{9}{\pi}+\frac{3}{\pi}\right]\) 4
= π \(\left[\frac{9+3}{\pi}\right]\) 4
= 12 × 4 = 48 cm²
Hence, Curved surface area of frustum = 48 cm².

Question 3.
A fez, the cap used by the Turks, is shaded like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Solution:

Radius of lower end of frustum (R) = 10 cm
Radius of upper end of frustum (r) = 4 cm
Slant height of frustum (l) = 15 cm
Curved surface area of frustum = πL [R+ r]
= \(\frac{22}{7}\) × 15 [10 + 4]
= \(\frac{22}{7}\) × 15 × 14
= 22 × 15 × 2 = 660 cm²
Area of the closed side = πr² = \(\frac{22}{7}\) × (4)²
= \(\frac{22}{7}\) × 4 × 4 = \(\frac{352}{7}\) cm²
Total area of the material used = Curved surface area of frustum + Area of the closed side
= 660 + 50.28 = 710.28 cm²
Hence, Total material used = 710.28 cm².

Question 4.
A container opened from the top is made up of a metal sheet ¡s in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate, of ₹ 20 per litre. Also fmd the cost of metal sheet used to make the container, If it costs ₹ 8 per 100 cm². (Take π = 3:14.)
Solution:

Radius of upper end of container (R) = 20 cm
Radius of lower end of container (r) = 8 cm
Height of container (H) = 16 cm
Slant height (l) = \(\sqrt{\mathrm{H}^{2}+(\mathrm{R}-r)^{2}}\)
= \(\sqrt{(16)^{2}+(20-8)^{2}}\) = \(\sqrt{256+144}\)
Slant height (l) = \(\sqrt{400}=\sqrt{20 \times 20}\) = 20 cm

Capacity of the container = \(\frac{1}{3}\) πH[R² + r² + Rr]
= \(\frac{1}{3}\) × 3.14 × 16 [(20)² + (8)² + 20 × 8]
= \(\frac{3.14 \times 16}{3}\) [400 + 64 + 100]
= 3.14 × 16 × 624 = 10449.92 cm³
Milk in the container = 10449.92 cm³
= \(\frac{10449.92}{1000}\) litres [∵ 1 cm³ = \(\frac{1}{1000}\) litres]
∴ Milk in the container = 10.45 litres
Cost of 1 it. milk = ₹ 20
∴ Cost of 10.45 litre = ₹ 20 × 10.45
Total cost of milk = ₹ 209
Curved surface area of frustum = πL [R + r]
= 3.14 × 20[20 + 8]
= 3.14 × 20 × 28 cm² = 1758.4 cm²
Area of base of container = πr²
= 3.14 × (8)π = 3.14 × 64 = 200.96 cm²

Total metal used to make coniainer = curved surface area of frustum + area of base
= (1758. 4 + 200.96) cm² = 1959.36 cm²
Cost of 100 cm² metal sheet used = ₹ 8
Cost of 1 cm² metal sheet used = ₹ \(\frac{8}{100}\)
Cost of 1959.36 cm² metal sheet used = ₹ \(\frac{8}{100}\) × 1959.36
= ₹ 156.748 = ₹ 156.75
Hence, Total cost of sheet used = ₹ 156.75
and Total cost of milk is ₹ 209.

Question 5.
A metallic right circular cone 20 cm high and whose vertical angle Ls 600 is cut into two parts at the middle of its height by a plane parallel to Its base. If the frustum so obtained be drawn into a wire of diameter \(\frac{4}{4}\) cm, find the length of the wire.
Solution:
Vertical angle of cone = 60°
Altitude of cone divide vertical angle ∠EOF = 30°

In ∆ODB,
\(\frac{\mathrm{BD}}{\mathrm{OD}}\) = tan 30°

\(\frac{r}{10}=\frac{1}{\sqrt{3}}\)

r = \(\frac{10}{\sqrt{3}}\) cm

In ∆OEF,

\(\frac{E F}{O E}\) = tan 30°

\(\frac{\mathrm{R}}{20}=\frac{1}{\sqrt{3}}\)

R = \(\frac{20}{\sqrt{3}}\) cm

Volume of frustum = \(\frac{\pi}{3}\)h[R² + r² + Rr]
= \(\frac{22}{73} \times \frac{10}{3}\left[\left(\frac{20}{\sqrt{3}}\right)^{2}+\left(\frac{10}{\sqrt{3}}\right)^{2}+\frac{20}{\sqrt{3}} \times \frac{10}{\sqrt{3}}\right]\)

= \(\frac{22}{7} \times \frac{10}{3}\left[\frac{400}{3}+\frac{100}{3}+\frac{200}{3}\right]\)

= \(\frac{22}{7} \times \frac{10}{3}\left[\frac{400+100+200}{3}\right]\)

Volume of frustum = \(\frac{22}{7} \times 10 \times \frac{700}{9}\) cm³

= \(\frac{22}{7} \times \frac{7000}{9}\) cm³
Frustum is made into wiie, which is in shape of cylinder having diameter \(\frac{1}{16}\) cm
∴ Radius of cylinderical wire (r₁) = \(\frac{1}{2} \times \frac{1}{16} \mathrm{~cm}=\frac{1}{32} \mathrm{~cm}\)

Let height of cylinder so formed be H cm
On recasting volume remain same
Volume of frustum = Volume of cylindrical wire

\(\frac{22}{7} \times \frac{7000}{9}\) = πr₁²H

\(\frac{22}{7} \times \frac{7000}{9}=\frac{22}{7} \times\left(\frac{1}{32}\right)^{2} \times \mathrm{H}\)

H = \(\frac{\frac{22}{7} \times \frac{7000}{9}}{\frac{22}{7} \times \frac{1}{32} \times \frac{1}{32}}\)

= \(\frac{7000}{9}\) × 32 × 32

H = 796444.44 cm

H = \(\) = 7964.44 m
Hence, Length of cylindrical wire (H) = 7964.44 m

Punjab State Board PSEB 6th Class Punjabi Book Solutions Chapter 1 ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Punjabi Chapter 1 ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ

ਪਾਠ-ਅਭਿਆਸ ਪ੍ਰਸ਼ਨ-ਉੱਤਰ

1. ਛੋਟੇ ਉੱਤਰ ਵਾਲੇ ਪ੍ਰਸ਼ਨ

ਪ੍ਰਸ਼ਨ 1.
‘ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਕਵਿਤਾ ਵਿਚ ਕਿਸ ਪੁੱਤ ਦਾ ਜ਼ਿਕਰ ਹੈ ?
ਉੱਤਰ :
ਪੰਜਾਬ ਦਾ ।

ਪ੍ਰਸ਼ਨ 2.
“ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਕਵਿਤਾ ਦਾ ਕਵੀ ਕੌਣ ਹੈ ?
ਉੱਤਰ :
ਸਰਦਾਰ ਅੰਜੁਮ ।

2. ਸੰਖੇਪ ਉੱਤਰ ਵਾਲੇ ਪ੍ਰਸ਼ਨ

ਪ੍ਰਸ਼ਨ 1.
ਹੇਠ ਲਿਖੀ ਕਾਵਿ-ਸਤਰ ਦਾ ਭਾਵ-ਅਰਥ ਲਿਖੋ :
‘ਇਹ ਮਿੱਟੀ ’ਤੇ ਘਰ-ਘਰ ਜਾ ਕੇ ਪਿਆਰ ਦਾ ਬੂਟਾ ਲਾਵੇ ।
ਉੱਤਰ :
ਪੰਜਾਬੀ ਲੋਕਾਂ ਦਾ ਮੁਢਲਾ ਸੁਭਾ ਸਾਰੀ ਮਨੁੱਖਤਾ ਵਿਚ ਪਿਆਰ ਦਾ ਪਸਾਰ ਕਰਨ ਵਾਲਾ ਹੈ ।

ਪ੍ਰਸ਼ਨ 2.
‘ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਕਵਿਤਾ ਦੀਆਂ ਚਾਰ ਸਤਰਾਂ ਜ਼ਬਾਨੀ ਲਿਖੋ ।
ਉੱਤਰ :
ਇਹਦਾ ਰੰਗ ਸੰਧੂਰੀ ਹੈ, ਇਹ ਗੋਰੀ ਚਿੱਟੀ ਹੈ ।
ਇਹਨੂੰ ਮੈਲੀ ਨਾ ਕਰਨਾ, ਮੇਰੇ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਹੈ ।
ਇਸ ਮਿੱਟੀ ਵਿਚ ਕੋਈ ਕੁੜੱਤਣ, ਕਿੱਦਾਂ ਕੋਈ ਉਗਾਵੇ ।
ਇਹ ਮਿੱਟੀ ‘ਤੇ ਘਰ-ਘਰ ਜਾ ਕੇ ਪਿਆਰ ਦਾ ਬੂਟਾ ਲਾਵੇ ।

ਪ੍ਰਸ਼ਨ 3.
“ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਕਵਿਤਾ ਨੂੰ ਟੋਲੀ ਬਣਾ ਕੇ ਆਪਣੀ ਜਮਾਤ ਵਿਚ ਗਾਓ ।
ਉੱਤਰ:
(ਨੋਟ :-ਵਿਦਿਆਰਥੀ ਆਪ ਕਰਨ ।)

3. ਅਧਿਆਪਕ ਲਈ

ਪ੍ਰਸ਼ਨ 1.
ਪੰਜਾਬ ਦੀ ਮਹਾਨਤਾ ਬਾਰੇ ਬੱਚਿਆਂ ਨੂੰ ਹੋਰ ਜਾਣਕਾਰੀ ਦਿੱਤੀ ਜਾਵੇ ।
ਉੱਤਰ :
ਪੰਜਾਬ ਪੰਜਾਂ ਦਰਿਆਵਾਂ ਦੀ ਧਰਤੀ ਹੈ । 1947 ਵਿਚ ਇਹ ਭਾਰਤੀ ਪੰਜਾਬ ਤੇ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਪਾਕਿਸਤਾਨੀ ਪੰਜਾਬ ਵਿਚ ਵੰਡੀ ਗਈ ਹੈ , ਜਿਸ ਕਰਕੇ ਹੁਣ ਢਾਈ ਦਰਿਆ, ਸਤਲੁਜ, ਬਿਆਸ ਤੇ ਅੱਧਾ ਰਾਵੀ ਇਧਰ ਰਹਿ ਗਏ ਹਨ ਤੇ ਢਾਈ ਦਰਿਆ ਚਨਾਬ, ਜਿਹਲਮ ਤੇ ਅੱਧਾ ਰਾਵੀ ਉਧਰ । ਇਹ ਦਸ ਗੁਰੂ ਸਾਹਿਬਾਨ ਤੇ ਸੂਫ਼ੀ ਫ਼ਕੀਰਾਂ ਦੀ ਧਰਤੀ ਹੈ । ਇਸੇ ਧਰਤੀ ਉੱਤੇ ਸਿੱਖ ਧਰਮ ਦਾ ਜਨਮ ਹੋਇਆ ਤੇ ਸ੍ਰੀ ਗੁਰੂ ਗ੍ਰੰਥ ਸਾਹਿਬ ਦੀ ਰਚਨਾ ਹੋਈ । ਸੰਸਾਰ ਦੇ ਸਭ ਤੋਂ ਪੁਰਾਤਨ ਧਰਮ ਗੰਥ ਰਿਗਵੇਦ ਦੀ ਰਚਨਾ ਇੱਥੇ ਹੀ ਹੋਈ, । ਮਹਾਂਭਾਰਤ ਦਾ ਯੁੱਧ ਤੇ ਸੀ । ਮਦ ਭਗਵਦ ਗੀਤਾ ਦੀ ਰਚਨਾ ਵੀ ਪੁਰਾਤਨ ਪੰਜਾਬ ਦੀ ਧਰਤੀ ‘ਤੇ ਹੀ ਹੋਈ ।ਇਸਦੀ ਬੋਲੀ ਪੰਜਾਬੀ ਹੈ । ਅਜੋਕੇ ਪੰਜਾਬ ਦੇ ਪੱਛਮੀ ਪਾਸੇ ਪਾਕਿਸਤਾਨ ਲਗਦਾ ਹੈ । ਉੱਤਰੀ ਪਾਸੇ ਕਸ਼ਮੀਰ, ਪੂਰਬੀ ਪਾਸੇ ਹਿਮਾਚਲ, ਦੱਖਣ-ਪੂਰਬੀ ਪਾਸੇ ਹਰਿਆਣਾ ਤੇ ਦੱਖਣ-ਪੱਛਮੀ ਪਾਸੇ ਰਾਜਸਥਾਨ ਲਗਦੇ ਹਨ । ਇਸਦਾ ਖੇਤਰਫਲ 50,362 ਵਰਗ ਕਿਲੋਮੀਟਰ ਹੈ । ਕਣਕ ਤੇ ਝੋਨਾ ਇਸਦੀਆਂ ਮੁੱਖ ਫ਼ਸਲਾਂ ਹਨ । ਇਸਦੇ 22 ਜ਼ਿਲ੍ਹੇ ਹਨ । ਇਸਦੀ ਆਬਾਦੀ 2 ਕਰੋੜ 80 ਲੱਖ ਹੈ । ਪੰਜਾਬੀ ਲੋਕ ਆਪਣੇ ਖੁੱਲ੍ਹੇ-ਡੁੱਲੇ, ਮਿਹਨਤੀ ਤੇ ਅਣਖੀਲੇ ਸੁਭਾ ਕਰਕੇ ਸੰਸਾਰ ਭਰ ਵਿਚ ਪ੍ਰਸਿੱਧ ਹਨ । ਉਹ ਕਿਰਤ ਕਰਨ, ਨਾਮ ਜਪਣ ਤੇ ਵੰਡ ਛਕਣ ਵਿਚ ਵਿਸ਼ਵਾਸ ਰੱਖਦੇ ਹਨ । ਅਫ਼ਸੋਸ ਕਿ ਅੱਜ ਦੀ ਨੌਜਵਾਨ ਪੀੜੀ ਖੇਡਾਂ ਵਿਚ ਨਾਮਣਾ ਕਮਾਉਣ ਤੇ ਸਰੀਰ ਪਾਲਣ ਦੇ ਸ਼ੌਕ ਛੱਡ ਕੇ ਨਸ਼ਿਆਂ ਵਿਚ ਗ਼ਰਕ ਹੋ ਚੁੱਕੀ ਹੈ ।

 

ਪਸ਼ਨ 1.
ਹੇਠ ਲਿਖੇ ਕਾਵਿ-ਟੋਟੇ ਦੇ ਸਰਲ ਅਰਥ ਕਰੋ :
(ਉ) ਇਹਦਾ ਰੰਗ ਸੰਧੂਰੀ ਹੈ ਇਹ ਗੋਰੀ ਚਿੱਟੀ ਹੈ ।
ਇਹਨੂੰ ਮੈਲੀ ਨਾ ਕਰਨਾ, ਮੇਰੇ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਹੈ ।
ਉੱਤਰ :
ਕਵੀ ਅੰਜੁਮ ਕਹਿੰਦਾ ਹੈ ਕਿ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਦਾ ਰੰਗ ਦਿਲ-ਖਿਚਵਾਂ ਸੰਧੂਰੀ ਹੈ । ਇਹ ਮਿੱਟੀ ਗੋਰੀ-ਚਿੱਟੀ ਅਰਥਾਤ ਸਾਫ਼-ਸੁਥਰੀ ਹੈ । ਸਾਡਾ ਸਭ ਦਾ ਫ਼ਰਜ਼ ਹੈ ਕਿ ਪੰਜਾਬ ਦੀ ਇਸ ਸਾਫ਼-ਸੁਥਰੀ ਮਿੱਟੀ ਨੂੰ ਕਿਸੇ ਤਰ੍ਹਾਂ ਵੀ ਮੈਲੀ ਨਾ ਕਰੀਏ ।

ਔਖੇ ਸ਼ਬਦਾਂ ਦੇ ਅਰਥ-ਸੰਧੂਰੀ = ਸੰਧੂਰ ਦੇ ਰੰਗ ਵਰਗੀ ।

ਪ੍ਰਸ਼ਨ 2.
ਹੇਠ ਲਿਖੇ ਕਾਵਿ-ਟੋਟੇ ਦੇ ਸਰਲ ਅਰਥ ਕਰੋ :
(ਅ) ਇਸ ਮਿੱਟੀ ਵਿਚ ਕੋਈ ਕੁੜੱਤਣ, ਕਿੱਦਾਂ ਕੋਈ ਉਗਾਵੇ ।
ਇਹ ਮਿੱਟੀ ’ਤੇ ਘਰ-ਘਰ ਜਾ ਕੇ, ਪਿਆਰ ਦਾ ਬੂਟਾ ਲਾਵੇ ।
ਉੱਤਰ :
ਕਵੀ ਅੰਜੁਮ ਕਹਿੰਦਾ ਹੈ ਕਿ ਇਹ ਕਿਵੇਂ ਹੋ ਸਕਦਾ ਹੈ ਕਿ ਕੋਈ ਮੇਰੇ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਵਿਚ ਫ਼ਿਰਕੂਪੁਣੇ ਦੀ ਕੁੜੱਤਣ ਪੈਦਾ ਕਰ ਦੇਵੇ, ਜਦਕਿ ਇਹ ਮਿੱਟੀ ਤਾਂ ਘਰ-ਘਰ ਜਾ ਕੇ ਆਪਸੀ ਪਿਆਰ ਦਾ ਬੂਟਾ ਲਾਉਣ ਦਾ ਕੰਮ ਕਰਦੀ ਹੈ, ਅਰਥਾਤ ਪੰਜਾਬੀ ਲੋਕਾਂ ਦਾ ਮੁਢਲਾ ਸੁਭਾ ਸਾਰੀ ਮਨੁੱਖਤਾ ਵਿਚ ਆਪਸੀ ਪਿਆਰ ਦਾ ਪਸਾਰਾ ਕਰਨ ਵਾਲਾ ਹੈ ।

ਪ੍ਰਸ਼ਨ 3.
ਹੇਠ ਲਿਖੇ ਕਾਵਿ-ਟੋਟੇ ਦੇ ਸਰਲ ਅਰਥ ਕਰੋ :
(ਈ) ਪੁੱਛ ਲਓ ਫ਼ਕੀਰਾਂ ਤੋਂ, ਇਹਦੀ ਬਾਣੀ ਮਿੱਠੀ ਹੈ ।
ਇਹਨੂੰ ਮੈਲੀ ਨਾ ਕਰਨਾ, ਮੇਰੇ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਹੈ ।
ਉੱਤਰ :
ਕਵੀ ਅੰਜੁਮ ਕਹਿੰਦਾ ਹੈ ਕਿ ਇਹ ਗੱਲ ਸੂਫ਼ੀ ਫ਼ਕੀਰਾਂ ਤੇ ਗੁਰੂ ਸਾਹਿਬਾਂ ਤੋਂ ਪੁੱਛ ਕੇ ਸਮਝੀ ਜਾ ਸਕਦੀ ਹੈ ਕਿ ਪੰਜਾਬ ਦੀ ਬੋਲੀ ਬਹੁਤ ਮਿੱਠੀ ਹੈ, ਇਸੇ ਕਰਕੇ ਹੀ ਉਨ੍ਹਾਂ ਨੇ ਆਪਣੇ ਭਾਵਾਂ ਤੇ ਵਿਚਾਰਾਂ ਦਾ ਪ੍ਰਗਟਾਵਾ ਇਸ ਬੋਲੀ ਵਿਚ ਕੀਤਾ ਹੈ । ਜਿਸ ਪੰਜਾਬ ਦੀ ਧਰਤੀ ਉੱਤੇ ਇਹ ਬੋਲੀ ਬੋਲੀ ਜਾਂਦੀ ਹੈ, ਉਸਦੀ ਮਿੱਟੀ ਨੂੰ ਕਿਸੇ ਤਰ੍ਹਾਂ ਵੀ ਮੈਲੀ ਨਹੀਂ ਕਰਨਾ ਚਾਹੀਦਾ

ਔਖੇ ਸ਼ਬਦਾਂ ਦੇ ਅਰਥ-ਬਾਣੀ = ਬੋਲੀ ।

ਪ੍ਰਸ਼ਨ 4.
ਹੇਠ ਲਿਖੇ ਕਾਵਿ-ਟੋਟੇ ਦੇ ਸਰਲ ਅਰਥ ਕਰੋ :
(ਸ) ਕਰੋ ਦੁਆ ਕਦੇ ਇਸ ਮਿੱਟੀ ’ਤੇ ਉਹ ਮੌਸਮ ਨਾ ਆਵੇ।
ਇਸ ਤੇ ਉੱਗਿਆ ਹਰ ਕੋਈ ਸੂਰਜ, ਕਾਲਖ਼ ਹੀ ਬਣ ਜਾਵੇ ।
ਉੱਤਰ :
ਕਵੀ ਅੰਜੁਮ ਸਾਨੂੰ ਸਭ ਨੂੰ ਰੱਬ ਅੱਗੇ ਇਹ ਅਰਦਾਸ ਕਰਨ ਲਈ ਕਹਿੰਦਾ ਹੈ ਕਿ ਪੰਜਾਬ ਦੀ ਧਰਤੀ ਦੀ ਮਿੱਟੀ ਉੱਤੇ ਕਦੇ ਵੀ ਫ਼ਿਰਕੂ ਜ਼ਹਿਰ ਨਾਲ ਭਰਿਆ ਉਹ ਮੌਸਮ ਨਾ ਆਵੇ ਕਿ ਇਸ ਉੱਤੇ ਉੱਗਿਆ ਹਰ ਇਕ ਸੁਰਜ ਭਾਵ ਇਸ ਉੱਤੇ ਚੜ੍ਹਨ ਵਾਲਾ ਹਰ ਦਿਨ ਚਾਨਣ ਦੀ ਥਾਂ ਨਫ਼ਰਤ ਦੀ ਕਾਲਖ਼ ਦਾ ਪਸਾਰ ਕਰ ਦੇਵੇ ।

ਔਖੇ ਸ਼ਬਦਾਂ ਦੇ ਅਰਥ-ਦੁਆ = ਬੇਨਤੀ, ਅਰਦਾਸ ।

ਪ੍ਰਸ਼ਨ 5.
ਹੇਠ ਲਿਖੇ ਕਾਵਿ-ਟੋਟੇ ਦੇ ਸਰਲ ਅਰਥ ਕਰੋ :
(ਹ) ਰੱਬ ਦੇ ਘਰ ਤੋਂ ਅੰਜੁਮ, ਆਈ ਇਹ ਚਿੱਠੀ ਹੈ ।
ਇਹਨੂੰ ਮੈਲੀ ਨਾ ਕਰਨਾ, ਮੇਰੇ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਹੈ ।
ਉੱਤਰ :
ਕਵੀ ਅੰਜੁਮ ਕਹਿੰਦਾ ਹੈ ਕਿ ਰੱਬ ਦੇ ਘਰ ਤੋਂ ਵੀ ਸਾਨੂੰ ਚਿੱਠੀ ਰਾਹੀਂ ਇਹ ਸੰਦੇਸ਼ ਪੁੱਜਾ ਹੈ ਕਿ ਅਸੀਂ ਆਪਣੇ ਪੰਜਾਬ ਦੀ ਸਾਫ਼-ਸੁਥਰੀ ਮਿੱਟੀ ਨੂੰ ਨਫ਼ਰਤ ਦੀ ਜ਼ਹਿਰ ਭਰ ਕੇ ਇਸਨੂੰ ਮੈਲੀ ਨਾ ਕਰੀਏ, ਸਗੋਂ ਸਾਫ਼-ਸੁਥਰੀ ਹੀ ਰਹਿਣ ਦੇਈਏ ।. ਔਖੇ ਸ਼ਬਦਾਂ ਦੇ ਅਰਥ-ਅੰਜੁਮ = ਕਵੀ ਦਾ ਨਾਂ, ਸਰਦਾਰ ਅੰਜੁਮ ॥

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

Question 1.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of cylinder of radius 6 cm. Find the height of the cylinder.
Solution:

Radius of sphere (r) = 4.2 cm
Radius of cylinder (R) = 6 cm
Let height of cylinder be H cm
On recasting volume remain same
Volume of sphere = Volume of cylinder
\(\frac{4}{3}\) πR³ = πR²H

\(\frac{4}{3}\) × \(\frac{22}{7}\) × 4.2 × 4.2 × 4.2 = \(\frac{22}{7}\) × 6 × 6 × H
∴ H = \(\frac{\frac{4}{3} \times \frac{22}{7} \times 4.2 \times 4.2 \times 4.2}{\frac{22}{7} \times 6 \times 6}\)

= \(\frac{4}{3} \times \frac{42}{10} \times \frac{42}{10} \times \frac{42}{10} \times \frac{1}{6 \times 6}\)

= \(\frac{2744}{1000}\) = 2.744 cm

Hence, Height of cylinder (H) = 2.744 cm.

Question 2.
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:

Radius of first sphere (r₁) = 6 cm
Radius of second sphere (r₂) = 8 cm
Radius of third sphere (r₃) = 10 cm
Let radius of new sphere formed be R cm
On recasting volume remain same
Volume of all three spheres = Volume of big sphere

\(\frac{4}{3}\) πr₁³ + \(\frac{4}{3}\) πr₂³ + \(\frac{4}{3}\) πr₃³ = \(\frac{4}{3}\) πR³

\(\frac{4}{3}\) π[(6)³ + (8)³ + (10)³] = \(\frac{4}{3}\) πR³

R³ = \(\frac{\frac{4}{3} \pi[216+512+1000]}{\frac{4}{3} \pi}\)

R³ = 1728
R = \(\sqrt[3]{1728}\) = \(\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3}\)

= 2 × 2 × 3
R= 12 cm
Hence, Radius of sphere = 12 cm.

Question 3.
A 20 m deep well with-diameter 7 m is dug and the earth from digging is evenly spread out to form of platform 22 m by 14 m. Find the height of the platform.
Solution:
Diameter of well = 7 m
Radius of well (cylinder) R = \(\frac{7}{2}\) m
Height of well (H) = 20 m
Length of Platform (L) =22 m
Width of Platform (B) = 14 m
Let height of Platform be H m

Volume of earth dug out from well = Volume of platform formed

πR²H = L × B × H
× × × 20 = 22 × 14 × H
∴ H = \(\frac{\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20}{22 \times 14}\)
H = 2.5 cm
Hence, Height of Platform H = 2.5 cm.

Question 4.
A well of diameter 3 m is dug 14 m deep. The earth taken out of ¡t has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution:
Depth of well (h) = 14 m
Radius of well (r) = \(\frac{3}{2}\) m

Embankment is in the shape of hollow cylinder whose inner radius is same as radius of well and width of embañkment 4 m
Timer radius of embankment = Radius of well(r) = \(\frac{3}{2}\) m
Outer radius of embankment (R) = (\(\frac{3}{2}\) + 4) m
R = \(\frac{11}{2}\) = 5.5 m
Volume of earth dug out = Volume of embankment (so formed)
πR²h = volume of outer cylinder – volume of inner cylinder
πr²h = πR²H – πr²H
= πH[R² – r²]

\(\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times 14\) = \(\frac{22}{7}\) × H[(5.5)² – (1.5)²]

H = \(\frac{\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times 14}{\frac{22}{7} \times(5.5-1.5)(5.5+1.5)}\)

= \(\frac{1.5 \times 1.5 \times 14}{4 \times 7}\) = 1.125 m.

Hence, Height of embankment H = 1.125 m.

Question 5.
A container shaped like a right circular cylinder having 4iameter 12 cm and height 15 cm Is full of ice-cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be tilled with ice-cream.

Dinieter of cvlrnder (D) = 12 cm
.. Radius of cylinder (R) = 6 cm
Height of cylinder (H) = 15 cm
Diameter of cone = 6 cm
Radius of cone (r) = 3 cm
Radius of hemisphere (r) = 3 cm
Height of cone (h) = 12 cm
Let us suppose number of cones used to fill the ice-cream = n
Volume of ice cream in container = n [Volume of ice cream in one cone]

n = 10
Hence, Number of cones formed = 10.

Question 6.
How many silver coins 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions
5.5 cm × 10 cm × 3.5 cm.
Solution:
Silver coin is in the form of cylinder
Diameter of silver coin = 1.75 cm
∴ Radius of silver coin (r) = \(\frac{1.75}{2}\) cm
Thickness of silver coin = Height of cylinder (H) = 2 mm

i.e., h = \(\frac{2}{10}\) cm.
Length of cuboid (L) = 5.5 cm
Width of cuboid (B) = 10 cm
Height of cuboid (H) = 3.5 cm
Let number of coins melted to form cuboid = n
∴ Volume of cuboid = n [volume of one silver coin]
= n[πr²h]
5.5 × 10 × 3.5 = n × \(\frac{22}{7} \times \frac{1.75}{2} \times \frac{1.75}{2} \times \frac{2}{10}\)

\(\frac{\frac{55}{10} \times 10 \times \frac{35}{10}}{\frac{22}{7} \times \frac{175}{200} \times \frac{175}{200} \times \frac{2}{10}}\) = n
n = 400
Hence, Number of corns so formed = 400.

Question 7.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:

Radius of cylindrical bucket (R) = 18 cm
Height of cylindrical bucket (H) = 32 cm
Height of conical heap (h) = 24 cm
Let ‘r’ cm and ‘l’ cm be the radius and slant height of cone
Volume of sand in bucket = Volume of sand in cone
πR²H = \(\frac{1}{3}\) πr²h

\(\frac{22}{7}\) × 18 × 18 × 32 = \(\frac{1}{3}\) × \(\frac{22}{7}\) × r² × 24
\(\frac{\frac{22}{7} \times 18 \times 18 \times 32}{\frac{1}{3} \times \frac{22}{7} \times 24}\) = r²
r² = \(\frac{18 \times 18 \times 32}{8}\)
r² = 1296
r = \(\sqrt{1296}\)
r = 36
∴ Radius of cone (r) = 36 cm
As we know,
(Slant height)² = (Radius)² + (Height)²
l = r² + h²
l = \(\sqrt{(36)^{2}+(24)^{2}}\)

= \(\sqrt{1296+576}\) = \(\sqrt{1872}\)

= \(\sqrt{12 \times 12 \times 13}\)

l = 12\(\sqrt{13}\) cm

∴ Slant height of cone (1) = 12\(\sqrt{13}\) cm.

Question 8.
Water in a canal 6m wide and 1.5m deep is flowing with a speed of 10 km/k How much area will it irrigate in 30 minutes, If 8 cm of standing water is needed?
Solution:
Width of canal = 6 m
Depth of water in canal = 1.5 m
Velocity at which water is flowing = 10km/hr
Volume of water discharge in one hour = (Area of cross section) velocity
= (6 × 1.5m2) × 10 km
= 6 × 1.5 × 10 × 6 × 1.5 × 10 × 1000 × 1000 m³
∴ Volume of water discharge in \(\frac{1}{2}\) hour = \(\frac{1}{2}\) × \(\frac{6 \times 15}{10}\) × 1000 = 45000 m³
Let us suppose area to be irrigate = (x) m²

According to question, 8 cm standing water is required in field
∴ Volume of water discharge by canal in \(\frac{1}{2}\) hours = Volume of water in field 45000 m³ = (Area of field) × Height of water
45000 m³ = x × (\(\frac{8}{100}\) m)
\(\frac{4500}{8}\) × 100 = x
x = 562500 m²
x = \(\frac{562500}{10000}\) hectares
[1 m² = \(\frac{1}{10000}\) hectares]
x = 56.25 hectares
Hence, Area of field = 56.25 hectares.

Question 9.
A farmer connects a pipe of Internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate 0f 3km/h, in how much time will the tank be filled?
Solution:
Velocity of water = 3 km/hr
Diameter of pipe = 20 cm
Radius of pipe (r) = 10 cm = \(\frac{10}{100}\) m = \(\frac{1}{10}\) m
Diameter of tank = 10 m
Radius of tank (R) = 5 m
Depth of tank (H) = 2 m
Let us suppose pipe filled a tank in n minutes
Volume of water tank = Volume of water through the pipe in n minutes
πR²H = n[Area of cross section × Velocity of water]

πR²H = n[(πr²) × 3 km/h]

\(\frac{22}{7}\) × (5)² × 2 = n[latex]\frac{22}{7} \times \frac{1}{10} \times \frac{1}{10} \times \frac{3 \times 1000}{60}[/latex]

25 × 2 × \(\frac{22}{7}\) = \(\frac{11}{7}\) n

n = \(\frac{25 \times 2 \times 22 \times 7}{11 \times 7}\)
n = 100 minutes
Hence, Time taken to fill the tank = 100 minutes.

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

PSEB Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2

In Question 1 to 3, choose the corred option and give justification.

Question 1.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Solution:
A circle with centre O from a point, Q the length of the tangent to a circle is 24 cm and distance of Q from the centre is 25 cm.

∴ ∠QPO = 90°
Now, in right angled ∠OPQ,
OQ² = PQ² + OP²
(25)² = (24)² + OP²
Or 625 = 576 + OP²
Or OP² = 625 – 576
Or OP² = 49 = (7)²
Or OP = 7 cm
∴ Option (A) is correct.

Question 2.
In Fig., if TP and TQ and tangents to a circle with centre O so ∠POQ = 110°, then ∠PTQ is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°

Solution:
In figure, OP is radius and PT is tangent to circle.
∠OPT = 90°
Similarly, ∠OQT = 90° and ∠POQ = 110° (Given)
Now, POQT is a Quadrilateral.
∴ ∠POQ + ∠OQT + ∠PTQ + ∠TPO = 360°
110° + 90° + ∠PTQ + 90° = 360°
Or ∠PTQ + 290° = 360°
Or ∠PTQ = 360° – 290°
Or ∠PTQ = 70°
∴ Option (B) is correct.

Question 3.
In tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 800, then LPOA is equal to
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Solution:
In given firgure, OA is radius and AP is a tangent to the circle.

∴ ∠OAP = 90°
Similarly, ∠OBP = 90°
Now, in right angled ∆PAO and ∆PBO
∠PAO = ∠PBO = 90°
OP = OP (Common side)
OA = OB (radii of same Circle)
∴ ∆PAO ≅ ∆PBO [RHS congruence]
∴ ∠AOP = ∠BOP [CPCT]
Or ∠AOP =∠BOP = \(\frac{1}{2}\) ∠AOB
Also, In Quad. OAPB,
∠OBP + ∠BPA + ∠PAO + ∠AOP = 360°
90° +80° +90° + ∠AOB = 360°
∠AOB = 360° – 260°
∠AOB = 100°
Form (1) and (2), we get
∠AOP = ∠BOP = \(\frac{1}{2}\) × 100° = 50°

∴ Option (A) is correct.

Question 4.
Prove that, the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
Given: A circle with center O and AB as its diameter l and m are tangents at points A and B.
To Prove: l || m
Proof: OA is the radius and l is the tangent to the circle.

∴ ∠1 = 90°
Similarly, ∠2 = 90°
Or ∠1 = ∠2 = 90°
But these are alternate angles between two lines, when one transversal cuts them.
∴ l || m
Hence, tangents drawn at the ends of a diameter of a circle are parallel.

Question 5.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution:
Given. A circle with centre O. AB its tangent meet circle at P.
i.e., P is the point of contact.

To Prove: Perpendicular at the point of contact to the tangent to a circle passes through the centre.
Construction: Join OP.
Proof: The perpendicular to a tangent line AB through the point of contact passes through the centre of the circle because only one perpendicular, OP can be drawn to the line AB through the point P.
Hence, perpendicular at the point of contact to the tangent to a circle passes through the centre.

Question 6.
The length of a tangent from a point A at a distance 5 cm. from the centre of the circle is 4 cm. Find the radius of the circle.
Solution:
A circle with centre ‘O’ A is any point outside the circle at a distance of 5 cm from the centre.

Length of tangent = PA = 4 cm
Since, OP is the radius and PA is the tangent to the circle.
∠OPA = 90°
Now, in right angled ∠OPA.
Using Pythagoras Theorem.
OA² = OP² + PA²
(5)² = OP² + (4)²
Or OP² = 25 – 16
Or OP² = 9 = (3)²
Or OP = 3 cm.
Hence, radius of the circle is 3 cm.

Question 7.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of Ihe chord of the larger circle which touches the smaller circle.
Solution:
Two concentric circles having same centre O, and radii 5 cm and 3 cm respectively. Let PQ be the chord of larger circle
but tangent to the smaller circle.

Since, OM be the radius of smaller circle and PMQ be the tangent.

∴ ∠OMP = ∠OMQ = 90°
Consider, right angled triangles OMP and OMQ,
∠OMP = ∠OMQ = 90°
OP = OQ [radii of same circle]
OM = OM [common side]
∴ ∆OMP ≅ ∆OMQ [RHS congurence]
∴ PM = MQ [CPCT]
Or PQ = 2PM = 2MQ
Now, in right angled ∆OMQ.
Using Pythagoras Theorem,
OQ² = OM² + MQ²
(5)² = (3)² + (MQ)²
Or MQ = 25 – 9
Or MQ² = 16 = (4)²
Or MQ = 4 cm
Length of chord PQ = 2 MQ = 2 (4) cm = 8 cm
Hence, length of required chord is 8 cm.

Question 8.
A quadrilateral ABCD is drawn to the circumscribe a circle (see figure). Prove that AB + CD = AD + BC.

Solution:
Given: A Quadrilateral ABCD is drawn to circumscribe a circle.
To Prove: AB + CD = AD + BC
Proof: Since the lengths of tangents drawn from an external point to a circle are equal. Now, B is any point outside the circle and BP : BQ are tangents to the circle.
∴ BP = BQ ……………(1)
Similarly, AP = AS ………….(2)
and CR = CQ …………..(3)
Also, DR = DS ………….(4)
Adding (1), (2), (3) and (4), we get
(BP + AP) + (CR + DR) = (BQ + CQ) + (AS + DS)
AB + CD = BC + AD
is the required result.

Question 9.
In figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C
intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

Solution:
Given: A circle with centre O having two parallel tangents XY and X’Y’ and= another tangent AB with point of contact C
intersecting XY at A and X’Y’ at B.
To Prove: ∠AOB = 90°
Contruction: Join OC, OA and OB.
Proof: Since the lengths of tangents drawn from an external point to a circle are equal.
Now, A is any point outside the circle from two tangents PA and AC are drawn to the circle.
∴ PA = AC
Also, in ∆ POA and ∆ AOC,
PA = AC (Proved)
OA = OA (common side)
OP = OC (radii of same circle)
∴ ∆POA ≅ ∆AOC [SSS congruence]
and ∠PAO = ∠CAO [CPCT]
Or ∠PAC = 2 ∠PAO = 2 ∠CAO ……………(1)
Similarly, ∠QBC = 2∠OBC = 2 ∠OBQ ………………(2)
Now, ∠PAC + ∠QBC = 180°
[Sum of the interior angles on the same side of transversal is 180°]
Or 2∠CAO + 2∠OBC = 180° [Using (1) & (2)]
Or ∠CAO + ∠OBC = 180 = 90° …(3)
Now, in ∆OAB,
∠CAO + ∠OBC + ∠AOB = 180°
90°+ ∠AOB = 180° [Using (3)]
Or ∠AOB = 180° – 90° = 90°
Hence, ∠AOB = 90°.

Question 10.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre. [Pb. 20191
Solution:
Given. A circle with centre O. P is any point outside the circle PQ and PR are the tangents to the given circle from point P.

To Prove. ∠ROQ + ∠QPR = 180°
Proof. OQ is the radius and PQ is tangent from point P to the given circle.
∴ ∠OQP = 90° ………..(1)
[∵ The tangent at any point of a circle is perpendicular to the radius through the point of contact]
Similarly, ∠ORP = 90°
Now, in quadrilateral ROQP,
∠ROQ + ∠PRO + ∠OQP + ∠QPR = 360°
Or ∠ROQ + 90° + 90° + ∠QPR = 360° [Using (1) & (2)]
Or ∠ROQ + ∠QPR + 180 = 360°
Or ∠ROQ + ∠QPR = 360° – 180°
Or ∠ROQ + ∠QPR = 180°
Hence, the angle between the two tangents drawn from and external point to a circle is supplementary to angle subtended by the line segment joining the pnts of contact at the centre.

Question 11.
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:
Given: A parallelogram ABCD circumscribed a circle with centre O.
To Prove: ABCD is a rhombus.
Proof: Since the lengths of tangents drawn from an external point to a circle are equal. Now, B is any point outside thé circle and BE; BF are tangents to the circle.

To Prove: ABCD is a rhombus.
Proof: Since the lengths of tangents drawn from an external point to a circle are equal. Now, B is any point outside thè circle and BE; BF are tangents to the circle.
∴ BE = BF
Similarly AE = AH ……………(2)
and CG = CF
Also DG = DH
Adding (1), (2), (3) and (4), we get
(BE + AE) + (CG + DG) = (BF + CF) – (AH + DH)
Or AB + CD = BC +AD ……….(5)
Now, ABCD is a parallelogram, (Given)
∴ AB = CD and BC = AD …………(6)
From (5) and (6), we get
AB + AB = BC + BC
Or 2AB = 2BC or AB = BC
Or AB = BC = CD = AD
∴ ABCD is a rhombus.
Hence, parallelogram circumscribing a circle is a rhombus.

Question 12.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig). Find the sides AB and AC.

Solution:
A triangle ABC is drawn to circumscribe a circle of radius 4 cm and the sides BC, CA, AB of ∆ABC touch the circle at
D, E, F respectively. Since the lengths of tangents drawn from an external point to a circle are equal.
∴ AE = AF = x cm(say)
CE = CD = 6 cm (Given)
and BF = BD = 5 cm
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ OD ⊥ BC; OE ⊥ AC and OF ⊥ AB.
Also, OE = OD = OF = 4 cm.
Consider, ∆ABC
a = AC = (x + 6) cm ;
b = CB = (6 + 8) cm = 14 cm
c = BA = (8 + x) cm
S = \(\frac{a+b+c}{2}\)
∴ S = \(\frac{x+6+14+8+x}{2}\) = \(\frac{2 x+28}{2}\) = (x + 14)

area (∆ABC)= \(\sqrt{\mathrm{S}(\mathrm{S}-a)(\mathrm{S}-b)(\mathrm{S}-c)}\)

= \(\sqrt{\begin{array}{r}
(x+14)(x+14-\overline{x+6}) \\
(x+14-14)(x+14-\overline{8+x})
\end{array}}\)

= \(\sqrt{(x+14)(8)(x)(6)}\)

= \(\sqrt{48 x^{2}+672 x}\) cm² ………………(1)

area (∆OBC) = \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × 14 × 4 = 28cm² …………….(2)

area(∆BOA)= \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × (8 + x) × 4 = 28cm²…………….(3)

area (∆AOC) = \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × (6 + x) × 4 = 28cm² …………….(4)

From the figure, by addition of areas, we have
Or (∆ABC) = ar (∆OBC) + ar (∆BOA) + ar (∆AOC)
\(\sqrt{48 x^{2}+672 x}\) = 28 + 16 + 2x + 12 + 2x
Or \(\sqrt{48 x^{2}+672 x}\) = 4x + 56
Or 48x² + 672x =4[x+ 14]
Squaring both sides, we get
Or 48x² + 672x = 16 (x + 14)²
Or 48x (x + 14) = 16(x + 14)²
Or 3x = x + 14
Or 2x = 14
Or x = \(\frac{14}{2}\) = 7
∴ AC = (x + 6) cm = (7 + 6) cm= 13 cm
and AB = (x + 8) cm = (7 + 8) cm = 15 cm
Hence, AB = 15 cm and AC = 13 cm.

Question 13.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
Given:
A quadrilateral PQRS circumscribing a circle having centre O. Sides PQ, QR, RS and SP touches the circles at L, M, N, T
respectively.

To Prove:
∠POQ + ∠SOR = 180°
and ∠SOP + ∠ROQ = 180°
Construction:
Join OP, OL, OQ, 0M. OR, ON, OS, OT
Proof: Since the two tangents drawn from an external point subtend equal angles at the centre.
∴ ∠2 = ∠3; ∠4 = ∠5 ; ∠6 = ∠7; ∠8 = ∠1
But, sum of all angles around a point is 360°
∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
Or ∠1 + ∠2 + ∠2 + ∠5 +∠5 + ∠6 + ∠6 + ∠1 = 360°
Or 2(∠1 + ∠2 + ∠5 + ∠6) = 360°
Or (∠1 + ∠2) + (∠5 + ∠6) = \(\frac{360^{\circ}}{2}\) = 180°
Or ∠POQ + ∠SOR = 180°
Similarly, ∠SOP + ∠ROQ = 180°
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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

PSEB Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1

Question 1.
How many tangents can a circle have?
Solution:
Since at any point on a circle, there can be one and only one tangeni. But circle is a collection of infinite points, so we can draw infinite number of tangents to a circle.

Question 2.
Fill in the blanks:
(i) A tangent to a circle intersects it in ………………. point(s).
Solution:
one

(ii) A line intersecting a circle in two points is called a ………………..
Solution:
secant.

(iii) A circle can have ……………. parallel tangents at the most.
Solution:
two

(iv) The common point of a lingent h, circle and the circit is called ………………
Solution:
point of contact.

Question 3.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is:
(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D) \(\sqrt{119}\) cm.
Solution:
According to given information we draw the figure such that,

OP = 5 cm and OQ = 12 cm
∵ PQ is a tangent and OP is the radius
∵ ∠OPQ = 90°
Now, In right angled ∆OPQ.
By Pythagoras Theorem,
OQ² = OP² + QP²
Or (12)² = (5)² + QP²
Or QP² = (12)² – (5)²
Or QP² = 144 – 25 = 119
Or QP = \(\sqrt{119}\) cm.
Hence, option (D) is correct.

Question 4.
Draw a circle and two lines parallel to given line such that one is a tangent and other a secant to the circle.
Solution:
According to thc given information we draw a circle having O as centre and l is the given line.

Now, m and n be two lines parallel to a given line l such that m is tangent as well as parallel to l and n is secant to the circle as well as parallel to l.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 9 Some Applications of Trigonometry Ex 9.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1

Question 1.
A circus artist is climbing 220 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, ¡f the angle made by the rope with the ground level is 30° (see fig.).

Solution:
Let AB be the heignt of pole;
AC = 20 m be the length of rope.
The angle of elevation in this situation is 30°.
Various arrangements are as shown in figure.

In right angled ∆ABC,
\(\frac{\mathrm{AB}}{\mathrm{AC}}\) = sin 30°

or \(\frac{\mathrm{AB}}{20}=\frac{1}{2}\)

or AB = \(\frac{1}{2}\) × 20 = 10
Hence, height of pole is 10 m.

Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 300 with it. The distance between the foot of the tree to the point where the top touches the ground is 8 rn Find the height of the tree.
Solution:
Let BD be length of tree before storm.
After storm AD = AC = length of broken part of tree.
The angle of elevation in this situation is 30°.
Various arrangements are as shown in the figure.

In right angled ∆ABC,

\(\frac{\mathrm{AB}}{\mathrm{AC}}\) = tan 30°

or \(\frac{h_{1}}{8}=\frac{1}{\sqrt{3}}\)
or h₁ = \(\frac{8}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{8}{3} \sqrt{3}\) m ……….(1)

\(\frac{\mathrm{BC}}{\mathrm{AC}}\) = cos 30°

or \(\frac{8}{h_{2}}=\frac{\sqrt{3}}{2}\)

or \(h_{2}=\frac{8 \times 2}{\sqrt{3}}=\frac{16}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

h₂ = \(\frac{16}{3}\) √3 …………..(2)

Total height of the tree = h₁ + h₂
= \(\frac{8}{3}\) √3 + \(\frac{16}{3}\) √3 [Using (1) & (2)]

= \(\left(\frac{8+16}{3}\right) \sqrt{3}=\frac{24}{3} \sqrt{3}\) = 8√3 m.
Hence, height of the tree is 8√3 m.

Question 3.
A contractor plants to install two slides for the children to play in a park. For the children below the age of 5 years, she
prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Solution:
Case I:
For children below 5 years.
Let AC = l₁ m denote the length of slide and BC = 1.5 m be the height of slide. The angle of elevation is 30°.
Various arrangements are shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{BC}}{\mathrm{AC}}\) = sin 30°

or \(\frac{1 \cdot 5}{l_{1}}=\frac{1}{2}\)

or l₁ = 1.5 × 2 = 3 m.

Case II:
For Elder children
Let AC = 12 m represent the length of slide and BC = 3 m be the height of slide. The angle of elevation is 60°. Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{BC}}{\mathrm{AC}}\) = sin 60°

or \(\frac{3}{l_{2}}=\frac{\sqrt{3}}{2}\)

or l₂ = \(\frac{3 \times 2}{\sqrt{3}}=\frac{6}{\sqrt{3}}\)

= \(\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{6 \sqrt{3}}{3}\)

= 2√3 m.

Hence, length of slides for children below 5 years and elder children are 3 m and 2 m.

Question 4.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Solution:
Let BC = h m be the height of tower and AB = 30 m be the distance at ground level. Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{BC}}{\mathrm{AB}}\) = tan 30°

or \(\frac{h}{30}=\frac{1}{\sqrt{3}}\)

or h = \(\frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{30 \sqrt{3}}{3}\)

= 10√3 = 10 × 1.732
h = 17.32 (approx).
Hence, height of tower is 17.32 m.

Question 5.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Solution:
Let us suppose position of the kite is at point CAC = l m be length of string with which kite is attached. The angle of elevation for this situation be 60°. Various arrangements are as shown in the figure.

In right angled ∆ABC,

\(\frac{\mathrm{CB}}{\mathrm{AB}}\) = sin 60°

or \(\frac{60}{l}=\frac{\sqrt{3}}{2}\)

or l = \(\frac{60 \times 2}{\sqrt{3}}=\frac{120}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

= \(\frac{120 \sqrt{3}}{3}\) = 40√3 m.
Hence, length of the string be 40√3 m.

Question 6.
A 15 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution. Let ED = 30 m be the height of building and EC = l5 m be the height of boy.
The angle of elevation at different situation are 30° and 60° respectively.
Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{DC}}{\mathrm{AC}}\) = tan 30°

or \(\frac{28 \cdot 5}{x+y}=\frac{1}{\sqrt{3}}\)

or x + y = 28.5 × √3 m ………………(1)

Now, in right angled ∆BCD,

\(\frac{\mathrm{DC}}{\mathrm{BC}}\) = tan 60°

or \(\frac{28 \cdot 5}{y}=\sqrt{3}\)

or y = \(\frac{28 \cdot 5}{\sqrt{3}}\)

or y = \(\frac{28 \cdot 5}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{28 \cdot 5 \times \sqrt{3}}{3}\) ……….(2)

Distance covered towards building = x = (x + y) – y
= (28.5 × √3) – (\(\frac{28.5}{3}\) × √3) m [sing (1) and (2)]

= 28.5 (1 – \(\frac{1}{3}\)) √3 m

= 28.5 (\(\frac{3-1}{4}\)) √3 m

= [28.5 × \(\frac{2}{3}\)]√3 m = 19√3 m.

Hence, distance covered by boy towards the building is 19√3 m.

Question 7.
From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Solution:
Let BC = 20 m be the height of building and DC = h m be the height of transmission tower. The angle of elevation of
the bottom and top of a transmission tower are 45° and 60° respectively.
Various arrangements are as shown in the figure.

In right angled ∆ABC,
\(\frac{A B}{B C}\) = cot 45°

or \(\frac{\mathrm{AB}}{20}\) = 1
or AB = 20 m ………………..(1)
Also, in right angled ∆ABD,
\(\frac{A B}{B C}\) = cot 60°

or \(\frac{\mathrm{AB}}{20+h}=\frac{1}{\sqrt{3}}\)

AB = \(\frac{(20+h)}{\sqrt{3}}\) ………….(2)

From (1) and (2), we get

20 = \(\frac{(20+h)}{\sqrt{3}}\)
or 20√3 = 20 + h
or h = 20√3 – 20
or h = 20 (√3 – 1) m
= 20 (1.732 – 1) m
= 20 × 0.732 = 14.64 m.

Hence, height of the tower is 14.64 m.

Question 8.
A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Solution. Let BC = h m be the height of Pedestal and CD = 1.6 m be the height of statue.
The angle of elevation of top of statue and top of pedestal are 60° and 45° respectively. Various arrangements are as shown in the figure.

In right angled ∆ABC,
\(\frac{A B}{B C}\) = cot 45°

or \(\frac{A B}{h}\) = 1

or AB = h m ………….(1)

In right angled ∆ABC,
\(\frac{\mathrm{AB}}{\mathrm{BD}}\) = cot 60°

or \(\frac{\mathrm{AB}}{h+1.6}=\frac{1}{\sqrt{3}}\)

or AB = \(\frac{h+1.6}{\sqrt{3}}\) ……….(2)

From (1) and (2), we get
h = \(\frac{h+1.6}{\sqrt{3}}\)
or √3h = h + 1.6
or (√3 – 1) h = 1.6
or (1.732 – 1) h = 16
or (0.732) h = 1.6
or h = \(\frac{1.6}{0.732}\) = 2.1857923
= 2.20 m (approx.)
Hence, height of pedestal is 2.20 m.

Question 9.
The angle of elevation of the top of a building from the foot of the tower is 300 and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the
building.
Solution:
Let BC = 50 m be height of tower and AD = h m be height of building. The angle of elevation of the top of a building from the foot of tower and top of tower from foot of the building are 30° and 60° respectively. Various arrangement are as shown in figure.

In right angled ∆ABC,
\(\frac{A B}{B C}\) = cot 60°

or \(\frac{\mathrm{AB}}{50}=\frac{1}{\sqrt{3}}\)

or AB = \(\frac{50}{\sqrt{3}}\) …………(1)

Also, in right angled ∆DAB,
\(\frac{\mathrm{AB}}{\mathrm{DA}}\) = cot 30°

or \(\frac{A B}{h}\) = √3
or AB = h√3 ……………(2)

From (1) and (2), we get
\(\frac{50}{\sqrt{3}}\) = h√3

or \(\frac{50}{\sqrt{3}} \times \frac{1}{\sqrt{3}}\) = h

or h = \(\frac{50}{3}\) = 16.6666

or h = 16.70 m (approx).
Hence, height of building is 16.70 m.

Question 10.
Two poles of equal heights are tanding opposite each other on either side of he road, which is 80 m wide. From a point
between them on the road the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
Solution:
Let BC = DE = h m he height of two equal poles and point A be the required position where the angle of elevations of top of two poles are 30° and 60° respectively. Various arrangement are as shown in the figure.

In right angled ∆ADE,

\(\frac{E D}{D A}\) = tan 30°

or \(\frac{h}{x}=\frac{1}{\sqrt{3}}\)

or h = \(\frac{x}{\sqrt{3}}\) ……………(1)

In right angled ∆ABC,

\(\frac{\mathrm{BC}}{\mathrm{AB}}\) = tan 60°

or \(\frac{h}{80-x}\) = √3

or h = (80 – x) √3 …………(2)

From (1) and (2), we get
\(\frac{x}{\sqrt{3}}\) = (80 – x)
or x = (80 – x) √3 × √3
or x = (80 – x) 3
or x = 240 – 3x
or 4x = 240
or x = \(\frac{240}{4}\) = 60
Substitute this value of x in (I), we get
h = \(\frac{60}{\sqrt{3}}=\frac{60^{\circ}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

= \(\frac{60 \sqrt{3}}{3}=20 \sqrt{3}\)

= (20 × 1.732) m = 34.64 m
DA = x = 60 m
and AB = 80 – x = (80 – 60) m = 20 m.
Hence, heigth of the poles are 3464 m and the distances of the point from the poles are 20 m and 60 m respectively.

Question 11.
A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From a point 20 m away from this point on the same bank, the angle of elevation of the top of the tower is 30° (see fig.). Find the height of the tower and the width of the canal.

Solution:
Let BC = x m be the width of canal and CD = h m be height of TV tower. The angles of elevation of top of tower at different position are 30° and 60° respectively. Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{AB}}{\mathrm{BC}}\) = tan 60°

or \(\frac{h}{x}\) = √3
or h = √3x …………..(1)

Also, in right angled ∆ABD,
\(\frac{\mathrm{AB}}{\mathrm{BD}}\) = tan 30°

or \(\frac{h}{20+x}=\frac{1}{\sqrt{3}}\)

or h = \(\frac{20+x}{\sqrt{3}}\) ……………….(2)

From (1) and (2), we get

√3x = \(\frac{20+x}{\sqrt{3}}\)
or √3(√3x) = 20 + x
or 3x = 20 + x
or 2x = 20
or x = \(\frac{20}{2}\) = 10

Substitute this value of x in (1), we get
h = 10(√3)
= 10 × 1.732
h = 17.32 m
Hence, height of TV tower is 17.32 m and. width of the canal is 10 m.

Question 12.
From the top of a 7m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Solution:
Let BD = hm be the height of cable tower and AE = 7 m be the height of building. The angle of elevation of the top of a cable tower and angle of depression of its foot from top of a building are 60° and 45° respectively.
Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{AB}}{\mathrm{AE}}\) = cot 45°

or \(\frac{\mathrm{AB}}{7}\) = 1

or AB = 7 m. ……………..(1)

Also, in right angled ∆DCE,

\(\) = cot 60°
or \(\frac{\mathrm{EC}}{h-7}=\frac{1}{\sqrt{3}}\)

or EC = \(\frac{h-7}{\sqrt{3}}\) ……………..(2)

But AB = EC ………….(Given)
7 = \(\frac{h-7}{\sqrt{3}}\) [Using (1) and (2)]
or 7√3 = h – 7
h = 7√3 + 7 = 7 (√3 + 1)
or h = 7 (1.732 + 1) = 7(2.732)
or h = 19.124
or h = 19.20 m (approx.)
Hence, height of the tower is 19.20 m.

Question 13.
As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Solution:

Let CD = 75 m be the height of light house and point D be top of light house from w’here angles of depression of two ships are 30° and 45° respectively. Various arrangements are as shown in the figure.

In right angled ∆BCD,
\(\frac{\mathrm{BC}}{\mathrm{CD}}\) = cot 45°

or \(\frac{y}{75}\) = 1
or y = 75 m ……………(1)

Also, in right angled ∆ACD
\(\frac{\mathrm{AC}}{\mathrm{CD}}\) = cot 30°

or \(\frac{x+y}{75}\) = √3
or x + y = 75√3
or x + 75 = 75√3 [using (1)]
or x = 75√3 – 75
= 75 (√3 – 1)
= 75( 1.732 – 1)
= 75 (0.732)
or x = 54.90
Hence, distance between the two ships is 54.90 m.

Question 14.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant ¡s 60°. After some time, the angle of elevation reduces to 30° (see fig.). Find the distance travelled by the balloon during the interval.

Solution:
Let ‘AB’ be the position of 1.2 m tall girl, at the point of the angles of elevation of balloon at
different distances are 30° and 60° respectively. Various arrangements are as shwon in th figure.
According to question,
FG = ED = CE – CD
= 88.2 m – 1.2 m
= 87 m

In right angled ∆AGF,
\(\frac{A G}{G F}\) = cot 60°

or \(\frac{x}{87}=\frac{1}{\sqrt{3}}\)

or x = \(\frac{87}{\sqrt{3}}\) m.

Also, in right angled ∆ADE,
\(\frac{A D}{E D}\) = cot 30°

or \(\frac{x+y}{87}\) = √3

or x + y = 87√3
or \(\frac{87}{\sqrt{3}}\) + y = 87√3
or y = 87√3 – \(\frac{87}{\sqrt{3}}\)

or y = 87√3 – \(\frac{1}{\sqrt{3}}\)

or y = 87 \(\frac{3-1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

or y = \(\frac{87 \times 2 \times \sqrt{3}}{3}\)

or y = 58√3
or y = 58(1.732) = 100.456
or y = 100.456 m.
Hence, distance travelled by the balloon during the interval is 100.46 m.

Question 15.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the further t(me taken by the car to reach the foot of the tower.
Solution:
Let CD = h m. be the tower of height.
Let A be initial position of the car and after six seconds the car be at 13. The angles of depression at A and B are 30° and 60° respectively. Various arrangements are as shown in figure.

Let speed of the car be υ metre per second using formula, Distance = Speed x Time
AB = Distance covered by car in 6 seconds
AB = 6υ metre
Also, time taken by car to reach the tower be ‘n’ seconds.
∴ BC = nυ metre
In right angled ∆ACD.
\(\frac{\mathrm{CD}}{\mathrm{AC}}\) = tan 30°

or \(\frac{h}{6 v+n v}=\frac{1}{\sqrt{3}}\)

or h = \(\frac{6 v+n v}{\sqrt{3}}\) ……………….(1)

Also, in right angled ∆BCD,
\(\frac{C D}{B C}\) = tan 60°

or \(\frac{h}{n v}\) = √3
h = nv (√3) ……….(2)

From (1) and (2), we get
\(\frac{6 v+n v}{\sqrt{3}}\) = nυ(√3)
or 6υ + nυ = nυ(√3)
or 6υ + nυ = 3nυ
or 6υ = 2nυ
or n = \(\frac{6 v}{2 v}\) = 3
Hence, time taken by car to reach the foot of tower is 3 seconds.

Question 16.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Solution:
Let CD = h m be the height of tower and B ; A be the required points which are at a distance of 4 m and 9 m from the tower respectively. Various arrangements are as shown in the figure.

In right angled ∆BCD
\(\frac{\mathrm{CD}}{\mathrm{BC}}\) = tan θ

or \(\frac{h}{4}\) = tan θ ………….(1)

Also, in right angled ∆ACD,
\(\frac{C D}{A C}\) = tan (90 – θ)

or \(\frac{h}{9}\) = cot θ

Multiplying (1) and (2), we get
\(\frac{h}{4} \times \frac{h}{9}\) = tan θ cot θ

or \(\frac{h^{2}}{36}=\tan \theta \times \frac{1}{\tan \theta}\)

or h² = 36 = (6)²
or h = 6
Hence, height of the tower is 6 m.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4

Question 1.
Express the trigonometric ratios of sin A, sec A and tan A in terms of cot A.
Solution:
By using Identity,
cosec² A – cot² A = 1
⇒ cosec² A = 1 + cot² A
⇒ (cosec A)² = cot² A + 1
⇒ \(\left(\frac{1}{\sin A}\right)^{2}\) = cot² A + 1
⇒ (sin A)² = \(\frac{1}{\cot ^{2} \mathrm{~A}+1}\)
⇒ sin A = ± \(\frac{1}{\sqrt{\cot ^{2} \mathrm{~A}+1}}\)
We reject negative values of sin A for acute angle A.
Therefore, sin A = \(\frac{1}{\sqrt{\cot ^{2} A+1}}\)
By using identity,
sec² A – tan² A = 1
⇒ sec² A = 1 + tan² A
= 1 + \(\frac{1}{\cot ^{2} A}\)
= \(\frac{\cot ^{2} A+1}{\cot ^{2} A}\)

⇒ sec A = \(\sqrt{\frac{\cot ^{2} A+1}{\cot ^{2} A}}\)

tan A = \(\frac{1}{\cot A}\).

Question 2.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution:
By using Identity
sin² A + cos² A = 1
⇒ sin² A = 1 – cos² A
= 1 – \(\frac{1}{\sec ^{2} \cdot A}\) = \(\frac{\sec ^{2} A-1}{\sec ^{2} A}\)

⇒ (sin A)² = \(\frac{\sec ^{2} A-1}{\sec ^{2} A}\)

⇒ sin A = ± \(\sqrt{\frac{\sec ^{2} A-1}{\sec ^{2} A}}\)

[Reject – ve sign for acute angle A]
⇒ sin A = ± \(\sqrt{\frac{\sec ^{2} A-1}{\sec ^{2} A}}\)
cos A = \(\frac{1}{\sec A}\)
1 + tan² A = sec² A
tan² A = sec² A – 1
(tan A)² = sec² A – 1
⇒ tan A = ± \(\sqrt{\sec ^{2} A-1}\)
[Reject – ve sign for acute angle A]
i.e., tan A = \(\sqrt{\sec ^{2} A-1}\)
cosec A = \(\frac{1}{\sin A}=\frac{1}{\sqrt{\sec ^{2} A-1}}\)

= \(\frac{\sec A}{\sqrt{\sec ^{2} A-1}}\)

cot A = \(\frac{1}{\tan A}=\frac{1}{\sqrt{\sec ^{2} A-1}}\).

Question 3.
Evaluate:
(i) \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}\)

(ii) sin 25° cos 65° + cos 25° sin 65°.
Solution:
(i) \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}\)

= \(\frac{\left\{\sin \left(90^{\circ}-27^{\circ}\right)+\sin ^{2} 27^{\circ}\right\}}{\cos ^{2} 17^{\circ}+\left\{\cos \left(90^{\circ}-17^{\circ}\right)\right\}^{2}}\)
[∵ sin(90 – θ) = cos θ and cos (90 – θ) = sin θ]

= \(\frac{\left\{\cos 27^{\circ}\right\}^{2}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\left\{\sin 17^{\circ}\right\}^{2}}\)

= \(\frac{\cos ^{2} 27^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\sin ^{2} 17^{\circ}}\)
= \(\frac{1}{1}\) = 1.

(ii) sin 25° cos 65° + cos 25° sin 65°
= sin 25° × cos (90° – 25°)
+ cos 25° × sin (90° – 25°)
[∵ cos (90° – θ) = sin θ
sin(90° – θ) = cos θ].
= sin 25° × sin 25° + cos 25° × cos 25°
= sin² 25° + cos² 25° = 1.

Question 4.
Choose the correct option. Justify your choice:
(i) 9 sec² A – 9 tan² A =
(A) 1
(B) 9
(C) 8
(D) 0.

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =
(A) θ
(B) 1
(C) 2
(D) – 1.

(iii) (sec A + tan A) (1 – sin A) =
(A) sec A
(B) sin A
(C) cosec A
(D) cos A.

(iv) \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\) =
(A) sec²A
(B) – 1
(C) cot² A
(D) tan² A.

Solution:
(i) Consider, 9 sec² A – 9 tan² A
= 9 (sec² A – tan² A)
= 9 × 1 = 9.
Option (B) is correct.

(ii) Consider, (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
= \(\left\{1+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right\} \times\left\{1+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}\right\}\)

= \(\left\{\frac{\cos \theta+\sin \theta+1}{\cos \theta}\right\} \times\left\{\frac{\sin \theta+\cos \theta+1}{\sin \theta}\right\}\)

= \(\begin{array}{r}
\{(\cos \theta+\sin \theta)+1\} \\
\times\{(\cos \theta+\sin \theta)-1\} \\
\hline \cos \theta \times \sin \theta
\end{array}\)

= \(\frac{(\cos \theta+\sin \theta)^{2}-(1)^{2}}{\cos \theta \times \sin \theta}\)

[∵ (a + b) (a – b) = a² – b²]

= \(\frac{\cos ^{2} \theta+\sin ^{2} \theta+2 \cos \theta \sin \theta-1}{\cos \theta \times \sin \theta}\)

= \(\frac{1+2 \cos \theta \sin \theta-1}{\cos \theta \sin \theta}\) = 2.

Option (C) is correct.

(iii) Consider, (sec A + tan A) (1 – sin A)
= \(\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)\) × (1 – sin A)

= \(\frac{(1+\sin A)}{\cos A}\) × (1 – sin A)

= \(\frac{(1+\sin A)(1-\sin A)}{\cos A}\)

= \(\frac{(1)^{2}-(\sin A)^{2}}{\cos A}=\frac{1-\sin ^{2} A}{\cos A}=\frac{\cos ^{2} A}{\cos A}\)
[∵ cos² A = 1 – sin² A]
= cos A.
Option (D) is correct.

(iv) Consider, \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\)

= tan² A.
Option (D) is correct.

Question 5.
Prove the following Identities, where the angles involved are acute angles for which the expressions are defined.
(i) (cosec θ – cot θ) = \(\frac{1-\cos \theta}{1+\cos \theta}\)

(ii) \(\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}\) = 2 sec A.

(iii) \(\frac{\tan \theta}{1+\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ cosec θ
[Hint : Write the expression in terms of sin θ and cos θ]

(iv) \(\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}\)
[Hint: Simplify L.H.S. and R.H.S. separately]

(v) \(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}\) using the identity cosec² A = 1 + cot² A.

(vi) \(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = sec A + tan A

(vii) \(\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}\) = tan θ

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A.

(ix) (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\)
[Hint : Simplify L.H.S. and R.H.S. separately]

(x) \(\left(\frac{1+\tan ^{2} A}{1+\cot A^{2}}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}\) = tan² A.

Solution:
(i) L.H.S. = (cosec θ – cot θ)²
= \(\left\{\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right\}^{2}\)

= \(\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2}=\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta}\)
Using identity, sin² θ + cos² θ = 1
⇒ sin² θ = 1 – cos² θ
= \(\frac{(1-\cos \theta)^{2}}{1-\cos ^{2} \theta}\)
= \(\)
[∵ a² – b² = (a + b) (a – b)]

= \(\)

∴ L.H.S. = R.H.S.
Hence, (cosec θ – cot θ)² = \(\frac{1-\cos \theta}{1+\cos \theta}\)

(ii) L.H.S. = \(\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}\)

= \(\frac{2}{\cos A}\) = cos A
∴L.H.S. = R.H.S.
Hence, \(\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}\) = 2 sec A.

(iii) L.H.S. = \(\frac{\tan \theta}{1+\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ cosec θ

= \(\frac{\left(\frac{\sin \theta}{\cos \theta}\right)}{\left(1-\frac{\cos \theta}{\sin \theta}\right)}+\frac{\left(\frac{\cos \theta}{\sin \theta}\right)}{\left(1-\frac{\sin \theta}{\cos \theta}\right)}\)

= \(\frac{1+\sin \theta \cos \theta}{\cos \theta \sin \theta}=\frac{1}{\cos \theta \sin \theta}+1\)

= 1 + \(\left(\frac{1}{\cos \theta}\right)\left(\frac{1}{\sin \theta}\right)\) = 1 + sec θ cosec θ
∴L.H.S. = R.H.S.
Hence, \(\frac{\tan \theta}{1+\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ cosec θ

(iv) L.H.S. = \(\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}\)
= \(\frac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}\)
= 1 + cos A …………….(1)
R.H.S = \(\frac{\sin ^{2} A}{1-\cos A}\)
(∵ 1 – cos² A = sin² A.)
= \(\frac{1-\cos ^{2} A}{1-\cos A}\)

= \(\frac{(1+\cos A)(1-\cos A)}{(1-\cos A)}\)

= 1 + cos A. …………….(2)
From (1) and (2) it is clear that
∴ L.H.S. = R.H.S.
Hence, \(\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}\).

(v) L.H.S. = \(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}\) using the identity cosec² A

= cosec A + cot A
= R.H.S
∴ L.H.S. = R.H.S.
Hence, \(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}\) using the identity cosec² A = 1 + cot² A.

(vi) L.H.S. = \(\sqrt{\frac{1+\sin A}{1-\sin A}}\)

= \(\sqrt{\frac{(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)}}\)

= \(\sqrt{\frac{(1+\sin A)^{2}}{(1)^{2}-(\sin A)^{2}}}\)

= \(\sqrt{\frac{(1+\sin A)^{2}}{1-\sin ^{2} A}}=\sqrt{\frac{(1+\sin A)^{2}}{\cos ^{2} A}}\)

= \(\frac{1+\sin A}{\cos A}=\frac{1}{\cos A}+\frac{\sin A}{\cos A}\)
= sec A + tan A
∴ L.H.S. = R.H.S.
Hence, \(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = sec A + tan A.

(vii) L.H.S. = \(\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}\)

∴ L.H.S. = R.H.S.
Hence, \(\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}\) = tan θ

(viii) L.H.S. = (sin A + cosec A)² + (cos A + sec A)²
= (sin² A + cosec² A + 2 sin A × cosec A) + {cos² A + sec² A
+ 2 cos A × sec A)
= [sin² A + co² A + 2sin A × \(\frac{1}{\sin A}\)] + [cos² A + sec² A + 2 cosA × \(\frac{1}{\cos A}\)]
= (sin² A + cosec² A + 2) + (cos² A + sec² A + 2)
= 2 + 2 + (sin² A + cos² A) + sec² A + cosec² A
= 2 + 2 + 1 + 1 + tan² A + 1 + cot² A
= 7 tan² A + cot² A
∴ L.H.S. = R.H.S.
Hence, (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A.

(ix) L.H.S. = (cosec A – sin A) (sec A – cos A)

From (1) and (2), it is clear that
L.H.S. = R.H.S.
Hence, (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\)

(x) \(\left(\frac{1+\tan ^{2} A}{1+\cot A^{2}}\right)\)
(∵ 1 + tan² A = sec² A
and 1 + cot² A = cosec² A)

From (1) and (2), it is clear that
LH.S. = R.H.S.
Hence, \(\left(\frac{1+\tan ^{2} A}{1+\cot A^{2}}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}\) = tan² A.