Class 10

RBSE Class 10 Science Notes Chapter 11 Work, Energy and Power are part of RBSE Solutions for Class 10 Science Notes. Here we have given Rajasthan Board RBSE Class 10 Science Notes Chapter 11 Work, Energy and Power.

Rajasthan Board RBSE Class 10 Science Notes Chapter 11 Work, Energy and Power

Displacement Calculator is really a nice calculator that will help you solve a lot of your problems related to displacement.

Work:
The product of force and displacement (in direction of force) is called work.
Work = Force x Displacement in direction of force
W = F x s

If direction of force is different from direction of displacement then we can find the work done by component of force in direction of displacement.

Let us assume that direction of force makes an angle θ with direction of displacement. Then, component of force in direction of displacement is;
F.cosθ

Work done
W = component of force in direction of displacement x displacement
Or   W = F.cosθ x s = F s cosθ = F.S
Work is a scalar quantity, i.e. its value can be positive or negative. If force or component of force is in the direction of displacement then work done is positive. But if force or component of force is in opposite direction of displacement then work done is negative.

Unit of Work:
If force is expressed in Newton and displacement is expressed in meter, then unit of work is Joule.
Work = Newton × meter = Joule
If force is expressed in Dyne and displacement is expressed in centimeter then unit of force is erg.
Work = Dyne×cm = erg 1 Joule = 1 N x 1 m = 10^s Dyne x 10²cm = 10⁷erg

Energy:
The capacity to do work is called energy. Thus, work is the measure of energy and hence unit of energy is same as that of work. Energy too is a scalar quantity. If 1 Joule of work is to be done then required energy for this would also be 1 Joule.

Types of Energy:
Some major forms of energy are as follows:

• Mechanical Energy: The energy in an object because of movement or position, or both is called mechanical energy.
• Heat Energy: The mobile energy in micro particles due to heat is called heat energy. These micro particles transfer energy from high temperature to low temperature,
• Chemical Energy: Energy obtained from chemical reactions is called chemical energy. For example, a battery provides energy because of chemical reaction. Similarly, energy from LPG or coal is obtained because of chemical reaction.
• Electrical Energy: Energy produced by electric charge is called electrical energy. Most of the household appliances work because of electrical energy. Gravitational Energy: Energy in an object due to gravitational force is called gravitational energy. Gravitational energy is responsible for falling water in waterfalls.
• Nuclear Energy: The energy produced by nuclear fission or fusion is called nuclear energy.

Mechanical Energy:
The sum of kinetic energy and potential energy in an object gives the mechanical energy of that object. Mechanical energy gives an object the capacity to do work.

Kinetic Energy:
The energy in an object due to its motion is called kinetic energy. Let us assume that an object of mass m is moving at uniform velocity μ and a force F is applied on it in the direction of motion then the object is displaced to distance s. Let us assume that the work done on the object increases its speed to ν and acceleration of object becomes. Then work done is equal to change in kinetic energy of object.

If initial velocity of the object is zero, i.e. the object starts from rest and obtains the velocity v then;

It can be said that the kinetic energy of an object of mass m and velocity v is;

Potential Energy :
The energy in an object due to the position of the object is called potential energy. Potential energy gives the capacity to do work to an object. The work done in putting an object in a specific position from normal position, gives the measure of its potential energy.

Potential Energy in Gravitational Field :
When an object is lifted up to a height then some work is done against acceleration due to gravity. The work done on object increases the energy of object. This energy gets stored in the object in the form of potential energy. If an object with mass m is lifted to height h from earth then the minimum force required to do this should be equal to its weight, i.e. w = mg.
Potential energy of object at height h = work done against gravitational force
= Force × Displacement = mgh

Potential Energy of a Simple Pendulum :
When a simple pendulum is displaced from its mean position to a particular side, its centre of gravity is raised. The work done on pendulum in this gets stored as potential energy in pendulum.

The pendulum moves ahead from its mean position due to this kinetic energy. While doing so, its kinetic energy reduces to zero and potential energy increases to maximum when pendulum reaches another extreme (point B). The pendulum swings back to its mean position due to the potential energy.

If mass of pendulum is m and pendulum is suspended by a thread of length / then potential energy of pendulum for displacement x can be obtained as follows:

Similarly, when a spring with spring constant k is displaced to x distance from its mean position (under elastic limits) then potential energy of spring is;

Potential Energy Calculator are physic/math calculator to find Mass, Grav. acceleration, Height or Potential energy fast and easy.

Electrical Energy
The energy in charged particles is called electrical energy. When a particle becomes charged then an electrical field is created around it. This electrical field applies a force on charged particles nearby and makes them move. Thus, transmission of energy takes place through particles in motion.

• The electrical field around a positively charged particle repels other positively charged particles but attracts negatively charged particles.
• Positively charged particles possess potential energy because of their position. When electrical field applies a force on these particles, they move in a certain direction. For example; we need to apply an external force in order to move a positively charged particle from negatively charged source. This process would increase the potential energy of positively charged particles. When the external force is removed, the positively charged particle would move from area of higher potential energy to area of lower potential energy. Thus, a positively charged particle would naturally move towards negatively charged source. While doing so, potential energy of charged particles changes into kinetic energy. This kinetic energy is obtained in the form of electrical energy.

Different types of electric plants are used for production of electricity. Some of them are as follows.

Thermal Power Plant: In a thermal power plant, heat energy is obtained from chemical energy of coal. The heat energy is utilized to convert highly pure water into steam. The steam is utilized to rotate the turbine and thus electricity is produced by generator which is attached with turbine.

Nuclear Power Plant:
Nuclear fission is carried out in a nuclear power plant. The heat generated from nuclear fission is utilized to convert water into steam. The steam is utilized to rotate the turbine and thus electricity is produced by generator which is attached with turbine.

• Hydel Power Plant: In a hydel power plant, potential energy of water is increased by making a dam. When water is released from the dam, potential energy of water changes to kinetic energy. This kinetic energy is utilized to rotate the turbine and thus electricity is produced by generator which is attached with turbine.
• Wind Power Plant: In a windmill, kinetic energy of air is utilized to rotate the turbine and thus electricity is produced by generator which is attached with turbine. Wind energy is a renewable energy and hence is environment friendly.
• Solar Power Plant: Energy from sun is converged at a point with the help of lens and mirrors and is changed into heat energy. This heat is utilized to produce steam which is utilized to rotate the turbine and thus electricity is produced by generator which is attached with turbine.
• Photovoltaic Power Plant: In a photovoltaic power plant, solar panels are installed at open spaces or on rooftops. These panels contain photovoltaic cells. When sunlight falls on these cells, the cells gain photons from light and bring the electrons in excited state. These charged particles provide electricity to circuit in the form of electric current.

Conservation of Energy :
As per law of conservation of energy, the total energy of an isolated system always remains constant. Energy can neither be created nor be destroyed. We can only change the form of energy.
As per law of conservation of mechanical energy, the mechanical energy of a system remains conserved. If kinetic energy of the system is increased then potential energy decreases, and vice- versa is also true.
If change in potential energy and kinetic energy is respectively ΔE_P and ΔEK then;
^ΔE_P^=- Δ Ek
Or,  ΔE_p+ΔE_k = 0
Or, Total mechanical energy E_m = E_p+E_k
In reality, there is some fall in mechanical energy during the whole cycle. This reduction is because of action of some non-conserving forces, like friction, viscosity, etc. These forces convert some of the energy into sound, heat, etc. Thus form of energy of the system changes but total energy of the system remains constant.

Dissipation of Energy:
When energy changes from one form to another then some of the energy is dissipated in the form of heat, light, sound, etc. Dissipation of energy means that energy is converted into some forms which are undesirable and which are not required by us. The total energy remains conserved but because of useless dissipation of energy it is impossible to make a cent per cent efficient system.

Power:
Power is defined as the rate of doing a work. It is a scalar quantity.

Unit of Power:
The unit of power is Watt which is named in honour of James Watt; the inventor of steam engine.

Another unit of power is horse power which is equal to 746 W.

Electrical Power:
The rate of transfer of electrical energy in electric circuit is called electric power. On a commercial scale, the electricity distribution company measures the consumption of electricity in the form of kilo Watt Hour. One kilowatt hour is called one electric unit which is read by electric meter.

This means that when a bulb of 1 KW is used for 1 hour, it will consume 1 unit of electricity

 

Rajasthan Board RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables Important Questions and Answers.

RBSE Class 10 Maths Chapter 3 Important Questions Pair of Linear Equations in Two Variables

Objective Type Questions—

Question 1.
If 2A + y = 6 then the pair satisfying it is—
(A) (1, 2)
(B) (2, 1)
(C) (2, 2)
(D) (1, 1)
Answer:
(C) (2, 2)

Question 2.
If \(\frac{4}{x}\) + 5y = 7 and x = \(\) then the value of y will be—
(A) \(\frac{37}{15}\)
(B) 2
(C) \(\frac{1}{2}\)
(D) \(\frac{1}{3}\)
Answer:
(B) 2

Question 3.
In the equation \(\frac{y-3}{7}\) – \(\frac{x}{2}\) = 1 if y = 10, then A is equal to—
(A) 0
(B) 1
(C) – 2
(D) 2
Answer:
(A) 0

Question 4.
The age of the father is three times the age of the son, if the age of father is A years, then age of the son after 5 years will be—
(A) 3A + 5
(B) A + 5
(C) \(\frac{x}{3}\) + 5
(D) \(\frac{x+5}{3}\)
Answer:
(C) \(\frac{x}{3}\) + 5

Question 5.
The point on A-axis is—
(A) (2, 3)
(B) (2, 0)
(C) (0, 2)
(D) (2, 2)
Answer:
(B) (2, 0)

Question 6.
The quadrant, in which the point P(3, – 4) lies, is—
(A) first
(B) second
(C) third
(D) fourth
Answer:
(D) fourth

Question 7.
If in any number the unit’s digit is a and the ten’s digit is b, then that number is—
(A) 10a + b
(B) a + 10b
(C) a + b
(D) ab
Answer:
(B) a + 10b

Very Short Answer Type Questions—

Question 1.
For what value of K there exists no solution of the pair of equation
2x + Ky = 1,
3x – 5y = 7
Solution:
For no solution

Hence, for K = \(\frac{-10}{3}\), the system will have no solution.

Question 2.
On comparing the ratios \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) and \(\frac{c_{1}}{c_{2}}\), find out whether the pair of linear equations 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0 are consistent, or inconsistent.
Solution:
Comparing the equations 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0 with a₁x + b₁y + c₁x = 0 and a₂x + b₂y + c₂ = 0

Hence the pair of equations is Inconsistent.

Question 3.
Show that the lines x – 4y + 5 = 0 and 3r – 12y +8 = 0 are parallel.
Solution:
The equations of the given lines are
x – 4y + 5 = 0
and 3x – 12y +8 = 0
Comparing the above pair of equations with general pair of equations
a₁ = 1, b₁ = – 4, c₁ = 5
a₂ = 3, b₂ = – 12, c₂ = 8

Hence the given pair of equations is inconsistent. Hence the given lines are parallel.

Question 4.
Twice a number x is 24 more than y. Write an equation representing this statement.
Solution:
2x – y = 24

Question 5.
The age of Ram is x years and the age of Shyam is y years. Five years ago the age of Ram was twice the age of Shyam. Write the equation representing this statement in the form of ax + by + c = 0.
Solution:
x – 2y + 5 = 0

Question 6.
Write the methods of representing and presenting the solution of pair of linear equations in two variables.
Solution:

1. Graphical Method,
2. Algebraic Method.

Question 7.
What do you understand by the inconsistent pair of linear equations?
Answer:
If both the lines are parallel, then there exists no solution of this pair of linear equations. In this case the pair of linear equations is called an inconsistent pair.

Question 8.
Find the nature of solution of the following system of equations—
2x + 4y = 7, 3x + 6y = 10
Solution:
2x + 4y – 7 = 0
3x + 6y – 10 = 0
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0.

Hence the pair of equations is inconsistent and the system has no solution.

Question 9.
If we add 1 in the numerator and the denominator each of a fraction, then the value of the fraction becomes \(\frac{1}{2}\). How will you write it in the form of an equation.
Solution:
Let the fraction be \(\frac{x}{y}\)
Then, \(\frac{x+1}{y+1}\) = \(\frac{1}{2}\)

Solve for x means find the value of x that would make the equation you see true.

Question 10.
Find the value of x in the following system of equations—
2x + 3y = 4
3x + 4y = 5
Solution:
From given equations
2x + 3y – 4 = 0 …. (1)
3x + 4y – 5 = 0 …. (2)
Multiplying equation (1) by 4 and equation (2) by 3

Question 11.
Solving the equations x + y = 2xy and \(\frac{1}{x}+\frac{2}{y}\) = 10, find the value of y only.
Solution:
x + y = 2xy
Dividing both sides by xy
\(\frac{x}{x y}+\frac{y}{x y}\) = \(\frac{2 x y}{x y}\)
\(\frac{1}{y}+\frac{1}{x}\) = 2 …(1)
From second equation
\(\frac{1}{x}+\frac{2}{y}\) = 10 …(2)
Subtracting equation (2) from equation (1)
\(-\frac{1}{y}\) = -8
∴ y = \(\frac{1}{8}\)

Question 12.
Write the names of the methods of solving pair of linear equations in algebraic form.
Solution:

1. Substitution method
2. Elimination method
3. Cross-multiplication method.

Question 13.
In the linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, if \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\), then explain the meaning of this situation.
Solution:
If \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\), then the pair of linear equations is inconsistent.
In this case lines are parallel and the pair of equation has no solution.

Question 14.
In the equation 5y – 3x – 10 = 0, express y in terms of x. Find the point where the line represented by the equation 5y – 3x – 10 = 0 intersects the y- axis.
Solution:
The given equation is 5y – 3x – 10=0
or 5y = 3x + 10
∴ y = \(\frac{3 x+10}{5}\)
The line represented by the equation 5y – 3x – 10 = 0 will intersect y-axis when x = 0. So y = \(\frac{3 \times 0+10}{5}\) = 2
Hence that point will be (0, 2)

Question 15.
In the linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0. If \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\), then explain the meaning of this situation.
Solution:
If \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\), then the pair of linear equations is consistent.

Question 16.
Write the solution of the pair of linear equations
\(\sqrt{2}\)x + \(\sqrt{3}\)y = 0 and \(\sqrt{3}\)x – \(\sqrt{2}\)y = 0
Solution:
The given pair of linear equations is
\(\sqrt{2}\)x + \(\sqrt{3}\)y = 0 ….(1)
\(\sqrt{3}\)x – \(\sqrt{2}\)y = 0 ….(2)
From equation (2)
\(\sqrt{3}\)x = \(\sqrt{2}\)y
x = \(\frac{\sqrt{2}}{\sqrt{3}}\)y
Putting this value of x in equation (1)

Hence, x = 0, y = 0

Question 17.
The total cost of 7 pencils and 5 pens is Rs. 29. Write it in algebraic form.
Solution:
Let the cost of 1 pencil = Rs. x
and the cost of 1 pen = Rs. y
Then according to the question 7x + 5y = 29

Question 18.
Write the solution of the pair of linear equations 3x + 4y = 0
and 2x – y = 0
Solution:
x = 0 and y = 0

Question 19.
Write the solution of the pair of linear equations 4x + 2y = 5 and x – 2y = 0
Solution:
Pair of linear equations
4x + 2y = 5 …. (1)
and x – 2y = 0 …. (2)
From equation (2) we get
x = 2y
putting this value x = 2y in equation (1) then we get
4(2y) + 2y = 5
⇒ 8y + 2y = 5 ∴10y = 5
or y = \(\frac{5}{10}\) = \(\frac{1}{2}\)
putting this value y =1/2 in equation (2) then we get
x – 2 × \(\frac{1}{2}\) = 0 ∴ x – 1 = 0
or x = 1
so the required solution of the pair of linear equations is
x = 1 and y = \(\frac{1}{2}\)

Short Answer Type Questions—

Question 1.
A two-digit number in such that the product of its digits is 12. When 36 is added to this number, then the digits of the number are interchanged. Find the number.
Solution:
Let the ten’s digit and the unit’s digit be x and y respectively. So the number = 10x + y
According to the question
x × y = 12 ….(1)
and 10x + y + 36 = 10y + x ….(2)
or 9x – 9y = – 36
⇒ x – y = – 4 …(3)
From equation (3)
(x – y)² = x² + y² – 2xy
⇒ (-4)² = x² + y² – 2xy
⇒ x² + y² = 16 + 24 = 40
Again (x + y)² = x² + y² + 2xy
= 40 + 2 × 12
⇒ (x + y)² – 64
⇒ x + y = \(\sqrt{64}\) = ± 8 ….(4)
From equation (3) and (4)
x = 2 and y = 6
Hence the required number will be 26.

Question 2.
The sum of the digits of a two-digit number is 9. Four times this number is seven times the number formed by reversing the digits of the number. Find that number.
Solution:
Let the ten’s digit be x and the unit’s digit be y
Then required number = 10x + y
According to the question
x + y = 9 ….(1)
Again according to the question
4(10x + y) = 7(10y + x)
⇒ 40x + 4y = 70y + 7x
⇒ 33x – 66y = 0
⇒ x – 2y = 0 ….(2)
Subtracting equation (2) from equation (1)

Putting the value of y in equation (2)
x – 2 × 3 = o
⇒ x = 6
Hence required number = 10x + y
= 10 × 6 + 3 = 60 + 3
= 63

Question 3.
The number of teachers in a government college and a private college in a town is 210. When 10 teachers from private college resign and join government college then the number of teachers in both the colleges becomes the same. Find the number of teachers in each college.
Solution:
Let the number of teachers in government college = x
and the number of teachers in private college = y
Then according to the question
x + y = 210 …(1)
and (x + 10) = (y – 10)
⇒ x – y = – 20 ….(2)
Adding equations (1) and (2)
2x = 190
⇒ x= \(\frac{190}{2}\) = 95
Putting the value of x in equation (1)
93 + y = 210
⇒ y = 210 – 95 = 115
Hence the number of teachers in government college is 95 and in private college is 115.

Question 4.
The sum of the digits of a two-digit number is 7. On reversing the order of the digits the number obtained is a more than the original number. Find that number.
Solution:
Let in the number the ten’s digit = x
and the unit’s digit = y
Then the given number = (10x + y)
The number on reversing the order of the digits = (10y + x)
According to the question
x + y = 7
or x + y – 7 = 0
and (10x + y) + 9 = 10y
or 10x + y + 9 = 10y
or 9x + 9 = 9y
or x – y + 1 = 0
Solving equations (1) and (2)

Putting the value y = 4 in equation (1)
x + 4 – 7 = 0
⇒ x – 3 = 0
⇒ x = 3
∴ Required number = 10x + y
= 10 × 3 + 4
= 30 + 4 = 34

Question, 5.
6 years hence the age of a man will be 3 times the age of his son and 3 years ago he was 9 times the age of his son. Find their present ages.
Solution:
Let the present ages of the man and his son be reprectively x years and y years.
Age of father 6 years hence = (x + 6) years
and age of son 6 years hence = (y + 6) years
Hence according to the question
x + 6 = 3(y + 6)
⇒ x + 6 = 3y + 18
or x – 3y = 12 ….(1)
Again 3 years ago the ages of the father and the son will be respectively (x – 3) years and (y – 3) years.
Hence according to the question x – 3 = 9(y – 3)
⇒ x – 3 = 9y – 27
⇒ x – 9y = – 24 ….(2)
Now writing both the equations

Putting the value of y in equations (1)
x – 3 × 6 = 12
⇒ x – 18 = 12
⇒ x = 12 + 18 = 30
Hence age of father = 30 years
and age of son = 6 years

Question 6.
Aftab says to his daughter, ‘seven years ago I was of an age seven times of you. 3 years hence from now I shall be of an age only three times of you.’ (Is it interesting?) Express this situation in algebraic and graphical form.
Solution:
Let the age (in years) of Aftab and his daughter be respectively s and t. Then the pair of linear equations are
s – 7 = 7 (t – 7)
i.e., s – 7t + 42 = 0 ….(1)
and s + 3 = 3 (t + 3)
i.e., s – 3t = 6 …(2)
Using equation (2)
s = 3t + 6
Putting the value of s in equation (1)
(31 + 6) – 7t + 42 = 0
i.e., 4t = 48, so that t = 12
Putting this value of t in equation (2)
s – 3 × 12 = 6
⇒ s – 36 = 6
∴ s = 6 + 36 = 42
Hence, Aftab and his daughter are of age 42 years and 12 years respectively.

Question 7.
Two rails are represented by the equation x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Will the rails intersect each other?
Solution:
According to the question the linear equations are
x + 2y – 4 = 0 ….(1)
2x + 4y – 1 2 = 0 …(2)
From equation (1) expressing x in terms of y
x = 4 – 2y
Now, substituting this value of x in equation (2)
2 (4 – 2y) + 4y – 12 = 0
or 8 – 12 = 0
or -4 = 0
Which is a false statement.
Hence, there is no common solution of the given equations. Therefore, both the rails will not intersect each other.

Question 8.
For what values of p, the following pair of equations has a unique solution
4x + py + 8 = 0
2x + 2y + 2 = 0
Solution:
Here a₁ = 4. a₂ = 2, b₁ = p, b₂ = 2
Now for a unique solution of the given pair

Hence, for each value of p except 4, the given pair of equations will have a unique solution.

Question 9.
Show graphically that the following pair of equations 2x + 4y = 10; 3x + 6y = 12 has no solution.
Solution:
From equation
2x + 4y= 10
4y= 10 – 2x
or y = \(\frac{10-2 x}{4}\)

x	1	– 3	3
y	2	4	1

From equation
3x + 6y = 12
y = \(\frac{12-3 x}{6}\)

x	0	2	– 2
y	2	1	3

Plotting the points (1, 2), (- 3, 4) and (3, 1) and joining we get the graph AB of the equation 2x + 4y =10 and plotting the points (0, 2), (2, 1) and (-2, 3) and joining we get the graph CD of the equation 3x + 6y= 12. These two lines are mutually parallel. Hence the given pair of equations is inconsistent and it has no solution.

Question 10.
For what value of k will the following pair of linear equations have infinitely many solutions?
kx + 3y = k – 3
12x + ky = k
Solution:
For infinitely many solutions

⇒ 3 = k – 3
⇒ k = 6
Hence k = 6

Long Answer Type Questions—

Question 1.
Check graphically whether the pair of equations
x + 3y = 6
and 2x – 3y = 12
are consistent. It in then, solve graphically.
OR
Solve the following linear equations graphically
x + 3y = 6
2x – 3y = 12
Solution:
Let us form tables from the first equation
x + 3y = 6
or 3y = 6 – x
∴ y = \(\frac{6-x}{3}\)

x	0	6
y	2	0

From second equation
2x – 3y = 12
or 2x – 12 = 3y
or y = \(\frac{2x-12}{3}\)

x	0	3
y	– 4	_ 2

Let us plot the points A(0, 2), B(6, 0), P(0, – 4) and Q(3, – 2) on a graph paper and joining the points lines AB and PQ are drawn as shown in the figure.
We see that the lines AB and PQ have a common point B(6, 0). Hence a solution of the pair of linear equations is x = 6,
y = 0, i.e., the pair of equations is consistent.

Question 2.
Find the solution of the pair of equations 2x – y = 1; x + 2y = 8 by graphical method and find the coordinates of these points where their corresponding lines meet y-axis.
Solution:
Equation 2x – y = 1
So y = 2x – 1

x	1	2	3
y	1	3	5

Equation x + 2y = 8
So y = \(\frac{8-x}{2}\)

x	2	4	6
y	3	2	1

Drawing the graph with the points (1, 1) (2, 3) (3, 5) and the points (2, 3) (4, 2) (6, 1) we find that the point (2, 3) is the point where both the graphs intersect. This is the required solution of the equations. Hence x = 2,y = 3.
The first line intersects the y-axis at point (0, – 1) and the second line intersects the y-axis at the point (0, 4).

Question 3.
The sum of a two digit number and the number formed by reversing the digits is 66. If the difference of the digits of the number is 2, then find the number. How many such numbers are there?
Solution:
Let the ten’s digit and the unit’s digit of the first number be A and y respectively. Therefore, the expanded form of the first number is 10x + y. When the digits are reversed, then x becomes the unit’s digit and y the ten’s digit. This number in the expanded form is 10y + x.
According to the given conditions
(10x + y) + (10y + x) = 66
or 11 (x + y) = 66
or x + y = 6 ….(1)
We are also given that the difference of the digits is 2.
Therefore,
either x – y = 2 …(2)
or y – x = 2 …(3)
If x – y = 2, then solving (1) and (2) by elimination method, we get A = 4 and y = 2. In this case, we get the number 42.
If y – x = 2, then solving (1) and (3) by elimination method, we get A = 2 and y = 4. In this case, we get the number 24.
Thus there are two such numbers 42 and 24.
Verification :
Here 42 + 24 = 66 and 4 – 2 – 2
and 24 + 42 = 66 and 4 – 2 = 2

Question 4.
For what value of k, will there be infinitely many solutions of the following pair of linear equations?
kx + 3y – (k – 3) = 0
12A + ky – k = 0
Solution:

For infinitely many solutions of the pair of linear equations, we should have

Which gives
3k = k² – 3k
i.e., 6k = k²
which means k = 0 or k = 6
Therefore the value of k which satisfies both the conditions is k = 6. For this value there are infinitely many solutions of the pair of equations.

Question 5.
From a bus stand of Bangalore if we purchase two tickets of Malleshwaram and 3 tickets of Yashwantpur, then the total cost is Rs. 46. But if we purchase 3 tickets of Malleshwaram and 5 ticket of Yaswantpur, then the total cost is Rs. 74. Find the fare from Bus stand to Malleshwaram and from Bus stand to Yashwantpur.
Solution:
Let the fare from bus stand of Bangalore to Malleshwaram be Rs. x and to Yashwantpur be Rs. y.
From the given informations
2x + 3y = 46
i.e., 2x + 3y – 46 = 0 ….(1)
3x + 5y = 74
i.e., 3x + 5y – 74 = 0 ….(2)
Solving these equations by cross-multiplication method


i.e., x = 8 and y = 10
Hence, from bus stand of Bangalore, the fare of Malleshwaram is Rs. 8 and the fare of Yashwantpur is Rs. 10.

Question 6.
Solve the following pair of equations by converting into pair of linear equations

Solution:
Putting \(\frac{1}{x-1}\) = p and \(\frac{1}{y-1}\) = q the given equations become

5p + q = 2 …(3)
6p – 3q = 1 …(4)
Equations (3) and (4) in general form a pair of linear equations. Now to solve these equations, we can use any method. We find,
p = \(\frac{1}{3}\) and q = \(\frac{1}{3}\)
Now substituting \(\frac{1}{x-1}\) for p

⇒ y – 2 = 3
⇒ y = 5
Hence, the required solution of the given pair of equations is x = 4, y = 5.

Question 7.
The ratio of the incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each person saves ₹ 2000 each month, then find their monthly incomes.
Solution:
Let the monthly incomes of both the persons be ₹ 9x and ₹ 7x respectively and their monthly expenditures be ₹ 4y and ₹ 3y respectively. Then, in that
situtation the equations formed are
9x – 4y = 2000 ….(1)
and 7x – 3y = 2000 …(2)
Step 1 : To make the coefficients of y equal multiplying equation (1) by 3 and equation (2) by 4
27x – 12y = 6000 ….(3)
28x – 12y = 8000 …(4)

Step 2 : To eliminate y subtracting equation (3) from equation (4) since the coefficients of y are equal, we get
(28x – 27x) – (12y- 12y) = 8000 – 6000
⇒ x = 2000

Step 3 : Substituting the value of x in (1)
9 (2000) – 4y= 2000
⇒ y = 4000
Hence the solution of the pair of equations is x = 2000, y = 4000. Therefore, the monthly incomes of the persons are respectively ₹ 18,000 and ₹ 14,000.

Question 8.
A boat gives 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.

Solution:
Let the speed of the boat in still water be x km/h and speed of the stream be y km/h. Tfyen the speed of the boat downstream = (x + v) km/h and the speed of the boat upstream = (x – y) km/h
Also, time = \(\frac{\text { distance }}{\text { speed }}\)
In the first case, when the boat goes 30 km upstream. Let the time taken in hours be t₁. Then,
t₁ = \(\frac{30}{(x-y)}\)
Let t₂ be the time in hours taken by the boat to go 44 km downstream. Then,
t₂ = \(\frac{44}{(x+y)}\)
The total time taken t₁ + t₂ = 10 hours
Therefore, \(\frac{30}{(x-y)}\) + \(\frac{44}{(x+y)}\) = 10 ….(1)
In the second case, in 13 hours it can go 40 km upstream and 55 km downstream. Then
\(\frac{40}{(x-y)}\) + \(\frac{55}{(x+y)}\) = 13 ….(2)
Putting \(\frac{1}{(x-y)}\) = u and \(\frac{1}{(x+y)}\) = v …(3)
Substituting these values in equations (1) and (2)
30u + 44v = 10
or 30u + 44v – 10 = 0 ….(4)
40u + 55v = 13
or 40u + 55v – 13 = 0 ….(5)
Using cross-multiplication method,

or x – y = 5 and x + y = 11 …(6)
Adding these equations
2x = 16
or x = 8
Subtracting equation given in (6)
2y = 6
or y = 3
Hence, the speed of the boat in still water in 8 km/h and the speed of the stream water is 3 km/h.

Question 9.
The coach of a cricket team buys one bat and 3 balls for ₹ 300. Later, she buys another 2 bats of the samekind and 3 balls for ₹ 525. Represent this situation algebraically and solve it by graphical method. Also find for how many rupees can be coach buy one bat and one ball.
Solution:
Let the cost of one bat = ₹ x
and the cost of one ball = ₹ y
Algebraic Form—
According to the first condition of the « problem
x + 2y = 300
According to the second condition of the problem
2x + 3y = 525
∴ Pair of linear equations in two variables
x + 2y = 300
and 2x + 3 y = 525
Graphical Form—
x + 2y = 300
⇒ x = 300 – 2y ….(1)
Substituting y = 0 in (1)
x = 300 – 2 × 0
⇒ x – 300
Substituting y = 75 in (1)
x = 300 – 2 × 75
= 300 – 150
⇒ x = 150
Substituting y = 100 in (2)
x = 300 – 2 × 100
= 300 – 200
⇒ x = 100

x	300	150	100
y	0	75	100

2x + 3y = 525 …(2)
2x = 525 – 3y
x = \(\frac{525-3 y}{2}\)
Substituting y = 25 in (2)
x = \(\frac{525-3 \times 25}{2}\)

x	225	150	75
y	25	75	125

On solving graphically x = 150 and y = 75 and the coach will be able to buy a bat and a ball for
= 150 + 75
= ₹ 225

Question 10.
Ashok scored 65 marks in a test, getting 5 marks for each right answer and losing 2 marks for each wrong answer. That 3 marks been awarded for each correct answer and 1 mark been deducted for each incorrect answer, then Ashok would have scored 40 marks. Represent this situation algebraically and solve it graphically. How many questions were there in the test?
Solution:
Let the right answer question be x and the wrong answer questions be y.
Then, according to the first condition of the problem
5x – 2y = 65 ….(1)

According to the second condition of the problem
3x – y = 40 ….(2)
From equation (1) 5x = 65 + 2y
∴ x = \(\frac{65+2 y}{5}\)

x	13	15	11
y	0	5	-5

Putting y = 0 x = \(\frac{65+0}{5}\) = 13
Putting y = – 5 x = \(\frac{65+10}{5}\) = 15
Putting y = -5 x = \(\frac{65-10}{5}\) = 11
From equation (2)
3x – y = 40
⇒ -y = 40 – 3x
⇒ y = 3x – 40

x	13	15	20
y	-10	5	20

Putting x = 10 y – 30 – 40 = – 10
Putting x = 15 y = 45 – 40 = 5
Putting x = 20 y = 60 – 40 = 20
∵ The point of intersection of both the graphs is (15, 5)
So total number of questions = 15 + 5 = 20

Question 11.
The cost of 5 apples and 3 oranges is Rs. 35 and the cost of 2 apples and 4 oranges is Rs. 28. For mulati the problem algebraically and solve it graphically.
Solution:
Let the number of apples denote by x and the number of oranges denote by y according to 1st condition of question.

5x + 3y = 35
3y = 35 – 3x
∴ y = \(\frac{35-5 x}{3}\) …(1)

Apples (x)	1	4	– 2	7
Oranges (y)	10	5	15	0

According to Ilnd condition of question
2x + 4y – 28
4y = 28 – 2x
∴ y = \(\frac{28-2 x}{4}\) …(2)

Apples (x)	0	14	4	-2
Oranges (y)	7	0	5	8

Drawing the graph with the points (1, 10), (4, 5), (-2, 15), (7, 0) and the points (0, 7), (14, 0) (4, 5) and (-2, 8). We find that the point (4, 5) in the point where both the graphs intersect. This is the required solution of the equations.
Hence x = 4 Apples and y = 5 Oranges.

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 11 Constructions Exercise 11.2

In each of the following, give also the justification of the construction :

Question 1.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution:
Steps of Construction :
1. First of all taking O as centre and 6 cm as radius draw a circle.
2. Locate a point P at a distance of 10 cm from 0.
3. Now join OP and bisect it. Denote its mid-point by M.
4. Taking M as centre and MP as radius, draw a circle which intersects the above circle at Q and R.
5. Join PQ and PR
Then, PQ and PR are the required tangent lines on measurement.
PQ = PR = 8 cm.

Justification—
Join OQ. ∠PQO = 90°, since angle ∠PQO is an angle formed in a semi-circle.
∴ PQ ⊥ OQ
Since OQ is a radius of the given circle, therefore PQ will be the tangent line to the circle.
Similarly, PR will also be a tangent line to the circle.

Question 2.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Solution:
Steps of Construction :
1. First of all taking O as centre, draw two concentric circles of radii 4 cm and 6 cm.
2. Locate a point P on the circum-ference of the larger circle.
3. Now join OP and bisect it. Take its mid-point as M.
4. Taking M as centre and MP as radius, draw a circle which intersects the above circle at Q and R.
5. Join PQ and PR.
Then PQ and PR are the required tangent lines.
On measurement
PQ = PR = 4.5 cm. (approx.)

Justification—
Join OQ, ∠PQO = 90°, since ∠PQO is an angle formed in a semi-circle.
∴ PQ ⊥ OQ
Since OQ is a radius of the given circle, therefore PQ will be a tangent line to the circle.
Similarly, PR will also be a tangent line to the circle.
On measuring the length of the tangent line
PQ = PR = 4.5 cm.
Calculation—
Tangent line PQ = \(\sqrt{(O P)^{2}-(O Q)^{2}}\)
= \(\sqrt{(6)^{2}-(4)^{2}}\)
= \(\sqrt{36-16}\)
= \(\sqrt{20}\) = 4.47 cm.
≃ 4.5 cm.

Question 3.
Draw a circle of radius of 3 cm. Take two points P and Q on one of its extended diameter each as a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Solution:
Steps of Construction :
1. First of all take a point O and taking it as centre draw a circle of radius 3 cm.
2. Take two points P and Q on its diameter produced. Thus OP = OQ = 7 cm.
3. Now bisect OP and OQ. Take their mid-points as M₁ and M₂ respectively.
4. Taking M₁ as centre and M₁P as radius, draw a circle which intersects the above circle at T₁ and T₂.
5. Join PT₁ and PT₂.
Then PT₁ and PT₂ are the required tangent lines.
Similarly the tangent lines QT₃ and QT₄ can be obtained.

Justification—
Joining OT₁ ∠PT₁O = 90°, since it is an angle formed in a semi-circle.
∴ PT₁ ⊥ OT₁
Since OT₁ is a radius of the given circle, therefore PT₁ will be a tangent line to the circle.
Similarly, PT₂, QT₃ and QT₄ will also be the tangent lines to the circle.

Question 4.
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.

Solution:
Steps of Construction :
1. First of all draw a circle with centre O and radius 5 cm.
2. Draw any diameter AOC.
3. Now draw radius OL such that DCOL = 60° (i.e., given angle)
4. At L drawU LM ^ OL.
5. At A drawU AN ^ OA.
6. Both of these perpendiculars intersect at P.
Then PA and PB will be the required tangent lines.
Justification—
Since OA is a radius, therefore PA will be a tangent line to the circle. Similarly, PL will also be a tangent line to the circle.
∠APL = 360° – ∠OAP – ∠OLP – ∠AOL
= 360° – 90° – 90° – (180° – 60°)
= 360° – 360° + 60° = 60°
Thus, the tangent lines PA and PL will be inclined at an angle of 60° with each other.

Question 5.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Solution:
Steps of Construction :
1. First of all draw a line segment AB = 8 cm.
2. Now taking A as centre and 4 cm as radius draw a circle and taking B as centre and 3 cm as radius draw a circle. Bisect AB and find its mid-point O.
3. Clearly O is the mid-point of AB. With centre O draw a circle of radius OA or OB which intersects the circle with centre B at T₁ and T₂ and the circle with centre A at T₃ and T₄.
4. Join AT₁, AT₂, BT₃ and BT₄.
Then there will be the required tangents.
Justification—
Joining BT₁ ∠BT₁A = 90°, since ∠BT₁A is an angle situated in a semi-circle.

∴ AT₁ ⊥ BT₁
Since BT₁ is a radius of the given circle, therefore AT₁ will be a tangent line to the circle.
Similarly, AT₂, BT₃ and BT₄ will also be the tangent lines.

Question 6.
Let ABC be a right triangle in which AB = 6 cm., BC = 8 cm. and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D drawn. Construct the tangents from A to this circle.
Solution:
Steps of Construction :
1. First of all construct a DABC with given values in which AB = 6 cm., BC = 8 cm. and DB = 90°.
2. Draw BD ^ AC.
3. Now draw a circle with D as centre and DA, DB or DC as radius.
4. Bisect BC and BD. Let their bisectors meet at O.

5. With O as centre draw a circle with centre OB. This circle passes through the points B, C and D.
6. Join OA.
7. Draw perpendicular bisector of OA. This perpendicular bisector is obtained at point M.
8. Taking M as centre draw a circle with radius MA which intersects the above circle at points P and Q.
9. Join AP and AQ which are the required tangent lines to this circle from A.

Question 7.
Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Solution:
Steps of Construction :
1. First of all draw a circle with the help of a bangle.

2. Draw a secant from an external point A. Produce RA to C such that AR = AC.
3. Now taking CS as diameter, draw a semi-circle.
4. At point A draw AB ⊥ AS which intersects the semi-circle at B.
5. Draw an arc with centre A and radius as AB which cuts the given circle at T and T’. Join A and AT’.
Then, AT and AT’ will be the required tangent lines.

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 11 Constructions Exercise 11.1

In each of the following constructions, give the justification of the construction also :

Question 1.
Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Solution:
Steps of Construction :

1. First of all draw a line segment AB = 7.6 cm.
2. Now draw a ray AC which makes an acute angle with AB.
3. Locate 5 + 8 = 13 equal line segments on ray AC starting from A : AA₁, A₁A₂, A₂A₃, A₃A₄, A₄A₅, A₅A₆, A₆A₇, A₇A₈, A₈A₉, A₉A₁₀, A₁₀A₁₁, A₁₁A₁₂ and A₁₂A₁₃
4. Now join A₁₃B
5. Through A₅ draw A₅P || A₁₃B which meets AB at R
6. Thus, P divides AB in the ratio 5 : 8.
On measuring the two parts AP = 2.9 cm. and PB = 4.7 cm.
Justification—
In △ABA₁₃
PA₅ || BA₁₃
∴ △APA₅ ~ △ABA₁₃
⇒ \(\frac{AP}{PB}\) = \(\frac{\mathrm{AA}_{5}}{\mathbf{A}_{5} \mathbf{A}_{13}}\) = \(\frac{5}{8}\)
⇒ AP : PB = 5 : 8

Question 2.
Construct a triangle of sides 4 cm., 5 cm. and 6 cm. and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.
Solution:
Steps of Construction :
1. First of all draw a line segment BC = 7 cm.
2. Now taking B as centre and radius of 5 cm, draw an arc.
3. With C as centre and radius 4 cm, draw an arc to intersect the former arc at A.
4. Join AB and AC.
Then DABC is the required triangle.
5. Make an acute angle below BC.
6. On BX, mark three points B₁, B₂ and B₃ such that
BB₁ = B₁B₂ = B₂B₃
7. Now join B₃C
8. Through B₂ draw B₂D || B₃C which meets BC at D.
9. Through D draw ED || AC which meets BA at E.
Thus △EBD is the required triangle whose sides are \(\frac{2}{3}\) times the corresponding sides of △ABC.

Justification—Since DE || CA
∴ △ABC ~ △EBD
and \(\frac{EB}{AB}\) = \(\frac{BD}{BC}\) = \(\frac{DE}{CA}\) = \(\frac{2}{3}\)
Thus, we get a new triangle which is similar to the given triangle, whose sides are \(\frac{2}{3}\) times the corresponding sides of △ABC.

Question 3.
Construct a triangle with sides 5 cm., 6 cm. and 7 cm. and then another triangle where sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.
Solution:
Steps of Construction :
1. First of all construct a triangle ABC with given values in which BC = 7 cm., CA = 6 cm. and AB = 5 cm.
2. Now below BC make an acute angle CBX,
3. Where 7 points on BX :
B₁,B₂, B₃, B₄, B₅, B₆ and B₇ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇
4. Now join B₅C.
5. Through B₇ draw B₇D || B₅C. Join BC producing upto D.
6. Through D draw DE || CA, which meets BA produced at E.
Thus △EBD is the required triangle whose sides are \(\frac{7}{5}\) times the corresponding sides of △ABC.

Justification—
Since DE || CA
∴ △ABC ~ △EBD
and \(\frac{EB}{AB}\) = \(\frac{BD}{BC}\) = \(\frac{DE}{CA}\) = \(\frac{7}{5}\)
Thus we get a new triangle similar to the given triangle, where sides are \(\frac{7}{5}\) times the
corresponding side of △ABC.

Question 4.
Construct an isosceles triangle where base is 8 cm and altitude 4 cm and then another triangle whose sides are 1\(\frac{1}{2}\) times the corresponding sides of the isosceles triangle.
Solution:
Steps of Construction :
1. First of all draw BC = 8 cm.
2. Draw the perpendicular bisector PQ of line segment BC which meets BC at M.
3. From MP, cut MA = 4 cm.
4. Join BA and CA.
Now the DABC obtained is the required triangle.
5. Produce BC to D such that
BD = 12 cm.
6. Draw DE || CA which meets BA produced at E.
Then, △EBD will be the required triangle.

Justification—Since DE || CA
∴ △ABC ~ △EBD
and, \(\frac{EB}{AB}\) = \(\frac{DE}{CA}\) = \(\frac{BD}{BC}\) = \(\frac{12}{8}\) = \(\frac{3}{2}\)
Thus, we get a new triangle similar to the given triangle, whose sides are \(\frac{3}{2}\) times, i.e. 1\(\frac{1}{2}\) the corresponding sides of △ABC.

Question 5.
Draw a triangle ABC with side BC = 6 cm., AB = 5 cm. and ∠ABC = 60°. Then construct a triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of the triangle ABC.
Solution:
Steps of Construction :
1. First of all construct DABC with given values in which BC = 6 cm., DABC = 60° and AB = 5 cm.
2. Now make an acute angle DCBX below
3. Locate four points B₁, B₂, B₃ and B₄ on BX. Thus BB₁ = B₁B₂ = B₂B₃ = B₃B₄
4. Then join B₄C and draw B₃D || B₄C.
5. Now through D, draw ED || AC which meets BA at E.
Then △EBD will be the required triangle whose sides will be \(\frac{3}{4}\) times the corresponding sides of △ABC.

Justification—
Since DE || CA
∴ △ABC ~ △EBD
and, \(\frac{EB}{AB}\) = \(\frac{BD}{BC}\) = \(\frac{DE}{CA}\) = \(\frac{3}{2}\)
Thus, we get a new triangle similar to the given triangle, where sides are \(\frac{3}{2}\) times the sides of △ABC.

Question 6.
Draw a triangle ABC with side BC = 7 cm., ∠B = 45°, ∠A = 105. Then, construct a triangle whose sides are \(\frac{4}{3}\) times the corresponding sides of △ABC.
Solution:
Steps of Construction :
1. First of all construct △ABC with the given values in which BC = 7 cm.,
∠B = 45°, ∠C = 180° – (∠A + ∠B)
⇒ ∠C = 180° – (105° + 45°)
= 180° – 150° = 30°
2. Now make aft acute angle ∠CBX below BC.
3. Locate four points on BX :
B₁, B₂, B₃ and B₄
Thus BB₁ = B₁B₂ = B₂B₃ = B₃B₄
4. Then join B₃C.
5. Through B₄, draw B₄D || B₃C which meets BC produced at D.
6. Through D, draw ED || AC which meets BA produced at E.
Then △EBD will be the required triangle where sides will be \(\frac{4}{3}\) times the corresponding sides of △ABC.

Justification—
Since, DE || CA
∴ △ABC ~ △EBD
and, \(\frac{EB}{AB}\) = \(\frac{BD}{BC}\) = \(\frac{DE}{CA}\) = \(\frac{4}{3}\)
Thus, we get a new triangle similar to the given triangle, whose sides are \(\frac{4}{3}\) times the corresponding sides of △ABC.

Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle where sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.
Solution:
Steps of Construction :
1. First of all construct △ABC with given values in which BC = 4 cm., ∠B = 90° and BA = 3 cm.
2. Now make an acute angle CBX below BC.
3. Locate five points on BX :
B₁, B₂, B₃, B₄ and B₅ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅
4. Join B₃C.
5. Through B₅, draw B₅D || B₃C which meets BC produced at D.
6. Through D, draw ED || AC which meets BA produced at E. Thus, △EBD is the required triangle, whose sides are \(\frac{5}{3}\) times the corresponding sides of △ABC.

Justification—
Since, DE || CA
∴ △ABC ~ △EBD
and, \(\frac{EB}{AB}\) = \(\frac{BD}{BC}\) = \(\frac{DE}{CA}\) = \(\frac{5}{3}\)
Thus, we get a new triangle similar to the given triangle, whose sides are \(\frac{5}{3}\) times the corresponding sides of △ABC.