Category: Class 10

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 4 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 4 Geometric Constructions.

Problem Set 4 Geometry 10th Std Maths Part 2 Answers Chapter 4 Geometric Constructions

Question 1.
Select the correct alternative for each of the following questions.

i. The number of tangents that can be drawn to a circle at a point on the circle is ______
(A) 3
(B) 2
(C) 1
(D) 0
Answer:
(C)

ii. The maximum number of tangents that can be drawn to a circle from a point outside it is ______
(A) 2
(B) 1
(C) one and only one
(D) 0
Answer:
(A)

iii. If ∆ABC ~ ∆PQR and \(\frac { AB }{ PQ } \) = \(\frac { 7 }{ 5 } \), then ______
(A) AABC is bigger.
(B) APQR is bigger.
(C) both triangles will be equal
(D) can not be decided
Answer:
(A)

Question 2.
Draw a circle with centre O and radius 3.5 cm. Take point P at a distance 5.7 cm from the centre. Draw tangents to the circle from point P.
Solution:
Analysis:
As shown in the figure, let P be a point in the exterior of circle at a distance of 5.7 cm.
Let PQ and PR be the tangents to the circle at points Q and R respectively.
∴ seg OQ ⊥ tangent PQ …[Tangent is perpendicular to radius]
∴ ∠OQP = 90°

∴ point Q is on the circle having OP as diameter. …[Angle inscribed in a semicircle is a right angle]
Similarly, point R also lies on the circle having OP as diameter.
∴ Points Q and R lie on the circle with OP as diameter.
On drawing a circle with OP as diameter, the points where it intersects the circle with centre O, will be the positions of points Q and R respectively.

Question 3.
Draw any circle. Take any point A on it and construct tangent at A without using the centre of the circle.
Solution:
Analysis:
As shown in the figure, line l is a tangent to the circle at point A.
seg BA is a chord of the circle and ∠BCA is an inscribed angle.
By tangent secant angle theorem,
∠BCA = ∠BAR

By converse of tangent secant angle theorem,
If we draw ∠BAR such that ∠BAR = ∠BCA, then ray AR (i.e. line l) is a tangent at point A.

Question 4.
Draw a circle of diameter 6.4 cm. Take a point R at a distance equal to its diameter from the centre. Draw tangents from point R.
Solution:
Diameter of circle = 6.4 cm 6.4
Radius of circle = \(\frac { 6.4 }{ 2 } \) = 3.2 cm
Analysis:
As shown in the figure, let R be a point in the exterior of circle at a distance of 6.4 cm.
Let RQ and RS be the tangents to the circle at points Q and S respectively.

∴ seg PQ ⊥tangent RQ …[Tangent is perpendicular to radius]
∴ ∠PQR = 90°
∴ point Q is on the circle having PR as diameter. …[Angle inscribed in a semicircle is a right angle]
Similarly,
Point S also lies on the circle having PR as diameter.
∴ Points Q and S lie on the circle with PR as diameter.
On drawing a circle with PR as diameter, the points where it intersects the circle with centre P, will be the positions of points Q and S respectively.

Question 5.
Draw a circle with centre P. Draw an arc AB of 100° measure. Draw tangents to the circle at point A and point B.
Solution:
m(arc AB) = ∠APB = 100°
Analysis:
seg PA ⊥ line l
seg PB ⊥line m … [Tangent is perpendicular to radius]

The perpendicular to seg PA and seg PB at points A and B respectively will give the required tangents at A and B.

Steps of construction:
i. With centre P, draw a circle of any radius and take any point A on it.
ii. Draw ray PA.
iii. Draw ray PB such that ∠APB = 100°.
iv. Draw line l ⊥ray PA at point A.
v. Draw line m ⊥ ray PB at point B.
Lines l and m are tangents at points A and B to the circle.

Question 6.
Draw a circle of radius 3.4 cm and centre E. Take a point F on the circle. Take another point A such that E – F – A and FA = 4.1 cm. Draw tangents to the circle from point A.
Solution:
Analysis:
Draw a circle of radius 3.4 cm
As shown in the figure, let A be a point in the exterior of circle at a distance of (3.4 + 4.1) = 7.5 cm.
Let AP and AQ be the tangents to the circle at points P and Q respectively.
∴ seg EP ⊥ tangent PA … [Tangent is perpendicular to radius]

∴ ∠EPA = 90°
∴ point P is on the circle having EA as diameter. …[Angle inscribed in a semicircle is a right angle]
Similarly, point Q also lies on the circle having EA as diameter.
∴ Points P and Q lie on the circle with EA as diameter.
On drawing a circle with EA as diameter, the points where it intersects the circle with centre E, will be the positions of points P and Q respectively.

Question 7.
∆ABC ~ ∆LBN. In ∆ABC, AB = 5.1 cm, ∠B = 40°, BC = 4.8 cm, \(\frac { AC }{ LN } \) = \(\frac { 4 }{ 7 } \). Construct ∆ABC and ∆LBN.
Solution:
Analysis:
As shown in the figure,
Let B – C – N and B – A – L.

∆ABC ~ ∆LBN …[Given]
∴ ∠ABC ≅ ∠LBN …[Corresponding angles of similar triangles]
\(\frac { AB }{ LB } \) = \(\frac { BC }{ BN } \) = \(\frac { AC }{ LN } \) …(i)[Corresponding sides of similar triangles]
But. \(\frac { AC }{ LN } \) = \(\frac { 4 }{ 7 } \) …(ii)[Given]
∴ \(\frac { AB }{ LB } \) = \(\frac { BC }{ BN } \) = \(\frac { AC }{ LN } \) = \(\frac { 4 }{ 7 } \) …[From(i)and(ii)]
∴ sides of ∆LBN are longer than corresponding sides of ∆ABC.
∴ If seg BC is divided into 4 equal parts, then seg BN will be 7 times each part of seg BC.
So, if we construct ∆ABC, point N will be on side BC, at a distance equal to 7 parts from B.
Now, point L is the point of intersection of ray BA and a line through N, parallel to AC.
∆LBN is the required triangle similar to ∆ABC.

Steps of construction:
i. Draw ∆ABC of given measure. Draw ray BD making an acute angle with side BC.
ii. Taking convenient distance on compass, mark 7 points B₁, B₂, B₃, B₄, B₅, B₆ and B₇ such that
BB₁ = B₁B₂ = B₂B₃ B₃= B4₄ = B₄B₅ = B₅B₆ = B₆B₇.
iii. Join B₄C. Draw line parallel to B₄C through B₇ to intersects ray BC at N.
iv. Draw a line parallel to side AC through N. Name the point of intersection of this line and ray BA as L.
∆LBN is the required triangle similar to ∆ABC.

Question 8.
Construct ∆PYQ such that, PY = 6.3 cm, YQ = 7.2 cm, PQ = 5.8 cm. If = \(\frac { YZ }{ YQ } \) = \(\frac { 6 }{ 5 } \) then construct ∆XYZ similar to ∆PYQ.
Solution:

Analysis:
As shown in the figure,
Let Y – Q – Z and Y – P – X.
∆XYZ ~ ∆PYQ …[Given]
∴ ∠XYZ ≅ ∠PYQ …[Corresponding angles of similar triangles]
\(\frac { XY }{ PY } \) = \(\frac { YZ }{ YQ } \) = \(\frac { XZ }{ PQ } \) …(i)[Corresponding sides of similar triangles]
But, \(\frac { YZ }{ YQ } \) = \(\frac { 6 }{ 5 } \) ,..(ii)[Given]
∴ \(\frac { XY }{ PY } \) = \(\frac { YZ }{ YQ } \) = \(\frac { XZ }{ PQ } \) = \(\frac { 6 }{ 5 } \) …[From (i) and (ii)]
∴ sides of ∆XYZ are longer than corresponding sides of ∆PYQ.
∴ If seg YQ is divided into 5 equal parts, then seg YZ will be 6 times each part of seg YQ.
So, if we construct ∆PYQ, point Z will be on side YQ, at a distance equal to 6 parts from Y.
Now, point X is the point of intersection of ray YP and a line through Z, parallel to PQ.
∆XYZ is the required triangle similar to ∆PYQ.

Steps of construction:
i. Draw ∆ PYQ of given measure. Draw ray YT making an acute angle with side YQ.
ii. Taking convenient distance on compass, mark 6 points Y₁, Y₂, Y₃, Y₄, Y₅ and Y₆ such that
YY₁ = Y₁Y₂ = Y₂Y₃ = Y₃Y₄ = Y₄Y₅ = Y₅Y₆.
iii. Join Y₅Q. Draw line parallel to Y₅Q through Y₆ to intersects ray YQ at Z.
iv. Draw a line parallel to side PQ through Z. Name the point of intersection of this line and ray YP as X.
∆XYZ is the required triangle similar to ∆PYQ.

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.2 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

Practice Set 2.2 Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations

Question 1.
Solve the following quadratic equations by factorisation.
i. x² – 15x + 54 = 0
ii. x² + x – 20 = 0
iii. 2y² + 27y + 13 = 0
iv. 5m² = 22m + 15
v. 2x² – 2x + \(\frac { 1 }{ 2 } \) = 0
vi. 6x – \(\frac { 2 }{ x } \) = 1
vii. √2x² + 7x + 5√2 = 0 to solve this quadratic equation by factorisation complete the following activity
viii. 3x² – 2√6x + 2 = 0
ix. 2m(m – 24) = 50
x. 252 = 9
xi. 7m² = 21 m
xii. m² – 11 = 0
Solution:

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x – 9 = 0 or x – 6 = 0
∴ x = 9 or x = 6
∴ The roots of the given quadratic equation are 9 and 6.

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 5 = 0 or x – 4 = 0
∴ x = -5 or x = 4
∴ The roots of the given quadratic equation are -5 and 4.

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ y + 13 = 0 or 2y + 1 = 0
∴ y = – 13 or 2y = -1
∴ y = -13 or y = –\(\frac { 1 }{ 2 } \)
∴ The roots of the given quadratic equation are -13 and – \(\frac { 1 }{ 2 } \)

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ m – 5 = 0 or 5m + 3 = 0
∴ m = 5 or 5m = -3
∴ m = 5 or m = \(\frac { -3 }{ 5 } \)
∴ The roots of the given quadratic equation are 5 and – \(\frac { 3 }{ 5 } \)

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ 3x – 2 = 0 or 2x + 1 = 0
∴ 3x = 2 or 2x = -1
∴ x = \(\frac { 2 }{ 3 } \) or 2x = -1
∴ The roots of the given quadratic equation are \(\frac { 2 }{ 3 } \) and \(\frac { -1 }{ 2 } \).

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

ix. 2m (m – 24) = 50
∴ 2m² – 48m = 50
∴ 2m² – 48m – 50 = 0
∴m² – 24m – 25 = 0 …[Dividing both sides by 2]

∴ m – 25 = 0 or m + 1 = 0
∴ m = 25 or m = -1
∴ The roots of thes given quadratic equation are 25 and -1.

x. 25m² = 9
∴ 25m² – 9 = 0
∴ (5m)² – (3)² = 0
∴ (5m + 3) (5m – 3) = 0
…. [∵a² – b² = (a + b) (a – b)]
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ 5m + 3 = 0 or 5m – 3 = 0
∴ 5m = -3 or 5m = 3
∴ m = \(\frac { -3 }{ 5 } \) or m = \(\frac { 3 }{ 5 } \)
∴ The roots of the given quadratic equation are \(\frac { -3 }{ 5 } \) and \(\frac { 3 }{ 5 } \).

xi. 7m² = 21m
∴ 7m² – 21m = 0
∴ m² – 3m = 0 …[Dividing both sides by 7]
∴ m(m – 3) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ m = 0 or m – 3 = 0
∴ m = 0 or m = 3
∴ The roots of the given quadratic equation are 0 and 3.

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ m + √11 = 0 or m – √11 = 0
∴ m = -√11 or m = √11
∴ The roots of the given quadratic equation are – √11 and √11

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.4 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

Practice Set 2.4 Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations

Question 1.
Compare the given quadratic equations to the general form and write values of a, b, c.
i. x² – 7x + 5 = 0
ii. 2m² = 5m – 5
iii. y² = 7y
Solution:
i. x² – 7x + 5 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = -7, c = 5

ii. 2m² = 5m – 5
∴ 2m² – 5m + 5 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 2, b = -5, c = 5

iii. y² = 7y
∴ y² – 7y + 0 = 0
Comparing the above equation with
ay² + by + c = 0, we get
a = 1, b = -7, c = 0

Question 2.
Solve using formula.
i. x² + 6x + 5 = 0
ii. x² – 3x – 2 = 0
iii. 3m² + 2m – 7 = 0
iv. 5m² – 4m – 2 = 0
v. y² + \(\frac { 1 }{ 3 } \) y = 2
vi. 5x² + 13x + 8 = 0
Solution:
i. x² + 6x + 5 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = 6, c = 5
∴ b² – 4ac = (6)2 – 4 × 1 × 5
= 36 – 20 = 16

∴ x = -3 + 2 or x = -3 -2
∴ x = -1 or x = -5
∴ The roots of the given quadratic equation are -1 and -5.

ii. x² – 3x – 2 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = -3, c = -2
∴ b² – 4ac = (-3)2 – 4 × 1 × (-2)
= 9 + 8 = 17

iii. 3m² + 2m – 7 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 3, b = 2, c = -7
∴ b² – 4ac = (2)² – 4 × 3 × ( -7)
= 4 + 84 = 88

iv. 5m² – 4m – 2 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 5, b = -4, c = -2
∴ b² – 4ac = (-4)² – 4 × 5 × (-2)
= 16 + 40 = 56

v. y² + \(\frac { 1 }{ 3 } \)y = 2
∴ 3y² + y = 6 …(Multiplying both sides by 3]
∴ 3y² + y – 6 = 0
Comparing the above equation with
ay² + by + c = 0, we get
a = 3, b = 1, c = -6
∴ b² – 4ac = (1)² – 4 × 3 × (-6)
= 1 + 72 = 73

vi. 5x² + 13x + 8 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 5, b = 13, c = 8
∴ b² – 4ac = (13)2 – 4 × 5 × 8
= 169 – 160 = 9


The roots of the given quadratic equation are -1 and \(\frac { -8 }{ 5 } \).

Question 3.
With the help of the flow chart given below solve the equation x² + 2√3 x + 3 = 0 using the formula.

Solution:
i. x² + 2√3 x + 3 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = 2√3 ,c = 3

ii. b² – 4ac = (2√3)2 -4 × 1 × 3
= 12 – 12
= 0

Question 1.
Solve the equation 2x² + 13x + 15 = 0 by factorisation method, by completing the square method and by using the formula. Verify that you will get the same roots every time. (Textbook pg. no. 43)
Solution:

By using the property, if the product of two numbers is zero, then at least zero, we get
∴ x + 5 = 0 or 2x + 3 = 0
∴ x + -5 = 0 or 2x = -3 = 0
∴ x + -5 = or x = \(\frac { -3 }{ 2 } \)
∴ The roots of the given quadratic equation are \(\frac { -3 }{ 2 } \) and -5.

ii. Completing the square method:
2x² + 13x + 15 = 0


∴ The roots of the given quadratic equation are \(\frac { -3 }{ 2 } \) and -5.

iii. Formula method:
2x² + 13x + 15 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 2, b = 13, c = 15
∴ b² – 4ac = (13)2 – 4 × 2 × 15
= 169 – 120 = 49

∴ The roots of the given quadratic equation are \(\frac { -3 }{ 2 } \) and -5.
∴ By all the above three methods, we get the same roots of the given quadratic equation.

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

Practice Set 2.1 Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations

Question 1.
Write any two quadratic equations.
Solution:
i. y² – 7y + 12 = 0
ii. x² – 8 = 0

Question 2.
Decide which of the following are quadratic
i. x² – 7y + 2 = 0
ii. y² = 5y – 10
iii. y² + \(\frac { 1 }{ y } \) = 2
iv. x + \(\frac { 1 }{ x } \) = -2
v. (m + 2) (m – 5) = 03
vi. m³ + 3m² – 2 = 3m³
Solution:
i. The given equation is x² + 5x – 2 = 0
Here, x is the only variable and maximum index of the variable is 2.
a = 1, b = 5, c = -2 are real numbers and a ≠ 0.
∴ The given equation is a quadratic equation.

ii. The given equation is
y² = 5y – 10
∴ y² – 5y + 10 = 0
Here, y is the only variable and maximum index of the variable is 2.
a = 1, b = -5, c = 10 are real numbers and a ≠ 0.
∴ The given equation is a quadratic equation.

iii. The given equation is
y² + \(\frac { 1 }{ y } \) = 2
∴ y³ + 1 = 2y …[Multiplying both sides by y]
∴ y³ – 2y + 1 = 0
Here, y is the only variable and maximum index of the variable is not 2.
∴ The given equation is not a quadratic equation.

iv. The given equation is
x + \(\frac { 1 }{ x } \) = -2
∴ x² + 1 = -2x …[Multiplying both sides by x]
∴ x² + 2x+ 1 = 0
Here, x is the only variable and maximum index of the variable is 2.
a = 1, b = 2, c = 1 are real numbers and a ≠ 0.
∴ The given equation is a quadratic equation.

v. The given equation is
(m + 2) (m – 5) = 0
∴ m(m – 5) + 2(m – 5) = 0
∴ m² – 5m + 2m – 10 = 0
∴ m² – 3m – 10 = 0
Here, m is the only variable and maximum index of the variable is 2.
a = 1, b = -3, c = -10 are real numbers and a ≠ 0.
∴ The given equation is a quadratic equation.

vi. The given equation is
m³ + 3m² – 2 = 3m³
∴ 3m³ – m³ – 3m² + 2 = 0
∴ 2m³ – 3m² + 2 = 0
Here, m is the only variable and maximum
index of the variable is not 2.
∴ The given equation is not a quadratic equation.

Question 3.
Write the following equations in the form ax² + bx + c = 0, then write the values of a, b, c for each equation.
i. 2y = 10 – y²
ii. (x – 1)² = 2x + 3
iii. x² + 5x = – (3 – x)
iv. 3m² = 2m² – 9
v. P (3 + 6p) = – 5
vi. x² – 9 = 13
Solution:
i. 2y – 10 – y²
∴ y² + 2y – 10 = 0
Comparing the above equation with
ay² + by + c = 0, we get
a = 1, b = 2, c = -10

ii. (x – 1)² = 2x + 3
∴ x² – 2x + 12x + 3
x² – 2x + 1 – 2x – 30
∴ x² – 4x – 2 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = -4, c = -2

iii. x² + 5x = – (3 – x)
∴ x² + 5x = -3 + x
∴ x² + 5x – x + 3 = 0
∴ x² + 4x + 3 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = 4, c = 3

iv. 3m² = 2m² – 9
∴ 3m² – 2m² + 9 = 0
∴ m² + 9 = 0
∴ m² + 0m + 9 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 1, b = 0, c = 9

v. p (3 + 6p) = – 5
∴ 3p + 6p² = -5
∴ 6p² + 3p + 5 = 0
Comparing the above equation with
ap² + bp + c = 0, we get
a = 6, b = 3, c = 5

vi. x² – 9 = 13
∴ x² – 9 – 13 = 0
∴ x² – 22 = 0
∴ x² + 0x – 22 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = 0, c = -22

Question 4.
Determine whether the values given against each of the quadratic equation are the roots of the equation.
i. x² + 4x – 5 = 0; x = 1,-1
ii. 2m² – 5m = 0; m = 2, \(\frac { 5 }{ 2 } \)
Solution:
i. The given equation is
x² + 4x – 5 = 0 …(i)
Putting x = 1 in L.H.S. of equation (i), we get
L.H.S. = (1)² + 4(1) – 5 = 1 + 4 – 5 = 0
∴ L.H.S. = R.H.S.
∴ x = 1 is the root of the given quadratic equation.
Putting x = -1 in L.H.S. of equation (i), we get
L.H.S. = (-1)² + 4(-1) – 5 = 1 – 4 – 5 = -8
∴ LH.S. ≠ R.H.S.
∴ x = -1 ¡s not the root of the given quadratic equation.

ii. The given equation is
2m² – 5m = 0 …(i)
Putting m = 2 in L.H.S. of equation (i), we get
L.H.S. = 2(2)² – 5(2) = 2(4) -10 = 8 – 10 = -2
∴ L.H.S. ≠ R.H.S.
∴ m = 2 is not the root of the given quadratic equation.
Putting m = \(\frac { 5 }{ 2 } \) in L.H.S. of equation (i), we get

Question 5.
Find k if x = 3 is a root of equation kx² – 10x + 3 = 0.
Solution:
x = 3 is the root of the equation kx² – 10x + 3 = 0.
Putting x = 3 in the given equation, we get
k(3)² – 10(3) + 3 = 0
∴ 9k – 30 +3 = 0
∴ 9k – 27 = 0
∴ 9k = 27
∴ k = \(\frac { 27 }{ 9 } \)
∴ k = 3

Question 6.
One of the roots of equation 5m² + 2m + k = 0 is \(\frac { -7 }{ 5 } \) Complete the following activity to find the value of ‘k’.
Solution:

Question 1.
x² + 3x – 5, 3x² – 5x, 5x²; Write the polynomials In the index form. Observe the coefficients and fill in the boxes. (Textbook p. no. 31)
Answer:
Index form of the given polynomials:
x² + 3x – 5, 3x² – 5x + 0, 5x² + 0x + 0
i. Coefficients of x2 are [1], [3] and [5] respectively, and these coefficients are non zero.
ii. Coefficients of x are 3, [-5] and [0] respectively.
iii. Constant terms are [-5], [0] and [0] respectively.
Here, constant terms of second and third polynomial is zero.

Question 2.
Complete the following table (Textbook p. no. 31)

Answer:

Question 3.
Decide which of the following are quadratic equations? (Textbook pg. no. 31)
i. 9y² + 5 = 0
ii. m³ – 5m² + 4 = 0
iii. (l + 2)(l – 5) = 0
Solution:
i. In the equation 9y² + 5 = 0, [y] is the only variable and maximum index of the variable is [2].
∴ It [is] a quadratic equation.

ii. In the equation m³ – 5m² + 4 = 0, [m] is the only variable and maximum index of the variable is not 2.
∴ It [is not] a quadratic equation.

iii. (l + 2)(l – 5) = 0
∴ l(l – 5) + 2(l – 5) = 0
∴ l² – 5l + 2l – 10 = 0
∴ l² – 3l – 10 = 0.
In this equation [l] is the only variable and maximum index of the variable is [2]
∴ it [is] a quadratic equation.

Question 4.
If x = 5 is a root of equation kx² – 14x – 5 = 0, then find the value of k by completing the following activity. (Textbook pg, no. 33)
Solution:

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.5 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 1 Linear Equations in Two Variables.

Practice Set 1.5 Algebra 10th Std Maths Part 1 Answers Chapter 1 Linear Equations in Two Variables

Question 1.
Two numbers differ by 3. The sum of twice the smaller number and thrice the greater number is 19. Find the numbers.
Solution:
Let the greater number be x and the smaller number be y.
According to the first condition, x – y = 3 …(i)
According to the second condition,
3x + 2y = 19 …(ii)
Multiplying equation (i) by 2, we get
2x – 2y = 6 …(iii)
Adding equations (ii) and (iii), we get

Substituting x = 5 in equation (i), we get
5 – y = 3
∴ 5 – 3 = y
∴ y = 2
∴ The required numbers are 5 and 2.

Question 2.
Complete the following.

Solution:
Opposite sides of a rectangle are equal.
∴ 2x + y + 8 = 4x – y
∴ 8 = 4x – 2x – y – y
∴ 2x – 2y = 8
∴ x – y = 4 …(i)[Dividingboth sides by 2]
Also, x + 4= 2y
∴ x – 2y = -4 …(ii)
Subtracting equation (ii) from (i), we get

Substituting y = 8 in equation (i), we get
x – 8 = 4
∴ x = 4 + 8
∴ x = 12
Now, length of rectangle = 4x – y
= 4(12) – 8
= 48 – 8
∴ Length of rectangle = 40
Breadth of rectangle = 2y = 2(8) = 16
Perimeter of rectangle = 2(length + breadth)
= 2(40 + 16)
= 2(56)
∴ Perimeter of rectangle =112 units
Area of rectangle = length × breadth
= 40 × 16
∴ Area of rectangle = 640 sq. units
∴ x = 12 and y = 8, Perimeter of rectangle is 112 units and area of rectangle is 640 sq. units.

Question 3.
The sum of father’s age and twice the age of his son is 70. If we double the age of the father and add it to the age of his son the sum is 95. Find their present ages.
Solution:
Let the present ages of father and son be x years and y years respectively.
According to the first condition,
x + 2y = 70 …(i)
According to the second condition,
2x + y = 95 …(ii)
Multiplying equation (i) by 2, we get
2x + 4y = 140 …(iii)
Subtracting equation (ii) from (iii), we get

Substituting y = 15 in equation (i), we get
x + 2(15) = 7O
⇒ x + 30 = 70
⇒ x = 70 – 30
∴ x = 40
∴ The present ages of father and son are 40 years and 15 years respectively.

Question 4.
The denominator of a fraction is 4 more than twice its numerator. Denominator becomes 12 times the numerator, if both the numerator and the denominator are reduced by 6. Find the fraction.
Solution:
Let the numerator of the fraction be x and the denominator be y.
∴ Fraction = \(\frac { x }{ y } \)
According to the first condition,
y = 2x + 4
∴ 2x – y = -4 …(i)
According to the second condition,
(y – 6)= 12(x – 6)
∴ y – 6 = 12x – 72
∴ 12x – y = 72 – 6
∴ 12x – y = 66 …(ii)
Subtracting equation (i) from (ii), we get

Question 5.
Two types of boxes A, B ,are to be placed in a truck having capacity of 10 tons. When 150 boxes of type A and 100 boxes of type B are loaded in the truck, it weights 10 tons. But when 260 boxes of type A are loaded in the truck, it can still accommodate 40 boxes of type B, so that it is fully loaded. Find the weight of each type of box.
Solution:
Let the weights of box of type A be x kg and that of box of type B be y kg.
1 ton = 1000 kg
∴ 10 tons = 10000 kg
According to the first condition,
150x + 100y = 10000
∴ 3x + 2y = 200 …(i) [Dividing both sides by 50]
According to the second condition,
260x + 40y = 10000
∴ 13x + 2y = 500 …(ii) [Dividing both sides by 20]
Subtracting equation (i) from (ii), we get

∴ The weights of box of type A is 30 kg and that of box of type B is 55 kg.

Question 6.
Out of 1900 km, Vishal travelled some distance by bus and some by aeroplane. Bus travels with average speed 60 km/hr and the average speed of aeroplane is 700 km/hr. It takes 5 hours to complete the journey. Find the distance Vishal travelled by bus.
Solution:
Let the distance Vishal travelled by bus be x km and by aeroplane be y km.
According to the first condition,
x + y = 1900 …(i)
\(\text { Time }=\frac{\text { Distance }}{\text { Speed }} \)
∴ Time required to cover x km by bus = \(\frac { x }{ 60 } \) hr
Time required to cover y km by aeroplane
= \(\frac { y }{ 700 } \) hr
According to the second condition,

Multiplying equation (i) by 6, we get
6x + 6y= 11400 …(iii)
Subtracting equation (iii) from (ii), we get

∴ The distance Vishal travelled by bus is 150 km.

Question 1.
There are some instructions given below. Frame the equations from the information and write them in the blank boxes shown by arrows. (Textbook pg. no. 20)
Answer:

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 1 Linear Equations in Two Variables.

Problem Set 1 Algebra 10th Std Maths Part 1 Answers Chapter 1 Linear Equations in Two Variables

Choose correct alternative for each of the following questions.

Question 1.
To draw graph of 4x + 5y = 19, find y when x = 1.
(a) 4
(b) 3
(c) 2
(d) -3
Answer:
(b)

Question 2.
For simultaneous equations in variables x and y, D_x = 49, Dy = – 63, D = 7 then what is x?
(a) 7
(b) -7
(c) \(\frac { 1 }{ 7 } \)
(d) \(\frac { -1 }{ 7 } \)
Answer:
(a)

Question 3.
Find the value of

(a) -1
(b) -41
(c) 41
(d) 1
Answer:
(d)

Question 4.
To solvex + y = 3; 3x – 2y – 4 = 0 by determinant method find D.
(a) 5
(b) 1
(c) -5
(d) -1
Answer:
(c)

Question 5.
ax + by = c and mx + n y = d and an ≠ bm then these simultaneous equations have-
(a) Only one common solution
(b) No solution
(c) Infinite number of solutions
(d) Only two solutions.
Answer:
(a)

Question 2.
Complete the following table to draw the graph of 2x – 6y = 3.
Answer:

Question 3.
Solve the following simultaneous equations graphically.
i. 2x + 3y = 12 ; x – y = 1
ii. x – 3y = 1 ; 3x – 2y + 4 = 0
iii. 5x – 6y + 30 = 0; 5x + 4y – 20 = 0
iv. 3x – y – 2 = 0 ; 2x + y = 8
v. 3x + y= 10 ; x – y = 2
Answer:
i. The given simultaneous equations are

The two lines interest at point (3,2).
∴ x = 3 and y = 2 is the solution of the simultaneous equations 2x + 3y = 12 and x – y = 1.

ii. The given simultaneous equations are

The two lines intersect at point (-2, -1).
∴ x = -2 and y = -1 is the solution of the simultaneous equations x – 3y = 1 and 3x – 2p + 4 = 0.

iii. The given simultaneous equations are

The two lines intersect at point (0, 5).
∴ x = 0 and y = 5 is the solution of the simultaneous equations 5x – 6y + 30 = 0 and 5x + 4y – 20 = 0.

iv. The given simultaneous equations are


The two lines intersect at point (2, 4).
∴ x = 2 and y = 4 is the solution of the simultaneous equations 3x – y – 2 = 0 and 2x + y = 8.

v. The given simultaneous equations are

The two lines intersect at point (3, 1).
∴ x = 3 and y = 1 is the solution of the simultaneous equations 3x + y = 10 and x – y = 2.

Question 4.
Find the values of each of the following determinants.

Solution:

Question 5.
Solve the following equations by Cramer’s method.

Solution:
i. The given simultaneous equations are
6x – 3y = -10 …(i)
3x + 5y – 8 = 0
∴ 3x + 5y = 8 …(ii)
Equations (i) and (ii) are in ax + by = c form. Comparing the given equations with a₁x + b₁y = c₁ and a₂x + b₂y = c₂, we get
a₁ = 6, b₁ = -3, c₁ = 10 and
a₂ = 3, b₂ = 5, c₂ = 8

ii. The given simultaneous equations are
4m – 2n = -4 …(i)
4m + 3n = 16 …(ii)
Equations (i) and (ii) are in am + bn = c form.
Comparing the given equations with a₁m + b₁n = c₁ and a₂m + b₂n = c₂, we get
a₁ = 4, b₁ = -2, c₁ = -4 and
a₂ = 4, b₂ = 3, c₂ = 16


∴ (m, n) = (1, 4) is the solution of the given simultaneous equations.

iii. The given simultaneous equations are

iv. The given simultaneous equations are
7x + 3y = 15 …(i)
12y – 5x = 39
i.e. -5x + 12y = 39 …(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a₁x + b₁y = c₁ and a₂x + b₂y = c₂, we get
a₁ = 7, b₁ = 3, c₁ = 15 and
a₂ = -5, b₂ = 12, c₂ = 39

v. The given simultaneous equations are

∴ 4(x + y – 8) = 2(3x – y)
∴ 4x + 4y – 32 = 6x – 2y
∴ 6x – 4x – 2y – 4y = -32
∴ 2x – 6y = -32
∴ x – 3y = -16 …(ii)[Dividing both sides by 2]
Equations (i) and (ii) are in ax + by = c form. Comparing the given equations with
a₁x + b₁y = c₁ and a₂x + b₂y = c₂, we get
a₁ = 1, b₁ = -1, c₁ = -4 and
a₂ = 1, b₂ = -3, c₂ = -16

∴ (x, y) = (2, 6) is the solution of the given simultaneous equations.

Question 6.
Solve the following simultaneous equations:

Answer:
i. The given simultaneous equations are

Subtracting equation (iv) from (iii), we get

∴ (x, y) = (6, – 4) is the solution of the given simultaneous equations.

ii. The given simultaneous equations are

Adding equations (v) and (vi), we get

iii. The given simultaneous equations are


∴ (x, y) = (1, 2) is the solution of the given simultaneous equations.

iv. The given simultaneous equations are

∴ Equations (i) and (ii) become 7q – 2p = 5 …(iii)
8q + 7p = 15 …(iv)
Multiplying equation (iii) by 7, we get
49q – 14p = 35 …(v)
Multiplying equation (iv) by 2, we get
16q + 14p = 30 …(vi)
Adding equations (v) and (vi), we get

Substituting q = 1 in equation (iv), we get
8(1) + 7p = 15
∴ 8 + 7p = 15
∴ 7p = 15 – 8
∴ 7p = 7
∴ p = \(\frac { 7 }{ 7 } \) = 1
∴ (P, q) = (1,1)
Resubstituting the values of p and q, we get
1 = \(\frac { 1 }{ x } \) and 1 = \(\frac { 1 }{ y } \)
∴ x = 1 and y = 1
∴ (x, y) = (1, 1) is the solution of the given simultaneous equations.

v. The given simultaneous equations are

Resubstituting the values of p and q, we get

∴ 3x + 4y = 10 …(v)
and 2x – 3y = 1 …(vi)
Multiplying equation (v) by 3, we get
9x + 12y = 30 …(vii)
Multiplying equation (vi) by 4, we get
8x – 12y = 4 …(viii)
Adding equations (vii) and (viii), we get

Substituting x = 2 in equation (v), we get
3(2) + 4y = 10
⇒ 6 + 4y = 10
⇒ 4y = 10 – 6
⇒ y = 4/4 = 1
∴ y = 1
∴ (x, y) = (2, 1) is the solution of the given simultaneous equations.

Question 7.
Solve the following word problems, i. A two digit number and the number with digits interchanged add up to 143. In the given number the digit in unit’s place is 3 more than the digit in the ten’s place. Find the original number.
Solution:
Let the digit in unit’s place be x
and that in the ten’s place be y.

ii. Kantabai bought 1 \(\frac { 1 }{ 2 } \) kg tea and 5 kg sugar from a shop. She paid ₹ 50 as return fare for rickshaw. Total expense was ₹ 700. Then she realised that by ordering online the goods can be bought with free home delivery at the same price. So, next month she placed the order online for 2 kg tea and 7 kg sugar. She paid ₹ 880 for that. Find the rate of sugar and tea per kg.
Solution:
Let the rate of tea be ₹ x per kg and that of sugar be ₹ y per kg.
According to the first condition,
cost of 1 \(\frac { 1 }{ 2 } \) kg tea + cost of 5 kg sugar + fare for rickshaw = total expense

According to the second condition,
cost of 2 kg tea + cost of 7 kg sugar = total expense
2x + 7y = 880 …(ii)
Multiplying equation (i) by 2, we get
6x + 20y = 2600 …(iii)
Multiplying equation (ii) by 3, we get
6x + 21y = 2640 …(iv)
Subtracting equation (iii) from (iv), we get

∴ The rate of tea is ₹ 300 per kg and that of sugar is ₹ 40 per kg.

iii. To find number of notes that Anushka had, complete the following activity.

Solution:
Anushka had x notes of ₹ 100 and y notes of ₹ 50.
According to the first condition,
100x + 50y = 2500
∴ 2x + y = 50 …(i) [Dividing both sides by 50]
According to the second condition,
100y + 50x = 2000
∴ 2y + x = 40 … [Dividing both sides by 50]
i.e. x + 2y = 40
∴ 2x + 4y = 80 …(ii) [Multiplying both sides by 2]
Subtracting equation (i) from (ii), we get

∴ Anushka had 20 notes of ₹ 100 and 10 notes of ₹ 50.

iv. Sum of the present ages of Manish and Savita is 31, Manish’s age 3 years ago was 4 times the age of Savita. Find their present ages.
Solution:
Let the present ages of Manish and Savita be x years and y years respectively.
According to the first condition,
x + y = 31 …(i)
3 years ago,
Manish’s age = (x – 3) years
Savita’s age = (y – 3) years
According to the second condition,
(x – 3) = 4 (y – 3)
∴ x – 3 = 4y – 12
∴ x – 4y = -12 + 3
∴ x – 4y = -9 …(ii)
Subtracting equation (ii) from (i), we get

∴ x = 31 – 8
∴ x = 23
The present ages of Manish and Savita are 23 years and 8 years respectively.

v. In a factory the ratio of salary of skilled and unskilled workers is 5 : 3. Total salary of one day of both of them is ₹ 720. Find daily wages of skilled and unskilled workers.
Solution:
Let the daily wages of skilled workers be ₹ x
that of unskilled workers be ₹ y.
According to the first condition,

∴ The daily wages of skilled workers is ₹ 450 and that of unskilled workers is ₹ 270.

vi. Places A and B are 30 km apart and they are on a straight road. Hamid travels from A to B on bike. At the same time Joseph starts from B on bike, travels towards A. They meet each other after 20 minutes. If Joseph would have started from B at the same time but in the opposite direction (instead of towards A), Hamid would have caught him after 3 hours. Find the speed of Hamid and Joseph.
Solution:
Let the speeds of Hamid and Joseph be x km/hr andy km/hr respectively.
Distance travelled by Hamid in 20 minutes

According to the first condition,

∴ The speeds of Hamid and Joseph 50 km/hr and 40 km/hr respectively.

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.4 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 1 Linear Equations in Two Variables.

Practice Set 1.4 Algebra 10th Std Maths Part 1 Answers Chapter 1 Linear Equations in Two Variables

Question 1.
Solve the following simultaneous equations.

Solution:
i. The given simultaneous equations are

∴ Equations (i) and (ii) become
2p – 3q = 15 …(iii)
8p + 5q = 77 …(iv)
Multiplying equation (iii) by 4, we get
8p – 12q = 60 …(v)
Subtracting equation (v) from (iv), we get

ii. The given simultaneous equations are



Substituting x = 3 in equation (vi), we get
3 + y = 5
∴ y = 5 – 3 = 2
∴ (x, y) = (3, 2) is the solution of the given simultaneous equations.

iii. The given simultaneous equations are

∴ Equations (i) and (ii) become
27p + 31q = 85 …(iii)
31p + 27q = 89 …(iv)
Adding equations (iii) and (iv), we get

iv. The given simultaneous equations are


Substituting x = 1 in equation (vi), we get
3(1) + y = 4
∴ 3 + y = 4
∴ y = 4 – 3 = 1
∴ (x, y) = (1, 1) is the solution of the given simultaneous equations.

Question 1.
Complete the following table. (Textbook pg. no. 16)
Solution:

Question 2.
In the above table the equations are not linear. Can you convert the equations into linear equations? (Textbook pg. no. 17)
Answer:
Yes, the above given simultaneous equations can be converted to a pair of linear equations by making suitable substitutions.

Steps for solving equations reducible to a pair of linear equations.

• Step 1: Select suitable variables other than those which are in the equations.
• Step 2: Replace the given variables with new variables such that the given equations become linear equations in two variables.
• Step 3: Solve the new simultaneous equations and find the values of the new variables.
• Step 4: By resubstituting the value(s) of the new variables, find the replaced variables which are to be determined.

Question 3.
To solve given equations fill the below boxes suitably. (Text book pg.no. 19)
Answer:

Question 4.
The examples on textbook pg. no. 17 and 18 obtained by transformation are solved by elimination method. If you solve these equations by graphical method and by Cramer’s rule will you get the same answers? Solve and check it. (Textbook pg. no. 18)

Solution:

The two lines intersect at point (1,-1).
∴ p = 1 and q = -1 is the solution of the simultaneous equations 4p + q = 3 and 2p – 3q = 5.
Re substituting the values of p and q, we get

The two lines intersect at point (0, -1).
∴ x = 0 and y = -1 is the solution of the simultaneous equations x – y = 1 and x + y = -1.
∴ (x, y) = (0, -1) is the solution of the given simultaneous equations.

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 3.2 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 3 Arithmetic Progression.

Practice Set 3.2 Algebra 10th Std Maths Part 1 Answers Chapter 3 Arithmetic Progression

Question 1.
Write the correct number in the given boxes from the following A.P.

Solution:

Question 2.
Decide whether following sequence is an A.P., if so find the 20^th term of the progression.
-12, -5, 2, 9,16, 23,30,…
Solution:
i. The given sequence is
-12, -5,2, 9, 16, 23,30,…
Here, t₁ = -1₂, t₂ = -5, t₃ = 2, t₄ = 9
∴ t₂ – t₁ – 5 – (-12) – 5 + 12 = 7
t₃ – t₂ = 2 – (-5) = 2 + 5 = 7
∴ t₄ – t₃ – 9 – 2 = 7
∴ t₂ – t₁ = t₃ – t₂ = … = 7 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P.

ii. t_n = a + (n – 1)d
∴ t₂₀ = -12 + (20 – 1)7 …[∵a = -12, d = 7]
= -12 + 19 × 7
= -12 + 133
∴ t₂₀ = 121
∴ 20^th term of the given A.P. is 121.

Question 3.
Given Arithmetic Progression is 12, 16, 20, 24, … Find the 24^th term of this progression.
Solution:
The given A.P. is 12, 16, 20, 24,…
Here, a = 12, d = 16 – 12 = 4 Since,
t_n = a + (n – 1)d
∴ t₂₄ = 12 + (24 – 1)4
= 12 + 23 × 4
= 12 + 92
∴ t₂₄ = 104
∴ 24th term of the given A.P. is 104.

Question 4.
Find the 19th term of the following A.P. 7,13,19,25…..
Solution:
The given A.P. is 7, 13, 19, 25,…
Here, a = 7, d = 13 – 7 = 6
Since, t_n = a + (n – 1)d
∴ t₁₉ = 7 + (19 – 1)6
= 7 + 18 × 6
= 7 + 108
∴ t₁₉ = 115
∴ 19^th term of the given A.P. is 115.

Question 5.
Find the 27^th term of the following A.P. 9,4,-1,-6,-11,…
Solution:
The given A.P. is 9, 4, -1, -6, -11,…
Here, a = 9, d = 4- 9 = -5
Since, t_n = a + (n – 1)d
∴ t₂₇ = 9 + (27 – 1)(-5)
= 9 + 26 × (-5)
= 9 – 130
∴ t₂₇ = -121
∴ 27^th term of the given A.P. is -121.

Question 6.
Find how many three digit natural numbers are divisible by 5.
Solution:
The three digit natural numbers divisible by
5 are 100, 105, 110, …,995
The above sequence is an A.P.
∴ a = 100, d = 105 – 100 = 5
Let the number of terms in the A.P. be n.
Then, t_n = 995
Since, t_n = a + (n – 1)d
∴ 995 = 100 +(n – 1)5
∴ 995 – 100 = (n – 1)5
∴ 895 = (n – 1)5
∴ n – 1 = \(\frac { 895 }{ 5 } \)
∴ n – 1 = 179
∴ n = 179 + 1 = 180
∴ There are 180 three digit natural numbers which are divisible by 5.

Question 7.
The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41st term of that A.P.
Solution:
Bor an A.P., let a be the first term and d be the common difference,
t₁₁ = 16, t₂₁ = 29 …[Given]
t_n = a + (n – 1)d
∴ t₁₁, = a + (11 – 1)d
∴ 16 = a + 10d
i.e. a + 10d = 16 …(i)
Also, t₂₁ = a + (21 – 1)d
∴ 29 = a + 20d
i.e. a + 20d = 29 …(ii)
Subtracting equation (i) from (ii), we get a

Question 8.
8. 11, 8, 5, 2, … In this A.P. which term is number-151?
Solution:
The given A.P. is 11, 8, 5, 2,…
Here, a = 11, d = 8 – 11 = -3
Let the n^th term of the given A.P. be -151.
Then, t_n = – 151
Since, t_n = a + (n – 1)d
∴ -151= 11 + (n – 1)(-3)
∴ -151 – 11 =(n – 1)(-3)
∴ -162 = (n – 1)(-3)
∴ n – 1 = \(\frac { -162 }{ -3 } \)
∴ n – 1 = 54
∴ n = 54 + 1 = 55
∴ 55th term of the given A.P. is -151.

Question 9.
In the natural numbers from 10 to 250, how many are divisible by 4?
Solution:
The natural numbers from 10 to 250 divisible
by 4 are 12, 16, 20, …,248
The above sequence is an A.P.
∴ a = 12, d = 16 – 12 = 4
Let the number of terms in the A.P. be n.
Then, t_n = 248
Since, t_n = a + (n – 1)d
∴ 248 = 12 + (n – 1)4
∴ 248 – 12 = (n – 1)4
∴ 236 = (n – 1)4
∴ n – 1 = \(\frac { 236 }{ 4 } \)
∴ n – 1 = 59
∴ n = 59 + 1 = 60
∴ There are 60 natural numbers from 10 to 250 which are divisible by 4.

Question 10.
In an A.P. 17^th term is 7 more than its 10th term. Find the common difference.
Solution:
For an A.P., let a be the first term and d be the common difference.
According to the given condition,
t₁₇ = t₁₀ + 7
∴ a + (17 – 1)d = a + (10 – 1)d + 7 …[∵ t_n = a + (n – 1)d]
∴ a + 16d = a + 9d + 7
∴ a + 16d – a – 9d = 7
∴ 7d = 7
∴ d = \(\frac { 7 }{ 7 } \) = 1
∴ The common difference is 1.

Question 1.
Kabir’s mother keeps a record of his height on each birthday. When he was one year old, his height was 70 cm, at 2 years he was 80 cm tall and 3 years he was 90 cm tall. His aunt Meera was studying in the 10th class. She said, “it seems like Kabir’s height grows in Arithmetic Progression”. Assuming this, she calculated how tall Kabir will be at the age of 15 years when he is in 10th! She was shocked to find it. You too assume that Kabir grows in A.P. and find out his height at the age of 15 years. (Textbook pg. no. 63)
Solution:
Height of Kabir when he was 1 year old = 70 cm Height of Kabir when he was 2 years old = 80 cm
Height of Kabir when he was 3 years old = 90 cm The heights of Kabir form an A.P.
Here, a = 70, d = 80 – 70 = 10
We have to find height of Kabir at the age of 15years i.e. t₁₅.
Now, tn = a + (n – 1)d
∴ t₁₅ = 70 + (15 – 1)10
= 70 + 14 × 10 = 70 + 140
∴ t₁₅ = 210
∴ The height of Kabir at the age of 15 years will be 210 cm.

Question 2.
Is 5, 8, 11, 14, …. an A.P.? If so then what will be the 100^th term? Check whether 92 is in this A.P.? Is number 61 in this A.P.? (Textbook pg. no, 62)
Solution:
i. The given sequence is
5, 8,11,14,…
Here, t₁ = 5, t₂ = 8, t₃ = 11, t₄ = 14
∴ t₂ – t₁ = 8 – 5 = 3
t₃ – t₂ = 11 – 8 = 3
t₄ – t₃ = 14 – 11 = 3
∴ t₂ – t₁ = t₃ – t₂ = t₄ – t₃ = 3 = d = constant
The difference between two consecutive terms is constant
∴ The given sequence is an A.P.

ii. t_n = a + (n – 1)d
∴ t₁₀₀ = 5 + (100 – 1)3 …[∵ a = 5, d = 3]
= 5 + 99 × 3
= 5 + 297
∴ t₁₀₀ = 302
∴ 100^th term of the given A.P. is 302.

iii. To check whether 92 is in given A.P., let t_n = 92
∴ t_n = a + (n – 1)d
∴ 92 = 5 + (n – 1)3
∴ 92 = 5 + 3n – 3
∴ 92 = 2 + 3n
∴ 90 = 3n
∴ n = \(\frac { 90 }{ 3 } \) = 30
∴ 92 is the 30th term of given A.P.

iv. To check whether 61 is in given A.P., let t_n = 61
61 = 5 + (n – 1)3
∴ 61 = 5 + 3n – 3
∴ 61 = 2 + 3n
∴ 61 – 2 = 3n
∴ 59 = 3n
∴ n = \(\frac { 59 }{ 3 } \)
But, n is natural number 59
∴ n ≠ \(\frac { 59 }{ 3 } \)
∴ 61 is not in given A.P.

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Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 4.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 4 Geometric Constructions.

Practice Set 4.2 Geometry 10th Std Maths Part 2 Answers Chapter 4 Geometric Constructions

Question 1.
Construct a tangent to a circle with centre P and radius 3.2 cm at any point M on it.
Solution:
Analysis:
seg PM ⊥ line l ….[Tangent is perpendicular to radius]

The perpendicular to seg PM at point M will give the required tangent at M.

Question 2.
Draw a circle of radius 2.7 cm. Draw a tangent to the circle at any point on it.
Solution:
Analysis:
seg OM ⊥ line l …[Tangent is perpendicular to radius]

The perpendicular to seg OM at point M will give the required tangent at M.

Question 3.
Draw a circle of radius 3.6 cm. Draw a tangent to the circle at any point on it without using the centre.
Solution:
Analysis:
As shown in the figure, line lis a tangent to the circle at point K.
seg BK is a chord of the circle and LBAK is an inscribed angle.
By tangent secant angle theorem,
∠BAK = ∠BKR

By converse of tangent secant angle theorem,
If we draw ∠BKR such that ∠BKR = ∠BAK, then ray KR
i.e. (line l) is a tangent at point K.

Question 4.
Draw a circle of radius 3.3 cm. Draw a chord PQ of length 6.6 cm. Draw tangents to the circle at points P and Q. Write your observation about the tangents.
Solution:
Analysis:
seg OP ⊥ line l …[Tangent is perpendicular to radius]
seg OQ ⊥ line m

The perpendicular to seg OP and seg OQ at points P and Q
respectively will give the required tangents at P and Q.

Radius = 3.3 cm
∴ Diameter = 2 × 3.3 = 6.6 cm
∴ Chord PQ is the diameter of the circle.
∴ The tangents through points P and Q (endpoints of diameter) are parallel to each other.

Question 5.
Draw a circle with radius 3.4 cm. Draw a chord MN of length 5.7 cm in it. Construct tangents at points M and N to the circle.
Solution:
Analysis:
seg ON ⊥ linel l
seg OM ⊥ Iine m …….[Tangent is perpendicular to radius]

The perpendicular to seg ON and seg 0M at points N and M respectively will give the required tangents at N and M.

Question 6.
Draw a circle with centre P and radius 3.4 cm. Take point Q at a distance 5.5 cm from the centre. Construct tangents to the circle from point Q.
Solution:
Analysis:
As shown in the figure, let Q be a point in the exterior of circle at a distance of 5.5 cm.
Let QR and QS be the tangents to the circle at points R and S respectively.

∴ seg PR ⊥ tangent QR …[Tangent is perpendicular to radius]
∴ ∠PRQ = 90°
∴ point R is on the circle having PQ as diameter. …[Angle inscribed in a semicircle is a right angle]
Similarly, point S also lies on the circle having PQ as diameter.
∴ Points R and S lie on the circle with PQ as diameter.
On drawing a circle with PQ as diameter, the points where it intersects the circle with centre P, will be the positions of points R and S respectively.
Ray QR and QS are the required tangents to the circle from point Q.

Question 7.
Draw a circle with radius 4.1 cm. Construct tangents to the circle from a point at a distance 7.3 cm from the centre.
Solution:
Analysis:
As shown in the figure, let Q be a point in the exterior of circle at a distance of 5.5 cm.
Let QR and QS be the tangents to the circle at points R and S respectively.
∴ seg PR ⊥ tangent QR …[Tangent is perpendicular to radius]
∴ ∠PRQ = 90°

∴ point R is on the circle having PQ as diameter. …[Angle inscribed in a semicircle is a right angle]
Similarly, point S also lies on the circle having PQ as diameter.
∴ Points R and S lie on the circle with PQ as diameter.
On drawing a circle with PQ as diameter, the points where it intersects the circle with centre P, will be the positions of points R and S respectively.
Ray QR and QS are the required tangents to the circle from point Q.

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 4.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 4 Geometric Constructions.

Practice Set 4.1 Geometry 10th Std Maths Part 2 Answers Chapter 4 Geometric Constructions

Question 1.
∆ABC ~ ∆LMN. In ∆ABC, AB = 5.5 cm, BC = 6 cm, CA = 4.5 cm. Construct ∆ABC and ∆LMN such that \(\frac { BC }{ MN } \) = \(\frac { 5 }{ 4 } \)
Solution:
Analysis:

Question 2.
∆PQR ~ ∆LTR. In ∆PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct ∆PQR and ∆LTR, such that \(\frac { PQ }{ LT } \) = \(\frac { 3 }{ 4 } \)
Solution:
Analysis:
As shown in the figure, Let R – P – L and R – Q – T.
∆PQR ~ ∆LTR … [Given]
∴ ∠PRQ ≅ ∠LRT … [Corresponding angles of similar triangles]

\(\frac { PQ }{ LT } \) = \(\frac { QR }{ TR } \) = \(\frac { PR }{ LR } \) …(i)[Corresponding sides of similar triangles]
But, \(\frac { PQ }{ LT } \) = \(\frac { 3 }{ 4 } \) ….(ii) [Given]
∴ \(\frac { PQ }{ LT } \) = \(\frac { QR }{ TR } \) = \(\frac { PR }{ LR } \) = \(\frac { 3 }{ 4 } \) …[From (i) and (ii)]
∴ sides of LTR are longer than corresponding sides of ∆PQR.
If seg QR is divided into 3 equal parts, then seg TR will be 4 times each part of seg QR.
So, if we construct ∆PQR, point T will be on side RQ, at a distance equal to 4 parts from R.
Now, point L is the point of intersection of ray RP and a line through T, parallel to PQ.
∆LTR is the required triangle similar to ∆PQR.

Steps of construction:
i. Draw ∆PQR of given measure. Draw ray RS making an acute angle with side RQ.
ii. Taking convenient distance on the compass, mark 4 points R₁, R₂, R₃, and R₄, such that RR₁ = R₁R₂ = R₂R₃ = R₃R₄.
iii. Join R₃Q. Draw line parallel to R₃Q through R₄ to intersects ray RQ at T.
iv. Draw a line parallel to side PQ through T. Name the point of intersection of this line and ray RP as L.
∆LTR is the required triangle similar to ∆PQR.

Question 3.
∆RST ~ ∆XYZ. In ∆RST, RS = 4.5 cm, ∠RST = 40°, ST = 5.7 cm. Construct ∆RST and ∆XYZ, such that \(\frac { RS }{ XY } \) = \(\frac { 3 }{ 5 } \).
Solution:
Analysis:
∆RST ~ ∆XYZ … [Given]
∴ ∠RST ≅ ∠XYZ = 40° … [Corresponding angles of similar triangles]

Question 4.
∆AMT ~ ∆ANE. In ∆AMT, AM = 6.3 cm, ∠TAM = 500, AT = 5.6 cm. \(\frac { AM }{ AH } \) = \(\frac { 7 }{ 5 } \) Construct ∆AHE.
Solution:
Analysis:
As shown in the figure,
Let A – H – M and A – E – T.
∆AMT ~ ∆AHE … [Given]
∴ ∠TAM ≅ ∠EAH … [Corresponding angles of similar triangles]

\(\frac { AM }{ AH } \) = \(\frac { MT }{ HE } \) = \(\frac { AT }{ AE } \) ….. (i)[Corresponding sides of similar triangles]
But, \(\frac { AM }{ AH } \) = \(\frac { 7 }{ 5 } \) …(ii)[Given]
∴ \(\frac { AM }{ AH } \) = \(\frac { MT }{ HE } \) = \(\frac { AT }{ AH } \) = \(\frac { 7 }{ 5 } \) …[From (i) and (ii)]
∴ Sides of AAMT are longer than corresponding sides of ∆AHE.
∴ The length of side AH will be equal to 5 parts out of 7 equal parts of side AM.
So, if we construct AAMT, point H will be on side AM, at a distance equal to 5 parts from A.
Now, point E is the point of intersection of ray AT and a line through H, parallel to MT.
∆AHE is the required triangle similar to ∆AMT.

Steps of construction:
i. Draw ∆AMT of given measure. Draw ray AB making an acute angle with side AM.
ii. Taking convenient distance on the compass, mark 7 points A1, A2, A3, A4, A5, Ag and A7, such that
AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇.
iii. Join A₇M. Draw line parallel to A₇M through A₅ to intersects seg AM at H.
iv. Draw a line parallel to side TM through H. Name the point of intersection of this line and seg AT as E.
∆AHE is the required triangle similar to ∆AMT.

Maharashtra Board Class 10 Maths Chapter 4 Geometric Constructions Intext Questions and Activities

Question 1.
If length of side AB is \(\frac { 11.6 }{ 2 } \) cm, then by dividing the line segment of length 11.6 cm in three equal parts, draw segment AB. (Textbook pg. no. 93)
Solution:

Steps of construction:
i. Draw seg AD of 11.6 cm.
ii. Draw ray AX such that ∠DAX is an acute angle.
iii. Locate points A₁, A₂ and A₃ on ray AX such that AA₁ = A₁A₂ = A₂A₃
iv. Join A₃D.
v. Through A₁, A₂ draw lines parallel to A₃D intersecting AD at B and C, wherein
AB = \(\frac { 11.6 }{ 3 } \) cm

Question 2.
Construct any ∆ABC. Construct ∆ A’BC’ such that AB : A’B = 5:3 and ∆ ABC ~ ∆ A’BC’. (Textbook pg. no. 93)
Analysis:
As shown in the figure,
Let B – A’ – A and B – C’ -C
∆ ABC – A’BC’ … [Given]
∴ ∠ABC ≅ ∠A’BC’ …[Corresponding angles of similar trianglesi


∴ Sides of ∆ABC are longer than corresponding sides of ∆A’BC’. Rough figure
∴ the length of side BC’ will be equal to 3 parts out of 5 equal parts of side BC.
So if we construct ∆ABC, point C’ will be on side BC, at a distance equal to 3 parts from B.
Now A’ is the point of intersection of AB and a line through C’, parallel to CA.
Solution:
Let ∆ABC be any triangle constructed such that AB = 7 cm, BC = 7 cm and AC = 6 cm.

Question 3.
Construct any ∆ABC. Construct ∆A’BC’ such that AB: A’B = 5:3 and ∆ABC ~ ∆A’BC’.
∆A’BC’ can also be constructed as shown in the adjoining figure. What changes do we have to make in steps of construction in that case? (Textbook pg. no. 94)

Solution:
Let ∆ABC be any triangle constructed such that AB = 5cm,
BC = 5.5 cm and AC = 6 cm.

i. Steps of construction:
Construct ∆ABC, extend rays AB and CB.
Draw line BM making an acute angle with side AB.
Mark 5 points B₁, B₂, B₃, B₄, B₅ starting from B at equal distance.
Join B₃C” (ie 3rd part)
Draw a line parallel to AB₅ through B₃ to intersect line AB at C”
Draw a line parallel to AC through C” to intersect line BC at A”
ii. Extra construction:
With radius BC” cut an arc on extended ray CB at C’ [C’ – B – C]
With radius BA” cut an arc on extended ray AB at A’ [A’ – B – A]
∆A’BC’ is the required triangle.