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10th Standard Maths 2 Practice Set 7.2 Chapter 7 Mensuration Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 7.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 7 Mensuration.

Class 10 Maths Part 2 Practice Set 7.2 Chapter 7 Mensuration Questions With Answers Maharashtra Board

Question 1.
The radii of two circular ends of frustum shaped bucket are 14 cm and 7 cm. Height of the bucket is 30 cm. How many litres of water it can hold? (1 litre = 1000 cm³)
Given: Radii (r₁) = 14 cm, and (r₂) = 7 cm,
height (h) = 30 cm
To find: Amount of water the bucket can hold.
Solution:
Volume of frustum = \(\frac { 1 }{ 3 } \) πh (r₁² + r₂² + r₁ × r₂)

∴ The bucket can hold 10.78 litres of water.

Question 2.
The radii of ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its
i. curved surface area,
ii. total surface area,
iii. volume, (π = 3.14)
Given: Radii (r₁) = 14 cm, and (r₂) = 6 cm,
height (h) = 6 cm
Solution:

i. Curved surface area of frustum
= πl (r₁ + r₂)
= 3.14 × 10(14 + 6)
= 3.14 × 10 × 20 = 628 cm²
∴ The curved surface area of the frustum is 628 cm².

ii. Total surface area of frustum
= πl (r₁+ r₂) + πr₁² + πr₂²
= 628 + 3.14 × (14)² + 3.14 × (6)²
= 628 + 3.14 × 196 + 3.14 × 36
= 628 + 3.14(196 + 36)
= 628 + 3.14 × 232
= 628 + 728.48
= 1356.48 cm²
∴ The total surface area of the frustum is 1356.48 cm².

iii. Volume of frustum
= \(\frac { 1 }{ 3 } \) πth(r₁² +r₂² + r₁ × r₂)
= \(\frac { 1 }{ 3 } \) × 3.14 × 6(14² + 6² + 14 × 6)
= 3.14 × 2(196 + 36 + 84)
= 3.14 × 2 × 316
= 1984.48 cm³
∴ The volume of the frustum is 1984.48 cm³.

Question 3.
The circumferences of circular faces of a frustum are 132 cm and 88 cm and its height is 24 cm. To find the curved surface area of frustum, complete the following activity. (π = \(\frac { 22 }{ 7 } \))
Solution:
Circumference1 = 27πr₁ = 132 cm

Curved surface area of frustum = π (r₁ + r₂) l
= π (21 + 14) × 25
=π × 35 × 35
= \(\frac { 22 }{ 7 } \) × 35 × 25
= 2750 cm²

Maharashtra State Board Class 10 Maths Solutions Part 2

• Co-ordinate Geometry Practice Set 5.1 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.2 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.3 Class 10 Maths Solutions
• Co-ordinate Geometry Problem Set 5 Class 10 Maths Solutions
• Trigonometry Practice Set 6.1 Class 10 Maths Solutions
• Trigonometry Practice Set 6.2 Class 10 Maths Solutions
• Trigonometry Problem Set 6 Class 10 Maths Solutions
• Mensuration Practice Set 7.1 Class 10 Maths Solutions
• Mensuration Practice Set 7.2 Class 10 Maths Solutions
• Mensuration Practice Set 7.3 Class 10 Maths Solutions
• Mensuration Practice Set 7.4 Class 10 Maths Solutions
• Mensuration Problem Set 7 Class 10 Maths Solutions

10th Standard Maths 2 Practice Set 7.1 Chapter 7 Mensuration Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 7.1  Geometry 10th Class Maths Part 2 Answers Solutions Chapter 7 Mensuration.

Class 10 Maths Part 2 Practice Set 7.1 Chapter 7  Mensuration Questions With Answers Maharashtra Board

Practice Set 7.1 Geometry 10th Question 1. Find the volume of a cone if the radius of its base is 1.5 cm and its perpendicular height is 5 cm.
Given: For the cone,
radius (r) = 1.5 cm,
perpendicular height (h) = 5 cm
To find: Volume of the cone.
Solution:
Volume of cone = \(\frac { 1 }{ 3 } \) πr²h

∴ The volume of the cone is 11.79 cm³.

Mensuration Practice Set 7.1 Question 2. Find the volume of a sphere of diameter 6 cm. [π = 3.14]
Given: For the sphere, diameter (d) = 6 cm
To find: Volume of the sphere.
Solution:
Radius (r) = \(\frac { d }{ 2 } \) = \(\frac { 6 }{ 2 } \) = 3 cm
Volume of sphere = \(\frac { 4 }{ 3 } \) πr²
= \(\frac { 4 }{ 3 } \) × 3.14 × (3)³
= 4 × 3.14 × 3 × 3
= 113.04 cm³
∴ The volume of the sphere is 113.04 cm³.

Practice Set 7.1 Geometry Class 10 Question 3. Find the total surface area of a cylinder if the radius of its base is 5 cm and height is 40 cm. [π = 3.14]
Given: For the cylinder,
radius (r) = 5 cm,
height (h) = 40 cm
To find: Total surface area of the cylinder.
Solution:
Total surface area of cylinder = 2πr (r + h)
= 2 × 3.14 × 5 (5 + 40)
= 2 × 3.14 × 5 × 45
= 1413 cm²
The total surface area of the cylinder is 1413 cm².

Practice Set 7.1 Geometry Question 4. Find the surface area of a sphere of radius 7 cm.
Given: For the sphere, radius (r) = 7 cm
To find: Surface area of the sphere.
Solution:
Surface area of sphere = Aπr²
= 4 × \(\frac { 22 }{ 7 } \) × (7)²
= 88 × 7
= 616 cm²
∴ The surface area of the sphere is 616 cm².

Practice Set 7.1 Question 5. The dimensions of a cuboid are 44 cm, 21 cm, 12 cm. It is melted and a cone of height 24 cm is made. Find the radius of its base.
Given: For the cuboid,
length (l) = 44 cm, breadth (b) = 21 cm,
height (h) = 12 cm
For the cone, height (H) = 24 cm
To find: Radius of base of the cone (r).
Solution:
Volume of cuboid = l × b × h
= 44 × 21 × 12 cm³
Volume of cone = \(\frac { 1 }{ 3 } \) πr²H
= \(\frac { 1 }{ 3 } \) × \(\frac { 22 }{ 7 } \) × r² × 24 cm³
Since the cuboid is melted to form a cone,
∴ volume of cuboid = volume of cone

∴ r² = 21 × 21
∴ r = 21 cm …[Taking square root of both sides]
∴ The radius of the base of the cone is 21 cm.

10th Class Geometry Practice Set 7.1 Question 6. Observe the measures of pots in the given figures. How many jugs of water can the cylindrical pot hold?

Given: For the conical water jug,
radius (r) = 3.5 cm, height (h) = 10 cm
For the cylindrical water pot,
radius (R) = 7 cm, height (H) = 10 cm
To find: Number of jugs of water the cylindrical pot can hold.
Solution:
Volume of conical jug = \(\frac { 1 }{ 3 } \) πr²h

∴ The cylindrical pot can hold 12 jugs of water.

Mensuration Class 10 Practice Set 7.1 Question 7. A cylinder and a cone have equal bases. The height of the cylinder is 3 cm and the area of its base is 100 cm². The cone is placed up on the cylinder. Volume of the solid figure so formed is 500 cm³. Find the total height of the figure

Given: For the cylindrical part,
height (h) = 3 cm,
area of the base (πr²)= 100 cm²
Volume of the entire figure = 500 cm³
To find: Total height of the figure.
Solution:
A cylinder and a cone have equal bases.
∴ they have equal radii.
radius of cylinder = radius of cone = r
Area of base = 100 cm²
∴ πr² =100 …(i)
Let the height of the conical part be H.
Volume of the entire figure
= Volume of the entire + Volume of cone

∴ Height of conical part (H) =6 cm
Total height of the figure = h + H
= 3 + 6
= 9 cm
∴ The total height of the figure is 9 cm.

10th Geometry Practice Set 7.1 Question 8. In the given figure, a toy made from a hemisphere, a cylinder and a cone is shown. Find the total area of the toy.

Given: For the conical Part,
height (h) = 4 cm, radius (r) = 3 cm
For the cylindrical part,
height (H) = 40 cm, radius (r) = 3 cm
For the hemispherical part,
radius (r) = 3 cm
To find: Total area of the toy.
Solution:
Slant height of cone (l) = \(\sqrt{\mathrm{h}^{2}+\mathrm{r}^{2}}\)
= \(\sqrt{\mathrm{4}^{2}+\mathrm{3}^{2}}\)
= \(\sqrt { 16+9 }\)
= \(\sqrt { 25 }\) = 5 cm
Curved surface area of cone = πrl
= π × 3 × 5
= 15π cm²
Curved surface area of cylinder = 2πrH
= 2 × π × 3 × 40
= 240π cm²
Curved surface area of hemisphere = 2πr²
= 2 × π × 3²
= 18π cm²
Total area of the toy
= Curved surface area of cone + Curved surface area of cylinder + Curved surface area of hemisphere
= 15π + 240π + 18π
= 2737 π cm²
∴ The total area of the toy is 273π cm².

7.1.8 Practice Question 9. In the given figure, a cylindrical wrapper of flat tablets is shown. The radius of a tablet is 7 mm and its thickness is 5 mm. How many such tablets are wrapped in the wrapper?

Given: For the cylindrical tablets,
radius (r) = 7 mm,
thickness = height(h) = 5 mm
For the cylindrical wrapper,
diameter (D) = 14 mm, height (H) = 10 cm
To find: Number of tablets that can be wrapped.
Solution:
Radius of wrapper (R) = \(\frac { Diameter }{ 2 } \)
= \(\frac { 14 }{ 2 } \) = 7 mm
Height of wrapper (H) = 10 cm
= 10 × 10 mm
= 100 mm
Volume of a cylindrical wrapper = πR²H
= π(7)² × 100
= 4900π mm³
Volume of a cylindrical tablet = πr²h
= π(7)² × 5
= 245 π mm³
No. of tablets that can be wrapped

∴ 20 tables can be wrapped in the wrapper

Class 10 Maths 7.1 Question 10. The given figure shows a toy. Its lower part is a hemisphere and the upper part is a cone. Find the volume and the surface area of the toy from the measures shown in the figure.
(π = 3.14)
Given: For the conical part,
height (h) = 4 cm, radius (r) = 3 cm
For the hemispherical part,
radius (r) = 3 cm
To find: Volume and surface area of the toy.
Solution:

Now, volume of the toy
= Volume of cone + volume of hemisphere
= 12π + 18π
= 30π
= 30 × 3.14
= 94.20 cm³
Also, surface area of the toy
= Curved surface area of cone + Curved surface area of hemisphere
= 15π + 18π
= 33π
= 33 × 3.14
= 103.62 cm²
∴ The volume and surface area of the toy are 94.20 cm³ and 103.62 cm² respectively.

Question 11.
Find the surface area and the volume of a beach ball shown in the figure.

Given: For the spherical ball,
diameter (d) = 42 cm
To find: Surface area and volume of the beach ball.
Solution:
Radius (r) = \(\frac { d }{ 2 } \) = \(\frac { 42 }{ 2 } \) = 21 cm
Surface area of sphere= 4πr²
= 4 × 3.14 × (21)²
= 4 × 3.14 × 21 × 21
= 5538.96 cm²
Volume of sphere = \(\frac { 4 }{ 3 } \) πr³
= \(\frac { 4 }{ 3 } \) × 3.14 × (21)³
= 4 × 3.14 × 7 × 21 × 21
= 38772.72 cm³
∴ The surface area and the volume of the beach ball are 5538.96 cm² and 38772.72 cm³ respectively.

Question 12.
As shown in the figure, a cylindrical glass contains water. A metal sphere of diameter 2 cm is immersed in it. Find the volume of the water.

Given: For the metal sphere,
diameter (d) = 2 cm
For the cylindrical glass, diameter (D) =14 cm,
height of water in the glass (H) = 30 cm
To find: Volume of water in the glass.
Solution:
Let the radii of the sphere and glass be r and R respectively.

Volume of water with sphere in it = πR²H
= π × (7)² × 30
= 1470π cm³
Volume of water in the glass
= Volume of water with sphere in it – Volume of sphere

∴ The volume of the water in the glass is 1468.67 π cm³ (i.e. 4615.80 cm³).

Maharashtra Board Class 10 Maths Chapter 7 Mensuration Intext Questions and Activities

Question 1.
The length, breadth and height of an oil can are 20 cm, 20 cm and 30 cm respectively as shown in the adjacent figure. How much oil will it contain? (1 litre = 1000 cm³) (Textbook pg. no.141)

Given: For the cuboidal can,
length (l) = 20 cm,
breadth (b) = 20 cm,
height (h) = 30 cm
To find: Oil that can be contained in the can.
Solution:
Volume of cuboid = l × b × h
= 20 × 20 × 30
= 12000 cm³
= \(\frac { 12000 }{ 1000 } \) litres
= 12 litres
∴ The oil can will contain 12 litres of oil.

Question 2.
The adjoining figure shows the measures of a Joker’s dap. How much cloth is needed to make such a cap? (Textbook pg. no. 141)

Given: For the conical cap,
radius (r) = 10 cm,
slant height (l) = 21 cm
To find: Cloth required to make the cap.
Solution:
Cloth required to make the cap
= Curved surface area of the conical cap
= πrl = \(\frac { 22 }{ 7 } \) × 10 × 21
=22 × 10 × 3
= 660 cm²
∴ 660 cm² of cloth will be required to make the cap.

Question 3.
As shown in the adjacent figure, a sphere is placed in a cylinder. It touches the top, bottom and the curved surface of the cylinder. If radius of the base of the cylinder is ‘r’,

i. what is the ratio of the radii of the sphere and the cylinder ?
ii. what is the ratio of the curved surface area of the cylinder and the surface area of the sphere?
iii. what is the ratio of the volumes of the cylinder and the sphere? (Textbook pg. no. 141)
Solution:
Radius of base of cylinder = radius of sphere
∴ Radius of sphere = r
Also, height of cylinder = diameter of sphere
∴ h = d
∴ h = 2r …(i)

∴ radius of sphere : radius of cylinder = 1 : 1.

∴ curved surface area of cylinder : surface area of sphere = 1:1.

∴ volume of cylinder : volume of sphere = 3 : 2.

Question 4.
Finding volume of a sphere using cylindrical beaker and water. (Textbook, pg. no. 142)

i. Take a ball and a beaker of the same radius.
ii. Cut a strip of paper of length equal to the diameter of the beaker.
iii. Draw two lines on the strip dividing it into three equal parts.
iv. Stick this strip on the beaker straight up from the bottom.
v. Fill the water in the beaker upto the first mark of the strip from bottom.
vi. Push the ball in the beaker so that it touches the bottom.
Observe how much water level rises.
You will notice that the water level has risen exactly upto the total height of the strip. Try to obtain the formula for volume of sphere using the volume of the cylindrical beaker.
Solution:
Suppose volume of beaker upto height 2r is V.
V = πr² h
∴ V = πr²(2r) …[∵ h = 2r]
∴ V = 2πr³
But, V = volume of the ball + volume of water in the beaker
∴ 2πr³ = Volume of the ball + \(\frac { 1 }{ 3 } \) × 2πr³
∴ Volume of the ball = 2πr³ – \(\frac { 2 }{ 3 } \) πr³
= \(\frac{6 \pi r^{3}-2 \pi r^{3}}{3}\)
∴ Volume of the ball = \(\frac { 4 }{ 3 } \) πr³
∴ Volume of a sphere = \(\frac { 4 }{ 3 } \) πr³

Maharashtra State Board Class 10 Maths Solutions Part 2

• Co-ordinate Geometry Practice Set 5.1 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.2 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.3 Class 10 Maths Solutions
• Co-ordinate Geometry Problem Set 5 Class 10 Maths Solutions
• Trigonometry Practice Set 6.1 Class 10 Maths Solutions
• Trigonometry Practice Set 6.2 Class 10 Maths Solutions
• Trigonometry Problem Set 6 Class 10 Maths Solutions
• Mensuration Practice Set 7.1 Class 10 Maths Solutions
• Mensuration Practice Set 7.2 Class 10 Maths Solutions
• Mensuration Practice Set 7.3 Class 10 Maths Solutions
• Mensuration Practice Set 7.4 Class 10 Maths Solutions
• Mensuration Problem Set 7 Class 10 Maths Solutions

10th Standard Maths 2 Practice Set 5.3 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.3 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

Class 10 Maths Part 2 Practice Set 5.3 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Practice Set 5.3 Geometry Class 10 Question 1. Angles made by the line with the positive direction of X-axis are given. Find the slope of these lines.
i. 45°
ii. 60°
iii. 90°
Solution:
i. Angle made with the positive direction of
X-axis (θ) = 45°
Slope of the line (m) = tan θ
∴ m = tan 45° = 1
∴ The slope of the line is 1.

ii. Angle made with the positive direction of X-axis (θ) = 60°
Slope of the line (m) = tan θ
∴ m = tan 60° = \(\sqrt { 3 }\)
∴ The slope of the line is \(\sqrt { 3 }\).

iii. Angle made with the positive direction of
X-axis (θ) = 90°
Slope of the line (m) = tan θ
∴ m = tan 90°
But, the value of tan 90° is not defined.
∴ The slope of the line cannot be determined.

Practice Set 5.3 Geometry Question 2. Find the slopes of the lines passing through the given points.
i. A (2, 3), B (4, 7)
ii. P(-3, 1), Q (5, -2)
iii. C (5, -2), D (7, 3)
iv. L (-2, -3), M (-6, -8)
v. E (-4, -2), F (6, 3)
vi. T (0, -3), s (0,4)
Solution:
i. A (x₁, y₁) = A (2, 3) and B (x₂, y₂) = B (4, 7)
Here, x₁ = 2, x₂ = 4, y₁ = 3, y₂ = 7

∴ The slope of line AB is 2.

ii. P (x₁, y₁) = P (-3, 1) and Q (x₂, y₂) = Q (5, -2)
Here, x₁ = -3, x₂ = 5, y₁ = 1, y₂ = -2

∴ The slope of line PQ is \(\frac { -3 }{ 8 } \)

iii. C (x₁, y₁) = C (5, -2) and D (x₂, y₂) = D (7, 3)
Here, x₁ = 5, x₂ = 7, y₁ = -2, y₂ = 3

∴ The slope of line CD is \(\frac { 5 }{ 2 } \)

iv. L (x₁, y₁) = L (-2, -3) and M (x₂,y₂) = M (-6, -8)
Here, x₁ = -2, x₂ = – 6, y₁ = – 3, y₂ = – 8

∴ The slope of line LM is \(\frac { 5 }{ 4 } \)

v. E (x₁, y₁) = E (-4, -2) and F (x₂, y₂) = F (6, 3)
Here,x₁ = -4, x₂ = 6, y₁ = -2, y₂ = 3

∴ The slope of line EF is \(\frac { 1 }{ 2 } \).

vi. T (x₁, y₁) = T (0, -3) and S (x₂, y₂) = S (0, 4)
Here, x₁ = 0, x₂ = 0, y₁ = -3, y₂ = 4

∴ The slope of line TS cannot be determined.

5.3.5 Practice Question 3. Determine whether the following points are collinear.
i. A (-1, -1), B (0, 1), C (1, 3)
ii. D (- 2, -3), E (1, 0), F (2, 1)
iii. L (2, 5), M (3, 3), N (5, 1)
iv. P (2, -5), Q (1, -3), R (-2, 3)
v. R (1, -4), S (-2, 2), T (-3,4)
vi. A(-4,4),K[-2,\(\frac { 5 }{ 2 } \)], N (4,-2)
Solution:


∴ slope of line AB = slope of line BC
∴ line AB || line BC
Also, point B is common to both the lines.
∴ Both lines are the same.
∴ Points A, B and C are collinear.

∴ slope of line DE = slope of line EF
∴ line DE || line EF
Also, point E is common to both the lines.
∴ Both lines are the same.
∴ Points D, E and F are collinear.

∴ slope of line LM ≠ slope of line MN
∴ Points L, M and N are not collinear.

∴ slope of line PQ = slope of line QR
∴ line PQ || line QR
Also, point Q is common to both the lines.
∴ Both lines are the same.
∴ Points P, Q and R are collinear.

∴ slope of line RS = slope of line ST
∴ line RS || line ST
Also, point S is common to both the lines.
∴ Both lines are the same.
∴ Points R, S and T are collinear.

∴ slope of line AK = slope of line KN
∴ line AK || line KN
Also, point K is common to both the lines.
∴ Both lines are the same.
∴ Points A, K and N are collinear.

Practice Set 5.3 Geometry 9th Standard Question 4. If A (1, -1), B (0,4), C (-5,3) are vertices of a triangle, then find the slope of each side.
Solution:

∴ The slopes of the sides AB, BC and AC are -5, \(\frac { 1 }{ 5 } \) and \(\frac { -2 }{ 3 } \) respectively.

Geometry 5.3 Question 5. Show that A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are the vertices of a parallelogram.
Proof:


∴ Slope of side AB = Slope of side CD … [From (i) and (iii)]
∴ side AB || side CD
Slope of side BC = Slope of side AD … [From (ii) and (iv)]
∴ side BC || side AD
Both the pairs of opposite sides of ꠸ABCD are parallel.
꠸ABCD is a parallelogram.
Points A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are the vertices of a parallelogram.

Question 6.
Find k, if R (1, -1), S (-2, k) and slope of line RS is -2.
Solution:
R(x₁, y₁) = R (1, -1), S (x₂, y₂) = S (-2, k)
Here, x₁ = 1, x₂ = -2, y₁ = -1, y₂ = k

But, slope of line RS is -2. … [Given]
∴ -2 = \(\frac { k+1 }{ -3 } \)
∴ k + 1 = 6
∴ k = 6 – 1
∴ k = 5

5.3 Class 10 Question 7. Find k, if B (k, -5), C (1, 2) and slope of the line is 7.
Solution:
B(x₁, y₁) = B (k, -5), C (x₂, y₂) = C (1, 2)
Here, x₁ = k, x₂ = 1, y₁ = -5, y₂ = 2

But, slope of line BC is 7. …[Given]
∴ 7 = \(\frac { 7 }{ 1-k } \)
∴ 7(1 – k) = 7
∴ 1 – k = \(\frac { 7 }{ 7 } \)
∴ 1 – k = 1
∴ k = 0

Question 8.
Find k, if PQ || RS and P (2, 4), Q (3, 6), R (3,1), S (5, k).
Solution:

But, line PQ || line RS … [Given]
∴ Slope of line PQ = Slope of line RS
∴ 2 = \(\frac { k-1 }{ 2 } \)
∴ 4 = k – 1
∴ k = 4 + 1
∴ k = 5

Maharashtra State Board Class 10 Maths Solutions Part 2

• Co-ordinate Geometry Practice Set 5.1 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.2 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.3 Class 10 Maths Solutions
• Co-ordinate Geometry Problem Set 5 Class 10 Maths Solutions
• Trigonometry Practice Set 6.1 Class 10 Maths Solutions
• Trigonometry Practice Set 6.2 Class 10 Maths Solutions
• Trigonometry Problem Set 6 Class 10 Maths Solutions
• Mensuration Practice Set 7.1 Class 10 Maths Solutions
• Mensuration Practice Set 7.2 Class 10 Maths Solutions
• Mensuration Practice Set 7.3 Class 10 Maths Solutions
• Mensuration Practice Set 7.4 Class 10 Maths Solutions
• Mensuration Problem Set 7 Class 10 Maths Solutions

10th Standard Maths 2 Practice Set 5.2 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

Class 10 Maths Part 2 Practice Set 5.2 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Question 1.
Find the co-ordinates of point P if P divides the line segment joining the points A (-1, 7) and B (4, -3) in the ratio 2:3.
Solution:
Let the co-ordinates of point P be (x, y) and A (x₁, y₁) B (x₂, y₂) be the given points.
Here, x₁ = -1, y₁ = 7, x₂ = 4, y₂ = -3, m = 2, n = 3
∴ By section formula,

∴ The co-ordinates of point P are (1,3).

Question 2.
In each of the following examples find the co-ordinates of point A which divides segment PQ in the ratio a : b.
i. P (-3, 7), Q (1, -4), a : b = 2 : 1
ii. P (-2, -5), Q (4, 3), a : b = 3 : 4
iii. P (2, 6), Q (-4, 1), a : b = 1 : 2
Solution:
Let the co-ordinates of point A be (x, y).
i. Let P (x₁, y₁), Q (x₂, y₂) be the given points.
Here, x₁ = -3, y₁ = 7, x₂ = 1, y₂ = -4, a = 2, b = 1
∴ By section formula,

∴ The co-ordinates of point A are (\(\frac { -1 }{ 3 } \),\(\frac { -1 }{ 3 } \)).

ii. Let P (x₁,y₁), Q (x₂, y₂) be the given points.
Here, x₁ = -2, y₁ = -5, x₂ = 4, y₂ = 3, a = 3, b = 4
By section formula,

∴ The co-ordinates of point A are (\(\frac { 4 }{ 7 } \),\(\frac { -11 }{ 7 } \))

iii. Let P (x₁, y₁), Q (x₂, y₂) be the given points.
Here,x₁ = 2,y₁ = 6, x₂ = -4, y₂ = 1, a = 1,b = 2
∴ By section formula,

∴ The co-ordinates of point A are (0,\(\frac { 13 }{ 3 } \))

Question 3.
Find the ratio in which point T (-1, 6) divides the line segment joining the points P (-3,10) and Q (6, -8).
Solution:
Let P (x₁, y₁), Q (x₂, y₂) and T (x, y) be the given points.
Here, x₁ = -3, y₁ = 10, x₂ = 6, y₂ = -8, x = -1, y = 6
∴ By section formula,

∴ Point T divides seg PQ in the ratio 2 : 7.

Question 4.
Point P is the centre of the circle and AB is a diameter. Find the co-ordinates of point B if co-ordinates of point A and P are (2, -3) and (-2,0) respectively.
Solution:
Let A (x₁, y₁), B (x₂, y₂) and P (x, y) be the given points.
Here, x₁ = 2, y₁ =-3,
x = -2, y = 0

Point P is the midpoint of seg AB.
∴ By midpoint formula,

∴ The co-ordinates of point B are (-6,3).

Question 5.
Find the ratio in which point P (k, 7) divides the segment joining A (8, 9) and B (1,2). Also find k.
Solution:
Let A (x₁, y₁), B (x₂, y₂) and P (x, y) be the given points.
Here, x₁ = 8, y₁ = 9, x₂ = 1, y₂ = 2, x = k, y = 7
∴ By section formula,

∴ Point P divides seg AB in the ratio 2 : 5, and the value of k is 6.

Question 6.
Find the co-ordinates of midpoint of the segment joining the points (22, 20) and (0,16).
Solution:
Let A (x₁, y₁) = A (22, 20),
B (x₂,y₂) = B (0, 16)
Let the co-ordinates of the midpoint be P (x,y).
∴ By midpoint formula,

The co-ordinates of the midpoint of the segment joining (22, 20) and (0, 16) are (11,18).

Question 7.
Find the centroids of the triangles whose vertices are given below.
i. (-7, 6), (2,-2), (8, 5)
ii. (3, -5), (4, 3), (11,-4)
iii. (4, 7), (8, 4), (7, 11)
Solution:
i. Let A (x₁, y₁) = A (-7, 6),
B (x₂, y₂) = B (2, -2),
C (x₃, y₃) = C(8, 5)
∴ By centroid formula,

∴ The co-ordinates of the centroid are (1,3).

ii. Let A (x₁ y₁) = A (3, -5),
B (x₂, y₂) = B (4, 3),
C(x₃, y₃) = C(11,-4)
∴ By centroid formula,

∴ The co-ordinates of the centroid are (6, -2).

iii. Let A (x₁, y₁) = A (4, 7),
B (x₂, y₂) = B (8,4),
C (x₃, y₃) = C(7,11)
∴ By centroid formula,

∴ The co-ordinates of the centroid are (\(\frac { 19 }{ 3 } \),\(\frac { 22 }{ 3 } \))

Question 8.
In ∆ABC, G (-4, -7) is the centroid. If A (-14, -19) and B (3, 5), then find the co-ordinates of C.
Solution:
G (x, y) = G (-4, -7),
A (x₁, y₁) = A (-14, -19),
B(x₂, y₂) = B(3,5)
Let the co-ordinates of point C be (x₃, y₃).
G is the centroid.
By centroid formula,

∴ The co-ordinates of point C are (-1, – 7).

Question 9.
A (h, -6), B (2, 3) and C (-6, k) are the co-ordinates of vertices of a triangle whose centroid is G (1,5). Find h and k.
Solution:
A(x₁,y₁) = A(h, -6),
B (x₂, y₂) = B(2, 3),
C (x₃, y₃) = C (-6, k)
∴ centroid G (x, y) = G (1, 5)
G is the centroid.
By centroid formula,

∴ 3 = h – 4
∴ h = 3 + 4
∴ h = 7

∴ 15 = -3 + k
∴ k = 15 + 3
∴ k = 18
∴ h = 7 and k = 18

Question 10.
Find the co-ordinates of the points of trisection of the line segment AB with A (2,7) and B (-4, -8).
Solution:
A (2, 7), B H,-8)
Suppose the points P and Q trisect seg AB.
∴ AP = PQ = QB

∴ Point P divides seg AB in the ratio 1:2.
∴ By section formula,

Co-ordinates of P are (0, 2).
Point Q is the midpoint of PB.
By midpoint formula,

Co-ordinates of Q are (-2, -3).
∴ The co-ordinates of the points of trisection seg AB are (0,2) and (-2, -3).

Question 11.
If A (-14, -10), B (6, -2) are given, find the co-ordinates of the points which divide segment AB into four equal parts.
Solution:
Let the points C, D and E divide seg AB in four equal parts.

Point D is the midpoint of seg AB.
∴ By midpoint formula,

∴ Co-ordinates of D are (-4, -6).
Point C is the midpoint of seg AD.
∴ By midpoint formula,

∴ Co-ordinates of C are (-9, -8).
Point E is the midpoint of seg DB.
∴ By midpoint formula,


∴ Co-ordinates of E are (1,-4).
∴ The co-ordinates of the points dividing seg AB in four equal parts are C(-9, -8), D(-4, -6) and E(1, – 4).

Question 12.
If A (20, 10), B (0, 20) are given, find the co-ordinates of the points which divide segment AB into five congruent parts.
Solution:
Suppose the points C, D, E and F divide seg AB in five congruent parts.
∴ AC = CD = DE = EF = FB

∴ co-ordinates of C are (16, 12).
E is the midpoint of seg CB.
By midpoint formula,

∴ co-ordinates of E are (8, 16).
D is the midpoint of seg CE.

∴ co-ordinates of F are (4, 18).
∴ The co-ordinates of the points dividing seg AB in five congruent parts are C (16, 12), D (12, 14), E (8, 16) and F (4, 18).

Maharashtra Board Class 10 Maths Chapter 5 Co-ordinate Geometry Intext Questions and Activities

Question 1.
A (15, 5), B (9, 20) and A-P-B. Find the ratio in which point P (11, 15) divides segment AB. Find the ratio using x and y co-ordinates. Write the conclusion. (Textbook pg. no. 113)
Solution:
Suppose point P (11,15) divides segment AB in the ratio m : n.
By section formula,

∴ Point P divides seg AB in the ratio 2 : 1.
The ratio obtained by using x and y co-ordinates is the same.

Question 2.
External division: (Textbook pg. no. 115)
Suppose point R divides seg PQ externally in the ratio 3:1.

Let the common multiple be k.
Let PR = 3k and QR = k
Now, PR = PQ + QR … [P – Q – R]
∴ 3k = PQ + k
∴ \(\frac { PQ }{ QR } \) = \(\frac { 2k }{ k } \) = \(\frac { 2 }{ 1 } \)
∴ Point Q divides seg PR in the ratio 2 : 1 internally.
Thus, we can find the co-ordinates of point R, when co-ordinates of points P and Q are given.

Maharashtra State Board Class 10 Maths Solutions Part 2

• Co-ordinate Geometry Practice Set 5.1 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.2 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.3 Class 10 Maths Solutions
• Co-ordinate Geometry Problem Set 5 Class 10 Maths Solutions
• Trigonometry Practice Set 6.1 Class 10 Maths Solutions
• Trigonometry Practice Set 6.2 Class 10 Maths Solutions
• Trigonometry Problem Set 6 Class 10 Maths Solutions
• Mensuration Practice Set 7.1 Class 10 Maths Solutions
• Mensuration Practice Set 7.2 Class 10 Maths Solutions
• Mensuration Practice Set 7.3 Class 10 Maths Solutions
• Mensuration Practice Set 7.4 Class 10 Maths Solutions
• Mensuration Problem Set 7 Class 10 Maths Solutions

10th Standard Maths 2 Practice Set 6.2 Chapter 6 Trigonometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 6 Trigonometry.

Class 10 Maths Part 2 Practice Set 6.2 Chapter 6 Trigonometry Questions With Answers Maharashtra Board

Question 1.
A person is standing at a distance of 80 m from a church looking at its top. The angle of elevation is of 45°. Find the height of the church.
Solution:
Let AB represent the height of the church and point C represent the position of the person.

BC = 80 m
Angle of elevation = ∠ACB = 45°
In right angled ∆ABC,
tan 45° = \(\frac { AB }{ BC } \) … [By definition]
∴ 1 = \(\frac { AB }{ 80 } \)
∴ AB = 80m
∴ The height of the church is 80 m.

Question 2.
From the top of a lighthouse, an observer looking at a ship makes angle of depression of 60°. If the height of the lighthouse is 90 metre, then find how far the ship is from the lighthouse. ( \(\sqrt { 3 }\) = 1.73)
Solution:
Let AB represent the height of lighthouse and point C represent the position of the ship.
AB = 90 m

Angle of depression = ∠PAC = 60°
Now, ray AP || seg BC
∴ ∠ACB = ∠PAC … [Alternate angles]
∴ ∠ACB = 60°
In right angled ∆ABC,
tan 60° = \(\frac { AB }{ BC } \) … [By definition]

∴ The ship is 51.90 m away from the lighthouse.

Question 3.
Two buildings are facing each other on a road of width 12 metre. From the top of the first building, which is 10 metre high, the angle of elevation of the top of the second is found to be 60°. What is the height of the second building?
Solution:
Let AB and CD represent the heights of the two buildings, and BD represent the width of the road.

Draw seg AM ⊥ seg CD.
Angle of elevation = ∠CAM = 60°
AB = 10 m
BD= 12 m
In ꠸ABDM,
∠B = ∠D = 90°
∠M = 90° … [seg AM ⊥ seg CD]
∴ ∠A = 90° … [Remaining angle of ꠸ABDM]
∴ ꠸ABDM is a rectangle …. [Each angle is 90°]
∴ AM = BD = 12 m opposite sides
DM = AB = 10 m of a rectangle
In right angled ∆AMC,
tan 60° = \(\frac { CM }{ AM } \) …[By definition]
∴ \(\sqrt { 3 }\) = \(\frac { CM }{ 12 } \)
∴ CM = 12\(\sqrt { 3 }\) m
Now, CD = DM + CM … [C – M – D]
∴ CD = (10 + 12\(\sqrt { 3 }\))m
= 10 + 12 × 1.73
= 10 + 20.76 = 30.76
∴ The height of the second building is 30.76 m.

Question 4.
Two poles of heights 18 metre and 7 metre are erected on a ground. The length of the wire fastened at their tops is 22 metre. Find the angle made by the wire with the horizontal.
Solution:
Let AB and CD represent the heights of two poles, and AC represent the length of the wire.

Draw seg AM ⊥ seg CD.
Angle of elevation = ∠CAM = θ
AB = 7 m
CD = 18 m
AC = 22 m
In ꠸ABDM,
∠B = ∠D = 90°
∠M = 90° …[seg AM ⊥ seg CD]
∴ ∠A = 90° … [Remaining angle of ꠸ABDM]
∴ □ABDM is a rectangle. … [Each angle is 90°]
∴ DM = AB = 7 m … [Opposite sides of a rectangle]
Now, CD = CM + DM … [C – M – D]
∴ 18 = CM + 7
∴ CM = 18 – 7 = 11 m
In right angled ∆AMC,
sin θ = \(\frac { CM }{ AC } \) …..[By definition]
∴ sin θ = \(\frac { 11 }{ 22 } \) = \(\frac { 1 }{ 2 } \)
But, sin 30° = \(\frac { 1 }{ 2 } \)
∴ θ = 30°
∴ The angle made by the wire with the horizontal is 30°.

Question 5.
A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60° with the horizontal. Find the height of the tree.
Solution:
Let AB represent the height of the tree.
Suppose the tree broke at point C and its top touches the ground at D.
AC is the broken part of the tree which takes position CD such that ∠CDB = 60°
∴ AC = CD …(i)
BD = 20 m
In right angled ∆CBD,

tan 60° = \(\frac { BC }{ BD } \) … [By definition]
∴ \(\sqrt { 3 }\) = \(\frac { BC }{ 20 } \)
∴ BC = 20\(\sqrt { 3 }\) m
Also, cos 60° = \(\frac { BC }{ CD } \) … [By definition]
∴ \(\frac { 1 }{ 2 } \) = \(\frac { 20 }{ CD } \)
∴ CD = 20 × 2 = 40 m
∴ AC = 40 m …[From(i)]
Now, AB = AC + BC ….[A – C – B]
= 40 + 20\(\sqrt { 3 }\)
= 40 + 20 × 1.73
= 40 + 34.6
= 74.6
∴ The height of the tree is 74.6 m.

Question 6.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is tied at the ground. It makes an angle of 60° with the ground. Assuming that the string is straight, find the length of the string. (\(\sqrt { 3 }\) = 1.73)
Solution:
Let AB represent the height at which kite is flying and point C represent the point where the string is tied at the ground.
∠ACB is the angle made by the string with the ground.

∠ACB = 60°
AB = 60 m
In right angled ∆ABC,
sin 60° = \(\frac { AB }{ AC } \) … [By definition]

∴ AC = 40 \(\sqrt { 3 }\) = 40 × 1.73 = 69.20 m
∴ The length of the string is 69.20 m.

Maharashtra State Board Class 10 Maths Solutions Part 2

• Co-ordinate Geometry Practice Set 5.1 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.2 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.3 Class 10 Maths Solutions
• Co-ordinate Geometry Problem Set 5 Class 10 Maths Solutions
• Trigonometry Practice Set 6.1 Class 10 Maths Solutions
• Trigonometry Practice Set 6.2 Class 10 Maths Solutions
• Trigonometry Problem Set 6 Class 10 Maths Solutions
• Mensuration Practice Set 7.1 Class 10 Maths Solutions
• Mensuration Practice Set 7.2 Class 10 Maths Solutions
• Mensuration Practice Set 7.3 Class 10 Maths Solutions
• Mensuration Practice Set 7.4 Class 10 Maths Solutions
• Mensuration Problem Set 7 Class 10 Maths Solutions

9th Standard Maths 2 Practice Set 9.3 Chapter 9 Surface Area and Volume Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 9.3 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 9 Surface Area and Volume.

Class 9 Maths Part 2 Practice Set 9.3 Chapter 9 Surface Area and Volume Questions With Answers Maharashtra Board

Question 1.
Find the surface areas and volumes of spheres of the following radii
i. 4 cm
ii. 9 cm
iii. 3.5 cm (π = 3.14)
i. Given: Radius (r) = 4 cm
To find: Surface area and volume of sphere
Solution:
Surface area of sphere = 4πr²
= 4 x 3.14 x 4²
∴ Surface area of sphere = 200.96 sq.cm
Volume of sphere = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) x 3.14 x 4²
∴ Volume of sphere = 267.95 cubic cm

ii. Given: Radius (r) = 9 cm
To find: Surface area and volume of sphere
Solution:
Surface area of sphere = 4πr²
= 4 x 3.14 x 9²
∴ Surface area of sphere = 1017.36 sq.cm

∴ Volume of sphere = 3052.08 cubic cm

iii. Given: Radius (r) = 3.5 cm
To find: Surface area and volume of sphere
Solution:
Surface area of sphere = 4πr²
= 4 x 3.14 x (3.5)²
∴ Surface area of sphere = 153.86 sq.cm
Volume of sphere = \(\frac { 4 }{ 3 }\) πr³
= \(\frac { 4 }{ 3 }\) x 3.14 x (3.5)³
∴ Volume of sphere = 179.50 cubic cm

Question 2.
If the radius of a solid hemisphere is 5 cm, then find its curved surface area and total surface area, (π = 3.14)
Given: Radius (r) = 5 cm
To find: Curved surface area and total surface area of hemisphere
Solution:
i. Curved surface area of hemisphere = 2πr²
= 2 x 3.14 x 5²
= 2 x 3.14 x 25
= 50 x 3.14
= 157 sq.cm.

ii. Total surface area of hemisphere = 3πr²
= 3 x 3.14 x 5²
= 235.5 sq.cm.
∴ The curved surface area and totai surface area of hemisphere are 157 sq.cm, and 235.5 sq.cm, respectively.

Question 3.
If the surface area of a sphere is 2826 cm² then find its volume. (π = 3.14)
Given: Surface area of sphere = 2826 sq.cm.
To find: Volume of sphere
Solution:
i. Surface area of sphere = 4πr²
∴ 2826 = 4 x 3.14 x r²
2826 = 282600 = 900
∴ \( r^{2}=\frac{2826}{4 \times 3.14}=\frac{282600}{4 \times 314}=\frac{900}{4}\)
∴ r² = 225
∴ r = \(\sqrt { 225 }\) … [Taking square root on both sides]
= 15 cm

ii. Volume of sphere = \(\frac { 4 }{ 3 }\) πr³
= \(\frac { 4 }{ 3 }\) x 3.14 x 15³
= \(\frac { 4 }{ 3 }\) x 3.14 x 15 x 15 x 15
= 4 x 3.14 x 5 x 15 x 15
= 14130 cubic cm.
∴ The volume of the sphere is 14130 cubic cm.

Question 4.
Find the surface area of a sphere, if its volume is 38808 cubic cm. (π = \(\frac { 22 }{ 7 }\))
Given: Volume of sphere = 38808 cubic cm.
To find: Surface area of sphere
Solution:

∴ r³ = 441 x 21 = 21 x 21 x 21
∴ r = 21 cm … [Taking cube root on both sides]
ii. Surface area of sphere = 4πr²
= 4 x \(\frac { 22 }{ 7 }\) x 21
= 4 x \(\frac { 22 }{ 7 }\) x 21 x 21
= 4 x 22 x 3 x 21
= 5544 sq.cm.
∴ The surface area of sphere is 5544 sq.cm.

Question 5.
Volume of a hemisphere is 18000π cubic cm. Find its diameter.
Given: Volume of hemisphere = 1 8000π cubic cm.
To find: Diameter of the hemisphere
Solution:
i. Volume of hemisphere = \(\frac { 2 }{ 3 }\) πr³

= 9000 x 3
∴ r³ = 27000
∴ r = \(\sqrt [ 3 ]{ 27000 }\) … [Taking cube root on both sides]
= 30 cm

ii. Diameter = 2r
= 2 x 30 = 60 cm
∴ The diameter of the hemisphere is 60 cm.

Maharashtra Board Class 9 Maths Solutions

Maharashtra Board Class 9 Maths Solutions

• Surface Area and Volume Practice Set 9.1 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.2 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.3 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9 Class 9 Maths Solutions

9th Standard Maths 2 Practice Set 9 Chapter 9 Surface Area and Volume Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 9 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 9 Surface Area and Volume.

Class 9 Maths Part 2 Practice Set 9 Chapter 9 Surface Area and Volume Questions With Answers Maharashtra Board

Question 1.
If diameter of a road roller is 0.9 m and its length is 1.4 m, how much area of a field will be pressed in its 500 rotations? ( π = \(\frac { 22 }{ 7 }\))
Given: For road roller,
diameter (d) = 0.9 m, length (h) = 1.4 m
To find: Area of a field pressed in 500 rotations
Solution:
i. Since, area of field pressed in 1 rotation of road roller = curved surface area of road roller
∴ Curved surface area of the road roller = 2πrh
= πdh ,..[∵ d = 2r]
= \(\frac { 22 }{ 7 }\) x 0.9 x 1.4 7
= 22 x 0.9 x 0.2
= 3.96 sq.m.

ii. Area of land pressed in 1 rotation = 3.96 sq.m.
∴Area of land pressed in 500 rotations = 500 x 3.96
= 1980 sq.m.
∴ 1980 sq.m, land will be pressed in 500 rotations of the road roller.

Question 2.
To make an open fish tank, a glass sheet of 2 mm gauge is used. The outer length, breadth and height of the tank are 60.4 cm, 40.4 cm and 40.2 cm respectively. How much maximum volume of water will be contained in it ?
Given: Thickness of the glass = 2 mm,
outer length of the tank = 60.4 cm,
outer breadth of the tank = 40.4 cm,
outer height of the tank = 40.2 cm
To find: Volume of water fish tank contains
Solution:

i. Thickness oldie glass = 2 mm.
= \(\frac { 2 }{ 10 }\) cm
= 0.2 cm
Outerlengthofthetank = 60.4 cm
∴ Inner length oldie tank (l) = Outer length – thickness oldie glass on both sides
= 60.4 – 0.2 – 0.2
= 60cm
Outer breadth oldie tank = 40.4 cm
∴ Inner breadth of the tank (b) = 40.4 – 0.2 – 0.2
= 40 cm
Outer height of the tank = 40.2 cm
∴Inner height of the tank (h) = 40.2 – 0.2
= 40 cm

ii. Maximum volume of water that can be contained in the tank = volume of the tank
= l x b x h
= 60 x 40 x 40
= 96000 cubic cm.
∴ The fishtank can contain maximum of 96000 cubic cm. water in it.

Question 3.
If the ratio of radius of base and height of a cone is 5 : 12 and its volume is 314 cubic metre. Find its perpendicular height and slant height (π = 3.14).
Given: Ratio of radius of base and height of a cone = 5 : 12,
Volume = 314 cubic metre
To find: Perpendicular height (h) and slant height (l)
Solution:
i. The ratio of radius and height of cone is 5 : 12
Let the common multiple be x.
∴ Radius of base (r) = 5x
Perpendicular height (h) = 12x

∴ x³ = 1
∴ x = 1 … [Taking cube root on both sides]
∴ r = 5x = 5(1) = 5m
h = 12x = 12(1) = 12 m

ii. Now, l² = r² + h²
= 5² + 12²
= 25 + 144
∴l² = 169
∴ l = \(\sqrt { 169 }\) … [Taking square root on both sides]
= 13 m
The perpendicular height and slant height of the cone are 12 m and 13 m respectively.

Question 4.
Find the radius of a sphere if its volume is 904.32 cubic cm. (π = 3.14)
Given: Volume of sphere = 904.32 cubic cm.
To find: Radius of a sphere
Solution:
Volume of sphere = \(\frac { 4 }{ 3 }\) πr³
∴ 904.32 = \(\frac { 4 }{ 3 }\) x 3.14 x r³

= 216
∴ r = \(\sqrt [ 3 ]{ 216 }\) … [Taking cube root on both sides]
= 6 cm
∴ The radius of the sphere is 6 cm.

Question 5.
Total surface area of a cube is 864 sq.cm. Find its volume.
Given: Total surface area of cube = 864 sq. cm
To find: Volume of cube
Solution:
i. Total surface area of cube = 6l²
∴ 864 = 6l²
∴ l²= \(\sqrt [ 864 ]{ 6 }\)
∴ l² = 144
∴ l = \(\sqrt { 144 }\) … [Taking square root on both sides]
= 12 cm

ii. Volume of cube = l²
= 12³
= 1728 cubic cm.
∴ The volume of cube is 1728 cubic cm.

Question 6.
Find the volume of a sphere, if its surface area is 154 sq.cm.
Given: Surface area of sphere = 154 sq. cm.
To find: Volume of sphere
Solution:
i. Surface area of sphere = 4πr²


∴ The volume of sphere is 179.67 cubic cm.

Question 7.
Total surface area of a cone is 616 sq.cm. If the slant ‘height of the cone Is three times the radius of its base, find its slant height.
Given: Total surface area of a cone = 616 sq.cm., slant height of the cone is three times the radius of its base
To find: Slant height (l)
Solution:
i. Let the radius of base be r cm.
∴ Slant height (l) = 3r cm
Total surface area of cone = πr (l + r)
∴ 616 = πr(l + r)
∴ 616 = \(\sqrt [ 22 ]{ 7 }\) x r x (3r + r)
∴ 616 = \(\sqrt [ 22 ]{ 7 }\) x 4r²

∴ r² = 49
∴ r = \(\sqrt { 49 }\) … [Taking square root on both sides]
= 7

ii. Slant height (l) = 3r = 3 x 7 = 21 cm
∴ The slant height of the cone is 21 cm.

Question 8.
The inner diameter of a well is 4.20 metre and its depth is 10 metre. Find the inner surface area of the well. Find the cost of plastering it from inside at the rate ₹ 52 per sq.m.
Given: Inner diameter (d) = 4.2 m,
To find: depth (h) = 10 m,
rate of plastering = ₹ 52 per sq.m.
Inner surface area and total cost of plastering
Solution:
i. Inner curved surface area of the well = 2πrh
= πdh …[∵ d = 2r]
= \(\sqrt [ 22 ]{ 7 }\) x 4.2 x 10
= \(\sqrt [ 22 ]{ 7 }\) x 42
= 22 x 6
= 132 sq.m.

ii. Rate of plastering = ₹52 per sq.m.
∴ Total cost = Curved surface area x Rate of plastering
= 132 x 52 = ₹6864
∴ The cost of plastering the well from inside is ₹6864.

Question 9.
The length of a road roller is 2.1 m and its diameter is 1.4 m. For levelling a ground 500 rotations of the road roller were required. How much area of ground was levelled by the road roller? Find the cost of levelling at the rate of ₹ 7 per sq.m.
Given: For road roller,
diameter (d) = 1.4 m,
length (h) = 2.1 m
number of rotations required for levelling the ground = 500,
rate of levelling = ₹ 7 per sq. m.
To find: Area of ground leveled by the road roller and cost of levelling
Solution:
i. Since, area of ground levelled in 1 rotation of road roller = curved surface area of road roller
∴Curved surface area of the road roller = 2πrh
= πdh …[∵ d = 2r]
= \(\frac { 22 }{ 7 }\) x 1.4 x 2.1
= 22 x 0.2 x 2.1
= 9.24 sq.m.

ii. Area of ground levelled in 1 rotation = 9.24 sq.m.
∴Area of ground levelled in 500 rotations = 9.24 x 500
= 4620 sq.m.

iii. Rate of levelling ₹ 7 per sq.m.
∴Total cost = Area of ground levelled x Rate of levelling
= 4620 x 7
= ₹32340
∴ The road roller levels 4620 sq.m. land in 500 rotation, and the cost of levelling is ₹32340.

Maharashtra Board Class 9 Maths Chapter 9 Surface Area and Volume Practice Set 9 Intext Questions and Activities

Question 1.
Curved surface area of cone. (Textbook pg. no. 116)

Circumference of base of the cone = 2πr
As shown in the figure (c), make pieces of the net as small as possible. Join them as shown in the figure (d),. By joining the small pieces of net of the cone, we get a rectangle ABCD approximately.
Total length of AB and CD is 2πr.
∴ length of side AB of rectangle ABCD is πr and length of side CD is also πr.
Length of side BC of rectangle = slant height of cone = l.
Curved surface area of cone is equal to the area of the rectangle.
∴ curved surface area of cone = Area of rectangle = AB x BC = πr x l = πrl

Question 2.
Prepare a cylinder of a card sheet, keeping one of its faces open. Prepare an open cone of card sheet which will have the same base-radius and the same height as that of the cylinder. Pour fine sand in the cone till it just fills up the cone. Empty the cone in the cylinder. Repeat the procedure till the cylinder is just filled up with sand. Note how many coneful of sand is required to fill up the cylinder. (Textbook pg, no 117)

Answer:
To fill the cylinder, three coneful of sand is required.

Question 3.
Finding total surface area of sphere. (Textbook pg, no 120)

i. Take a sweet lime (Mosambe), Cut it into two equal parts.

ii. Take one of the parts. Place its circular face on a paper. Draw its circular border. Copy three more such circles. Again, cut each half of the sweet lime into two equal parts.

iii. Now you get 4 quarters of sweet lime. Separate the peel of a quarter part. Cut it into pieces as small as possible. Try to cover one o’f the circles drawn, by the small pieces. Observe that the circle gets nearly covered.
The activity suggests that,
Curved surface area of a sphere = 4πr²

∴ Curved surface area of a sphere = 4 x Area of a circle

Question 4.
Make a cone and a hemisphere of cardsheet such that radii of cone and hemisphere are equal and height of cone is equal to radius of the hemisphere.
Fill the cone with fine sand. Pour the sand in the hemisphere. How many cones are required to fill the hemisphere completely ? (Textbook pg. no. 121)

Answer:
To fill the hemisphere, two coneful of sand is required.

Maharashtra Board Class 9 Maths Solutions

• Surface Area and Volume Practice Set 9.1 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.2 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.3 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9 Class 9 Maths Solutions

10th Standard Maths 2 Practice Set 5.1 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

Class 10 Maths Part 2 Practice Set 5.1 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Practice Set 5.1 Geometry Class 10 Question 1. Find the distance between each of the following pairs of points.
i. A (2, 3), B (4,1)
ii. P (-5, 7), Q (-1, 3)
iii. R (0, -3), S (0,\(\frac { 5 }{ 2 } \))
iv. L (5, -8), M (-7, -3)
v. T (-3, 6), R (9, -10)
vi. W(\(\frac { -7 }{ 2 } \),4), X(11, 4)
Solution:
i. Let A (x₁, y₁) and B (x₂, y₂) be the given points.
∴ x₁ = 2, y₁ = 3, x₂ = 4, y₂ = 1
By distance formula,

∴ d(A, B) = 2\(\sqrt { 2 }\) units
∴ The distance between the points A and B is 2\(\sqrt { 2 }\) units.

ii. Let P (x₁, y₁ ) and Q (x₂, y₂) be the given points.
∴ x₁ = -5, y₁ = 7, x₂ = -1, y₂ = 3
By distance formula,

∴ d(P, Q) = 4\(\sqrt { 2 }\) units
∴ The distance between the points P and Q is 4\(\sqrt { 2 }\) units.

iii. Let R (x₁, y₁) and S (x₂, y₂) be the given points.
∴ x₁ = 0, y₁ = -3, x₂ = 0, y₂ = \(\frac { 5 }{ 2 } \)
By distance formula,

∴ d(R, S) = \(\frac { 11 }{ 2 } \) units
∴ The distance between the points R and S is \(\frac { 11 }{ 2 } \) units.

iv. Let L (x₁, y₁) and M (x₂, y₂) be the given points.
∴ x₁ = 5, y₁ = -8, x₂ = -7, y₂ = -3
By distance formula,

∴ d(L, M) = 13 units
∴ The distance between the points L and M is 13 units.

v. Let T (x₁,y₁) and R (x₂, y₂) be the given points.
∴ x₁ = -3, y₁ = 6,x₂ = 9,y₂ = -10
By distance formula,

∴ d(T, R) = 20 units
∴ The distance between the points T and R 20 units.

vi. Let W (x₁, y₁) and X (x₂, y₂) be the given points.

∴ d(W, X) = \(\frac { 29 }{ 2 } \) units
∴ The distance between the points W and X is \(\frac { 29 }{ 2 } \) units.

Practice Set 5.1 Geometry 10th Question 2. Determine whether the points are collinear.
i. A (1, -3), B (2, -5), C (-4, 7)
ii. L (-2, 3), M (1, -3), N (5, 4)
iii. R (0, 3), D (2, 1), S (3, -1)
iv. P (-2, 3), Q (1, 2), R (4, 1)
Solution:
i. By distance formula,


∴ d(A, B) = \(\sqrt { 5 }\) …(i)
On adding (i) and (iii),
d(A, B) + d(A, C)= \(\sqrt { 5 }\) + 5\(\sqrt { 5 }\) = 6\(\sqrt { 5 }\)
∴ d(A, B) + d(A, C) = d(B, C) … [From (ii)]
∴ Points A, B and C are collinear.

ii. By distance formula,

On adding (i) and (iii),
d(L, M) + d(L, N) = 3\(\sqrt { 5 }\) + 5\(\sqrt { 2 }\) ≠ \(\sqrt { 65 }\)
∴ d(L, M) + d(L, N) ≠ d(M, N) … [From (ii)]
∴ Points L, M and N are not collinear.

iii. By distance formula,

On adding (i) and (ii),
∴ d(R, D) + d(D, S) = \(\sqrt { 8 }\) + \(\sqrt { 5 }\) ≠ 5
∴ d(R, D) + d(D, S) ≠ d(R, S) … [From (iii)]
∴ Points R, D and S are not collinear.

iv. By distance formula,

On adding (i) and (ii),
d(P, Q) + d(Q, R) = \(\sqrt { 10 }\) + \(\sqrt { 10 }\) = 2\(\sqrt { 10 }\)
∴ d(P, Q) + d(Q, R) = d(P, R) … [From (iii)]
∴ Points P, Q and R are collinear.

Coordinate Geometry Class 10 Practice Set 5.1 Question 3. Find the point on the X-axis which is equidistant from A (-3,4) and B (1, -4).
Solution:
Let point C be on the X-axis which is equidistant from points A and B.
Point C lies on X-axis.
∴ its y co-ordinate is 0.
Let C = (x, 0)
C is equidistant from points A and B.
∴ AC = BC

∴ (x + 3)² + (-4)² = (x- 1)² + 4²
∴ x² + 6x + 9 + 16 = x² – 2x + 1 + 16
∴ 8x = – 8
∴ x = – \(\frac { 8 }{ 8 } \) = -1
∴ The point on X-axis which is equidistant from points A and B is (-1,0).

10th Geometry Practice Set 5.1 Question 4. Verify that points P (-2, 2), Q (2, 2) and R (2, 7) are vertices of a right angled triangle.
Solution:
Distance between two points

Consider, PQ² + QR² = 42 + 52 = 16 + 25 = 41 … [From (i) and (ii)]
∴ PR² = PQ² + QR² … [From (iii)]
∴ ∆PQR is a right angled triangle. … [Converse of Pythagoras theorem]
∴ Points P, Q and R are the vertices of a right angled triangle.

Question 5.
Show that points P (2, -2), Q (7, 3), R (11, -1) and S (6, -6) are vertices of a parallelogram.
Proof:
Distance between two points

PQ = RS … [From (i) and (iii)]
QR = PS … [From (ii) and (iv)]
A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent.
∴ □ PQRS is a parallelogram.
∴ Points P, Q, R and S are the vertices of a parallelogram.

Question 6.
Show that points A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are vertices of rhombus ABCD.
Proof:
Distance between two points


∴ AB = BC = CD = AD …[From (i), (ii), (iii) and (iv)]
In a quadrilateral, if all the sides are equal, then it is a rhombus.
∴ □ ABCD is a rhombus.
∴ Points A, B, C and D are the vertices of rhombus ABCD.

Practice Set 5.1 Question 7. Find x if distance between points L (x, 7) and M (1,15) is 10.
Solution:
X₁ = x, y₁ = 7, x₂ = 1, y₂ = 15
By distance formula,

∴ 1 – x = ± 6
∴ 1 – x = 6 or l – x = -6
∴ x = – 5 or x = 7
∴ The value of x is – 5 or 7.

Geometry 5.1 Question 8. Show that the points A (1, 2), B (1, 6), C (1 + 2\(\sqrt { 3 }\), 4) are vertices of an equilateral triangle.
Proof:
Distance between two points

∴ AB = BC = AC … [From (i), (ii) and (iii)]
∴ ∆ABC is an equilateral triangle.
∴ Points A, B and C are the vertices of an equilateral triangle.

Maharashtra Board Class 10 Maths Chapter 5 Coordinate Geometry Intext Questions and Activities

Question 1.
In the figure, seg AB || Y-axis and seg CB || X-axis. Co-ordinates of points A and C are given. To find AC, fill in the boxes given below. (Textbook pa. no. 102)

Solution:
In ∆ABC, ∠B = 900
∴ (AB)² + (BC)² = [(Ac)² …(i) … [Pythagoras theorem]
seg CB || X-axis
∴ y co-ordinate of B = 2
seg BA || Y-axis
∴ x co-ordinate of B = 2
∴ co-ordinate of B is (2, 2) = (x₁,y₁)
co-ordinate of A is (2, 3) = (x₂, Y₂)
Since, AB || to Y-axis,
d(A, B) = Y₂ – Y₁
d(A,B) = 3 – 2 = 1
co-ordinate of C is (-2,2) = (x₁,y₁)
co-ordinate of B is (2, 2) = (x₂, y₂)
Since, BC || to X-axis,
d(B, C) = x₂ – x₁
d(B,C) = 2 – -2 = 4
∴ AC² = 12 + 42 …[From (i)]
= 1 + 16 = 17
∴ AC = \(\sqrt { 17 }\) units …[Taking square root of both sides]

Maharashtra State Board Class 10 Maths Solutions Part 2

• Co-ordinate Geometry Practice Set 5.1 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.2 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.3 Class 10 Maths Solutions
• Co-ordinate Geometry Problem Set 5 Class 10 Maths Solutions
• Trigonometry Practice Set 6.1 Class 10 Maths Solutions
• Trigonometry Practice Set 6.2 Class 10 Maths Solutions
• Trigonometry Problem Set 6 Class 10 Maths Solutions
• Mensuration Practice Set 7.1 Class 10 Maths Solutions
• Mensuration Practice Set 7.2 Class 10 Maths Solutions
• Mensuration Practice Set 7.3 Class 10 Maths Solutions
• Mensuration Practice Set 7.4 Class 10 Maths Solutions
• Mensuration Problem Set 7 Class 10 Maths Solutions

10th Standard Maths 2 Problem Set 4 Chapter 4 Geometric Constructions Textbook Answers Maharashtra Board

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.

Class 10 Maths Part 2 Problem Set 4 Chapter 1 Geometric Constructions Questions With Answers Maharashtra Board

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.

Maharashtra State Board Class 10 Maths Solutions Part 2

• Circle Practice Set 3.1 Class 10 Maths Solutions
• Circle Practice Set 3.2 Class 10 Maths Solutions
• Circle Practice Set 3.3 Class 10 Maths Solutions
• Circle Practice Set 3.4 Class 10 Maths Solutions
• Circle Practice Set 3.5 Class 10 Maths Solutions
• Circle Problem Set 3 Class 10 Maths Solutions
• Geometric Constructions Practice Set 4.1 Class 10 Maths Solutions
• Geometric Constructions Practice Set 4.2 Class 10 Maths Solutions
• Geometric Constructions Problem Set 4 Class 10 Maths Solutions

10th Standard Maths 2 Practice Set 4.2 Chapter 4 Geometric Constructions Textbook Answers Maharashtra Board

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.

Class 10 Maths Part 2 Practice Set 4.2 Chapter 1 Geometric Constructions Questions With Answers Maharashtra Board

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.

Maharashtra State Board Class 10 Maths Solutions Part 2

• Circle Practice Set 3.1 Class 10 Maths Solutions
• Circle Practice Set 3.2 Class 10 Maths Solutions
• Circle Practice Set 3.3 Class 10 Maths Solutions
• Circle Practice Set 3.4 Class 10 Maths Solutions
• Circle Practice Set 3.5 Class 10 Maths Solutions
• Circle Problem Set 3 Class 10 Maths Solutions
• Geometric Constructions Practice Set 4.1 Class 10 Maths Solutions
• Geometric Constructions Practice Set 4.2 Class 10 Maths Solutions
• Geometric Constructions Problem Set 4 Class 10 Maths Solutions