Class 9

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 9.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 9 Surface Area and Volume.

Practice Set 9.2 Geometry 9th Std Maths Part 2 Answers Chapter 9 Surface Area and Volume

Question 1.
Perpendicular height of a cone is 12 cm and its slant height is 13 cm. Find the radius of the base of the cone.
Given: Height (h) = 12 cm, length (l) = 13 cm
To find: Radius of the base of the cone (r)
Solution:
l² = r² + h²
∴ 13² = r² + 12²
∴ 169 = r² + 144
∴169 – 144 = r²
∴ r² = 25
∴ r = √25 … [Taking square root on both sides]
= 5 cm
∴ The radius of base of the cone is 5 cm.

Question 2.
Find the volume of a cone, if its total surface area is 7128 sq.cm and radius of base is 28 cm. ( π = \(\frac { 22 }{ 7 }\))
Given: Radius (r) = 28 cm,
Total surface area of cone = 7128 sq.cm
To find: Volume of the cone
Solution:
i. Total surface area of cone = πr (l + r)
∴ 7128= y x 28 x (l + 28)
∴ 7128 = 22 x 4 x(l +28)
∴ l + 28 = \(\frac { 7128 }{ 22\times 4 }\)
∴ l + 28 = 81
∴ l = 81 – 28
∴ l = 53cm

ii. Now, l² = r² + h²
∴ 53² = 28²+ h²
∴ 2809 = 784 + h²
∴ 2809 – 784 = h²
∴ h² = 2025
∴ h = \(\sqrt { 2025 }\) …… [Taking square root on both sides]
= 45 cm

= 22 x 4 x 28 x 15
= 36960 cubic.cm
∴ The volume of the cone is 36960 cubic.cm.

Question 3.
Curved surface area of a cone is 251.2 cm² and radius of its base is 8 cm. Find its slant height and perpendicular height, (π = 3.14)
Given: Radius (r) = 8 cm, curved surface area
of cone = 251.2 cm²
To find: Slant height (l) and the perpendicular height (h) of the cone
Solution:
i. Curved surface area of cone = πrl
∴ 251.2 = 3.14 x 8 x l

∴ l= 10 cm

ii. Now, l² = r² + h²
∴ 10² = 8² + h²
∴ 100 = 64 + h²
∴ 100 – 64 = h²
∴ h² = 36
∴ h = √36 … [Taking square root on both sides]
= 6 cm
∴ The slant height and the perpendicular height of the cone are 10 cm and 6 cm respectively.

Question 4.
What will be the cost of making a closed cone of tin sheet having radius of base 6 m and slant height 8 m if the rate of making is ₹ 10 per sq.m?
Given: Radius (r) = 6 m, length (l) = 8 m
To find: Total cost of making the cone
Solution:
i. To find the total cost of making the cone of tin sheet, first we need to find the total surface area of the cone.
Total surface area of the cone = πr (l + r)
= \(\frac { 22 }{ 7 }\) x 6 x (8 + 6)
= \(\frac { 22 }{ 7 }\) x 6 x 14
= 22 x 6 x 2 = 264 sq.m

ii. Rate of making the cone = ₹ 10 per sq.m
∴ Total cost = Total surface area x Rate of making the cone
= 264 x 10
= ₹ 2640
∴ A The total cost of making the cone of tin sheet is ₹ 2640.

Question 5.
Volume of a cone is 6280 cubic cm and base radius of the cone is 20 cm. Find its perpendicular height, (π = 3.14)
Given: Radius (r) = 20 cm,
Volume of cone = 6280 cubic cm
To find: Perpendicular height (h) of the cone
Solution:

∴ The perpendicular height of the cone is 15 cm.

Question 6.
Surface area of a cone is 188.4 sq.cm and its slant height is 10 cm. Find its perpendicular height (π = 3.14).
Given: Length (l) =10 cm, curved surface area of the cone = 188.4 sq.cm
To find: Perpendicular height (h) of the cone
Solution:
i. Curved surface area of the cone = πrl
∴ 188.4 = 3.14 x r x 10

ii. Now, l² = r² + h²
∴ 10² = 6² + h²
∴ 100 = 36 + h²
∴ 100 – 36 = h²
∴ h² = 64
∴ h = \(\sqrt { 64 }\) … [Taking square root on both sides]
= 8 cm
∴ The perpendicular height of the cone is 8 cm.

Question 7.
Volume of a cone is 1232 cm³ and its height is 24 cm. Find the surface area of the cone. (π = \(\frac { 22 }{ 7 }\))
Given: Height (h) = 24 cm,
Volume of cone = 1232 cm³
To find: Surface area of the cone
Solution:

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

ii. Now, l² = r² + h²
∴ l² = 7² + 24²
= 49 + 576 = 625
∴ l = \(\sqrt { 625 }\) … [Taking square root on both sides]
= 25

iii. Curved surface area of cone = πrl
= \(\frac { 22 }{ 7 }\) x 7 x 25
= 22 x 25
= 550 sq.cm
∴The surface area of the cone is 550 sq.cm.

Question 8.
The curved surface area of a cone is 2200 sq.cm and its slant height is 50 cm. Find the total surface area of cone. (π = \(\frac { 22 }{ 7 }\))
Given: Length (l) = 50 cm, curved surface area of cone = 2200 sq.cm
To find: Total surface area of the cone
Solution:
i. Curved surface area of cone = πrl

ii. Total surface area of cone = πr (l + r)
= \(\frac { 22 }{ 7 }\) x 14 x (50 + 14)
= \(\frac { 22 }{ 7 }\) x 14 x 64
= 22 x 2 x 64
= 2816 sq.cm
∴ The total surface area of the cone is 2816 sq.cm.

Question 9.
There are 25 persons in a tent which is conical in shape. Every person needs an area of 4 sq.m, of the ground inside the tent. If height of the tent is 18 m, find the volume of the tent.
Given: For the tent,
height (h) = 18m,
number of people in the tent = 25,
area required for each person = 4 sq.m
To find: Volume of the tent
Solution:
i. Every person needs an area of 4 sq.m, of the ground inside the tent.
Surface area of the base of the tent = number of people in the tent × area required for each person
= 25 × 4
= 100 sq.m

ii. Surface area of the base of the tent = πr²
∴ 100 = πr²
∴ πr² = 100

iii. Volume of the tent= \(\frac { 1 }{ 3 }\) πr²h
= \(\frac { 1 }{ 3 }\) x 100 x 18 …….[∵ πr² = 100]
= 100 x 6
= 600 cubic metre
∴ The volume of the tent is 600 cubic metre.

Question 10.
In a field, dry fodder for the cattle is heaped in a conical shape. The height of the cone is 2.1 m and diameter of base is 7.2 m. Find the volume of the heap of the fodder. If it is to be covered by polythene in rainy se&son then how much minimum polythene
sheet is needed? (π = \(\frac { 22 }{ 7 }\) and \(\sqrt { 17.37 }\) = 4.17 ]
Given: Height of the heap (h) = 2.1 m.
diameter of the base (d) = 7.2 m
∴Radius of the base (r) = \(\frac { d }{ 2 }\) = \(\frac { 7.2 }{ 2 }\) = 3.6 m
To find: Volume of the heap of the fodder and polythene sheet required
Solution:
i. Volume of the heap of fodder = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) x \(\frac { 22 }{ 7 }\) x (3.6)² x 2.1
= \(\frac { 1 }{ 3 }\) x \(\frac { 22 }{ 7 }\) x 3.6 x 3.6 x 2.1
= 1 x 22 x 1.2 x 3.6 x 0.3
= 28.51 cubic metre

ii. Now, l² = r² + h²
= (3.6)² + (2.1)²
= 12.96 + 4.41
∴ l² =17.37
∴ l² = \(\sqrt { 17.37 }\) .. .[Taking square root on both sides]
= 4.17 m

iii. Area of the polythene sheet needed to cover the heap of the fodder = Curved surface area of the conical heap
= πrl
= \(\frac { 22 }{ 7 }\) x 3.6 x 4.17
= 47.18 sq.m
∴ The volume of the heap of the fodder is 28.51 cubic metre and a polythene sheet of 47.18 sq.m will be required to cover it.

Maharashtra Board Class 9 Maths Solutions

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.

Practice Set 9 Geometry 9th Std Maths Part 2 Answers Chapter 9 Surface Area and Volume

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.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 8 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 8 Trigonometry.

Problem Set 8 Geometry 9th Std Maths Part 2 Answers Chapter 8 Trigonometry

Question 1.
Choose the correct alternative answer for the following multiple choice questions.

i. Which of the following statements is true?
(A) sin θ = cos (90 – θ)
(B) cos θ = tan (90 – θ)
(C) sin θ = tan (90 – θ)
(D) tan θ = tan (90 – θ)
Answer:
(A) sin θ = cos (90 – θ)

ii. Which of the following is the value of sin 90°?
(A) \( \frac { \sqrt { 3 } }{ 2 }\)
(B) 0
(C) \(\frac { 1 }{ 2 }\)
(D) 1
Answer:
(D) 1

iii. 2 tan 45° + cos 45° – sin 45° = ?
(A) 0
(B) 1
(C) 2
(D) 3
Answer:
2 tan 45° + cos 45° – sin
\( =2(1)+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=2\)
(C) 2

iv. \( \frac{\cos 28^{\circ}}{\sin 62^{\circ}}\) =?
(A) 2
(B) -1
(C) 0
(D) 1
Answer:
\( \frac{\cos 28^{\circ}}{\sin 62^{\circ}}\)

(D) 1

Question 2.
In right angled ∆TSU, TS = 5, ∠S = 90°, SU = 12, then find sin T, cos T, tan T. Similarly find sin U, cos U, tan U.

Solution:
i. TS = 5, SU = 12 …[Given]
In ∆TSU, ∠S = 90° … [Given]
∴ TU² = TS² + SU² …[Pythagoras theorem]
= 5² + 12² = 25 + 144 = 169
∴ TU = \(\sqrt { 169 }\) .. .[Taking square root of both sides]
= 13

Question 3.
In right angled ∆YXZ, ∠X = 90°, XZ = 8 cm, YZ = 17 cm, find sin Y, cos Y, tan Y, sin Z, cos Z, tan Z.

Solution:
i. XZ = 8 cm, YZ = 17 cm …[Given]
In ∆YXZ, ∠X = 90° … [Given]
∴ YZ² = XY² + XZ² .. .[Pythagoras theorem]
∴ 17² = XY² + 8²
∴ 289 = XY² + 64
∴ XY² = 289 – 64
= 225
∴ x = \(\sqrt { 225 }\) .. .[Taking square root of both sides]
= 15

Question 4.
In right angled ∆LMN, if ∠N = θ, ∠M = 90°, cos θ = \(\frac { 24 }{ 25 }\), find sin θ and tan θ. Similarly, find (sin²θ) and (cos²θ).

Solution:
i. cos θ = \(\frac { 24 }{ 25 }\)
In ∆LMN, ∠M = 90°, ∠N = θ

Let the common multiple be k.
∴ MN = 24k and LN = 25k
Now, LN²= LM² + MN² … [Pythagoras theorem]
∴ (25k)² = LM² + (24k)²
∴ 625 k² = LM² + 576k²
∴ LM² = 625k² – 576k²
∴ LM² = 49k²
∴ LM = \(\sqrt { 49{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 7k

Question 5.
Fill in the blanks.
i. sin 20° = cos
ii. tan 30° x tan  = 1
iii. cos 40° = sin
Solution:
i. sin 20° = cos (90° – 20°) …..[∵ sin θ = cos (90 – θ)]
= cos 70°

ii. tan θ x tan (90 – θ) = 1
Substituting θ = 30°,
tan 30° x tan (90 – 30)° = 1
∴ tan 30° x tan 60° = 1

iii. cos 40° = sin (90° – 40°) …[∵ COS θ = sin (90 – θ)]
= sin 50°

Maharashtra Board Class 9 Maths Solutions Chapter 8 Trigonometry Problem Set 8

Question 1.
Measuring height of a tree using trigonometric ratios. (Textbook pg. no. 101)

This experiment can be conducted on a clear sunny day. Look at the figure given above. Height of the tree is QR, height of the stick is BC.
Thrust a stick in the ground as shown in the figure. Measure its height and length of its shadow. Also measure the length of the shadow of the tree. Using these values, how will you determine the height of the tree?
Solution:
Rays of sunlight are parallel.
So, ∆PQR and ∆ABC are equiangular i.e., similar triangles.
Sides of similar triangles are proportional.
∴ \(\frac { QR }{BC }\) = \(\frac { PR }{ AC }\)
∴ Height of the tree (QR) = \(\frac { BC }{ AC }\) x PR
Substituting the values of PR, BC and AC in the above equation, we can get length of QR i.e., the height of the tree.

Question 2.
It is convenient to do the above experiment between 11:30 am and 1:30 pm instead of doing it in the morning at 8’O clock. Can you tell why? (Textbook pg. no. 101)
Solution:
At 8’O clock in the morning, the sunlight is not very bright. At the same time, the sun is on the horizon and the shadow would by very long. It would be extremely difficult to measure shadow in that case.
Between 11:30 am and 1:30 pm, the sun is overhead and it would be easier to measure the length of shadow.

Question 3.
Conduct the above discussed activity and find the height of a tall tree in your surrounding. If there is no tree in the premises, then find the height of a pole. (Textbook pg. no. 101)

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 8.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 8 Trigonometry.

Practice Set 8.2 Geometry 9th Std Maths Part 2 Answers Chapter 8 Trigonometry

Question 1.
In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table.

Solution:
i. cos θ = \(\frac { 35 }{ 37 }\) …(i) )[Given]
In right angled ∆ABC,

∠C = θ.

Let the common multiple be k.
∴ BC = 35k and AC = 37k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (37k)² = AB²+ (35k)²
1369k² = AB² + 1225k²
AB2 = 1369k² – 1225k²
= 144k²
AB = 144k²
AB = \(\sqrt { 2ghK }\)² … [Taking square root of both sides]
= 12k

ii. sin θ = \(\frac { 11 }{ 61 }\) …..(i) [Given]
In right angled ∆ABC, ∠C = θ.


Let the common multiple be k.
AB = 11k and AC = 61k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (61k)² = (11k)² + BC²
∴ 3721k² = 121k² + BC²
∴ BC² = 3721k² – 121k² = 3600k²
BC = \(\sqrt { 3600{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 60k

iii. tan θ = 1 = \(\frac { 1 }{ 1 }\) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and BC = 1k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= K² + K²
= 2K²
∴ AC = \(\sqrt { 2{ k }\)

iv. sin θ = \(\frac { 1 }{ 2 }\) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and BC = 2k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ 2K² = K² + BC²
∴ 4K² = K² + BC²
∴ BC² = 4K² – K² = 3K²
∴ BC = \(\sqrt { 3{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= \(\sqrt { 3{ k }\)

v. cos θ = \(\frac { 1 }{ \sqrt { 3 } } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and BC = √3k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (√3K)² = AB² + K²
∴ 3K² = 3K² – K² = 2K²
∴ AB = \(\sqrt { 2{ k }^{ 2 } }\) .. .[Taking square root of both sides]
AB = √2K

vi. cos θ = \(\frac { 21 }{ \sqrt { 20 } } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 21k and BC = 20k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= (21)K² + (20K)²
= 441K² – 400²
= 841K²
∴ AB = \(\sqrt { 841{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 29K

vii. tan θ = \(\frac { 8 }{ 15 } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 8k and BC = 15k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= (8)K² + (15K)²
= 64K² – 225²
= 289K²
∴ AC = \(\sqrt { 289{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 17K

viii. sin θ = \(\frac { 3 }{ 5 } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.



Let the common multiple be k.
∴ AB = 3k and AC = 5k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (5)K²= (3)K² + BC²
∴ 25K² = 9K² – 225²
∴ BC² = 25K² – 9K²
∴ BC = \(\sqrt { 16{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 4K

ix. tan θ = \(\frac { 1 }{ 2\sqrt { 2 } }\) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and AC = 2√2 k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= K² + (2√2 k )²
= K² – 225²
= 25K² + 8K²
= 9K²
∴ AC = \(\sqrt { 9{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 3K

Question 2.
Find the values of:
i. 5 sin 30° + 3 tan 45°
ii. \(\frac { 4 }{ 5 }\)tan² 60° + 3 sin² 60°
iii. 2 sin 30° + cos 0° + 3 sin 90°
iv. \(\frac{\tan 60^{\circ}}{\sin 60^{\circ}+\cos 60^{\circ}}\)
v. cos² 45° + sin² 30°
vi. cos 60° x cos 30° + sin 60° x sin 30°
Solution:
i. sin 30° = \(\frac { 1 }{ 2 }\) and tan 45° = 1

ii. \(\frac { 4 }{ 5 }\)tan² 60° + 3 sin² 60°

iii. 2 sin 30° + cos 0° + 3 sin 90°
2 sin 30° + cos0° + 3 sin 90° = 2 (\(\frac { 1 }{ 2 }\)) + 1 + 3(1)
= 1 + 1 + 3
∴ 2 sin 30° + cos 0° + 3 sin 90° = 5

iv. \(\frac{\tan 60^{\circ}}{\sin 60^{\circ}+\cos 60^{\circ}}\)

v. cos² 45° + sin² 30°

vi. cos 60° x cos 30° + sin 60° x sin 30°

Question 3.
If sin θ = \(\frac { 4 }{ 5 }\) , then find cos θ.
Solution:
sin θ = \(\frac { 4 }{ 5 }\) .. .(i)[Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 4k and AC = 5k
Now, AC² = AB² + BC² … [Pythagoras theorem]
∴ (5 k)² = (4k)² + BC²
∴ 25k² = 16k² + BC²
∴ BC² = 25k² – 16k² = 9k²
∴ BC = \(\sqrt { 9{ k }^{ 2 } }\) . .[Taking square root of both sides]
= 3k

Question 4.
If cos θ = \(\frac { 15 }{ 17 }\) , then find sin θ.
Solution:
cos θ = \(\frac { 15 }{ 17 }\) .. .(i)[Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ BC = 15k and AC = 17k
Now, AC² = AB² + BC² … [Pythagoras theorem]
∴ (17 k)² = AB² + (15K)²
∴ 289k² = AB² + 225²
∴ AB² = 289k² – 225k²
= 64k²
∴ AB = \(\sqrt { 64{ k }^{ 2 } }\) . .[Taking square root of both sides]
= 8k

Maharashtra Board Class 9 Maths Chapter 8 Trigonometry Practice Set 8.2 Intext Questions and Activities

Question 1.
In right angled ∆PQR, ∠Q = 900. Therefore ∠P and ∠R are complementary angles of each other. Verify the following ratios.
i. sin θ = cos (90 – θ)
ii. cos θ = sin (90 – θ)
iii. sin 30° = cos (90° – 30°) = cos 60°
iv. cos 30° = sin (90° – 30°) = sin 60° (Textbook pg. no. 107)
Solution:
In ∆PQR, ∠Q = 90°, ∠P = θ
∴ ∠R = 90 – θ

i. sin θ = cos (90 – θ)

ii. cos θ = sin (90 – θ)

iii. Let ∠P = θ = 30°
∴ ∠R = 90° – 30°

sin 30° = cos (90° – 30°) … [From (i) and (ii)]
sin 30° = cos 60°

iv. cos 30° = sin (90° – 30°) = sin 60°

∴ cos 30° = sin (90° – 30°) .,.[From (i) and (ii)]
∴ cos 30° = sin 60°

Question 2.
In right angled ∆PQR, ∠Q = 90°, ∠R = θ and if sin θ = \(\frac { 5 }{ 13 }\), then find cos θ and tan θ. (Textbook pg. no. 110)
Solution:
i. Take the given trigonometric ratio as 13k equation (i).
sin θ = \(\frac { 5 }{ 13 }\) .. .(i)[Given]
By using the definition write the trigonometric ratio of sin O and take it as equation (ii).
In right angled ∆PQR, ∠R = θ


Let the common multiple be k.
∴ PQ = 5k and PR = 13k
Find QR by using Pythagoras theorem.
PR² = PQ² + QR² … [Pythagoras theorem]
∴ (13k)² = (5k)² + QR²
∴ 169k² = 25k² + QR²
∴ QR² = 169k² – 25k²
= 144k²
∴ QR = \(\sqrt { 144{ k }^{ 2 } }\) . . . [Taking square root of both sides]
= 12k

Question 3.
While solving the above Illustrative example, why the lengths of PQ and PR are taken 5k and 13k? (Textbook pg. no. 111)
Solution:
\(\frac { PQ }{ PR }\) = \(\frac { 5 }{ 13 }\) … [Given]
Here, the ratio of the lengths of sides PQ and PR is 5 : 13.
The actual lengths of the sides can be any multiple of the ratio. Hence, we consider the multiple k while solving.

Question 4.
While solving the above illustrative example, can we take the lengths of PQ and PR as 5 and 13? If so, then what changes are needed In the writing of the solution. (Tcxtbook pg. no. 111)
Solution:
Yes, we can take lengths of PQ and PR as 5 and 13.
In that case, we will have to take k = 1 and solve the problem accordingly.

Question 5.
Verify that the equation ‘sin² θ + cos² θ = 1’ is true when θ = 0° or θ = 90°.
(Textbook pg. no. 112)
Solution:
sin² θ + cos² θ = 1
i. lf θ = 0°,
LH.S. = sin² θ + cos² θ
= sin² 0° + cos² 0°
= 0 + 1 …[∵ sin 0° = 0, cos 0° = 1]
= R.H.S.
∴ sin² θ + cos² θ = 1

ii. If θ = 90°,
L.H.S.= sin² θ +cos² θ
= sin² 90° + cos² 90°
= 1 + 0 … [ ∵ sin 90° = 1, cos 90° = 0]
= 1
= R.H.S.
∴ sin² θ + cos² θ = 1

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 8.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 8 Trigonometry.

Practice Set 8.1 Geometry 9th Std Maths Part 2 Answers Chapter 8 Trigonometry

Question 1.
In the given figure, ∠R is the right angle of ∆PQR. Write the following ratios.

i. sin P
ii. cos Q
iii. tan P
iv. tan Q
Solution:

Question 2.
In the right angled ∆XYZ, ∠XYZ = 90° and a, b, c are the lengths of the sides as shown in the figure. Write the following ratios.

i. sin x
ii. tan z
iii. cos x
iv. tan x.
Solution:

Question 3.
In right angled ∆LMN, ∠LMN = 90°, ∠L = 50° and ∠N = 40°. Write the following ratios.

i. sin 50°
ii. cos 50°
iii. tan 40°
iv. cos 40°
Solution:

Question 4.
In the given figure, ∠PQR = 90°, ∠PQS = 90°, ∠PRQ = α and ∠QPS = θ. Write the following trigonometric ratios.
i. sin α, cos α , tan α
ii. sin θ, cos θ, tan θ

Solution:
i. In ∆PQR,

ii. In ∆PQS,

Maharashtra Board Class 9 Maths Chapter 8 Trigonometry Practice Set 8.1 Intext Questions and Activities

Question 1.
In the figure gIven below, ∆PQR is a right angled triangle. Write the names of sides opposite and adjacent to ∠P and ∠R. (Textbook pg no. 102)

Solution:
In right angled ∆PQR,
i. side opposite to ∠P = QR
ii. side opposite to ∠R = PQ
iii. side adjacent to ∠P = PQ
iv. side adjacent to ∠R = QR

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 7.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 7 Co-ordinate Geometry.

Problem Set 7.2 Geometry 9th Std Maths Part 2 Answers Chapter 7 Co-ordinate Geometry

Question 1.
Choose the correct alternative answer for the following questions.

i. What is the form of co-ordinates of a point on the X-axis?
(A) (b,b)
(B) (0, b)
(C) (a, 0)
(D) (a, a)
Answer:
(C) (a, 0)

ii. Any point on the line y = x is of the form _____.
(A) (a, a)
(B) (0, a)
(C) (a, 0)
(D) (a, -a)
Answer:
(A) (a, a)

iii. What is the equation of the X-axis ?
(A) x = 0
(B) y = 0
(C) x + y = 0
(D) x = y
Answer:
(B) y = 0

iv. In which quadrant does the point (-4, -3) lie ?
(A) First
(B) Second
(C) Third
(D) Fourth
Answer:
(C) Third

v. What is the nature of the line which includes the points (-5, 5), (6, 5), (-3, 5), (0, 5)?
(A) Passes through the origin
(B) Parallel to Y-axis
(C) Parallel to X-axis
(D) None of these
Answer:
The y co-ordinate of all the points is the same.
∴ The line which passes through the given points is parallel to X-axis.
(C) Parallel to X-axis

vi. Which of the points P(-1, 1), Q(3, -4), R( -1, -1), S(-2, -3), T (-4, 4) lie in the fourth quadrant?
(A) P and T
(B) Q and R
(C) only S
(D) P and R
Answer:
(B) Q and R

Question 2.
Some points are shown in the adjoining figure. With the help of it answer the following questions :

i. Write the co-ordinates of the points Q and R.
ii. Write the co-ordinates of the points T and M.
iii. Which point lies in the third quadrant ?
iv. Which are the points whose x and y co-ordinates are equal ?
Solution:
i. Q(-2, 2) and R(4, -1)
ii. T(0, -1) and M(3, 0)
iii. Point S lies in the third quadrant.
iv. The x and y co-ordinates of point O are equal.

Question 3.
Without plotting the points on a graph, state in which quadrant or on which axis do the following points lie.
i. (5, -3)
ii. (-7, -12)
iii. (-23, 4)
iv. (-9, 5)
v. (0, -3)
vi. (-6, 0)
Solution:

Question 4.
Plot the following points on one and the same co-ordinate system.
A(1, 3), B(-3, -1), C(1, -4), D(-2, 3), E(0, -8), F(1, 0)
Solution:

Question 5.
In the graph alongside, line LM is parallel to the Y-axis.

i. What is the distance of line LM from the Y-axis?
ii. Write the co-ordinates of the points P, Q and R.
iii. What is the difference between the x co-ordinates of the points L and M?
Solution:
i. Distance of line LM from the Y-axis is 3 units.
ii. P(3, 2), Q (3, -1), R(3, 0)
iii. x co-ordinate of point L = 3
x co-ordinate of point M = 3
∴ Difference between the x co-ordinates of the points L and M = 3 – 3
= 0

Question 6.
How many lines are there which are parallel to X-axis and having a distance 5 units?
Solution:
The equation of a line parallel to the X-axis is y = b.
There are 2 lines which are parallel to X-axis and at a distance of 5 units.
Their equations are y = 5 and y = -5.

Question 7.
If ‘a’ is a real number, what is the distance between the Y-axis and the line x = a?
Solution:
Equation of Y-axis is x = 0.
Since, ‘a’ is a real number, there are two possibilities.
Case I: a > 0
Case II: a < 0 ∴ Distance between the Y-axis and the line x = a = a-0 = a Since, |a| = a, a > 0
= – a, a < 0
∴ Distance between the Y-axis and the line x = a is |a|.

Maharashtra Board Class 9 Maths Chapter 7 Co-ordinate Geometry Problem Set 7 Intext Questions and Activities

Question 1.
As shown in the adjoining figure, ask girls to sit in lines so as to form the X-axis and Y-axis.
i. Ask some boys to sit at the positions marked by the coloured dots in the four quadrants.
i. Now, call the students turn by turn using the initial letter of each student’s name. As his or her initial is called, the student stands and gives his or her own co-ordinates. For example Rajendra (2, 2) and Kirti (-1, 0)
iii. Even as they have fun during this field activity, the students will leam how to state the position of a point in a plane. (Textbook pg. no. 92)

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 7 Co-ordinate Geometry.

Practice Set 7.2 Geometry 9th Std Maths Part 2 Answers Chapter 7 Co-ordinate Geometry

Question 1.
On a graph paper plot the points A(3, 0), B(3, 3), C(0, 3). Join A, B and B, C. What is the figure formed?
Soiution:

d(O, A) = 3 cm, d(A, B) = 3 cm, d(B, C) = 3 cm, d(O, C) = 3 cm and each angle of □ OABC is 90°
∴ □ OABC is a square.

Question 2.
Write the equation of the line parallel to the Y-axis at a distance of 7 units from it to its left.
Solution:
The equation of a line parallel to the Y-axis is x = a.
Since, the line is at a distance of 7 units to the left of Y-axis,
∴ a = -7
∴ x = -1 is the equation of the required line.

Question 3.
Write the equation of the line parallel to the X-axis at a distance of 5 units from it and below the X-axis.
Solution:
The equation of a line parallel to the X-axis is y = b.
Since, the line is at a distance of 5 units below the X-axis.
∴ b = -5
∴ y = -5 is the equation of the required line.

Question 4.
The point Q( -3, -2) lies on a line parallel to the Y-axis. Write the equation of the line and draw its graph.
Solution:
The equation of a line parallel to the Y-axis is x = a.
Here, a = -3
∴ x = -3 is the equation of the required line.

Question 5.
Y-axis and line x = – 4 are parallel lines. What is the distance between them?
Solution:
Equation of Y-axis is x = 0.
Equation of the line parallel to the Y-axis is x = – 4. … [Given]
∴ Distance between the Y-axis and the line x = – 4 is 0 – (- 4) … [0 > -4]
= 0 + 4 = 4 units
∴ The distance between the Y-axis and the line x = – 4 is 4 units.
[Note: The question is modified as X-axis cannot be parallel to the line x = – 4.]

Question 6.
Which of the equations given below have graphs parallel to the X-axis, and which ones have graphs parallel to the Y-axis? [1 Mark each]
i. x = 3
ii. y – 2 = 0
iii. x + 6 = 0
iv. y = -5
Solution:
i. The equation of a line parallel to the Y-axis is x = a.
∴ The line x = 3 is parallel to the Y-axis.

ii. y – 2 = 0
∴ y = 2
The equation of a line parallel to the X-axis is y = b.
∴ The line y – 2 = 0 is parallel to the X-axis.

iii. x + 6 = 0
∴ x = -6
The equation of a line parallel to the Y-axis is x = a.
∴ The line x + 6 = 0 is parallel to the Y-axis.

iv. The equation of a line parallel to the X-axis is y = b.
∴ The line y = – 5 is parallel to the X-axis.

Question 7.
On a graph paper, plot the points A(2, 3), B(6, -1) and C(0, 5). If these points are collinear, then draw the line which includes them. Write the co-ordinates of the points at which the line intersects the X-axis and the Y-axis.
Solution:

From the graph, the line drawn intersects the X-axis at D(5, 0) and the Y-axis at C(0, 5).

Question 8.
Draw the graphs of the following equations on the same system of co-ordinates. Write the co-ordinates of their points of intersection.
x + 4 = 0,
y – 1 = 0,
2x + 3 = 0,
3y – 15 = 0
Solution:
i. x + 4 = 0
∴ x = – 4

ii. y – 1 = 0
∴ y = 1

iii. 2x + 3 = 0
∴2x = -3
∴ x = \(\frac { -3 }{ 2 }\)
∴ x = -1.5

iv. 3y- 15 = 0
3y = 15
y = \(\frac { 15 }{ 3 }\)
∴ y = 5

The co-ordinates of the point of intersection of x + 4 = 0 and y – 1 = 0 are A(-4, 1).
The co-ordinates of the point of intersection ofy – 1 = 0 and 2x + 3 = 0 are B(-1.5, 1).
The co-ordinates of the point of intersection of 3y – 15 = 0 and 2x + 3 = 0 are C(-1.5, 5).
The co-ordinates of the point of intersection of x + 4 = 0 and 3y – 15 = 0 are D(-4, 5).

Question 9.
Draw the graphs of the equations given below.
i. x + y = 2
ii. 3x – y = 0
iii. 2x + y = 1
Solution:
i. x + y = 2
∴ y = 2 – x
When x = 0,
y = 2 – x
= 2 – 0
= 2
When x = 1,
y = 2 – x
= 2 – 1
= 1
When x = 2,
y = 2 – x
= 0

ii. 3x – y = 0
∴ y = 3x
When x = 0,
y = 3x
= 3(0)
= 0

When x = 1,
y = 3x
= 3(1)
= 3

When x = -1,
y = 3x
= 3(-1)
= -3

iii. 2x + y = 1
∴ y = 1 – 2x
When x = 0,
y = 1 – 2x
= 1 – 2(0)
= 1 – o
When x = 1,
y = 1 – 2x
= 1- 2(1)
= 1 – 2
= -1
When x = -1,
y = 1 – 2x
= 1 – 2(-1)
= 1 + 2
= 3

Maharashtra Board Class 9 Maths Chapter 7 Co-ordinate Geometry Practice Set 7.2 Intext Questions and Activities

Question 1.
i. Can we draw a line parallel to the X-axis at a distance of 6 unIts from It and below the X-axis?
ii. Will all of the points (-3,-6), (10,-6), ( \(\frac { 1 }{ 2 }\), -6) be on that line?
iii. What would be the equation of this line?(Textbook pg. no. 94)
Solution:

i. Yes.
This line will pass through the point (0,-6).

ii. Yes.
Here, y co-ordinate of the points (-3, -6), (10,-6), ( \(\frac { 1 }{ 2 }\), -6) is the same, which is -6.
∴ All the above points lie on the same line.

iii. Since, the line is at a distance of 6 units below the X-axis.
∴ b = -6
∴ Equation of the line is y = -6.

Question 2.
i. Can we draw a line parallel to the Y – axis at a distance of 2 units from ¡t and to its right?
ii. Will all of the points (2, 10), (2, 8), (2, -) be on that line?
iii. What would be the equation of this line? (Textbook pg. no. 95)
Solution:

i. Yes.
(2, 10)
This line will pass through the point (2, 0).
(2,8)
ii. Yes.
Here, x co-ordinate of the points (2, 10), (2, 8), (2,-\(\frac { 1 }{ 2 }\) ) is the same, which is 2.
∴ All the above points lie on the same line.

iii. Since, the line is at a distance of 2 units to the right of Y-axis.
a = 2
∴ Equation of the line is x = 2.

Question 3.
On a graph paper, plot the points (0, 1), (1, 3), (2, 5). Are they collinear? If so, draw the line that passes through them.
i. Through which quadrants does this line pass ?
ii. Write the co-ordinates of the point at which it intersects the Y-axis.
iii. Show any point in the third quadrant which lies on this line. Write the co-ordinates of the point. (Textbook pg. no. 96)
Solution:
i. The line passes through the quadrants I, II and III.
ii. The line intersects the Y-axis at (0, 1).
iii. (-1,-1)

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 6 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 6 Circle.

Problem Set 6 Geometry 9th Std Maths Part 2 Answers Chapter 6 Circle

Question 1.
Choose correct alternative answer and fill in the blanks.

i. Radius of a circle is 10 cm and distance of a chord from the centre is 6 cm. Hence, the length of the chord is ____.
(A) 16 cm
(B) 8 cm
(C) 12 cm
(D) 32 cm
Answer:

∴ OA² = AC² + OC²
∴ 10² = AC² + 6²
∴ AC² = 64
∴ AC = 8 cm
∴ AB = 2(AC)= 16 cm
(A) 16 cm

ii. The point of concurrence of all angle bisectors of a triangle is called the ____.
(A) centroid
(B) circumcentre
(C) incentre
(D) orthocentre
Answer:
(C) incentre

iii. The circle which passes through all the vertices of a triangle is called ____.
(A) circumcircle
(B) incircle
(C) congruent circle
(D) concentric circle
Answer:
(A) circumcircle

iv. Length of a chord of a circle is 24 cm. If distance of the chord from the centre is 5 cm, then the radius of that circle is ____.
(A) 12 cm
(B) 13 cm
(C) 14 cm
(D) 15 cm
Answer:

OA² = AC² + OC²
∴ OA² = 12² + 5²
∴ OA² = 169
∴ OA = 13 cm
(B) 13 cm

v. The length of the longest chord of the circle with radius 2.9 cm is ____.
(A) 3.5 cm
(B) 7 cm
(C) 10 cm
(D) 5.8 cm
Answer:
Longest chord of the circle = diameter = 2 x radius = 2 x 2.9 = 5.8 cm
(D) 5.8 cm

vi. Radius of a circle with centre O is 4 cm. If l(OP) = 4.2 cm, say where point P will lie ____.
(A) on the centre
(B) inside the circle
(C) outside the circle
(D) on the circle
Answer:
l(OP) > radius
∴Point P lies in the exterior of the circle.
(C) outside the circle

vii. The lengths of parallel chords which are on opposite sides of the centre of a circle are 6 cm and 8 cm. If radius of the circle is 5 cm, then the distance between these chords is _____.
(A) 2 cm
(B) 1 cm
(C) 8 cm
(D) 7 cm
Answer:

PQ = 8 cm, MN = 6 cm
∴ AQ = 4 cm, BN = 3 cm
∴ OQ² = OA² + AQ²
∴ 5² = OA² + 4²
∴ OA² = 25 – 16 = 9
∴ OA = 3 cm
Also, ON² = OB² + BN²
∴ 5² = OB² + 3²
∴ OB = 4 cm
Now, AB = OA + OB = 3 + 4 = 7 cm

Question 2.
Construct incircle and circumcircle of an equilateral ADSP with side 7.5 cm. Measure the radii of both the circles and find the ratio of radius of circumcircle to the radius of incircle.
Solution:


Steps of construction:
i. Construct ∆DPS of the given measurement.
ii. Draw the perpendicular bisectors of side DP and side PS of the triangle.
iii. Name the point of intersection of the perpendicular bisectors as point C.
iv. With C as centre and CM as radius, draw a circle which touches all the three sides of the triangle.
v. With C as centre and CP as radius, draw a circle which passes through the three vertices of the triangle.

Radius of incircle = 2.2 cm and Radius of circumcircle = 4.4 cm

Question 3.
Construct ∆NTS where NT = 5.7 cm. TS = 7.5 cm and ∠NTS = 110° and draw incircle and circumcircle of it.
Solution:


Steps of construction:
For incircle:
i. Construct ∆NTS of the given measurement.
ii. Draw the bisectors of ∠T and ∠S. Let these bisectors intersect at point I.
iii. Draw a perpendicular IM on side TS. Point M is the foot of the perpendicular.
iv. With I as centre and IM as radius, draw a circle which touches all the three sides of the triangle.
For circumcircle:
i. Draw the perpendicular bisectors of side NT and side TS of the triangle.
ii. Name the point of intersection of the perpendicular bisectors as point C.
iii. Join seg CN
iv. With C as centre and CN as radius, draw a circle which passes through the three vertices of the triangle.

Question 4.
In the adjoining figure, C is the centre of the circle, seg QT is a diameter, CT = 13, CP = 5. Find the length of chord RS.

Given: In a circle with centre C, QT is a diameter, CT = 13 units, CP = 5 units
To find: Length of chord RS
Construction: Join points R and C.
Solution:

i. CR = CT= 13 units …..(i) [Radii of the same circle]
In ∆CPR, ∠CPR = 90°
∴ CR² = CP² + RP² [Pythagoras theorem]
∴ 13² = 5² + RP² [From (i)]
∴ 169 = 25 + RP² [From (i)]
∴ RP² = 169 – 25 = 144
∴ RP = \(\sqrt { 144 }\) [Taking square root on both sides]
∴ RP = 12 cm ….(ii)

ii. Now, seg CP _L chord RS [Given]
∴ RP = \(\frac { 1 }{ 2 }\) RS [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ 12 = \(\frac { 1 }{ 2 }\) RS [From (ii)]
∴ RS = 2 x 12 = 24
∴ The length of chord RS is 24 units.

Question 5.
In the adjoining figure, P is the centre of the circle. Chord AB and chord CD intersect on the diameter at the point E. If ∠AEP ≅ ∠DEP, then prove that AB = CD.

Given: P is the centre of the circle.
Chord AB and chord CD intersect on the diameter at the point E. ∠AEP ≅ ∠DEP
To prove: AB = CD
Construction: Draw seg PM ⊥ chord AB, A-M-B
seg PN ⊥ chord CD, C-N-D
Proof:

∠AEP ≅ ∠DEP [Given]
∴ Seg ES is the bisector of ∠AED.
PoInt P is on the bisector of ∠AED.
∴ PM = PN [Every point on the bisector of an angle is equidistant from the sides of the angle.]
∴ chord AB ≅ chord CD [Chords which are equidistant from the centre are congruent.]
∴ AB = CD [Length of congruent segments]

Question 6.
In the adjoining figure, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to chord AB at point E. Show that ∆ABC is an isosceles triangle.

Given: O is the centre of the circle.
diameter CD ⊥ chord AB, A-E-B
To prove: ∆ABC is an isosceles triangle.
Proof:
diameter CD ⊥ chord AB [Given]
∴ seg OE ⊥ chord AB [C-O-E, O-E-D]
∴ seg AE ≅ seg BE ……(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord]
In ∆CEA and ∆CEB,
∠CEA ≅ ∠CEB [Each is of 90°]
seg AE ≅ seg BE [From (i)]
seg CE ≅ seg CE [Common side]
∴ ∆CEA ≅ ∆CEB [SAS test]
∴ seg AC ≅ seg BC [c. s. c. t.]
∴ ∆ABC is an isosceles triangle.

Maharashtra Board Class 9 Maths Chapter 6 Circle Problem Set 6 Intext Questions and Activities

Question 1.
Every student in the group should do this activity. Draw a circle in your notebook. Draw any chord of that circle. Draw perpendicular to the chord through the centre of the circle. Measure the lengths of the two parts of the chord. Group leader should prepare a table as shown below and ask other students to write their observations in it. Write the property which you have observed. (Textbook pg. no. 77)


Answer:
On completing the above table, you will observe that the perpendicular drawn from the centre of a circle on its chord bisects the chord.

Question 2.
Every student from the group should do this activity. Draw a circle in your notebook. Draw a chord of the circle. Join the midpoint of the chord and centre of the circle. Measure the angles made by the segment with the chord.
Discuss about the measures of the angles with your friends. Which property do the observations suggest ? (Textbook pg. no. 77)

Answer:
The meausure of the angles made by the drawn segment with the chord is 90°. Thus, we can conclude that, the segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord.

Question 3.
Draw circles of convenient radii. Draw two equal chords in each circle. Draw perpendicular to each chord from the centre. Measure the distance of each chord from the centre. What do you observe? (Textbook pg. no. 79)
Answer:
Congruent chords of a circle are equidistant from the centre.

Question 4.
Measure the lengths of the perpendiculars on chords in the following figures.

Did you find OL = OM in fig (i), PN = PT in fig (ii) and MA = MB in fig (iii)?
Write the property which you have noticed from this activity. (Textbook pg. no. 80)
Answer:
In each figure, the chords are equidistant from the centre. Also, we can see that the measures of the chords in each circle are equal.
Thus, we can conclude that chords of a circle equidistant from the centre of a circle are congruent.

Question 5.
Draw different triangles of different measures and draw in circles and circumcircles of them. Complete the table of observations and discuss. (Textbook pg. no. 85)
Answer:

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 6.3 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 6 Circle.

Practice Set 6.3 Geometry 9th Std Maths Part 2 Answers Chapter 6 Circle

Question 1.
Construct ∆ABC such that ∠B =100°, BC = 6.4 cm, ∠C = 50° and construct its incircle.
Solution:


Steps of construction:
i. Construct ∆ABC of the given measurement.
ii. Draw the bisectors of ∠B and ∠C. Let these bisectors intersect at point I.
iii. Draw a perpendicular IM on side BC. Point M is the foot of the perpendicular.
iv. With I as centre and IM as radius, draw a circle which touches all the three sides of the triangle.

Question 2.
Construct ∆PQR such that ∠P = 70°, ∠R = 50°, QR = 7.3 cm and construct its circumcircle.
Solution:
In ∆PQR,
m∠P + m∠Q + m∠R = 180° … [Sum of the measures of the angles of a triangle is 180°]
∴ 70° + m∠Q + 50° = 180°
∴ m∠Q = 180° – 70° + m∠Q + 50° = 180°
∴ m∠Q = 180° – 70° – 50°
∴ m∠Q = 60°


Steps of construction:
i. Construct A PQR of the given measurement.
ii. Draw the perpendicular bisectors of side PQ and side QR of the triangle.
iii. Name the point of intersection of the perpendicular bisectors as point C.
iv. Join seg CP
v. With C as centre and CP as radius, draw a circle which passes through the three vertices of the triangle.

Question 3.
Construct ∆XYZ such that XY = 6.7 cm, YZ = 5.8 cm, XZ = 6.9 cm. Construct its incircle.
Solution:


Steps of construction:
i. Construct ∆XYZ of the given measurement
ii. Draw the bisectors of ∠X and ∠Z. Let these bisectors intersect at point I.
iii. Draw a perpendicular IM on side XZ. Point M is the foot of the perpendicular.
iv. With I as centre and IM as radius, draw a circle which touches all the three sides of the triangle.

Question 4.
In ∆LMN, LM = 7.2 cm, ∠M = 105°, MN = 6.4 cm, then draw ∆LMN and construct its circumcircle.
Solution:


Steps of construction:
i. Construct ∆LMN of the given measurement.
ii. Draw the perpendicular bisectors of side MN and side ML of the triangle.
iii. Name the point of intersection of the perpendicular bisectors as point C.
iv. Join seg CM
v. With C as centre and CM as radius, draw a circle which passes through the three vertices of the triangle.

Question 5.
Construct ∆DEF such that DE = EF = 6 cm. ∠F = 45° and construct its circumcircle.
Solution:


Steps of construction:
i. Construct ∆DEF of the given measurement.
ii. Draw the perpendicular bisectors of side DE and side EF of the triangle.
iii. Name the point of intersection of perpendicular bisectors as point C.
iv. Join seg CE
v. With C as centre and CE as radius, draw a circle which passes through the three vertices of the triangle.

Maharashtra Board Class 9 Maths Chapter 6 Circle Practice Set 6.3 Intext Questions and Activities

Question 1.
Draw any equilateral triangle. Draw incircle and circumcircle of it. What did you observe while doing this activity? (Textbook pg. no. 85)
i. While drawing incircle and circumcircle, do the angle bisectors and perpendicular bisectors coincide with each other?
ii. Do the incentre and circumcenter coincide with each other? If so, what can be the reason of it?
iii. Measure the radii of incircle and circumcircle and write their ratio.
Solution:


Steps of construction:
i. Construct equilateral ∆XYZ of any measurement.
ii. Draw the perpendicular bisectors of side XY and side YZ of the triangle.
iii. Draw the bisectors of ∠X and ∠Z.
iv. Name the point of intersection of the perpendicular bisectors and angle bisectors as point I.
v. With I as centre and IM as radïus, draw a circle which touches all the three sides of the triangle.
vi. With I as centre and IZ as radius, draw a circle which passes through the three vertices of the triangle.
[Note: Here, point of intersection of perpendicular bisector and angle bisector is same.]

i. Yes.
ii. Yes.
The angle bisectors of the angles and the perpendicular bisectors of the sides of an equilateral triangle are coincedent. Hence, its incentre and circumcentre coincide.
iii. Radius of circumcircle = 3.6 cm,
Radius of incircle = 1.8 cm
\(\text { Ratio }=\frac{\text { Radius of circumcircle }}{\text { Radius of incircle }}=\frac{3.6}{1.8}=\frac{2}{1}=2 : 1\)

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 5 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 5 Quadrilaterals.

Problem Set 5 Geometry 9th Std Maths Part 2 Answers Chapter 5 Quadrilaterals

Question 1.
Choose the correct alternative answer and fill in the blanks.

i. If all pairs of adjacent sides of a quadrilateral are congruent, then it is called ____.
(A) rectangle
(B) parallelogram
(C) trapezium
(D) rhombus
Answer:
(D) rhombus

ii. If the diagonal of a square is 22√2 cm, then the perimeter of square is ____.
(A) 24 cm
(B) 24√2 cm
(C) 48 cm
(D) 48√2 cm
Answer:
In ∆ABC,
AC² = AB² + BC²

∴ (12² √2 )² = AB² + AB²
∴ \( A B^{2}=\frac{12^{2} \times 2}{2}=12^{2}\)
∴ AB = 12 cm
∴ Perimeter of □ABCD = 4 x 12 = 48 cm
(C) 48 cm

iii. If opposite angles of a rhombus are (2x)° and (3x – 40)°, then the value of x is ____.
(A) 100°
(B) 80°
(C) 160°
(D) 40°
Answer:
2x = 3x – 40 … [Pythagoras theorem]
∴ x = 40°
(D) 40°

Question 2.
Adjacent sides of a rectangle are 7 cm and 24 cm. Find the length of its diagonal.
Solution:

Let □ABCD be the rectangle.
AB = 7 cm, BC = 24 cm
In ∆ABC, ∠B = 90° [Angle of a rectangle]
AC² = AB² + BC² [Pythagoras theorem]
= 7² + 24²
= 49 + 576
= 625
AC = √625 [Taking square root of both sides]
= 25 cm
∴ The length of the diagonal of the rectangle is 25 cm.

Question 3.
If diagonal of a square is 13 cm, then find its side.
Solution:

Let □PQRS be the square of side x cm.
∴ PQ = QR = x cm …..(i) [Sides of a square]
∴ In ∆PQR, ∠Q = 90° [Angle of a square]
∴ PR² = PQ² + QR² [Pythagoras theorem]
∴ 13 = x + x [From (i)]
∴ 169 = 2x²

The length of the side of the square is 6.5√2 cm.

Question 4.
Ratio of two adjacent sides of a parallelogram is 3 : 4, and its perimeter is 112 cm. Find the length of its each side.
Solution:

Let □STUV be the parallelogram.
Ratio of two adjacent sides of a parallelogram is 3 : 4.
Let the common multiple be x.
ST = 3x cm and TU = 4x cm
∴ ST = UV = 3x cm
TU = SV = 4x cm …..(i) [Opposite sides of a parallelogram]
Perimeter of □STUV = 112 [Given]
∴ ST + TU + UV + SV = 112
∴ 3x + 4x + 3x + 4x = 112 [From (i)]
∴ 14x = 112
∴ x = \(\frac { 112 }{ 14 }\)
∴ x = 8
∴ ST = UV = 3x = 3 x 8 = 24 cm
∴ TU = SV = 4x = 4 x 8 = 32 cm [From (i)]
∴ The lengths of the sides of the parallelogram are 24 cm, 32 cm, 24 cm and 32 cm.

Question 5.
Diagonals PR and QS of a rhombus PQRS are 20 cm and 48 cm respectively. Find the length of side PQ.
Solution:

□PQRS is a rhombus. [Given]
PR = 20 cm and QS = 48 cm [Given]
∴ PT = \(\frac { 1 }{ 2 }\) PR [Diagonals of a rhombus bisect each other]
= \(\frac { 1 }{ 2 }\) x 20 = 10 cm
Also, QT = \(\frac { 1 }{ 2 }\) QS [Diagonals of a rhombus bisect each other]
= \(\frac { 1 }{ 2 }\) x 48 = 24 cm

ii. In ∆PQT, ∠PTQ = 90° [Diagonals of a rhombus are perpendicular to each other]
∴ PQ² = PT² + QT² [Pythagoras- theorem]
= 10² + 24²
= 100 + 576
∴ PQ² = 676
∴ PQ = \(\sqrt {676 }\) [Taking square root of both sides]
= 26 cm
∴ The length of side PQ is 26 cm.

Question 6.
Diagonals of a rectangle PQRS are intersecting in point M. If ∠QMR = 50°, then find the measure of ∠MPS.
Solution:

□PQRS is a rectangle.
∴ PM = \(\frac { 1 }{ 2 }\) PR …(i)
MS = \(\frac { 1 }{ 2 }\) QS …(ii) [Diagonals of a rectangle bisect each other]
Also, PR = QS …..(iii) [Diagonals of a rectangle are congruent]
∴ PM = MS ….(iv) [From (i), (ii) and (iii)]
In ∆PMS,
PM = MS [From (iv)]
∴ ∠MSP = ∠MPS = x° …..(v) [Isosceles triangle theorem]
∠PMS = ∠QMR = 50° ……(vi) [Vertically opposite angles]
In ∆MPS,
∠PMS + ∠MPS + ∠MSP = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 50° +x + x = 180° [From (v) and (vi)]
∴ 50° + 2x= 180
∴ 2x= 180-50
∴ 2x= 130
∴ x = \(\frac { 130 }{ 2 }\) = 65°
∴ ∠MPS = 65° [From (v)]

Question 7.
In the adjoining figure, if seg AB || seg PQ , seg AB ≅ seg PQ, seg AC || seg PR, seg AC ≅ seg PR, then prove that seg BC || seg QR and seg BC ≅ seg QR.

Solution:
Given: seg AB || seg PQ , seg AB ≅ seg PQ,
seg AC || seg PR, seg AC ≅ seg PR
To prove: seg BC || seg QR, seg BC ≅ seg QR
Proof:
Consider □ABQP,
seg AB || seg PQ [Given]
seg AB ≅ seg PQ [Given]
∴ □ABQP is a parallelogram. [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent]
∴ segAP || segBQ …..(i)
∴ seg AP ≅ seg BQ …..(ii) [Opposite sides of a parallelogram]
Consider □ACRP,
seg AC || seg PR [Given]
seg AC ≅ seg PR [Given]
∴ □ACRP is a parallelogram. [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent]
∴ seg AP || seg CR …(iii)
∴ seg AP ≅ seg CR …….(iv) [Opposite sides of a parallelogram]
Consider □BCRQ,
seg BQ || seg CR
seg BQ ≅ seg CR
∴ □BCRQ is a parallelogram. [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent]
∴ seg BC || seg QR
∴ seg BC ≅ seg QR [Opposite sides of a parallelogram]

Question 8.
In the adjoining figure, □ABCD is a trapezium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively. Then prove that PQ || AB and PQ = \(\frac { 1 }{ 2 }\) ( AB + DC).

Given : □ ABCD is a trapezium.
To prove:
Construction: Join points A and Q. Extend seg AQ and let it meet produced DC at R.
Proof:

seg AB || seg DC [Given]
and seg BC is their transversal.
∴ ∠ABC ≅ ∠RCB [Alternate angles]
∴ ∠ABQ ≅ ∠RCQ ….(i) [B-Q-C]
In ∆ABQ and ∆RCQ,
∠ABQ ≅∠RCQ [From (i)]

seg BQ ≅ seg CQ [Q is the midpoint of seg BC]
∠BQA ≅ ∠CQR [Vertically opposite angles]
∴ ∆ABQ ≅ ∆RCQ [ASA test]
seg AB ≅ seg CR …(ii) [c. s. c. t.]
seg AQ ≅ seg RQ [c. s. c. t.]
∴ Q is the midpoint of seg AR. ….(iii)

In ∆ADR,
Points P and Q are the midpoints of seg AD and seg AR respectively. [Given and from (iii)]

∴ seg PQ || seg DR [Midpoint theorem]
i.e. seg PQ || seg DC ……..(iv) [D-C-R]
But, seg AB || seg DC …….(v) [Given]
∴ seg PQ || seg AB [From (iv) and (v)]
In ∆ADR,

Question 9.
In the adjoining figure, □ABCD is a trapezium. AB || DC. Points M and N are midpoints of diagonals AC and DB respectively, then prove that MN || AB.

Solution:
Given: □ABCD is a trapezium. AB || DC.
Points M and N are midpoints of diagonals AC and DB respectively.
To prove: MN || AB
Construction: Join D and M. Extend seg DM to meet seg AB at point E such that A-E-B.
Proof:
seg AB || seg DC and seg AC is their transversal. [Given]
∴ ∠CAB ≅ ∠ACD [Alternate angles]
∴ ∠MAE ≅ ∠MCD ….(i) [C-M-A, A-E-B]
In ∆AME and ∆CMD,

∠AME ≅ ∠CMD [Vertically opposite angles]
seg AM ≅ seg CM [M is the midpoint of seg AC]
∠MAE ≅∠MCD [From (i)]
∴ ∆AME ≅ ∆CMD [ASA test]
∴ seg ME ≅ seg MD [c.s.c.t]
∴ Point M is the midpoint of seg DE. …(ii)
In ∆DEB,

Points M and N are the midpoints of seg DE and seg DB respectively. [Given and from (ii)]
∴ seg MN || seg EB [Midpoint theorem]
∴ seg MN || seg AB [A-E-B]

Maharashtra Board Class 9 Maths Chapter 5 Quadrilaterals Problem Set 5 Intext Questions and Activities

Question 1.
Draw five parallelograms by taking various measures of lengths and angles. (Textbook page no. 59)

Question 2.
Draw a parallelogram PQRS. Draw diagonals PR and QS. Denote the intersection of diagonals by letter O. Compare the two parts of each diagonal with a divider. What do you find? (Textbook page no. 60)

Answer:
seg OP = seg OR, and seg OQ = seg OS
Thus we can conclude that, point O divides the diagonals PR and QS in two equal parts.

Question 3.
To verify the different properties of quadrilaterals.

Material: A piece of plywood measuring about 15 cm x 10 cm, 15 thin screws, twine, scissor.
Note: On the plywood sheet, fix five screws in a horizontal row keeping a distance of 2 cm between any two adjacent screws. Similarly make two more rows of screws exactly below the first one. Take care that the vertical distance between any two adjacent screws is also 2 cm.
With the help of the screws, make different types of quadrilaterals of twine. Verify the properties of sides and angles of the quadrilaterals. (Textbook page no. 75)