Class 9

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

Note: Assume π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.
Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Answer:
For the given cone, diameter d = 10.5 cm.
Then, radius r = \(\frac{10.5}{2}\) cm and slant height
l = 10 cm.
Curved surface area of a cone
= πrl
= \(\frac{22}{7}\) × \(\frac{10.5}{2}\) × 10 cm²
= \(\frac{22}{7}\) × \(\frac{105}{2}\) × cm²
= 11 × 15 cm²
= 165 cm²
Thus, the curved surface area of the given cone is 165 cm².

Question 2.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Answer:
For the given cone, diameter d = 24 m.
Then, radius r = \(\frac{24}{2}\) = 12 m and slant height l = 21 m.
Total surface area of a cone
= πr (l + r)
= \(\frac{22}{7}\) × 12(21 + 12) m²
= \(\frac{22 \times 12 \times 33}{7}\) m²
= \(\frac{8712}{7}\) m²
= 1244.57 m²
Thus, the total surface area of the given cone is 1244.57 m².

Question 3.
Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find,
(i) radius of the base and
(ii) total surface area of the cone.
Answer:
For the given cone, slant height l = 14 cm and curved surface area = 308 cm²
(i) Curved surface area of a cone = πrl
∴ 308 cm² = \(\frac{22}{7}\) × r × 14 cm
∴ \(\frac{308 \times 7}{22 \times 14}\) cm = r
∴ r = 7 cm
Thus, the radius of the base of the cone is 7 cm.

(ii) Total surface area of a cone
= πrl + πr²
= 308 + \(\frac{22}{7}\) × 7 × 7 cm²
= 308 + 154 cm²
= 462 cm²
Thus, the total surface area of the cone is 462 cm².

Question 4.
A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is ₹ 70.
Answer:
For the conical tent,
radius r = 24 m and height h = 10 m.

(i) l = \(\sqrt{h^{2}+r^{2}}\)
= \(\sqrt{10^{2}+24^{2}}\)
= \(\sqrt{100+576}\)
= \(\sqrt{676}\)
∴ l = 26 m
Thus, the slant height of the tent is 26 m.

(ii) Area of the canvas used to make tent
= Curved surface area of the conical tent
= πrl
= \(\frac{22}{7}\) × 24 × 26 m²
= \(\frac{13728}{7}\) m²
Cost of 1 m² canvas = ₹ 70
∴ Cost of \(\frac{13728}{7}\) m² canvas
= ₹ \(\left(70 \times \frac{13728}{7}\right)\)
= ₹ 1,37,280
Thus, the cost of canvas required is ₹ 1,37,280.

Question 5.
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm.
(Use π = 3.14)
Answer:
For the conical tent to be made, radius r = 6 m and height h = 8 m.
l² = h² + r² = 8² + 6² = 64 + 36 = 100
∴ l = √100 = 10 m
Area of the tarpaulin used in making tent
= Curved surface area of conical tent
= πrl
= 3.14 × 6 × 10 m²
= 188.4 m²
Now, the width of the tarpaulin is 3 m.
∴ Length of tarpaulin required = \(\frac{188.4}{3}\) m
= 62.8 m
But, 20 cm, i.e., 0.2 m of tarpaulin is required more for margins and wastage.
∴ Total length of the tarpaulin required = 62.8 + 0.2 m = 63 m
Thus, total length of tarpaulin required is 63 m.

Question 6.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of ₹ 210 per 100 m².
Answer:
For the given conical tomb,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{14}{2}\) = 7 m
and slant height l = 25 m. .
Area of the region to be whitewashed
= Curved surface area of the conical tomb
= πrl
= \(\frac{22}{7}\) × 7 × 25 m²
= 550 m²
Cost of whitewashing 100 m² region = ₹ 210
∴ Cost of whitewashing 550 m² region
= ₹ \(\left(\frac{210 \times 550}{100}\right)\)
= ₹ 1155
Thus, the cost of whitewashing the curved surface of the tomb is ₹ 1155.

Question 7.
A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Answer:
For the conical cap, radius r = 7 cm and height h = 24 cm.
l = \(\sqrt{h^{2}+r^{2}}\)
= \(\sqrt{24^{2}+7^{2}}\)
= \(\sqrt{576+49}\)
= \(\sqrt{625}\) = 25 cm
Area of the sheet required to make 1 conical cap
= Curved surface area of the conical cap
= πrl
= \(\frac{22}{7}\) × 7 × 25 cm²
= 550 cm²
Area of sheet required to make 1 cap = 550 cm²
∴ Area of sheet required to make 10 caps
= 550 × 10 cm²
= 5500 cm²
Thus, the area of the sheet required to make 10 caps is 5500 cm²

Question 8.
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled carboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹ 12 per m², what will be the cost of painting all these cones? (Use π = 3.14 and take \(\sqrt{1.04}\) = 1.02)
Answer:
For the given cone,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{40}{2}\) = 20 cm = 0.2 m and
height h = 1 m.
l = \(\sqrt{h^{2}+r^{2}}\)
= \(\sqrt{1^{2}+0.2^{2}}\)
= \(\sqrt{1.04}\)
= 1.02 m
Curved surface area of a cone
= πrl
= 3.14 × 0.2 × 1.02 m²
∴ Curved surface area of 50 cones
= 50 × 3.14 × 0.2 × 1.02 m²
= 32.028 m²
Cost of painting 1 m² region = ₹ 12
∴ Cost of painting 32.028 m² region
= ₹ (12 × 32.028)
= ₹ 384.34 (approx.)
Thus, the cost of painting all the 50 cones is ₹ 384.34 (approx.)

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

Note: Assume π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.
The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder.
Answer:
Height of cylinder h = 14 cm.
Curved surface area of a cylinder = 2 πrh
∴ 88 cm² = 2 × r × 14cm
∴ \(\frac{88 \times 7}{2 \times 22 \times 14}\) cm = r
∴ r = 1 cm
Now, diameter of the cylinder = 2r = 2 × 1 cm
= 2 cm
Thus, the diameter of the base of the cylinder is 2 cm.

Question 2.
It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same ?
Answer:
Height of cylindrical tank h = 1 m
Diameter of the cylinder =140 cm
∴ Radius of the cylinder r = \(\frac{\text { diameter }}{2}\)
= \(\frac{140}{2}\) cm
= 70 cm
= 0.7 m
Total surface area of the closed cylindrical tank
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × 0.7 (0.7 + 1) m²
= 4.4 × 1.7 m²
= 7.48 m²
Thus, 7.48 m² sheet is required to make the closed cylindrical tank.

Question 3.
A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see the given figure). Find its

(i) inner curved surface area,
Answer:
For inner cylinder, diameter = 4 cm
∴ For inner cylinder,
radius r = \(\frac{\text { diameter }}{2}\) = 2 cm
and height (length) h = 77 cm.
Inner curved surface area of the pipe
= 2πrh
= 2 × \(\frac{22}{7}\) × 2 × 77 cm²
= 968 cm²
Thus, the inner curved surface area is 968 cm².

(ii) outer curved surface, area,
Answer:
For outer cylinder, diameter = 4.4 cm
∴ For outer cylinder,
radius R = \(\frac{\text { diameter }}{2}\) = \(\frac{4.4}{2}\) = 2.2
and height h = 77 cm.
Outer curved surface area of the pipe
= 2πRh
= 2 × \(\frac{22}{7}\) × 2 × 77 cm²
= 1064.8 cm²
Thus, the outer curved surface area is
1064.8 cm².

(iii) total surface area.
Answer:
Total surface area includes the area of two circular rings at the ends together with the inner and outer curved surface areas.
For each circular ring, outer radius R = 2.2 cm and inner radius r = 2 cm
Area of one circular ring
= π(R² – r²)
= \(\frac{22}{7}\)(2.2² – 2²)cm²
= \(\frac{22}{7}\) (4.84 – 4) cm²
= \(\frac{22}{7}\) × 0.84 cm²
= 2.64 cm²
∴ Area of two circular rings.
= 2 × 2.64 cm²
= 5.28 cm²
Now, total surface area of the pipe = Inner curved surface area + outer curved surface area + area of two circular rings
= 968 + 1064.8 + 5.28 cm²
= 2038.08 cm²

Question 4.
The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².
Answer:
For the cylindrical roller, diameter d = 84 cm and height (length) h = 120 cm.
Curved surface area of the cylindrical roller
= πdh
= \(\frac{22}{7}\) × 84 × 120 cm²
= 31680 cm²
= \(\frac{31680}{10000}\) m²
= 3.168 m²
Thus, the area of playground levelled in 1 complete revolution of the roller = 3.168 m²
∴ The area of playground levelled in 500 complete revolutions of the roller
= 3.168 × 500 m² = 1584 m²
Thus, the area of the playground is 1584 m².

Question 5.
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹ 12.50 per m².
Answer:
For the cylindrical pillar, diameter d = 50 cm = 0.5 m and height h = 3.5 m.
Curved surface area of the cylindrical pillar
= πdh
= \(\frac{22}{7}\) × 0.5 × 3.5 m²
= 5.5 m²
Cost of painting 1 m² area = ₹ 12.50
∴ Cost of painting 5.5 m2 area = ₹ (12.50 x 5.5)
= ₹ 68.75
Thus, the cost of painting the curved surface of the pillar is ₹ 68.75.

Question 6.
Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.
Answer:
For the given cylinder, radius r = 0.7 m and
curved surface area = 4.4 m².
Curved surface area of a cylinder = 2πrh
∴ 4.4 m² = 2 × \(\frac{22}{7}\) × 0.7m × h
∴ h = \(\frac{4.4 \times 7}{2 \times 22 \times 0.7}\)m
∴ h = 1 m
Thus, the height of the cylinder is 1 m.

Question 7.
The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of ₹ 40 per m².
Answer:
A circular well means a cylindrical well. For the cylindrical well, diameter d = 3.5 m and height (depth) h = 10 m.
(i) Curved surface area of the well
= πdh
= \(\frac{22}{7}\) × 3.5 × 10 m²
= 110 m²

(ii) Cost of plastering 1 m² region = ₹ 40
∴ Cost of plastering 110 m2 region
= ₹ (40 × 110)
= ₹ 4400

Question 8.
In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Answer:
For the cylindrical pipe, diameter d = 5 cm = 0.05 m and height (length) h = 28 m.
The radiation surface in the system is the •curved surface of the pipe.
Hence, we find the curved surface area of the cylindrical pipe.
Curved surface area of the cylindrical pipe
= πdh
= \(\frac{22}{7}\) × 0.05 × 28 m²
= 4.4 m²
Thus, the total radiating surface in the system is 4.4 m².

Question 9.
Find: (i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if \(\frac{1}{12}\) of the steel actually used was wasted in making the tank.
Answer:
For the closed cylindrical tank, diameter d = 4.2 m, hence radius
r = \(\frac{4.2}{2}\) = 2.1 m and height h = 4.5 m.

(i) Curved surface area of the cylindrical tank
= 2 πrh
= 2 × \(\frac{22}{7}\) × 2.1 × 4.5 m²
= 59.4 m²

(ii) Total surface area of the closed cylindrical tank
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × 2.1 (2.1 + 4.5) m²
= 13.2 × 6.6 m²
= 87.12 m²
Suppose, x m² steel was used for making the tank. But during production, \(\frac{1}{12}\) of the steel was wasted.
∴ Actual quantity of steel used = \(\frac{11}{12}\)x m².
Hence, \(\frac{11}{12}\)x = 87.12
∴ x = \(\frac{8712}{100} \times \frac{12}{11}\)
∴ x = 95.04 m²
Thus, the quantity of steel actually used during the preparation of the tank is 95.04 m².

Question 10.
In the given figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Answer:
The shape of the decorative cloth will be cylindrical.
For the cylinder of cloth, diameter d = 20 cm and height h = 30 cm + 2.5 cm + 2.5 cm = 35 cm.
Curved surface area of the cylinder of cloth
= πdh
= \(\frac{22}{7}\) × 20 × 35 cm²
Thus, 2200 cm² cloth is required for covering the lampshade.

Question 11.
The students of Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboards. If there were 35 competitors, how much cardboard was required to be bought for the competition ?
Answer:
The cylindrical penholders to be made have base but open at the top. Thus, to prepare a penholder, the area of the cardboard required will be given by the curved surface area of the cylinder and the area of base.

For cylindrical penholder, radius r = 3 cm and height h = 10.5 cm.
Area of cardboard required for 1 penholder
= Curved surface area of cylinder + Area of base
= 2πrh + πr²
= πr (2h + r)
= \(\frac{22}{7}\) × 3(2 × 10.5 + 3) cm2
= \(\frac{66 \times 24}{7}\) cm²
∴ Area of the cardboard required for 35 penholders
= 35 × \(\frac{66 \times 24}{7}\) cm²
= 7920 cm²
Thus, 7920 cm² cardboard was required to be bought for the competition.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

Question 1.
A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is open at the top. Ignoring the thickness of the plastic sheet, determine: (i) The area of the sheet required for making the box. (ii) The cost of sheet for it, if a sheet measuring 1 m² costs ₹ 20.
Answer:
The plastic box to be made is open at the top. Hence, the plastic sheet is required for the lateral surfaces and the base.
Here, for the box to be made,
length l = 1.5 m;
breadth b = 1.25 m and
height h = 65 cm = 0.65 m.
Area of the plastic sheet required for open box = Lateral surface area + Area of base
= 2 h(l + b) + l × b
= 2 × 0.65 (1.5 + 1.25) + 1.5 × 1.25 m²
= 1.3 × 2.75 + 1.875 m²
= 3.575 + 1.875 m²
= 5.45 m²
Cost of 1 m² sheet = ₹ 20
∴ Cost of 5.45 m² sheet = ₹ (5.45 × ₹ 20)
= ₹ 109

Question 2.
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹ 7.50 per m².
Answer:
For the given room, length 1 = 5 m; breadth b = 4 m and height h = 3 m.
Area of the region to be white washed
= Area of four walls + Area of ceiling
= 2 h(l + b) + l × b
= 2 × 3 (5 + 4) + 5 × 4 m²
= 54 + 20 m²
= 74 m²
Cost of white washing 1 m² region = ₹ 7.5
∴ Cost of white washing 74 m² region
= ₹ (74 × 7.5)
= ₹ 555

Question 3.
The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ₹ 10 per m² is ₹ 15,000, find the height of the hall. [Hint: Area of the four walls = Lateral surface area.]
Answer:
Area painted at the cost of ₹ 10 = 1 m²
∴ Area painted at the cost of ₹ 15,000
= \(\frac{15000}{10}\)
= 1500 m²
∴ Area of the four walls = 1500m²
∴ Lateral surface area = 1500 m²
∴ Perimeter Of the floor × Height = 1500 m²
∴ 250 m × Height = 1500 m²
∴ Height = \(\frac{15000}{250}\)
∴ Height = 6 m

Question 4.
The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimensions 22.5 cm × 10 cm× 7.5 cm can be painted out of this container?
Answer:
For each brick, length l = 22.5 cm; breadth b = 10 cm and height h = 7.5 cm.
Total surface area of one brick
= 2 (lb + bh + hl)
= 2 (22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm²
= 2 (225 + 75 + 168.75) cm²
= 2 (468.75) cm²
= 937.5 cm²
= \(\frac{937.5}{10000}\) m² = 0.09375 m²
No. of bricks that can be painted with paint sufficient to paint 0.09375 m² area = 1
∴ No. of bricks that can be painted with paint sufficient to paint 9.375 m² area 9.375
= \(\frac{9.375}{0.09375}\) = 100

Question 5.
A cubical box has each edge 10 cm and |> another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much ?
Answer:
For the cubical box, edge a = 10 cm and for the cuboidal box, length l = 12.5 cm; breadth b = 10 cm and height h = 8 cm
(i) Lateral surface area of cubical box
= 4a²
= 4 (10)² cm²
= 400 cm
Lateral surface area of cuboidal box
= 2h(l + b)
= 2 × 8(12.5 + 10) cm²
= 16 × 22.5 cm²
= 360 cm²
Thus, the lateral surface area of cubical box is greater by 40 cm² (400 – 360).

(ii) Total surface area of cubical box = 6a2
= 6 (10)² cm²
= 600 cm²
Total surface area of cuboidal box
= 2 (lb + bh + hl)
= 2(12.5 × 10 + 10 × 8 + 8 × 12.5) cm²
= 2 (125 + 80 + 100) cm²
= 2 (305) cm²
= 610 cm²
Thus, the total surface area of cubical box is smaller by 10 cm² (610 – 600).

Question 6.
A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass ?
(ii) How much of tape is needed for all the 12 edges ?
Answer:
(i) For the cuboidal greenhouse, length l = 30 cm; breadth fa = 25 cm and height h = 25 cm.
Area of glass used
= Total surface area of cuboid
= 2 (lb + bh + hl)
= 2 (30 × 25 + 25 × 25 + 25 × 30) cm²
= 2 (750 + 625 + 750) cm²
= 2 (2125) cm²
= 4250 cm²

(ii) 12 edges of the cuboidal greenhouse is made-up of 4 lengths, 4 breadths and 4 heights.
∴ Length of tape needed for 12 edges
= 4l + 4b + 4h
= 4 (l + b + h)
= 4 (30 + 25 + 25) cm
= 4 (80) cm
= 320 cm

Question 7.
Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5 % of the total surface area is required extra. If the cost of the cardboard is ₹ 4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.
Answer:
For bigger cuboidal boxes, length l = 25 cm;
breadth b = 20 cm and height h = 5 cm.
Total surface area of a bigger box
= 2 (lb + bh + hl)
= 2 (25 × 20 + 20 × 5 + 5 × 25) cm²
= 2 (500 + 100 + 125) cm²
= 1450 cm²
Area of cardboard required for overlap
= 5 % of 1450 cm²
= 72.5 cm²
Thus, the total area of cardboard required for 1 bigger box = 1450 + 72.5 cm²
= 1522.5 cm²
∴ The total area of cardboard required for 250 bigger boxes = (1522.5 × 250) cm²
For smaller cuboidal boxes, length l = 15 cm; breadth b = 12 cm and height h = 5 cm.
Total surface area of a smaller box
= 2 (lb + bh + hl)
= 2 (15 × 12 + 12 × 5 + 5 × 15) cm²
= 2(180 +60 + 75) cm²
= 2 (315) cm²
= 630 cm²
Area of cardboard required for overlap
= 5% of 630 cm²
= 31.5 cm²
Thus, the total area of cardboard required for 1 smaller box = 630 + 31.5 cm² = 661.5 cm²
∴ The total area of cardboard required for 250 smaller boxes = (661.5 × 250) cm²
Now, the total area of cardboard required for all the boxes
= (1522.5 × 250) + (661.5 × 250) cm²
= 250(1522.5 + 661.5) cm²
= 250 × 2184 cm²
Cost of 1000 cm² cardboard = ₹ 4
∴ Cost of 250 × 2184 cm² cardboard
= ₹ \(\left(\frac{4 \times 250 \times 2184}{1000}\right)\)
= ₹ 2184

Question 8.
Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m ?
Answer:
For the box-like structure without base, length
l = 4m; breadth b = 3m and height h = 2.5m.
Area of tarpaulin required
= Area of lateral surfaces + Area of top
= 2 h(l + b) + l × b
= 2 × 2.5 (4 + 3) + 4 × 3 m²
= 35 + 12 m²
= 47 m²

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 12 Heron’s Formula MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 12 Heron’s Formula MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
The sides of a triangle measure 8cm, 12cm and 6 cm. Then, the semiperimeter of the triangle is ……………………… cm.
A. 26
B. 52
C. 13
D. 6.5
Answer:
C. 13

Question 2.
Each side of an equilateral triangle measures 8 cm. Then, the semiperimeter of the triangle is ……………………….. cm.
A. 4
B. 24
C. 12
D.36
Answer:
C. 12

Question 3.
In a right angled triangle, the length of the hypotenuse is 15 cm and one of the sides forming right angle is 9 cm. Then, the semiperimeter of the triangle is ……………………….. cm.
A. 36
B. 18
C. 12
D. 15
Answer:
B. 18

Question 4.
The ratio of the measures of the sides of a triangle is 3:4:5. If the semiperimeter of the < triangle is 36 cm, the measure of the longest side of the triangle is ……………………. cm.
A. 12
B. 15
C. 20
D. 30
Answer:
D. 30

Question 5.
The area of a triangle is 48 cm2 and one of its sides measures 12 cm. Then, the length of the altitude corresponding to this side is …………………. cm.
A. 4
B. 8
C. 16
D. 6
Answer:
B. 8

Question 6.
The sides of a triangle measure 12 cm, 17 cm and 25 cm. Then, the area of the triangle is ……………………….. cm².
A. 54
B. 90
C. 180
D. 135
Answer:
B. 90

Question 7.
Two sides of a triangle measure 9 cm and 10 cm. If the perimeter of the triangle is 36cm, then its area is …………………. cm².
A. 17
B. 36
C. 72
D. 18
Answer:
B. 36

Question 8.
The area of an equilateral triangle with each side measuring 10 cm is ………………….. cm².
A. \(\frac{5 \sqrt{3}}{2}\)
B. 25√3
C. 5√3
D. 3√5
Answer:
B. 25√3

Question 9.
∆ ABC is an isosceles triangle in which BC = 8 cm and AB = AC = 5 cm. Then, area of ∆ ABC = ……………………….. cm².
A. 6
B. 12
C. 18
D. 24
Answer:
B. 12

Question 10.
ABCD is a parallelogram. If ar (ABC) = 18 cm², then ar(ABCD) = …………………. cm².
A. 18
B. 9
C. 36
D. 27
Answer:
C. 36

Question 11.
ABCD is a parallelogram. If ar (ABC) = 18 cm², then ar (ABCD) = …………………. cm².
A. 3.6
B. 7.2
C. 7.5
D. 6
Answer:
B. 7.2

Question 12.
In quadrilateral ABCD, AC = 10 cm. BM and DN are altitudes on AC from B and D respectively. If BM = 12cm and DN = 4 cm, then ar (ABCD) = …………………. cm².
A. 160
B. 80
C. 320
D. 480
Answer:
B. 80

Question 13.
The perimeter of rhombus ABCD is 40 cm and BD =16 cm. Then, ar (ABCD) = ……………………. cm².
A. 96
B. 48
C. 24
D. 72
Answer:
A. 96

Question 14.
The area of a rhombus is 72 cm² and one of its diagonals measures 16 cm. Then, the length of the other diagonal is ………………… cm.
A. 12
B. 9
C. 18
D. 15
Answer:
B. 9

Question 15.
PQRS is a square. If PQ = 10 cm, then PR = ……………………….. cm.
A. 10
B. 20
C. 10√2
D. 2√10
Answer:
C. 10√2

Punjab State Board PSEB 9th Class English Book Solutions English Reading Comprehension  Picture/Poster Based Exercise Questions and Answers, Notes.

PSEB 9th Class English Reading Comprehension Picture/Poster Based

Answers have been given at the end of this set.

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question.

Question 1.
Who did the fox invite to dinner ?
(a) The duck.
(b) The crane.
(c) The vixen.
(d) The deer.
Answer:
(b) The crane.

Question 2.
What did he serve his guest in the dinner ?
(a) Fruits.
(b) Meat.
(c) Soup.
(d) Eggs.
Answer:
(c) Soup.

Question 3.
The fox was very cunning. He placed a …….. before his guest.
(a) deep bowl
(b) flat dish
(c) narrow jar
(d) sound pitcher.
Answer:
(b) flat dish

Question 4.
What did the crane serve the fox when he invited the fox to dinner ?
(a) Cake.
(b) Milk.
(c) Rice.
(d) Chicken Curry
Answer:
(c) Rice.

Question 5.
The fox had to go hungry. Why ?
(a) Because the crane served the rice in a narrrow jar.
(b) Because the fox could not put his mouth in the narrow jar.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question.

Question 1.
An elephant and a ………. were very good friends.
(a) barber
(b) carpenter
(c) tailor
(d) cobbler
Answer:
(c) tailor

Question 2.
While going to the pond for water, the elephant would daily ………….
(a) stop at the tailor’s shop
(b) have a banana from the tailor
(c) both (a) and (b)
(d) neither (a) nor (b).
Answer:
(c) both (a) and (b)

Question 3.
One day when the elephant put his trunk into the shop,
(a) the tailor gave him a banana
(b) the tailor’s son gave him a banana
(c) the tailor pricked a needle into it
(d) the tailor’s son pricked a needle into it.
Answer:
(d) the tailor’s son pricked a needle into it.

Question 4.
The elephant had his revenge by ……….
(a) filling his trunk with muddy water
(b) throwing muddy water in the tailor’s shop
(c) both (a) and (b)
(d) neither (a) nor (b).
Answer:
(b) throwing muddy water in the tailor’s shop

Question 5.
The moral conveyed through these picture is ……….
(a) Might is right.
(b) Tit for tat.
(c) Do good have good.
(d) No pains no gains.
Answer:
(b) Tit for tat.

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question :

Question 1.
The man is this picture is the famous cricketer
(a) Virat Kohli
(b) Irfan Pathan
(c) Sachin Tendulkar
(d) Mahendra Singh Dhoni.
Answer:
(c) Sachin Tendulkar

Question 2.
He is popularly known as ……. of cricket.
(a) Master Bowler
(b) Master Blaster
(c) Master Batsman
(d) Master Crickter.
Answer:
(b) Master Blaster

Question 3.
At the age of sixteen, he made his international debut against
(a) England
(b) Australia
(c) Sri Lanka
(d) Pakistan.
Answer:
(d) Pakistan.

Question 4.
Sachin took retirement from cricket in
(a) 2005
(b) 2018
(c) 2020
(d) 2013.
Answer:
(d) 2013.

Question 5.
Sachin Tendulkar was honoured with many prestigious awards like
(a) Arjuna Award and Rajiv Khel Ratna
(b) Padma Shri and Bharat Ratna
(c) Both (a) and (b)
(d) Neither (a) nor (b)
Answer:
(c) Both (a) and (b)

Look at this poster and answer the questions given below :

Choose the correct option to answer each question :

Question 1.
For what purpose is this poster designed ?
(a) To promote education for boys.
(b) To promote education for girls.
(c) To provide employment for boys.
(d) To provide employment for girls.
Answer:
(b) To promote education for girls.

Question 2.
Girls and boys have ……….. to education.
(a) no right
(b) equal right
(c) no interest
(d) equal interest.
Answer:
(b) equal right

Question 3.
How can the nation’s progress be accelerated ?
(a) By educating the boys.
(b) By educating the girls
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 4.
We should not …….. the girls their rights.
(a) excuse
(b) give
(c) accept
(d) deny.
Answer:
(d) deny.

Question 5.
This poster teaches us to stop ……….
(a) the evil of dowry
(b) the evil of female foeticide
(c) the evil of bride burning
(d) the evil of gender discrimation.
Answer:
(b) the evil of female foeticide

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question :

Question 1.
Why did the fox jump into the well to drink water ?
(a) Because the water was very low.
(b) Because he wanted to eat the goat.
(c) Because he wanted to have a bath.
(d) None of these three.
Answer:
(a) Because the water was very low.

Question 2.
The fox drank water and ……….
(a) drenched his thirst
(b) quenched his thirst
(c) satisfied his thirst
(d) toasted his thirst.
Answer:
(b) quenched his thirst

Question 3.
How did the fox succeed in be fooling the goat who was passing that way ?
(a) He told the goat that it was very hot outside.
(b) He told the goat that it was very cold inside the well.
(c) He told that the water was very sweet.
(d) All of these three.
Answer:
(d) All of these three.

Question 4.
What happened when the foolished goat jumped into the well ?
(a) The fox at once climbed over her back.
(b) The fox jumped out of the well.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 5.
The moral of this story is
(a) As you sow so shall you reap
(b) Think before you speak.
(c) Look before you leap.
(d) All that glitters is not gold.
Answer:
(c) Look before you leap.

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question :

Question 1.
Once upon a time a shepherd-boy ……… the sheep of the villagers.
(a) looked into
(b) looked after
(c) looked at
(d) looked for.
Answer:
(b) looked after

Question 2.
What mischief did he make one day ?
(a) He climbed up a tree.
(b) He started crying, “Wolf! Wolf!’’
(c) He shouted for help.
(d) All of these three.
Answer
(d) All of these three.

Question 3.
Why did the villagers who came to help the boy become cross with him ?
(a) Because they found no wolf there.
(b) Because the boy told them that he had shouted in fun only.
(c) Both (a) and (b).
(d) Neithor (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 4.
What happened when a wolf really came there.
(a) The boy shouted for help but nobody came.
(b) The villagers didn’t believe the boy’s cries as he had be fooled them once.
(c) The wolf sprang upon the boy and. tore him to pieces.
(d) All of these three.
Answer:
(d) All of these three.

Question 5.
The moral of the story is —
(a) No pain no gain.
(b) Never give up hope in the hour of difficulty.
(c) Once a liar, always a liar
(d) Think before you speak.
Answer:
(c) Once a liar, always a liar

Look at these pictures and answer the questions given below :

Choose die correct option to answer each question:

Question 1.
An old farmer had three sons who were ………..
(a) very active
(b) very idle
(c) very naughty
(d) very hardworking.
Answer:
(b) very idle

Question 2.
What did the farmer said to sons before his death?
(a) He asked them to work hard in their uk.
(b) He asked them not to fight with each other after his death.
(c) He told them that there was a big treasure in his field.
(d) He asked them to live together happily.
Answer:
(c) He told them that there was a big treasure in his field.

Question 3.
What happened when the sons dig their field ?
(a) They found there a big treasure.
(b) They found there no treasure.
(c) They found there tools of farming.
(d) None of these three.
Answer:
(b) They found there no treasure.

Question 4.
What did the old man ask them to do?
(a) He asked them to sell that field to him.
(b) He asked them to sow seeds in their field.
(c) He asked them to dig their field more deeply.
(d) Any of these three.
Answer:
(b) He asked them to sow seeds in their field.

Question 5.
What happened when the Sons sow seeds in their field?
(a) There was a good crop that year.
(b) They got a lot of money for it.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 6.
The moral of this story is ……
(a) Do good find good.
(b) As you sow, so shall you reap.
(c) Hard work is the key to success.
(d) Hard work is man’s greatest treasure.
Answer:
(d) Hard work is man’s greatest treasure.

Look at these pictures and answer the questions given below:

Choose the correct option to answer each question :

Question 1.
What was the capseller doing in a forest ?
(a) He was passing through the forest to reach a village.
(b) He lay down under a tree to take some rest.
(c) He went there to sell his caps.
(d) Both (a) and (b).
Answer:
(d) Both (a) and (b).

Question 2.
What happened when the capseller fell asleep ?
(a) Some monkeys came there.
(b) The monkeys untied the bundle of caps.
(c) The monkeys took away all the caps.
(d) All of these three.
Answer:
(d) All of these three.

Question 3.
What did the capseller see when he woke up ?
(a) He found all his caps missing.
(b) He found the monkeys wearing his caps.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 4.
What did he do to recover his caps ?
(a) He took off his caps and threw it down.
(b) The monkeys imitated him.
(c) The monkey threw down their caps.
(d) All of these three.
Answer:
(d) All of these three.

Question 5.
The moral of the story is ………
(a) Tit for tat.
(b) A stitch in time saves nine.
(c) God helps those who help themselves.
(d) Never give up hope in the hour of difficulty.
Answer:
(d) Never give up hope in the hour of difficulty.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 11 Constructions Ex 11.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1

Question 1.
Construct an angle of 90° at the initial point of a given ray and justify the construction.
Answer:

Steps of construction :

1. Ray AB is given. Produce ray AB on the side of A to get line MAB.
2. Taking A as centre and some radius, draw an arc of a circle to intersect line MAB at X and Y.
3. Taking X and Y as centres and radius more than \(\frac{1}{2}\)XY, draw arcs to intersect at P on one side of line MAB.
4. Draw ray AC passing through E
   Thus, ∠CAB is the required angle of 90°.

Justification:
Draw PX and PY.
In ∆ PAX and ∆ PAY,
AX = AY (Radii of same arc)
PX = PY (Radii of congruent arcs)
PA = PA (Common)
∴ By SSS rule, ∆ PAX ≅ ∆ PAY
∴ ∠PAX = ∠PAY (CPCT)
But, ∠PAX + ∠PAY = 180° (Linear pair)
∴ ∠PAY = \(\frac{180^{\circ}}{2}\) = 90°
∴ ∠CAB = 90°

Question 2.
Construct an angle of 45° at the initial point of a given ray and justify the construction.
Answer:

Steps of construction:

1. Ray AB is given. Produce ray AB on the side of A to get line MAB.
2. Taking A as centre and some radius, draw an arc of a circle to intersect line MAB at X and Y.
3. Taking X and Y as centres and radius more than \(\frac{1}{2}\)XY, draw arcs to intersect at P on one side of line MAB.
4. Draw ray AC passing through E Thus, ∠CAB of 90° is received.
5. Name the point of intersection of the arc with centre A and ray AC as Z.
6. Taking Y and Z as centres and radius more than \(\frac{1}{2}\)YZ, draw arcs to intersect each other at Q.
7. Draw ray AQ.
   Thus, ∠QAB is the required angle of 45°.

Justification:
In example 1, we have already justified that ∠CAB = 90°. So, we do not repeat that part’ here.
Draw QZ and QY.
In ∆ AYQ and ∆ AZQ,
AY = AZ (Radii of same arc)
YQ = ZQ (Radii of congruent arcs)
AQ = AQ (Common)
∴ By SSS rule, ∆ AYQ ≅ ∆ AZQ
∴ ∠QAY = ∠QAZ (CPCT)
But, ∠QAY + ∠QAZ = ∠ZAY = ∠CAB = 90°
∴ ∠QAY = \(\frac{90^{\circ}}{2}\) = 45°
∴ ∠QAB = 45°

Question 3.
Construct the angles of the following measurements:
(i) 30°
Answer:

Steps of construction:

1. Draw any ray AB. With centre A and any radius, draw an arc to intersect AB at X.
2. With centre X and the same radius [as in step (1)], draw an arc to intersect the previous arc at Y. Draw ray AY. Then, ∠YAB = 60°.
3. Draw ray AT, the bisector of ∠YAB.
   Thus, ∠TAB is the required angle of 30°.

(ii) 22\(\frac{1}{2}\)°
Answer:

Steps of construction:

1. Draw any ray AB. Produce AB on the side of A to get line CAB.
2. Taking A as centre and any radius, draw an arc of a circle to intersect line CAB at X and Y.
3. Taking X and Y as centres and radius more than \(\frac{1}{2}\)XY, draw arcs to intersect each other at L on one side of line CAB. Draw ray AL. Then, ∠LAB = 90°.
4. Draw ray AM, the bisector of ∠LAB. Then, ∠MAB = 45°.
5. Draw ray AN, the bisector of ∠MAB. Then, ∠NAB = 22\(\frac{1}{2}\)°.
   Thus, ∠NAB is the required angle of 22\(\frac{1}{2}\)°.

(iii) 15°
Answer:

Steps of construction:

1. Draw any ray AB. Taking A as centre and any radius, draw an arc of a circle to intersect AB at X.
2. Taking X as centre and the same radius as before, draw an arc to intersect the previous arc at Y. Draw ray AY. Then, ∠YAB = 60°.
3. Draw ray AL, the bisector of ∠YAB. Then, ∠LAB = 30°.
4. Draw ray AM, the bisector of ∠LAB. Then, ∠MAB = 15°.
   Thus, ∠MAB is the required angle of 15°.

Question 4.
Construct the following angles and verify by measuring them by a protractor:
(i) 75° and (ii) 105°
Answer:

Steps of construction:

1. Draw any ray AB and produce it on the side of A to get line CAB. Taking A as centre and any radius draw an arc of a circle to intersect line CAB at X and Y.
2. Taking X and Y as centres and radius more than \(\frac{1}{2}\)XY, draw arcs to intersect each other at point L on one side of line CAB. Draw ray AL. Then, ∠LAB = 90°.
3. Taking X as centre and radius AX, draw an arc of a circle to Intersect the first arc (arc XY) with centre A at Z.
4. Draw ray AZ. Then, ∠ZAB = 60°.
5. Now, draw ray AM, the bisector of ∠LAZ. Then, ∠MAB = 75° and ∠MAC = 105°.
   Thus, ∠MAB and ∠MAC are the required angles of measure 75° and 105° respectively.

(iii) 135°
Answer:

Steps of construction:

1. Draw line CAB. Taking A as centre and any radius, draw an arc of a circle to Intersect line CAB at X and Y.
2. Taking X and Y as centres and radius more them \(\frac{1}{2}\)XY, draw arcs to intersect each other at P on one side of line CAB.
3. Draw ray AP Then, ∠PAB = ∠PAC = 90°.
4. Draw ray AQ, the bisector of ∠PAC.
5. Then, ∠QAB = 135°.
   Thus, ∠QAB is the required angle of 135°.

Question 5.
Construct an equilateral triangle, given its side and justify the construction.
Answer:
Line segment XY is given. We have to construct an equilateral triangle with each side being equal to XY.

Steps of construction:

1. Draw any ray BM.
2. With centre B and radius XY, draw an arc of a circle to intersect BM at C.
3. Taking B and C as centres and rhdius XY, draw arcs to intersect each other at A on one side of line AC.
4. Draw AB and AC.
   Thus, ∆ ABC is the required equilateral triangle with each side being equal to XY.

Justification:
The arc drawn with centre B and radius XY intersects ray BM at C. ∴ BC = XY. The arcs drawn with centres B and C and radius XY intersect at A.
∴ AB = XY and AC = XY.
Thus, in ∆ ABC, AB = BC = AC = XY.
Hence, ∆ ABC is an equilateral triangle in which all sides are equal to XY.
Note: If the measure of sides are given numerically, e.g., 4 cm, 5 cm, etc., then we have to use graduated scale instead of straight edge.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 10 Circles MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 10 Circles MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
In a circle with centre P, AB and CD are congruent chords. If ∠PAB = 40°, then ∠CPD = ………………..
A. 40°
B. 80°
C. 100°
D. 50°
Answer:
C. 100°

Question 2.
In a circle with radius 5 cm, the length of a chord lying at distance 4 cm from the centre is …………………. cm.
A. 3
B. 6
C. 12
D. 15
Answer:
B. 6

Question 3.
In a circle with radius 13 cm, the length of a chord is 24 cm. Then, the distance of the chord from the centre is ……………….. cm.
A. 10
B. 5
C. 12
D. 6.5
Answer:
B. 5

Question 4.
In a circle with radius 7 cm, the length of a minor arc is always less than ………………… cm.
A. 11
B. 22
C. 15
D. π
Answer:
B. 22

Question 5.
In a circle with centre P, AB is a minor arc. Point R is a point other than A and B on major arc AB. If ∠APB = 150°, then ∠ARB = …………… .
A. 150°
B. 75°
C. 50°
D. 100°
Answer:
B. 75°

Question 6.
In a circle with centre P, AB is a minor arc. Point R is a point other than A and B on major arc AB. If ∠ARB = 80°, then ∠APB = ……………. .
A. 40°
B. 80°
C. 160°
D. 60°
Answer:
C. 160°

Question 7.
In cyclic quadrilateral ABCD, ∠A – ∠C = 20°.
Then, ∠A = ………………. .
A. 20°
B. 80°
C. 100°
D. 50°
Answer:
C. 100°

Question 8.
In cyclic quadrilateral PQRS, 7∠P = 2∠R.
Then, ∠P = ………………….. .
A. 20°
B. 40°
C. 140°
D. 100°
Answer:
B. 40°

Question 9.
The measures of two angles of a cyclic quadrilateral are 40° and HOP. Then, the measures of other two angles of the quadrilateral are ……………….. .
A. 40° and 110°
B. 50° and 100°
C. 140° and 70°
D. 20° and 120°
Answer:
C. 140° and 70°

Question 10.
In cyclic quadrilateral PQRS, ∠SQR = 60° and ∠QPR = 20°. Then, ∠QRS = ……………… .
A. 40°
B. 60°
C. 80°
D. 100°
Answer:
D. 100°

Question 11.
In cyclic quadrilateral ABCD, ∠CAB = 30° and ∠ABC = 100°. Then, ∠ADB =
A. 50°
B. 100°
C. 75°
D. 60°
Answer:
A. 50°

Question 12.
Equilateral ∆ ABC is inscribed in a circle with centre P. Then, ∠BPC = ……………. .
A. 60°
B. 90°
C. 120°
D. 75°
Answer:
C. 120°

Question 13.
∆ ABC is inscribed in a circle with centre O and radius 5 cm and AC is a diameter of the circle. If AB = 8 cm, then BC = ………………… cm.
A. 10
B. 8
C. 6
D. 15
Answer:
C. 6

Question 14.
In cyclic quadrilateral ABCD, ∠A = 70° and ∠B + ∠C = 160°. Then, ∠B = ………………. .
A. 35°
B. 25°
C. 50°
D. 130°
Answer:
C. 50°

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 10 Circles Ex 10.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6

Question 1.
Prove that the line segment joining the centres of two intersecting circles subtends equal angles at the two points of intersection.
Answer:

Circles with centres O and P intersect each other at points A and B.
In ∆ OAP and ∆ OBR
OA = OB (Radii of circle with centre O)
PA = PB (Radii of circle with centre P)
OP = OP (Common)
∴ By SSS rule, ∆ OAP = ∆ OBP
∴ ∠OAP = ∠OBP (CPCT)
Thus, OP subtends equal angles at A and B. Hence, the line segment joining the centres of two intersecting circles subtends equal angles at the two points of intersection.

Question 2.
Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Answer:

Draw the perpendicular bisector of AB to intersect AB at M and draw the perpendicular bisector of CD to intersect CD at N.
Both these perpendicular bisectors pass through centre O and since AB || CD; M, O and N are collinear points.
Now, MB = \(\frac{1}{2}\)AB = \(\frac{5}{2}\) = 2.5 cm,
CN = \(\frac{1}{2}\)CD = \(\frac{11}{2}\) = 5.5 cm and MN = 6 cm.
Let ON = x cm s
∴ OM = MN – ON = (6 – x) cm
Suppose the radius of the circle is r cm.
∴ OB = OC = r cm
In ∆ OMB, ∠M = 90°
∴OB² = OM² + MB²
∴ r² = (6 – x)² + (2.5)²
∴ r² = 36 – 12x + x² + 6.25 ………….. (1)
In ∆ ONC, ∠N = 90°
∴ OC² = ON² + CN²
∴ r² = (x)² + (5.5)²
∴ r² = x² + 30.25 ………………. (2)
From (1) and (2),
36 – 12x + x² + 6.25 = x² + 30.25
∴ – 12x = 30.25 – 6.25 – 36
∴- 12x = – 12
∴x = 1
Now, r² = x² + 30.25
∴ r² = (1)2 + 30.25
∴ r² = 31.25
∴ r = √31.25 (Approximately 5.6)
Thus, the radius of the circle is √31.25 (approximately 5.6) cm.
Note: If the calculations are carried out in simple fractions, then MB = \(\frac{5}{2}\) cm, CN = \(\frac{11}{2}\) cm and radius is \(\frac{5 \sqrt{5}}{2}\) (approximately 5.6) cm.

Question 3.
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chords is at distance 4 cm from the cehtre, what is the distance of the other chord from the centre?
Answer:

In a circle with centre O, chord AB is parallel to chord CD, AB = 8 cm and CD = 6 cm.
Draw OM ⊥ AB, ON ⊥ CD, radius OB and radius OC.
Then, MB = \(\frac{1}{2}\)AB = \(\frac{1}{2}\) × 8 = 4 cm,
NC = \(\frac{1}{2}\)CD = \(\frac{1}{2}\) × 6 = 3cm and ON = 4cm.
In ∆ ONC, ∠N = 90°
∴ OC² = ON² + NC² = 4² + 3² = 16 + 9 = 25
∴ OC = 5 cm
∴ OB = 5 cm (OB = OC = Radius)
In ∆ OMB, ∠M = 90°
∴ OB² = OM² + MB²
∴ 5² = OM² + 4²
∴ 25 = OM² + 16
∴ OM² = 9
∴ OM = 3 cm
Thus, the distance of the other chord from the centre is 3 cm.

Question 4.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Answer:

In ∆ ABE, ∠AEC is an exterior angle.
∴ ∠AEC = ∠ABE + ∠BAE
∴ ∠ABE = ∠AEC – ∠BAE
∴ ∠ABC = ∠AEC – ∠DAE ……………. (1)
Now, ∠AEC = \(\frac{1}{2}\) ∠AOC (Theorem 10.8)
and ∠ DAE = \(\frac{1}{2}\) ∠DOE (Theorem 10.8)
Substituting above values in (1),
∠ABC = \(\frac{1}{2}\) ∠AOC – \(\frac{1}{2}\)∠DOE
∴ ∠ABC = \(\frac{1}{2}\) (∠AOC – ∠DOE)
Here, ∠AOC is the angle subtended by chord AC at the centre and ∠DOE is the angle subtended by chord DE at the centre.
Thus, ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Note: There is no need for chords AD and CE to be equal.

Question 5.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Answer:

ABCD is a rhombus and its diagonals intersect at M.
∴ ∠BMC is a right angle.
A circle is drawn with diameter BC.
There are three possibilities for point M:
(1) M lies in the interior of the circle,
(2) M lies in the exterior of the circle.
(3) M lies on the circle.
According to (1), if M lies in the interior of the circle, then BM produced will intersect the circle at E. Then, ∠BEC is an angle in a semicircle and hence a right angle, i.e.,
∠MEC = 90°.
In ∆ MEC, ∠ BMC is an exterior angle.
∴ ∠ BMC > ∠ MEC, i.e., ∠ BMC > 90°. In this situation, ∠ BMC is an obtuse angle which contradicts that ∠ BMC = 90°.
Similarly, according to (2), if M lies in the exterior of the circle, then ∠BMC is an acute angle which contradicts that ∠BMC 90°. Thus, possibilities (1) and (2) cannot be true.
Hence, only possibility (3) is true, i.e., M lies on the circle.
Thus, the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

Question 6.
ABCD is a parallelogram. The circle through A, B and C intersects CD (produced if necessary at) E. Prove that AE = AD.
Answer:

Here, the circle through A, B and C intersects CD at E.
∴ Quadrilateral ABCE is cyclic.
ABCD is a parallelogram.
∴ ∠ABC = ∠ADC
∴ ∠ABC = ∠ADE
In cyclic quadrilateral ABCE,
∠ABC + ∠AEC = 180°
∴ ∠ADE + ∠AEC = 180° ……………… (1)
Moreover, ∠AEC and ∠AED form a linear pair.
∴ ∠AED + ∠AEC = 180° ………………. (2)
From (1) and (2),
∠ADE + ∠AEC = ∠AED + ∠AEC
∴ ∠ ADE = ∠ AED
Thus, in ∆ AED, ∠ADE = ∠AED.
∴ AE = AD (Sides opposite to equal angles)
Note: If the circle intersect CD produced, l then also the result can be proved in similar way.

Question 7.
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.
Answer:

Chords AC and BD of a circle bisect each other at point O.
Hence, the diagonals of quadrilateral ABCD bisect each other.
∴ Quadrilateral ABCD Is a parallelogram.
∴ ∠BAC = ∠ACD (Alternate angles formed by transversal AC of AB || CD)
Moreover, ∠ACD = ∠ABD (Angles in same segment)
∴ ∠BAC = ∠ABD
∴ ∠BAO = ∠ABO
∴ In A OAB, OA = OB.
But, OA = OC and OB = OD
∴ OA = OB = OC = OD
∴ OA + OC = OB + OD
∴ AC = BD
Thus, the diagonals of parallelogram ABCD are equal.
∴ ABCD is a rectangle.
∴ ∠ABC = 90°
Hence, ∠ABC is an angle in a semicircle and AC is a diameter.
Similarly, ∠BAD = 90°.
Hence, ∠BAD is an angle in a semicircle and BD is a diameter.

Question 8.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90° – \(\frac{1}{2}\)A, 90° – \(\frac{1}{2}\)B and 90° – \(\frac{1}{2}\)C.
Answer:

The bisectors of ∠A, ∠B and ∠ C of ∆ ABC intersect the circumcircle of ∆ ABC at D, E and F respectively. .
∠FDE = ∠FDA + ∠EDA (Adjacent angles)
= ∠ FCA + ∠ EBA (Angles in same segment)
= \(\frac{1}{2}\)∠C + \(\frac{1}{2}\)∠B (Bisector of angles in ∆ ABC)
= \(\frac{1}{2}\)(∠ B + ∠ C)
= \(\frac{1}{2}\)(180° – ∠A) [∠A + ∠B + ∠C = 180°)
= 90° – \(\frac{1}{2}\) ∠A
Thus, ∠FDE = 90° – \(\frac{1}{2}\) ∠A.
Similarly, ∠ DEF = 90° – \(\frac{1}{2}\) ∠B and
∠ EFD = 90° – \(\frac{1}{2}\) ∠C.

Question 9.
Two congruent circles intersect each Other at points A and B. Through A any line segment PAQ is drawn so that P 9 lie-on. , the two circles. Prove that BP = BQ.
Answer:

Two congruent circles with centres X and Y intersect at A and B.
Hence, AB is their common chord.
In congruent circles, equal chords subtend equal angles at the centres.
∴ ∠AXB = ∠AYB
In the circle with centre X, ∠AXB = 2∠APB and in the circle with centre Y, ∠AYB = 2∠AQB.
∴ 2∠ APB = 2∠ AQB
∴ ∠APB = ∠AQB
∴ ∠QPB = ∠PQB
Thus, in ∆ BPQ, ∠QPB = ∠PQB
∴ QB = PB (Sides opposite to equal angles)
Hence, BP = BQ.

Question 10.
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Answer:

In ∆ ABC, the bisector of ∠A intersects the circumcircle of ∆ ABC at D.
∴∠BAD = ∠CAD
Aso, ∠BAD = ∠BCD and ∠CAD = ∠CBD (Angles in same segment)
∴ ∠BCD = ∠CBD
Thus, in ∆ BCD, ∠BCD = ∠CBD
∴BD = CD (Sides opposite to equal angles)
Thus, point D is equidistant from B and C.
Hence, D is a point on the perpendicular bisector of BC.
Thus, the bisector of ∠ A and the perpendicular bisector of side BC intersect at D and D is a point on the circumcircle of ∆ ABC.
Thus, in ∆ ABC, if the angle bisector of ∠A and the perpendicular bisector of side BC intersect, they intersect on the circumcircle of ∆ ABC.
Note: In ∆ ABC, if AB = AC, then the bisector of ∠A and the perpendicular bisector of side BC will coincide , and would not intersect in a single point.

Punjab State Board PSEB 9th Class English Book Solutions English E-mail Message Writing Exercise Questions and Answers, Notes.

PSEB 9th Class English E-mail Message Writing

Inviting friend to Watch a Play

Suppose you are Surjit Singh. Write an e-mail to your friend, Vipin Goyal, inviting him to watch a play.

Hi Vipin.
I am going to Government College for Women, Amritsar, to watch a play on 6 July, 20 – -.
Would you like to come? Let me know by Tuesday so that I can buy your ticket too.
Love
Surjit Singh.

House for Rent

Suppose you are Ramneek. Write an e-mail to your cousin, Darshan Pal, to put up a notice on his college notice-board to rent out your house.

Dear Pal
My father wants to rent our the Second floor of our house. There are two rooms, a kitchen and two attached bathrooms. He should like to have ₹ 2000 as rent. He will rake two months’ rent in advance. He warns to rent out the house to students. Please put up a notice on your college notice-board.
Regards
Ramneek.

Congratulation on Engagement

Suppose you are Shvinder Gill. Write an e-mail to your friend, Alok Wasn, congratulating him on his engagement.

Hi Alok
I have learnt that you are engaged. Congratulating! who is the lucky girl? who does she live and what does she do? Let me know when are you getting married? Is the dare fixed.
Love
Shivinder Gill.

Trip to the South

Suppose you are Varsha Gill. Write an e-mail to your friend, Asha Lakhpal, describing your visit to the south.

Hello Asha
Sorry, I couldn’t write to you earlier. I visited the south with my friend last month. We spent eight days there. We liked the Meenakshi Temple ar Madurai very much. The sunset at Kanyakumari was fascinating. We also went to Aurbindo Ashratn at Puducherry. It was very powerful there.
Love
Varsha.

E-mail (electronic mail) is the medium of communication that sends and receives messages through a specially designed computer network. With the revolution in information technology along with the rapid growth of the Internet, e-mail has become the most popular medium of communication. More and more people are using e-mail to send their messages. Due to its high speed, efficiency and low cost, e-mail has become one of the most important channels of communication. As e-mails are faster than letters, they are used for a quick transmission of all sorts of information.

Specimen of an Informal E-mail

Hi Paul
Sorry to say I’ll he a bit late for tonight’s rehearsal as something’s come up at home and I won’t be able to get away on time. I hope to make it by 7.15.
D. Paul.

Specimen of an Formal E-mail

Dear Ms. Maya
The books you ordered last week are now in stock and awaiting collection. I attach a list of the course books currently in stock at the bookshop.
Julie
Assistant Manage!

HFI Bookshop
Tel : 01123318301
Fax : 01123317931

Punjab State Board PSEB 9th Class English Book Solutions English Note-Making & Messages Exercise Questions and Answers, Notes.

PSEB 9th Class English Note-Making & Messages

Note-making

का अर्थ है किसी पैरे की मुख्य बातों को संक्षिप्त और साफ-सुथरे ढंग से प्रस्तुत करना। अच्छे Notes में निम्नलिखित विशेषताएं होती है –
1. वे संक्षिप्त होते हैं।

2. केवल प्रासंगिक बातें ही उनमें दी जाती हैं।

3. केवल शब्दों या वाक्यांशों का प्रयोग ही किया जाता है। पूरे वाक्यों की आमतौर पर आवश्यकता नहीं होती। अन्य शब्दों में हम कह सकते हैं कि Notes बनाते समय प्रयुक्त भाषा व्याकरण की दृष्टि से पूरी तरह सही नहीं भी हो सकती।

4. सूचना को सूचीबद्ध ढंग से प्रस्तुत किया जाता है। इसे विभाजित व उपविभाजित किया जाता है। विभाजन निम्न प्रकार से हो सकता है –
मुख्य खण्ड : 1, 2, 3, इत्यादि।
उपखण्ड : a, b, c, इत्यादि।

Passage 1:

If the young students in schools and colleges do not learn discipline, they will never be able to extract’ obedience from others in society. In fact, society will never accept them as persons fit for any responsible position in life. A school or college without discipline can never impart? suitable education to students.

Such a school or college is no better than a factory that turns out imperfect’ men and women. Sense of discipline plays a very important part in the playground and the battlefield. A disciplined team is likely to win the match in spite of its weakness but a very good team may not fare well for want of discipline. The rule of discipline equally applies to soldiers in the battlefield.

Read the above passage carefully and make notes on it in points only, using headings and sub-headings.
Answer:
1. Need for discipline in schools and colleges for good education.
2. Indisciplined students fail’ to win any respect or position later on in their life.
3. Importance of discipline, for players on the playground
4. Importance of discipline, for soldiers in the battlefield.

Passage 2

Early rising leads to health and happiness. The man who rises late, can have little rest in the course of the day. Anyone who lies in bed late is compelled to work till a late hour in the evening. He has to go without the morning exercise which is so necessary for his health. In spite of all efforts’, his work will not produce as good results as that of the early riser. The reason for this is that he cannot take advantage of the refreshing hours in the morning.

Some people say that the quiet hour of midnight is the best time for working. Several great thinkers say that they can write best only when they burn the midnight oil. Yet it is true to say that few men have a clear brain at midnight when the body needs rest and sleep. Those who work at that time soon ruin their health. Bad health must, in the long run, have a bad effect on the quality of their work.

Read the above passage carefully and make notes on it in points only, using headings and sub-headings.
Answer:
1. Advantages of early rising :
(i) health
(ii) Disadvantages

2. Disadvantages of late rising
(i) work till late in the evening
(ii) go without morning exercise
(iii) work not done properly.

(3) (i) Burning midnight oil bad for health.
(ii) Bad health, poor quality of our work.

Passage 3

Games, though essential, should not become the be-all and end-all of student life. Generally, the sportsmen waste too much time on them, and fail in their examinations. One must never devote more than an hour to sports and after that, one should not even think about them. Again, if a player plays a game rashly’, there is every danger of breaking bones.

If it is played without the spirit of sportsmanship, it can lead to bad blood and quarrels. In some of the colleges, there is a tradition that if the visiting team is winning a match, the home team plays foul, picks a quarrel and breaks the bones of the visitors. But in spite of these minor defects, sports are very useful in keeping the students busy and in developing their personalities.

India expects its citizens to have the qualities of true sportsmen. If we all acquire these qualities, there will be no narrow-mindedness, no corruption, and no injustice. There will be independence in the real sense of the word.

Read the given passage carefully and make notes on it in points only, using headings and sub-headings.

1. Sports essential for students, but not the be-all and end-all.
2. Wasting too much time → failure in examinations.
3. Playing rashly → breaking of bones.
4. Lack of sportsmanship → quarrels between teams.
5. True sportsmanship can end narrow – mindedness, corruption and injustice.

Passage 4

Of all amusements which can possibly be imagined for a hard-working man after his daily toil, there is nothing like reading an entertaining book. It calls for no bodily exertion of which he has had enough. It relieves his home of its dullness. It transports him to a livelier and more interesting scene, and while he enjoys himself there, he may forget the evils of the present moment.

It accompanies him to his next day’s work and if the book he has been reading be anything above the very idlest and the dullest, it gives him something to think about besides the drudgery of his everyday occupation. If I were to pray for a taste which should stand me in good stead under every variety of circumstances and be. a source of happiness and cheerfulness through life, it would be a taste for reading.

Give a man this taste,.and the means of gratifying it, and you can hardly fail to make him happy unless indeed you put into his hand a most perverse selection of books.

Read the above passage carefully and make notes on it in points only, using headings and sub-headings.
Answer:
1. Reading of an interesting book a good diversion after the day’s hard work.
2. (i) Removes dullness of home,
(ii) Transports one into a livelier world.
3. (i) A good book food for thought.
(ii) Stands in good srcad under every circumstance.
4. Taste for reading, a source of great happiness.

Passage -5

English is important not because a number of people know it in India, although it is a factor to be remembered. It is not important because it is the language of Milton and Shakespeare, although that has to be considered. English is important because it is the major window for us on the modern world. And we dare not close that window. If we close it, we imperil our future. We think of Industrialisation, scientific development, research and technology.

But every door of modern knowledge will be closed if we do not have one or more foreign languages. We need not have English : we can have Russian, French or German, if you like, but obviously it is infinitely simpler for us to deal with a language which we know than to shift over to Russian, French or German which will be a tremendous job.
Certainly, we want to learn foreign languages, because we deal with the people of those languages in business, trade and science. So in the present stage of our development, we cannot go ahead without English and other foreign languages.

Read the above passage carefully and make notes on it in points only, using headings and sub-headings.
Answer:
1. Mistakes do little harm

• when admitted.
• set tight before they a do any carnage

2. Delay in admitting mistakes

• harmful for the task in hand.
• harmful for the reputation.

3. Person who admits his mistakes

• is liked by everybody
• wins the confidence and respect of others.

4. Person who hides his mistakes

• is considered a fool.
• nobody likes him

Passage – 7

Teachers have a great responsibility at this time when our society is undergoing transformation. The future of the teaching profession in India will depend on the decision which the teachers take on vital questions relating to social change. In normal times, when society is comparatively more stable, the teachers’ primary task is transmitting culture. But in a period of transition, like the one through which we are passing, they have sometimes to set aside the culture in which they live, make a proper appraisal9 of it, pick out its salient features and reinterpret them for the new generation. The oncoming generations can rise to a high level of wisdom and cultivation only when teachers guide them carefully during this period of change.

Read the given passage carefully and make notes on it in points only, using headings and sub-headings.
Answer:
A. Teachers’ role in normal times :

• transmitting culture.

B. Modern rimes :

• not normal
• a period of transition.

C. teacher roles:

• proper appraisal of old culture
• pick out its salient features
•  re-interpreting them for future generations.

Passage – 8
Each one of us must realize that the only future for India and her people is one of tolerance and co-operation, which have been the basis of our culture for ages past. We have laid down in our constitution that Indians a Secular State. This does not mean irreligion. It means equal respect for all faiths and equal opportunities for those who profess different faiths. We have, therefore, always to keep in mind this vital aspect of our culture which is also of the highest importance in India today. Those who put up barriers between one Indian and another and who promote disruptive tendencies do not serve the cause of India and her culture. They weaken us at home and discredit us abroad.
Read the above passage carefully and make notes on it in points only, using headings and sub-headings.
Answer
(A) Tolerance and co-operation

• basis of our past culture
• pillars of our future

(B) Secularism

• equal respect for all faiths
• equal opportunities
• of highest importance in present-day India.

(C) Disruptive tendencies

• serve no cause
• discredit the country

Passage – 9

As a result of a long series of discoveries, mans life has been altered more radically and more rapidly during the last one hundred and fifty years than during the whole of the preceding two thousand years. In what ways does this alteration chiefly show itself ? In the first place, most of the external enemies to which our species in the past had been exposed are either overcome or are in a fair way to being overcome.

Look back over mans life in the past and you cannot but realize what sordid, meagre, frightened affair it must have been. His crops and, therefore, his livelihood have been at the mercy of forces which he could neither understand nor control; forces of fire and flood, of earthquake and drought; his communities were swept by pestilence and famine; and with the sweat of his brow, he wrung meager sustenance from nature. Today, thanks to science, all these enemies to man’s well-being have either disappeared or have been reduced to comparative impotence.

Read the above passage carefully and make notes on it in points only, using headings and sub-headings.
Answer
1. Discoveries of science during the
2. Man’s life completely changed.
3. External enemies overpowered :

• floods and fires
• earthquakes
• famines.
• pestilence

4. Now getting one’s livelihood not so difficult as it used to bo’