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Exercise 13.1

1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened 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 1m² costs Rs 20.

Answer

Length of plastic box (l) = 1.5 m
Width of plastic box (b) = 1.25 m
Depth of plastic box (h) = 0.65 m
(i) The area of sheet required to make the box is equal to the surface area of the box excluding the top.
Surface area of the box = Lateral surface area + Area of the base
= 2(l+b)×h + (l×b)
= 2[(1.5 + 1.25)×0.65] + (1.5 × 1.25) m² 
= (3.575 + 1.875) m² 
= 5.45 m² 
The sheet required required to make the box is 5.45 m² 

(ii) Cost of 1 m² of sheet = Rs 20
∴ Cost of 5.45 m² of sheet = Rs (20 × 5.45) = Rs 109

2. The length, breadth and height of a room are 5m, 4m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 per m².

Answer


length of the room = 5m
breadth of the room = 4m
height of the room = 3m
Area of four walls including the ceiling = 2(l+b)×h + (l×b)
 = 2(5+4)×3 + (5×4) m² 
= (54 + 20) m² 
= 74 m² Cost of white washing = ₹7.50 per m²
Total cost = ₹ (74×7.50) = ₹ 555

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 ₹15000, find the height of the hall.
[Hint : Area of the four walls = Lateral surface area.]

Answer

Perimeter of rectangular hall = 2(l + b) = 250 m
Total cost of painting = ₹15000
Rate per m² = ₹10
Area of four walls = 2(l + b) h m² = (250×h) m²
A/q,
(250×h)×10 = ₹15000
⇒ 2500×h = ₹15000 
⇒ h = 15000/2500 m
⇒ h = 6 m 
Thus the height of the hall is 6 m.

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

Volume of paint = 9.375 m² = 93750 cm²
Dimension of brick = 22.5 cm×10 cm×7.5 cm
Total surface area of a brick = 2(lb + bh + lh) cm² 
= 2(22.5×10 + 10×7.5 + 22.5×7.5) cm²
= 2(225 + 75 + 168.75) cm²
= 2×468.75 cm² = 937.5 cm²
Number of bricks can be painted = 93750/937.5 = 100

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

(i) Lateral surface area of cubical box of edge 10cm = 4×10² cm² = 400 cm²
Lateral surface area of cuboid box = 2(l+b)×h
= 2×(12.5+10)×8 cm²
= 2×22.5×8 cm² = 360 cm²
Thus, lateral surface area of the cubical box is greater by (400 – 360) cm² = 40 cm²

(ii) Total surface area of cubical box of edge 10 cm =6×102cm2=600cm2
Total surface area of cuboidal box = 2(lb + bh + lh)
= 2(12.5×10 + 10×8 + 8×12.5) cm²
= 2(125+80+100) cm²
= (2×305) cm² = 610 cm²
Thus, total surface area of cubical box is smaller by 10 cm²



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) Dimensions of greenhouse:
l = 30 cm, b = 25 cm, h = 25 cm
Total surface area of green house = 2(lb + bh + lh)
= 2(30×25 + 25×25 + 25×30) cm²
= 2(750 + 625 + 750) cm²
= 4250 cm²

(ii) Length of the tape needed = 4(l + b + h)
= 4(30 + 25 + 25) cm
= 4×80 cm = 320 cm

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

Dimension of bigger box = 25 cm × 20 cm × 5 cm
Total surface area of bigger box = 2(lb + bh + lh)
= 2(25×20 + 20×5 + 25×5) cm²
= 2(500 + 100 + 125) cm²
= 1450 cm²

Dimension of smaller box = 15 cm × 12 cm × 5 cm
Total surface area of smaller box = 2(lb + bh + lh)
= 2(15×12 + 12×5 + 15×5) cm²
= 2(180 + 60 + 75) cm²
= 630 cm²

Total surface area of 250 boxes of each type = 250(1450 + 630) cm²
= 250×2080 cm² = 520000 cm²
Extra area required = 5/100(1450 + 630) × 250 cm² = 26000 cm²

Total Cardboard required = 520000 + 26000 cm² = 546000 cm²
Total cost of cardboard sheet = ₹ (546000 × 4)/1000 = ₹ 2184

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 4m × 3m?

Answer

Dimensions of the box- like structure = 4m × 3m × 2.5
Tarpaulin only required for all the four sides and top.
Thus, Tarpaulin required = 2(l+b)×h + lb = [2(4+3)×2.5 + 4×3] m²
= (35+12) m²
= 47 m²

Exercise 13.2

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

Let r be the radius of the base and h = 14 cm be the height of the cylinder.
Curved surface area of cylinder = 2πrh = 88 cm²
⇒ 2 × 22/7 × r × 14 = 88
⇒ r = 88/ (2 × 22/7 × 14)
⇒ r = 1 cm
Thus, the diameter of the base = 2r = 2×1 = 2cm

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

Let r be the radius of the base and h be the height of the cylinder.
Base diameter = 140 cm and Height (h) = 1m
Radius of base (r) = 140/2 = 70 cm = 0.7 m
Metal sheet required to make a closed cylindrical tank = 2πr(h + r)
= (2 × 22/7 × 0.7) (1 + 0.7) m²
= (2 × 22 × 0.1 × 1.7) m²
=7.48 m²

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 Fig. 13.11). Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.

 

r = 4/2 cm = 2 cm
h = 77 cm
(i) Inner curved surface = 2πrh cm²
= 2 × 22/7 × 2 × 77cm²
= 968 cm²

(ii) Outer curved surface = 2πRh cm²
 = 2 × 22/7 × 2.2 × 77 cm²
= 1064.8 cm²

(iii) Total surface area of a pipe = Inner curved surface area + outer curved surface area + areas of two bases
= 2πrh + 2πRh + 2π(R^{2 –} r²)
= [968 + 1064.8 + (2 × 22/7) (4.84 – 4)] cm²
= (2032.8 + 44/7 × 0.84) cm²
= (2032.8 + 5.28) cm² = 2038.08 cm²

Page No: 217

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

Length of the roller (h) = 120 cm = 1.2 m
Radius of the cylinder = 84/2 cm = 42 cm = 0.42 m
Total no. of revolutions = 500
Distance covered by roller in one revolution = Curved surface area = 2πrh
= (2 × 22/7 × 0.42 × 1.2) m² = 3.168 m²
Area of the playground = (500 × 3.168) m² = 1584 m²

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

Radius of the pillar (r) = 50/2 cm = 25 cm = 0.25 m
Height of the pillar (h) = 3.5 m.
Rate of painting = ₹12.50 per m²
Curved surface = 2πrh
= (2 × 22/7 × 0.25 × 3.5) m²
=5.5 m²
Total cost of painting = (5.5 × 12.5) = ₹68.75

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

Let r be the radius of the base and h be the height of the cylinder.
Curved surface area = 2πrh = 4.4 m²
⇒ 2 × 22/7 × 0.7 × h = 4.4
⇒ h = 4.4/(2 × 22/7 × 0.7) = 1m
⇒ h = 1m

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

Radius of circular well (r) = 3.5/2 m = 1.75 m
Depth of the well (h) = 10 m
Rate of plastering = ₹40 per m²
(i) Curved surface = 2πrh
= (2 × 22/7 × 1.75 × 10) m²
= 110 m²

(ii) Cost of plastering = ₹(110 × 40) = ₹4400

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

Radius of the pipe (r) = 5/2 cm = 2.5 cm = 0.025 m
Length of the pipe (h) = 28/2 m = 14 m
Total radiating surface = Curved surface area of the pipe = 2πrh
= (2 × 22/7 × 0.025 × 28) m² = 4.4 m²

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 1/12 of the steel actually used was wasted in making the tank.

Answer

(i) Radius of the tank (r) = 4.2/2 m = 2.1 m
Height of the tank (h) = 4.5 m
Curved surface area = 2πrh m²
= (2 × 22/7 × 2.1 × 4.5) m²
= 59.4 m²

(ii) Total surface area of the tank = 2πr(r + h) m²
= [2 × 22/7 × 2.1 (2.1 + 4.5)] m²
= 87.12 m²

Let x be the actual steel used in making tank.
∴ (1 – 1/12) × x = 87.12
⇒ x = 87.12 × 12/11
⇒ x = 95.04 m²

10. In Fig. 13.12, 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

Radius of the frame (r) = 20/2 cm = 10 cm
Height of the frame (h) = 30 cm + 2×2.5 cm = 35 cm
2.5 cm of margin will be added both side in the height.
Cloth required for covering the lampshade = curved surface area = 2πrh
= (2 × 22/7 × 10 × 35)cm²
= 2200 cm²

11. The students of a 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 cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

Answer

Radius of the penholder (r) = 3cm
Height of the penholder (h) = 10.5cm
Cardboard required by 1 competitor = CSA of one penholder + area of the base
= 2πrh + πr²
= [(2 × 22/7 × 3 × 10.5) + 22/7 × 3²] cm²
= (198 + 198/7) cm²
= 1584/7 cm²
Cardboard required for 35 competitors = (35 × 1584/7) cm²
= 7920 cm²

Exercise 13.3

Exercise 13.4

1. Find the surface area of a sphere of radius:
(i) 10.5 cm 
(ii) 5.6 cm 
(iii) 14 cm

Answer

(i) Radius of the sphere (r) = 10.5 cm
Surface area = 4πr²
= (4 × 22/7 × 10.5 × 10.5) cm²
= 1386 cm²

(ii) Radius of the sphere (r) = 5.6 cm
Surface area = 4πr²
= (4 × 22/7 × 5.6 × 5.6) cm²
= 394.24 cm²

(iii) Radius of the sphere (r) = 14 cm
Surface area = 4πr²
= (4 × 22/7 × 14 × 14) cm²
= 2464 cm²

2. Find the surface area of a sphere of diameter:
(i) 14 cm 
(ii) 21 cm 
(iii) 3.5 m

Answer

(i) r = 14/2 cm = 7cm
Surface area = 4πr²
= (4 × 22/7 × 7 × 7)cm²
=616cm²

(ii) r = 21/2 cm = 10.5 cm
Surface area = 4πr²
= (4 × 22/7 × 10.5 × 10.5) cm²
= 1386 cm²

(iii) r = 3.5/2 m = 1.75 m
Surface area = 4πr² 
= (4 × 22/7 × 1.75 × 1.75) m²
= 38.5 m²

3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

Answer

r = 10 cm  
Total surface area of hemisphere = 3πr²
= (3 × 3.14 × 10 ×10) cm²
= 942 cm²

4. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Answer

Let r be the initial radius and R be the increased radius of balloons.
r = 7cm and R = 14cm
Ratio of the surface area =4πr²/4πR²
= r²/R²
= (7×7)/(14×14) = 1/4
Thus, the ratio of surface areas = 1 : 4

5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹16 per 100 cm².

Answer

Radius of the bowl (r) = 10.5/2 cm = 5.25 cm
Curved surface area of the hemispherical bowl = 2πr²
= (2 × 22/7 × 5.25 × 5.25) cm²
= 173.25 cm²
Rate of tin – plating is = ₹16 per 100 cm²
Therefor, cost of 1 cm² = ₹16/100 
Total cost of tin-plating the hemisphere bowl = 173.25 × 16/100
= ₹27.72

6. Find the radius of a sphere whose surface area is 154 cm².

Answer

Let r be the radius of the sphere.
Surface area = 154 cm²
⇒ 4πr² = 154
⇒ 4 × 22/7 × r² = 154
⇒ r² = 154/(4 × 22/7)
⇒ r² = 49/4
⇒ r = 7/2 = 3.5 cm

7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas. 

Answer

Let the diameter of earth be r and that of the moon will be r/4
Radius of the earth = r/2
Radius of the moon = r/8
Ratio of their surface area = 4π(r/8)²/4π(r/2)²
= (1/64)/(1/4)
= 4/64 = 1/16
Thus, the ratio of their surface areas is 1:16

8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Answer

Inner radius of the bowl (r) = 5 cm
Thickness of the steel = 0.25 cm
∴ outer radius (R) = (r + 0.25) cm
= (5 + 0.25) cm  = 5.25 cm
Outer curved surface = 2πR²
= (2 × 22/7 × 5.25 × 5.25) cm²
= 173.25 cm²

9. A right circular cylinder just encloses a sphere of radius r (see Fig. 13.22). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).

Answer

(i) The surface area of the sphere with raius r = 4πr²

(ii) The right circular cylinder just encloses a sphere of radius r.

Exercise 13.5

1. A matchbox measures 4cm × 2.5cm × 1.5cm. What will be the volume of a packet containing 12 such boxes?

Answer

Dimension of matchbox = 4cm × 2.5cm × 1.5cm
l = 4 cm, b = 2.5 cm and h = 1.5 cm
Volume of one matchbox = (l × b × h)
= (4 × 2.5 × 1.5)  cm³ = 15 cm³
Volume of a packet containing 12 such boxes = (12 × 15)  cm³ = 180 cm³

2. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1m³ = 1000 l)

Answer

Dimensions of water tank = 6m × 5m × 4.5m
l = 6m , b = 5m and h = 4.5m
Therefore Volume of the tank =ℓbh m³
=(6×5×4.5)m3=135 m³
Therefore , the tank can hold = 135 × 1000 litres  [Since 1m³=1000litres]
= 135000 litres of water.

3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?

Answer

Length = 10 m , Breadth = 8 m and Volume = 380 m³
Volume of cuboid = Length × Breadth × Height
⇒ Height = Volume of cuboid/(Length × Breadth) 
= 380/(10×8) m
= 4.75m

4. Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ₹30 per m³.

Answer

l = 8 m, b = 6 m and h = 3 m
Volume of the pit = lbh m³
= (8×6×3) m³
= 144 m³
Rate of digging = ₹30 per m³
Total cost of digging the pit = ₹(144 × 30) = ₹4320

5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.

Answer

length = 2.5 m, depth = 10 m and volume = 50000 litres
1m³ = 1000 litres

∴ 50000 litres = 50000/1000 m³ = 50 m³
Breadth = Volume of cuboid/(Length×Depth)
= 50/(2.5×10) m
= 2 m

6. A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20m × 15m × 6m. For how many days will the water of this tank last?

Answer

Dimension of tank = 20m × 15m × 6m
l = 20 m , b = 15 m and h = 6 m
Capacity of the tank = lbh m³
= (20×15×6) m³
= 1800 m³
Water requirement per person per day =150 litres
Water required for 4000 person per day = (4000×150) l
= (4000×150)/1000
= 600 m³
Number of days the water will last = Capacity of tank Total water required per day
=(1800/600) = 3
The water will last for 3 days.

7. A godown measures 40m × 25m × 15m. Find the maximum number of wooden crates each measuring 1.5m × 1.25m × 0.5m that can be stored in the godown.

Answer

Dimension of godown = 40 m × 25 m × 15 m
Volume of the godown = (40 × 25 × 15) m³ = 10000 m³
Dimension of crates = 1.5m × 1.25m × 0.5m 
Volume of 1 crates = (1.5 × 1.25 × 0.5) m³ = 0.9375 m³
Number of crates that can be stored =Volume of the godown/Volume of 1 crate
= 10000/0.9375 = 10666.66 = 10666

8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.

Answer

Edge of the cube = 12 cm. 
Volume of the cube = (edge)³ cm³
= (12 × 12 × 12) cm³
= 1728 cm³ 
Number of smaller cube = 8
Volume of the 1 smaller cube =1728/8 cm³ = 216 cm³
Side of the smaller cube = a
a³ = 216
⇒ a = 6 cm
Surface area of the cube = 6 (side)²
Ratio of their surface area = (6 × 12 × 12)/(6 × 6 × 6)
= 4/1 = 4:1

9. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?

Answer

Depth of river (h) = 3 m
Width of river (b) = 40 m
Rate of flow of water (l) = 2 km per hour = (2000/60) m per minute
= 100/3 m per minute 
Volume of water flowing into the sea in a minute = lbh m³
= (100/3 × 40 × 3) m³
= 4000 m³

Exercise 13.6

1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm 3 = 1l)

Answer

Let the base radius of the cylindrical vessel be ‘r’ cm.
∴ Circumference = 2πr
⇒ 2πr = 132     [∵ Circumference = 132 cm]
⇒ 2× 22/7 × r = 132 cm
R=(132×7)/(2×22) cm = 21 cm
∵ Height of the vessel = 25 cm
∴ Volume = πr² x h    [∵ Volume of a cylinder = πr²h]
= (22/7) (21)² × 25 cm³
= (22/7) × 21 × 21 × 25 cm³
= 22 × 3 × 21 × 25 cm³
= 34650 cm³ 
∵ Capacity of the vessel = Volume of the vessel
∴ Capacity of cylindrical vessel = 34650 cm³ 
Since 1000 cm³ = 1 litre
⇒ 34650 cm³ = (34650/1000) litres = 34.65 l

2. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ of wood has a mass of 0.6 g.
Answer

Here, Inner diameter of the cylindrical pipe = 24 cm
⇒ Inner radius of the pipe (r) = (24/2) cm = 12 cm
Outer diameter of the pipe = 28 cm
⇒ Outer radius of the pipe (R) = (28/2)cm = 14 cm
Length of the pipe (h) = 35 cm
∵ Inner volume of the pipe = πr²h
Outer volume of the pipe = πr²h
∴ Amount of wood (volume) in the pipe = Outer volume – Inner volume
= πR²h – πr²h
= πh(R²-r²)
= πh(R+r)(R-r) [ ∵ a² – b² = (a+b)(a-b)]
= 22/7 x 35 x (14+12) x (14-12) cm³
= 22 x 5 x 26 x 2 cm³ 

Mass of the wood in the pipe = [Mass of wood in 1 m³ of wood] x [Volume of wood in the pipe]
= [0.6g] x [22 x 5 x 26 x 2] cm³
= (6/10)x 22 x 10 x 26 g = 6 x 22 x 26 g
= 3432 g = (3432/1000)= 3.432 kg   [∵ 1000 g = 1 kg]
Thus, the required mass of the pipe is 3.432 kg.

3. A soft drink is available in two packs (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Answer

For rectangular pack: Length (l) = 5 cm
Breadth (b) = 4 cm Height (h) = 15 cm
∴ Volume = l x b x h = 5 x 4 x 15 cm³
= 300 cm³
⇒ Capacity of the rectangular pack = 300 cm³     …(1)
For cylindrical pack: Base diameter = 7 cm
⇒ Radius of the base (r) = (7/2)cm
Height (h) = 10 cm
∴ Volume = πr²h = (22/7) x (7/2)² x 10 cm³
= (22/7) x (7/2) x (7/2) x 10 cm³
= 11 x 7 x 5 cm³ = 385 cm³
⇒ Volume of the cylindrical pack = 385 cm³   …(2)
From (1) and (2),
we have 385 cm³ – 300 cm³ = 85 cm³
⇒ The cylindrical pack has the greater capacity by 85 cm³.

4. If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, then find: (i) radius of its base (ii) its volume. (Use π = 3.14)

Answer

(i) Since lateral surface of the cylinder = 2 πrh
But lateral surface of the cylinder = 94.2 cm²
 2πrh = 94.2
2×3.14×r×5 =942/10
⇒{(10×314)/100} × r = 942/10
⇒ r=(942/10)×{100/(10×314)}cm
⇒ r =471/157 cm 
Thus, the radius of the cylinder = 3 cm

(ii) Volume of a cylinder = πr²h
⇒ Volume of the given cylinder = 3.14 x (3)² x 5 cm³
=314×100×3×3×5cm³
=(157×3×3)/10
= 1413/10 = 141.3 cm³

Thus, the required volume = 141.3 cm³

5. It costs ₹ 2200 to paint the inner curved surface of cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m²; find: (i) inner curved surface of the vessel (ii) radius of the base (iii) capacity of the vessel.

Answer

(i) To find inner curved surface
Total cost of painting = ₹ 2200
Rate of painting = ₹ 20 per m²
∴  Area =cost/rate = 2200/20 = 110 m²   
⇒ Inner curved surface of the vessel = 110 m²

(ii) To find radius of the base Let the base radius of the cylindrical vessel.
∵ Curved surface of a cylinder = 2 πrh
∴  2πrh = 110
⇒ 2× 22/7 ×r×10 = 110 [∵Height=10 m ]
⇒ r= (110×7)/(2×22×10)m = 7/4 m
= 1.75 m
⇒ The required radius of the base = 1.75 m

(iii) To find the capacity of the vessel
Since, volume of a cylinder = πr²h
 ∴ Volume (capacity) of the vessel =22/7 × (7/2)² × 10 m³
= 22/7 × 7/4 × 7/4 × 10 m³ 
= (11×7×5)/4 m³ = 385/4 m3 = 96.25 m³

Since, 1 m³ = 1000000 cm³ = 1000 l = 1 kl
∴ 96.5 m³ = 96.5 kl
Thus, the required volume = 96.25 kl

6. The capacity of closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?

Answer

Capacity of the cylindrical vessel = 15.4 l
= 15.4×1000 cm³ 

⇒ Volume of the vessel = (15.4/1000)m³
Height of the vessel = 1m Let ‘r’ metres be the radius of the base of the vessel
∴ Volume = πr²h

⇒ πr²h= 15.4/1000

Thus, the required sheet = 0.4708 m²

7. A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Answer

Since, 10 mm = 1cm
∴ 1 mm = (1/10) cm
For graphite cylinder

Thus, the required volume of the graphite = 0.11 cm³ 
For the pencil Diameter of the pencil = 7 mm = (7/10)cm
∴ Radius of the pencil (R) = (7/20) cm

Height of the pencil (h) = 14 cm
Volume of the pencil = πr²h

Volume of the wood Volume of the wood = [Volume of the pencil] – [Volume of the graphite]
= 5.39 cm³ – 0.11 cm³ = 5.28 cm³ 
Thus, the required volume of the wood is 5.28 cm³.

8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

Answer

The bowl is cylindrical.
Diameter of the base = 7 cm
⇒ Radius of the base (r) = (7/3) cm
Height (h) = 4 cm
Volume of soup = πr²h

= 38500 / 100 liters             

Thus, the hospital needs to prepare 38.5 litres of soup daily for 250 patients.

Exercise 13.8

1. Find the volume of a sphere whose radius is
(i) 7 cm        (ii) 0.63 m

Answer

(i) Radius of the sphere(r) = 7 cm
Therefore, Volume of the sphere = 4/3 πr³
                                                     = (4/3 × 22/7 × 7 × 7 × 7) cm³
                                                     = 4312/3 cm³

(ii) Radius of the sphere(r) = 0.63 m
Volume of the sphere = 4/3 πr³
                                   = (4/3 × 22/7 × 0.63 × 0.63 × 0.63) m³
                                   = 1.05 m³

2. Find the amount of water displaced by a solid spherical ball of diameter.
(i) 28 cm                     (ii) 0.21 m

Answer

(i) Diameter of the spherical ball = 28 cm
Radius = 28/2 cm = 14 cm
Amount of water displaced by the spherical ball = Volume
                              = 4/3 πr³
                              = (4/3 × 22/7 × 14 × 14 × 14) cm³
                              = 34496/3 cm³

(ii) Diameter of the spherical ball = 0.21 m
Radius (r) = 0.21/2 m = 0.105 m
Amount of water displaced by the spherical ball = Volume
                              = 4/3 πr³
                              = (43×227×0.105×0.105×0.105) m³
                              = 0.004851 m³

3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³?

Answer

Diameter of the ball = 4.2 cm
Radius = (4.2/2) cm = 2.1 cm
Volume of the ball = 4/3 πr³
                               = (4/3 × 22/7 × 2.1 × 2.1 × 2.1) cm³
                               = 38.808 cm³
Density of the metal is 8.9g per cm3
Mass of the ball = (38.808 × 8.9) g = 345.3912 g

4. The diameter of the moon is approximately one-fourth of the diameter of the earth.What fraction of the volume of the earth is the volume of the moon? 

Answer

Let the diameter of the moon be r
Radius of the moon = r/2
A/q,
Diameter of the earth = 4r
Radius(r) = 4r/2 = 2r
Volume of the moon = v = 4/3 π(r/2)³
                                  = 4/3 πr³ × 1/8
⇒ 8v = 4/3 πr³ — (i)
Volume of the earth = r³ = 4/3 π(2r)³
                                = 4/3 πr³× 8
⇒ V/8 = 4/3 πr³ — (ii)
From (i) and (ii), we have
8v = V/8
⇒ v = 1/64 V
Thus, the volume of the moon is 1/64 of the volume of the earth.

5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Answer

Diameter of a hemispherical bowl = 10.5 cm
Radius(r) = (10.5/2) cm = 5.25cm
Volume of the bowl = 2/3 πr³
                                = (2/3 × 22/7 × 5.25 × 5.25 × 5.25) cm³
                                = 303.1875 cm³
Litres of milk bowl can hold = (303.1875/1000) litres
                                               = 0.3031875 litres (approx.)

6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Answer

Internal radius = r = 1m
External radius = R = (1 + 0.1) cm = 1.01 cm
Volume of iron used = External volume – Internal volume
                                  = 2/3 πR³ – 2/3 πr³
                                  = 2/3 π(R³ – r³)
                                  = 2/3 × 22/7 × [(1.01)³−(1)³] m³
                                  = 44/21 × (1.030301 – 1) m³
                                  = (44/21 × 0.030301) m³
                                  = 0.06348 m³(approx)

7. Find the volume of a sphere whose surface area is 154 cm².

Answer

Let r cm be the radius of the sphere
So, surface area = 154cm²
⇒ 4πr² = 154
⇒ 4 × 22/7 × r² = 154
⇒ r² = (154×7)/(4×22) = 12.25
⇒ r = 3.5 cm
Volume = 4/3 πr³
             = (4/3 × 22/7 × 3.5 × 3.5 × 3.5) cm³
             = 539/3 cm³

8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹498.96. If the cost of white-washing is ₹2.00 per square metre, find the
(i) inside surface area of the dome,      (ii) volume of the air inside the dome.

Answer

(i) Inside surface area of the dome =Total cost of white washing/Rate of white washing
                                                        =  (498.96/2.00) m² = 249.48 m² 

 (ii) Let r be the radius of the dome.
Surface area = 2πr²
⇒ 2 × 22/7 × r² = 249.48
⇒ r² = (249.48×7)/(2×22) = 39.69
⇒ r²=  39.69
⇒ r = 6.3m
Volume of the air inside the dome = Volume of the dome
                                                       = 2/3 πr³
                                                                    = (2/3 × 22/7 × 6.3 × 6.3 × 6.3) m³
                                                       = 523.9 m³ (approx.)

9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. Find the
(i) radius r′ of the new sphere,      (ii) ratio of S and S′. 

Answer

(i) Volume of 27 solid sphere of radius r = 27 × 4/3 πr³ — (i)
Volume of the new sphere of radius r′ = 4/3 πr’³ — (ii)
A/q,
4/3 πr’³= 27 × 4/3 πr³
⇒ r’³ = 27r³
⇒ r’³ = (3r)³
⇒ r′ = 3r
(ii) Required ratio = S/S′ = 4 πr²/4πr′² = r²/(3r)²
                              = r²/9r² = 1/9 = 1:9

10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule?

Answer

Diameter of the spherical capsule = 3.5 mm
Radius(r) = 3.52mm
                = 1.75mm
Medicine needed for its filling = Volume of spherical capsule
                                                  = 4/3 πr³
                                                  = (4/3 × 22/7 × 1.75 × 1.75 × 1.75) mm³
                                                  = 22.46 mm³ (approx.)

 Question 1.
A wooden bookshelf has external dimensions as follows : Height = 110cm, Depth = 25cm, Breadth = 85cm (see figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing-is 20 paise per cm² and the rate of pointing is 10 paise per cm², find the total expenses required for palishing and painting the surface of the bookshelf.

Solution:
Here, l = 85 cm b = 25 cm and h = 110 cm
Area of the bookshelf of outer surface = 2 lb + 2bh + hl
= [2(85 x 25)+ 2(110 x 25)+ 85×110] cm2 = (4250 + 5500 + 9350) cm² = 19100 cm²
Cost of polishing of the outer surface of bookshelf
= 19100 x 20100 = ₹ 3820
Thickness of the plank = 5 cm
Internal height of bookshelf = (110 – 2 x 5) = 100 cm
Internal depth of bookshelf = (25 – 5) = 20 cm
Internal breadth of bookshelf = 85 – 2 x 5 = 75 cm
Hence, area of the internal surface of bookshelf
= 2(75 x 20)+ 2(100x 20)+ 75×100
= 3000 + 4000 + 7500 = 14500 cm2
So, cost of painting of internal surface of bookshelf
= 14500 x 10100 = ₹1450
Hence, total costing of polishing and painting = 3820 + 1450= ₹ 5270

Question 2.
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in figure. Eight such spheres are-used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm² and black paint costs 5 paise per cm².

Solution:
It is obvious, we have to subtract the cost of the sphere that is resting on the supports while calculating the cost of silver paint.
Surface area to be silver paint

 Question 3.
The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?
Solution:
Let d be the diameter of the sphere

Exercise 12.1

1. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?

Answer

Length of the side of equilateral triangle = a
Perimeter of the signal board = 3a = 180 cm
∴ 3a = 180 cm ⇒ a = 60 cm
Semi perimeter of the signal board (s) = 3a/2
Using heron’s formula,
Area of the signal board = √s (s-a) (s-b) (s-c)
                                       = √(3a/2) (3a/2 – a) (3a/2 – a) (3a/2 – a)
                                       = √3a/2 × a/2 × a/2 × a/2
                                       = √3a⁴/16
                                       = √3a²/4
                                       = √3/4 × 60 × 60 = 900√3 cm²

2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig. 12.9). The advertisements yield an earning of ₹5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Answer

The sides of the triangle are 122 m, 22 m and 120 m.
Perimeter of the triangle is 122 + 22 + 120 = 264m
Semi perimeter of triangle (s) = 264/2 = 132 m
Using heron’s formula,
Area of the advertisement = √s (s-a) (s-b) (s-c)
                                       = √132(132 – 122) (132 – 22) (132 – 120) m²
                                       = √132 × 10 × 110 × 12 m²
                                       = 1320 m²
Rent of advertising per year = ₹ 5000 per m²
Rent of one wall for 3 months = ₹ (1320 × 5000 × 3)/12 = ₹ 1650000

Page No: 203

3. There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see Fig. 12.10 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Answer

Sides of the triangular wall are 15 m, 11 m and 6 m.
Semi perimeter of triangular wall (s) = (15 + 11 + 6)/2 m = 16 m
Using heron’s formula,
Area of the message = √s (s-a) (s-b) (s-c)
                                       = √16(16 – 15) (16 – 11) (16 – 6) m²
                                       = √16 × 1 × 5 × 10 m² = √800 m²
                                       = 20√2 m²

4. Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.

Answer

Two sides of the triangle = 18cm and 10cm
Perimeter of the triangle = 42cm
Third side of triangle = 42 – (18+10) cm = 14cm
Semi perimeter of triangle = 42/2 = 21cm
Using heron’s formula,
Area of the triangle = √s (s-a) (s-b) (s-c)
                                       = √21(21 – 18) (21 – 10) (21 – 14) cm²
                                       = √21 × 3 × 11 × 7 m²
                                       = 21√11 cm²

5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.

Answer

Ratio of the sides of the triangle = 12 : 17 : 25
Let the common ratio be x then sides are 12x, 17x and 25x
Perimeter of the triangle = 540cm
12x + 17x + 25x = 540 cm
⇒ 54x = 540cm
⇒ x = 10
Sides of triangle are,
12x = 12 × 10 = 120cm
17x = 17 × 10 = 170cm
25x = 25 × 10 = 250cm
Semi perimeter of triangle(s) = 540/2 = 270cm
Using heron’s formula,
Area of the triangle = √s (s-a) (s-b) (s-c)
                                       = √270(270 – 120) (270 – 170) (270 – 250)cm²
                                       = √270 × 150 × 100 × 20 cm²
                                       = 9000 cm²

6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

Answer

Length of the equal sides = 12cm
Perimeter of the triangle = 30cm
Length of the third side = 30 – (12+12) cm = 6cm
Semi perimeter of the triangle(s) = 30/2 cm = 15cm
Using heron’s formula,
Area of the triangle = √s (s-a) (s-b) (s-c)
                                       = √15(15 – 12) (15 – 12) (15 – 6)cm²
                                       = √15 × 3 × 3 × 9 cm²
                                       = 9√15 cm²

Exercise 11.1

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

Answer

Steps of construction:

Step 1: A ray YZ is drawn.
Step 2: With Y as a centre and any radius, an arc ABC is drawn cutting YZ at C.
Step 3: With C as a centre and the same radius, mark a point B on the arc ABC.

Step 4: With B as a centre and the same radius, mark a point A on the arc ABC.

Step 5: With A and B as centre, draw two arcs intersecting each other with the same radius at X.
Step 6: X and Y are joined and a ray XY making an angle 90° with YZ is formed.

Justification for construction:
We constructed ∠BYZ = 60° and also ∠AYB = 60°.
Thus, ∠AYZ = 120°.
Also, bisector of ∠AYB is constructed such that:
∠AYB = ∠XYA + ∠XYB
⇒ ∠XYB = 1/2∠AYB
⇒ ∠XYB = 1/2×60°
⇒ ∠XYB = 30°
Now,
∠XYZ = ∠BYZ + ∠XYB = 60° + 30° = 90°

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

Answer

Steps of construction:

Step 1: A ray OY is drawn.
Step 2: With O as a centre and any radius, an arc ABC is drawn cutting OY at A.
Step 3: With A as a centre and the same radius, mark a point B on the arc ABC.

Step 4: With B as a centre and the same radius, mark a point C on the arc ABC.
Step 5: With A and B as centre, draw two arcs intersecting each other with the same radius at X.
Step 6: X and Y are joined and a ray making an angle 90° with YZ is formed.
Step 7: With A and E as centres, two arcs are marked intersecting each other at D and the bisector of ∠XOY is drawn.

Justification for construction:
By construction,
∠XOY = 90°
We constructed the bisector of ∠XOY as DOY.
Thus,
∠DOY = 1/2 ∠XOY
∠DOY = 1/2×90° = 45°

3. Construct the angles of the following measurements:
(i) 30°        (ii) 22.5°        (iii) 15°

Answer

(i) 30°

Steps of constructions:
Step 1: A ray OY is drawn.
Step 2: With O as a centre and any radius, an arc AB is drawn cutting OY at A.
Step 3: With A and B as centres, two arcs are marked intersecting each other at X and the bisector of is drawn.

Thus, ∠XOY is the required angle making 30° with OY.

(ii) 22.5°

 

Steps of constructions:
Step 1: An angle ∠XOY = 90° is drawn.
Step 2: Bisector of ∠XOY is drawn such that ∠BOY = 45° is constructed.
Step 3: Again, ∠BOY is bisected such that ∠AOY is formed.
Thus, ∠AOY is the required angle making 22.5° with OY.

(iii) 15°

Steps of constructions:
Step 1: An angle ∠AOY = 60° is drawn.
Step 2: Bisector of ∠AOY is drawn such that ∠BOY = 30° is constructed.
Step 3: With C and D as centres, two arcs are marked intersecting each other at X and the bisector of ∠BOY is drawn.
Thus, ∠XOY is the required angle making 15° with OY.

4. Construct the following angles and verify by measuring them by a protractor:
(i) 75°         (ii) 105°        (iii) 135°

Answer

(i) 75°

Step 1: A ray OY is drawn.
Step 2: An arc BAE is drawn with O as a centre.
Step 3: With E as a centre, two arcs are A and C are made on the arc BAE.
Step 4: With A and B as centres, arcs are made to intersect at X and ∠XOY = 90° is made.
Step 5: With A and C as centres, arcs are made to intersect at D
Step 6: OD is joined and and ∠DOY = 75° is constructed.
Thus, ∠DOY is the required angle making 75° with OY.

(ii) 105°

 

Step 1: A ray OY is drawn.
Step 2: An arc ABC is drawn with O as a centre.
Step 3: With A as a centre, two arcs are B and C are made on the arc ABC.
Step 4: With B and C as centres, arcs are made to intersect at E and ∠EOY = 90° is made.
Step 5: With B and C as centres, arcs are made to intersect at X
Step 6: OX is joined and and ∠XOY = 105° is constructed.
Thus, ∠XOY is the required angle making 105° with OY.

Steps of constructions:Step 1: A ray DY is drawn.
Step 2: An arc ACD is drawn with O as a centre.
Step 3: With A as a centre, two arcs are B and C are made on the arc ACD.
Step 4: With B and C as centres, arcs are made to intersect at E and ∠EOY = 90° is made.
Step 5: With F and D as centres, arcs are made to intersect at X or bisector of ∠EOD is constructed.
Step 6: OX is joined and and ∠XOY = 135° is constructed.
Thus, ∠XOY is the required angle making 135° with DY.

5. Construct an equilateral triangle, given its side and justify the construction.

Answer

Steps of constructions:
Step 1: A line segment AB=4 cm is drawn.
Step 2: With A and B as centres, two arcs are made.
Step 4: With D and E as centres, arcs are made to cut the previous arc respectively and forming angle of 60° each.
Step 5: Lines from A and B are extended to meet each other at C.
Thus, ABC is the required triangle formed.

Justification:
By construction,
AB = 4 cm, ∠A = 60° and ∠B = 60°
We know that,
∠A + ∠B + ∠C = 180° (Sum of the angles of a triangle)
⇒ 60° + 60° + ∠C = 180°
⇒ 120° + ∠C = 180°
⇒ ∠C = 60°
BC = CA = 4 cm (Sides opposite to equal angles are equal)
AB = BC = CA = 4 cm
∠A = ∠B = ∠C = 60°

Exercise 10.1

1. Fill in the blanks:
(i) The centre of a circle lies in ____________ of the circle. (exterior/ interior)
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in __________ of the circle. (exterior/ interior)

(iii) The longest chord of a circle is a _____________ of the circle.

(iv) An arc is a ___________ when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and _____________ of the circle.
(vi) A circle divides the plane, on which it lies, in _____________ parts.

Answer

(i)   The centre of a circle lies in interior of the circle. (exterior/interior)
(ii)  A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle. (exterior/interior)
(iii) The longest chord of a circle is a diameter of the circle.
(iv) An arc is a semicircle when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and chord of the circle.
(vi) A circle divides the plane, on which it lies, in three parts.

2. Write True or False: Give reasons for your answers.

(i) Line segment joining the centre to any point on the circle is a radius of the circle.
(ii) A circle has only finite number of equal chords.
(iii) If a circle is divided into three equal arcs, each is a major arc.
(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
(v) Sector is the region between the chord and its corresponding arc.
(vi) A circle is a plane figure.

Answer

(i) True.
All the line segment from the centre to the circle is of equal length.
(ii) False.
We can draw infinite numbers of equal chords.
(iii) False.
We get major and minor arcs for unequal arcs. So, for equal arcs on circle we can’t say it is major arc or minor arc.
(iv) True.
A chord which is twice as long as radius must pass through the centre of the circle and is diameter to the circle.
(v) False.
Sector is the region between the arc and the two radii of the circle.
(vi) True.
A circle can be drawn on the plane.

Exercise 10.2

1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Answer

A circle is a collection of points whose every every point is equidistant from the centre. Thus, two circles can only be congruent when they the distance of every point of the both circle is equal from the centre.

Given,
AB = CD (Equal chords)
To prove,
∠AOB = ∠COD
Proof,
In ΔAOB and ΔCOD,
OA = OC (Radii)
OB = OD (Radii)
AB = CD (Given)
∴ ΔAOB ≅ ΔCOD (SSS congruence condition)
Thus, ∠AOB = ∠COD by CPCT.
Equal chords of congruent circles subtend equal angles at their centres.

2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Answer

Given,
∠AOB = ∠COD (Equal angles)
To prove,
AB = CD
Proof,
In ΔAOB and ΔCOD,
OA = OC (Radii)
∠AOB = ∠COD (Given)
OB = OD (Radii)
∴ ΔAOB ≅ ΔCOD (SAS congruence condition)
Thus, AB = CD by CPCT.
If chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Exercise 10.3

1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Answer

One point P is common.

One point P is common.

Two points P and Q are common.

No point is common.

2. Suppose you are given a circle. Give a construction to find its centre.

Answer

Steps of construction:
Step I: A circle is drawn.
Step II: Two chords AB and CD are drawn.
Step III: Perpendicular bisector of the chords AB and CD are drawn.
Step IV: Let these two perpendicular bisector meet at a point. The point of intersection of these two perpendicular bisector is the centre of the circle.

Given,
Two circles which intersect each other at P and Q.
To prove,
OO’ is perpendicular bisector of PQ.
Proof,
In ΔPOO’ and ΔQOO’,
OP = OQ (Radii)
OO’ = OO’ (Common)
O’P = OQ (Radii)
∴ ΔPOO’ ≅ ΔQOO’ (SSS congruence condition)
Thus,
∠POO’ = ∠QOO’ — (i)
In ΔPOR and ΔQOR,
OP = OQ (Radii)
∠POR = ∠QOR (from i)
OR = OR (Common)
∴ ΔPOR ≅ ΔQOR (SAS congruence condition)
Thus,
∠PRO = ∠QRO
also,
∠PRO + ∠QRO = 180°
⇒ ∠PRO = ∠QRO = 180°/2 = 90°
Hence,
OO’ is perpendicular bisector of PQ.

Exercise 10.4

1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

Answer

OP = 5cm, PS = 3cm and OS = 4cm.
also, PQ = 2PR
Let RS be x.

In ΔPOR,
OP² = OR² + PR²
⇒ 5² = (4-x)² + PR²
⇒ 25 = 16 + x² – 8x +  PR²
⇒ PR² = 9 – x² + 8x — (i)


In ΔPRS,
PS² = PR² + RS²
⇒ 3² = PR² + x²
⇒  PR² = 9 – x² — (ii)

Equating (i) and (ii),
9 – x² + 8x = 9 – x²
⇒ 8x = 0
⇒ x = 0
Putting the value of x in (i) we get,
PR² = 9 – 0²
⇒ PR = 3cm
Length of the cord PQ = 2PR = 2×3 = 6cm

2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Answer

Given,
AB and CD are chords intersecting at E.
AB = CD
To prove,
AE = DE and CE = BE
Construction,
OM ⊥ AB and ON ⊥ CD. OE is joined.

Proof,
OM bisects AB (OM ⊥ AB)
ON bisects CD (ON ⊥ CD)
As AB = CD thus,
AM = ND — (i)
and MB = CN — (ii)
In ΔOME and ΔONE,
∠OME = ∠ONE (Perpendiculars)
OE = OE (Common)
OM = ON (AB = CD and thus equidistant from the centre)
ΔOME ≅ ΔONE by RHS congruence condition.
ME = EN by CPCT — (iii)
From (i) and (ii) we get,
AM + ME = ND + EN
⇒ AE = ED
From (ii)  and (iii) we get,
MB – ME = CN – EN
⇒ EB = CE

3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Answer

Given,
AB and CD are chords intersecting at E.
AB = CD, PQ is the diameter.
To prove,
∠BEQ = ∠CEQ
Construction,
OM ⊥ AB and ON ⊥ CD. OE is joined.

In ΔOEM and ΔOEN,
OM = ON (Equal chords are equidistant from the centre)
OE = OE (Common)
∠OME = ∠ONE (Perpendicular)
ΔOEM ≅ ΔOEN by RHS congruence condition.
Thus,
∠MEO = ∠NEO by CPCT
⇒ ∠BEQ = ∠CEQ

4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 10.25).

Answer

OM ⊥ AD is drawn from O.
OM bisects AD as OM ⊥ AD.
⇒ AM = MD — (i)
also, OM bisects BC as OM ⊥ BC.
⇒ BM = MC — (ii)
From (i) and (ii),
AM – BM = MD – MC
⇒ AB = CD

5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?

Answer

 Let A, B and C represent the positions of Reshma, Salma and Mandip respectively.
AB = 6cm and BC = 6cm.
Radius OA = 5cm
BM ⊥ AC is drawn.
ABC is an isosceles triangle as AB = BC, M is mid-point of AC. BM is perpendicular bisector of AC and thus it passes through the centre of the circle.
Let AM = y and OM = x then BM = (5-x).
Applying Pythagoras theorem in ΔOAM,
OA² = OM² + AM²
⇒ 5² = x² + y² — (i)
Applying Pythagoras theorem in ΔAMB,
AB² = BM² + AM²
⇒ 6² = (5-x)² + y² — (ii)
Subtracting (i) from (ii), we get
36 – 25 = (5-x)² – x² –
⇒ 11 = 25 – 10x
⇒ 10x = 14 ⇒ x= 7/5
Substituting the value of x in (i), we get
y² + 49/25 = 25
⇒ y² = 25 – 49/25
⇒ y² = (625 – 49)/25
⇒ y² = 576/25
⇒ y = 24/5
Thus,
AC = 2×AM = 2×y = 2×(24/5) m = 48/5 m = 9.6 m
Distance between Reshma and Mandip is 9.6 m.

6. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Answer

Let A, B and C represent the positions of Ankur, Syed and David respectively. All three boys at equal distances thus ABC is an equilateral triangle.
AD ⊥ BC is drawn. Now, AD is median of ΔABC and it passes through the centre O.
Also, O is the centroid of the ΔABC. OA is the radius of the triangle.
OA = 2/3 AD
Let the side of a triangle a metres then BD = a/2 m.
Applying Pythagoras theorem in ΔABD,
AB² = BD² + AD²
⇒ AD² = AB² – BD² 
⇒ AD² = a² – (a/2)²
⇒ AD² = 3a²/4
⇒ AD = √3a/2
OA = 2/3 AD ⇒ 20 m = 2/3 × √3a/2
⇒ a = 20√3 m
Length of the string is 20√3 m.

 Question 1.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Solution:
Given: Two circles with centres O and O’ which intersect each other at C and D.

To prove: ∠OCO’ = ∠ODO’
Construction: Join OC, OD, O’C and O’D
Proof: In ∆ OCO’and ∆ODO’, we have
OC = OD (Radii of the same circle)
O’C = O’D (Radii of the same circle)
OO’ = OO’ (Common)
∴ By SSS criterion, we get
∆ OCO’ ≅ ∆ ODO’
Hence, ∠OCO’ = ∠ODO’ (By CPCT)

 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.
Solution:
Let O be the centre of the given circle and let its radius be cm.
Draw ON ⊥ AB and OM⊥ CD since, ON ⊥ AB, OM ⊥ CD and AB || CD, therefore points N, O, M are collinear.

Let ON = a cm
∴ OM = (6 – a) cm
Join OA and OC.
Then, OA = OC = b c m
Since, the perpendicular from the centre to a chord of the circle bisects the chord.
Therefore, AN = NB= 2.5 cm and OM = MD = 5.5 cm
In ∆OAN and ∆OCM, we get
OA² = ON² + AN²
OC² = OM² + CM²
⇒ b² = a² + (2.5)²
and, b² = (6-a)² + (5.5)² …(i)
So, a² + (2.5)² = (6 – a)² + (5.5)²
⇒ a² + 6.25= 36-12a + a² + 30.25
⇒ 12a = 60
⇒ a = 5
On putting a = 5 in Eq. (i), we get
b² = (5)² + (2.5)²
= 25 + 6.25 = 31.25
So, r = 31.25−−−−√ = 5.6cm (Approx.)

 Question 3.
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre ?
Solution:
Let PQ and RS be two parallel chords of a circle with centre O such that PQ = 6 cm and RS = 8 cm.
Let a be the radius of circle.

Draw ON ⊥ RS, OM ⊥ PQ. Since, PQ || RS and ON ⊥ RS, OM⊥ PQ, therefore points 0,N,M are collinear.
∵ OM = 4 cm and M and N are the mid-points of PQ and RS respectively.
PM = MQ = 12 PQ = 62 = 3 cm
and RN = NS = 12 RS = 82 = 4 cm
In ∆OPM, we have
OP² = OM² + PM²
⇒ a² =4² + 3² = 16 + 9 = 25
⇒ a = 5
In ∆ORN, we have
⇒ OR² = ON² + RN²
⇒ a² = ON² + (4)²
⇒ 25 = ON² + 16
⇒ ON² = 9
⇒ ON = 3cm
Hence, the distance of the chord PS 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.
Solution:
Since, an exterior angle of a triangle is equal to the sum of the interior opposite angles.

∴ In ∆BDC, we get
∠ADC = ∠DBC + ∠DCB …(i)
Since, angle at the centre is twice at a point on the remaining part of circle.
∴ ∠DCE = 12 ∠DOE
⇒ ∠DCB = 12 ∠DOE (∵ ∠DCE = ∠DCB)
∠ADC = 12 ∠AOC
∴ 12 ∠AOC = ∠ABC + 12 ∠DOE (∵ ∠DBC = ∠ABC)
∴ ∠ABC = 12 (∠AOC – ∠DOE)
Hence, ∠ABC is equal to half the difference of angles subtended by the chords AC and DE at the centre.

 Question 5.
Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
Solution:
Given: PQRS is a rhombus. PR and SQ are its two diagonals which bisect each other at right angles.
To prove: A circle drawn on PQ as diameter will pass through O.

Construction: Through O, draw MN || PS and EF || PQ.
Proof : ∵ PQ = SR ⇒ 12 PQ = 12 SR
So, PN = SM
Similarly, PE = ON
So, PN = ON = NQ
Therefore, a circle drawn with N as centre and radius PN passes through P, O, Q.

 Question 6.
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Solution:
Since, ABCE is a cyclic quadrilateral, therefore

∠AED+ ∠ABC= 180°
(∵ Sum of opposite angle of a cyclic quadrilateral is 180°) .. .(i)
∵ ∠ADE + ∠ADC = 180° (EDC is a straight line)
So, ∠ADE + ∠ABC = 180°
(∵ ∠ADC = ∠ABC opposite angle of a || gm).. .(ii)
From Eqs. (i) and (ii), we get
∠AED + ∠ABC = ∠ADE + ∠ABC
⇒ ∠AED = ∠ADE
∴ In ∆AED We have
∠AED = ∠ADE
So, AD = AE
(∵ Sides opposite to equal angles of a triangle are equal)

 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.
Solution:
(i) Let BD and AC be two chords of a circle bisect at P.

In ∆APB and ∆CPD, we get
PA = PC ( ∵ P is the mid-point of AC)
∠APB = ∠CPD (Vertically opposite angles)
and PB = PD (∵ P is the mid-point of BD)
∴ By SAS criterion
∆CPD ≅ ∆APB
∴ CD= AB (By CPCT) …(i)

∴ BD divides the circle into two equal parts. So, BD is a diameter.
Similarly, AC is a diameter.
(ii) Now, BD and AC bisect each other.
So, ABCD is a parallelogram.
Also, AC = BD
∴ ABCD is a rectangle.

Ex 10.6 Class 9 Maths Question 8.
Bisectors of angles A, B and C of a ∆ABC intersect its circumcircle at D, E and F, respectively. Prove that the angles of the ∆DEF are 90° – 12 A, 90° – 12 B and 90° – 12 C.
Solution:
∵ ∠EDF = ∠EDA + ∠ADF
∵ ∠EDA and ∠EBA are the angles in the same segment of the circle.
∴ ∠EDA = ∠EBA
and similarly ∠ADF and ∠FCA are the angles in the same segment and hence

Ex 10.6 Class 9 Maths Question 9.
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
Solution:
Let O’ and O be the centres of two congruent circles.

Since, AB is a common chord of these circles.
∴ ∠BPA = ∠BQA
(∵ Angle subtended by equal chords are equal)
⇒ BP = BQ

Ex 10.6 Class 9 Maths Question 10.
In any ∆ ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the ∆ABC.
Solution:
(i) Let bisector of ∠A meet the circumcircle of ∆ABC at M.
Join BM and CM.

∴ ∠MBC = ∠MAC (Angles in same segment)
and ∠BCM = ∠BAM (Angles in same segment)
But ∠BAM = ∠CAM (∵ AM is bisector of ∠A)…. .(i)
∴ ∠MBC = ∠BCM
So, MB = MC (Sides opposite to equal angles are equal)
So, M must lie on the perpendicular bisector of BC
(ii) Let M be a point on the perpendicular bisector of BC which lies on circumcircle of ∆ ABC.
Join AM.

Since, M lies on perpendicular bisector of BC.
∴ BM = CM
∠MBC = ∠MCB
But ∠MBC = ∠MAC (Angles in same segment)
and ∠MCB = ∠BAM (Angles in same segment)
So, from Eq. (i),
∠BAM = ∠CAM
AM is the bisector of A.
Hence, bisector of ∠A and perpendicular bisector of BC at M which lies on circumcircle of ∆ABC.

Exercise 9.1

1. Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.

Answer

(i) Trapezium ABCD and ΔPDC lie on the same DC and between the same parallel lines AB and DC.
(ii) Parallelogram PQRS and  trapezium SMNR lie on the same base SR but not between the same parallel lines.

Exercise 9.2

1. In Fig. 9.15, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Answer

Given,
AB = CD = 16 cm (Opposite sides of a parallelogram)
CF = 10 cm and AE = 8 cm
Now,
Area of parallelogram = Base × Altitude
= CD × AE = AD × CF
⇒ 16 × 8 = AD × 10
⇒ AD = 128/10 cm
⇒ AD = 12.8 cm

2. If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that
ar (EFGH) = 1/2 ar(ABCD).

Answer 

Given,
E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD.
To Prove,
ar (EFGH) = 1/2 ar(ABCD)
Construction,
H and F are joined.
Proof,
AD || BC and AD = BC (Opposite sides of a parallelogram)
⇒ 1/2 AD = 1/2 BC
Also,
AH || BF and and DH || CF
⇒ AH = BF and DH = CF (H and F are mid points)
Thus, ABFH and HFCD are parallelograms.
Now,
ΔEFH and ||gm ABFH lie on the same base FH and between the same parallel lines AB and HF.
∴ area of EFH = 1/2 area of ABFH — (i)
also, area of GHF = 1/2 area of HFCD — (ii)
Adding (i) and (ii),
area of ΔEFH + area of ΔGHF = 1/2 area of ABFH + 1/2 area of HFCD
⇒ area of EFGH = area of ABFH
⇒ ar (EFGH) = 1/2 ar(ABCD)

3. P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar(APB) = ar(BQC).

Answer

ΔAPB and ||gm ABCD are on the same base AB and between same parallel AB and DC.
Therefore,
ar(ΔAPB) = 1/2 ar(||gm ABCD) — (i)
Similarly,
ar(ΔBQC) = 1/2 ar(||gm ABCD) — (ii)
From (i) and (ii),
we have ar(ΔAPB) = ar(ΔBQC)

4. In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar(APB) + ar(PCD) = 1/2 ar(ABCD)
(ii) ar(APD) + ar(PBC) = ar(APB) + ar(PCD)
[Hint : Through P, draw a line parallel to AB.]

Answer 

(i) A line GH is drawn parallel to AB passing through P.
In a parallelogram,
AB || GH (by construction) — (i)
Thus,
AD || BC ⇒ AG || BH — (ii)
From equations (i) and (ii),
ABHG is a parallelogram.
Now,
In ΔAPB and parallelogram ABHG are lying on the same base AB and between the same parallel lines AB and GH.
∴ ar(ΔAPB) = 1/2 ar(ABHG) — (iii)
also,
In ΔPCD and parallelogram CDGH are lying on the same base CD and between the same parallel lines CD and GH.
∴ ar(ΔPCD) = 1/2 ar(CDGH) — (iv)
Adding equations (iii) and (iv),
ar(ΔAPB) + ar(ΔPCD) = 1/2 {ar(ABHG) + ar(CDGH)}
⇒ ar(APB) + ar(PCD) = 1/2 ar(ABCD)

(ii) A line EF is drawn parallel to AD passing through P.
In a parallelogram,
AD || EF (by construction) — (i)
Thus,
AB || CD ⇒ AE || DF — (ii)
From equations (i) and (ii),
AEDF is a parallelogram.
Now,
In ΔAPD and parallelogram AEFD are lying on the same base AD and between the same parallel lines AD and EF.
∴ ar(ΔAPD) = 1/2 ar(AEFD) — (iii)
also,
In ΔPBC and parallelogram BCFE are lying on the same base BC and between the same parallel lines BC and EF.
∴ ar(ΔPBC) = 1/2 ar(BCFE) — (iv)
Adding equations (iii) and (iv),
ar(ΔAPD) + ar(ΔPBC) = 1/2 {ar(AEFD) + ar(BCFE)}
⇒ ar(APD) + ar(PBC) = ar(APB) + ar(PCD)

5. In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = 1/2 ar (PQRS) 

 

∴ ar(PQRS) = ar(ABRS) — (i)
(ii) In ΔAXS and parallelogram ABRS are lying on the same base AS and between the same parallel lines AS and BR.
∴ ar(ΔAXS) = 1/2 ar(ABRS) — (ii)
From (i) and (ii),
ar(ΔAXS) = 1/2 ar(PQRS)

Page No: 106

6. A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

Answer

The field is divided into three parts. The three parts are in the shape of triangle. ΔPSA, ΔPAQ and ΔQAR.
Area of ΔPSA + ΔPAQ + ΔQAR = Area of PQRS — (i) 
Area of ΔPAQ = 1/2 area of PQRS — (ii)
Triangle and parallelogram on the same base and between the same parallel lines.
From (i) and (ii),
Area of ΔPSA + Area of ΔQAR = 1/2 area of PQRS — (iii)
Clearly from (ii) and (iii),
Farmer must sow wheat or pulses in ΔPAQ or either in both ΔPSA and ΔQAR.

Exercise 8.1

1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Answer

Let x be the common ratio between the angles.
Sum of the interior angles of the quadrilateral = 360°
Now,
3x + 5x + 9x + 13x = 360°
⇒ 30x = 360°
⇒ x = 12°
Angles of the quadrilateral are:
3x = 3×12° = 36°
5x = 5×12° = 60°
9x = 9×12° = 108°
13x = 13×12° = 156°

2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer

Given,
AC = BD
To show,
To show ABCD is a rectangle we have to prove that one of its interior angle is right angled.
Proof,
In ΔABC and ΔBAD,
BC = BA (Common)
AC = AD (Opposite sides of a parallelogram are equal)
AC = BD (Given)
Therefore, ΔABC ≅ ΔBAD by SSS congruence condition.
∠A = ∠B (by CPCT)
also,
∠A + ∠B = 180° (Sum of the angles on the same side of the transversal)
⇒ 2∠A = 180°
⇒ ∠A = 90° = ∠B
Thus ABCD is a rectangle.

3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

 Answer

Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.
Given,
OA = OC, OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°
To show,
ABCD is parallelogram and AB = BC = CD = AD
Proof,
In ΔAOB and ΔCOB,
OA = OC (Given)
∠AOB = ∠COB (Opposite sides of a parallelogram are equal)
OB = OB (Common)
Therefore, ΔAOB ≅ ΔCOB by SAS congruence condition.
Thus, AB = BC (by CPCT)
Similarly we can prove,
AB = BC = CD = AD
Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.
Thus, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.

4. Show that the diagonals of a square are equal and bisect each other at right angles.

Answer

Let ABCD be a square and its diagonals AC and BD intersect each other at O.
To show,
 AC = BD, AO = OC and ∠AOB = 90°
Proof,
In ΔABC and ΔBAD,
BC = BA (Common)
∠ABC = ∠BAD = 90°
AC = AD (Given)
Therefore, ΔABC ≅ ΔBAD by SAS congruence condition.
Thus, AC = BD by CPCT. Therefore, diagonals are equal.
Now,
In ΔAOB and ΔCOD,
∠BAO = ∠DCO (Alternate interior angles)
∠AOB = ∠COD (Vertically opposite)
AB = CD (Given)
Therefore, ΔAOB ≅ ΔCOD by AAS congruence condition.
Thus, AO = CO by CPCT. (Diagonal bisect each other.)
Now,
In ΔAOB and ΔCOB,
OB = OB (Given)
AO = CO (diagonals are bisected)
AB = CB (Sides of the square)
Therefore, ΔAOB ≅ ΔCOB by SSS congruence condition.
also, ∠AOB = ∠COB
∠AOB + ∠COB = 180° (Linear pair)
Thus, ∠AOB = ∠COB = 90° (Diagonals bisect each other at right angles)

5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Answer

Given,
Let ABCD be a quadrilateral in which diagonals AC and BD bisect each other at right angle at O.
To prove,
Quadrilateral ABCD is a square.
Proof,
In ΔAOB and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOB = ∠COD (Vertically opposite)
OB = OD (Diagonals bisect each other)
Therefore, ΔAOB ≅ ΔCOD by SAS congruence condition.
Thus, AB = CD by CPCT. — (i)
also,
∠OAB = ∠OCD (Alternate interior angles)
⇒ AB || CD
Now,
In ΔAOD and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOD = ∠COD (Vertically opposite)
OD = OD (Common)
Therefore, ΔAOD ≅ ΔCOD by SAS congruence condition.
Thus, AD = CD by CPCT. — (ii)
also,
AD = BC and AD = CD
⇒ AD = BC = CD = AB — (ii)
also,  ∠ADC = ∠BCD  by CPCT.
and ∠ADC + ∠BCD = 180° (co-interior angles)
⇒ 2∠ADC = 180°
⇒ ∠ADC = 90° — (iii)
One of the interior ang is right angle.
Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square.

6. Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.

Answer

(i)
In ΔADC and ΔCBA,
AD = CB (Opposite sides of a ||gm)
DC = BA (Opposite sides of a ||gm)
AC = CA (Common)
Therefore, ΔADC ≅ ΔCBA by SSS congruence condition.
Thus,
∠ACD = ∠CAB by CPCT
and ∠CAB = ∠CAD (Given)
⇒ ∠ACD = ∠BCA
Thus, AC bisects ∠C also.

(ii) ∠ACD = ∠CAD (Proved)
⇒ AD = CD (Opposite sides of equal angles of a triangle are equal)
Also, AB = BC = CD = DA (Opposite sides of a ||gm)
Thus, ABCD is a rhombus.

7. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Answer

Let ABCD is a rhombus and AC and BD are its diagonals.
Proof,
AD = CD (Sides of a rhombus)
∠DAC = ∠DCA (Angles opposite of equal sides of a triangle are equal.)
also, AB || CD
⇒ ∠DAC = ∠BCA (Alternate interior angles)
⇒ ∠DCA = ∠BCA
Therefore, AC bisects ∠C.
Similarly, we can prove that diagonal AC bisects ∠A.

Also, by preceding above method we can prove that diagonal BD bisects ∠B as well as ∠D.

8. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
(i) ABCD is a square
(ii) diagonal BD bisects ∠B as well as ∠D.

Answer

(i)∠DAC = ∠DCA (AC bisects ∠A as well as ∠C)
⇒ AD = CD (Sides opposite to equal angles of a triangle are equal)
also, CD = AB (Opposite sides of a rectangle)
Therefore, AB = BC = CD = AD
Thus, ABCD is a square.

(ii) In ΔBCD,
BC = CD
⇒ ∠CDB = ∠CBD (Angles opposite to equal sides are equal)
also, ∠CDB = ∠ABD (Alternate interior angles)
⇒ ∠CBD = ∠ABD
Thus, BD bisects ∠B
Now,
∠CBD = ∠ADB
⇒ ∠CDB = ∠ADB
Thus, BD bisects ∠D

Page No: 147

9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that:
(i) ΔAPD ≅ ΔCQB
(ii) AP = CQ
(iii) ΔAQB ≅ ΔCPD
(iv) AQ = CP
(v) APCQ is a parallelogram



Answer

(i) In ΔAPD and ΔCQB,
DP = BQ (Given)
∠ADP = ∠CBQ (Alternate interior angles)
AD = BC (Opposite sides of a ||gm)
Thus, ΔAPD ≅ ΔCQB by SAS congruence condition.

(ii) AP = CQ by CPCT as ΔAPD ≅ ΔCQB.

(iii) In ΔAQB and ΔCPD,
BQ = DP (Given)
∠ABQ = ∠CDP (Alternate interior angles)
AB = BCCD (Opposite sides of a ||gm)
Thus, ΔAQB ≅ ΔCPD by SAS congruence condition.

(iv) AQ = CP by CPCT as ΔAQB ≅ ΔCPD.

(v) From (ii)  and (iv), it is clear that APCQ has equal opposite sides also it has equal opposite angles. Thus, APCQ is a ||gm.

10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that
(i) ΔAPB ≅ ΔCQD
(ii) AP = CQ

Answer

(i) In ΔAPB and ΔCQD,
∠ABP = ∠CDQ (Alternate interior angles)
∠APB = ∠CQD (equal to right angles as AP and CQ are perpendiculars)
AB = CD (ABCD is a parallelogram)
Thus, ΔAPB ≅ ΔCQD by AAS congruence condition.

(ii) AP = CQ by CPCT as ΔAPB ≅ ΔCQD.

11. In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22).
Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iii) AD || CF and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ΔABC ≅ ΔDEF.

Answer

12. ABCD is a trapezium in which AB || CD and

AD = BC (see Fig. 8.23). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ΔABC ≅ ΔBAD
(iv) diagonal AC = diagonal BD

Answer

Construction: Draw a line through C parallel to DA intersecting AB produced at E.
(i) CE = AD (Opposite sides of a parallelogram)
AD = BC (Given)
Therefor, BC = CE
⇒ ∠CBE = ∠CEB
also,
∠A + ∠CBE = 180° (Angles on the same side of transversal and ∠CBE = ∠CEB)
∠B + ∠CBE = 180° (Linear pair)
⇒ ∠A = ∠B

(ii) ∠A + ∠D = ∠B + ∠C = 180° (Angles on the same side of transversal)
⇒ ∠A + ∠D = ∠A + ∠C (∠A = ∠B)
⇒ ∠D = ∠C

(iii) In ΔABC and ΔBAD,
AB = AB (Common)
∠DBA = ∠CBA
AD = BC (Given)
Thus, ΔABC ≅ ΔBAD by SAS congruence condition.

(iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.

Exercise 7.1

1. In quadrilateral ACBD, AC = AD and AB bisects ∠A (see Fig. 7.16). Show that ΔABC ≅ ΔABD. What can you say about BC and BD?

Answer

Given,
AC = AD and AB bisects ∠A
To prove,
ΔABC ≅ ΔABD
Proof,
In ΔABC and ΔABD,
AB = AB (Common)
AC = AD (Given)
∠CAB = ∠DAB (AB is bisector)
Therefore, ΔABC ≅ ΔABD by SAS congruence condition.
BC and BD are of equal length.

Page No: 119

2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Fig. 7.17). Prove that
(i) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.

Answer

Given,
AD = BC and ∠DAB = ∠CBA

(i) In ΔABD and ΔBAC,
AB = BA (Common)
∠DAB = ∠CBA (Given)
AD = BC (Given)
Therefore, ΔABD ≅ ΔBAC by SAS congruence condition.
(ii) Since, ΔABD ≅ ΔBAC
Therefore BD = AC by CPCT
(iii) Since, ΔABD ≅ ΔBAC
Therefore ∠ABD = ∠BAC by CPCT

3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.

Answer

Given,
AD and BC are equal perpendiculars to AB.
To prove,
CD bisects AB
Proof,
In ΔAOD and ΔBOC,
∠A = ∠B (Perpendicular)
∠AOD = ∠BOC (Vertically opposite angles)
AD = BC (Given)
Therefore, ΔAOD ≅ ΔBOC by AAS congruence condition.
Now,
AO = OB (CPCT). CD bisects AB.

4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ΔABC ≅ ΔCDA.

Answer

Given,
l || m and p || q
To prove,
ΔABC ≅ ΔCDA
Proof,
In ΔABC and ΔCDA,
∠BCA = ∠DAC (Alternate interior angles)
AC = CA (Common)
∠BAC = ∠DCA (Alternate interior angles)
Therefore, ΔABC ≅ ΔCDA by ASA congruence condition.

5. Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see Fig. 7.20). Show that:
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.

 Answer

Given,
l is the bisector of an angle ∠A.
BP and BQ are perpendiculars.

(i) In ΔAPB and ΔAQB,
∠P = ∠Q (Right angles)
∠BAP = ∠BAQ (l is bisector)
AB = AB (Common)
Therefore, ΔAPB ≅ ΔAQB by AAS congruence condition.
(ii) BP = BQ by CPCT. Therefore, B is equidistant from the arms of ∠A.

Page No: 120

6. In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Answer

Given,
AC = AE, AB = AD and ∠BAD = ∠EAC
To show,
BC = DE
Proof,
∠BAD = ∠EAC (Adding ∠DAC both sides)
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠EAD
In ΔABC and ΔADE,
AC = AE (Given)
∠BAC = ∠EAD
AB = AD (Given)
Therefore, ΔABC ≅ ΔADE by SAS congruence condition.
BC = DE by CPCT.

7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see Fig. 7.22). Show that
(i) ΔDAP ≅ ΔEBP
(ii) AD = BE

Answer

Given,
P is mid-point of AB.
∠BAD = ∠ABE and ∠EPA = ∠DPB

(i) ∠EPA = ∠DPB (Adding ∠DPE both sides)
∠EPA + ∠DPE = ∠DPB + ∠DPE
⇒ ∠DPA = ∠EPB
In ΔDAP ≅ ΔEBP,
∠DPA = ∠EPB
AP = BP (P is mid-point of AB)
∠BAD = ∠ABE (Given)
Therefore, ΔDAP ≅ ΔEBP by ASA congruence condition.
(ii) AD = BE by CPCT.

8. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see Fig. 7.23). Show that:
(i) ΔAMC ≅ ΔBMD
(ii) ∠DBC is a right angle.
(iii) ΔDBC ≅ ΔACB
(iv) CM = 1/2 AB

Answer

Given,
∠C = 90°, M is the mid-point of AB and DM = CM

(i) In ΔAMC and ΔBMD,
AM = BM (M is the mid-point)
∠CMA = ∠DMB (Vertically opposite angles)
CM = DM (Given)
Therefore, ΔAMC ≅ ΔBMD by SAS congruence condition.

(ii) ∠ACM = ∠BDM (by CPCT)
Therefore, AC || BD as alternate interior angles are equal.
Now,
∠ACB + ∠DBC = 180° (co-interiors angles)
⇒ 90° + ∠B = 180°
⇒ ∠DBC = 90°

(iii) In ΔDBC and ΔACB,
BC = CB (Common)
∠ACB = ∠DBC (Right angles)
DB = AC (byy CPCT, already proved)
Therefore, ΔDBC ≅ ΔACB by SAS congruence condition.

(iv)  DC = AB (ΔDBC ≅ ΔACB)
⇒ DM = CM = AM = BM (M is mid-point)
⇒ DM + CM = AM + BM
⇒ CM + CM = AB
⇒ CM = 1/2AB

Exercise 7.2

1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that :
(i) OB = OC                     (ii) AO bisects ∠A

Answer

Given,
AB = AC, the bisectors of ∠B and ∠C intersect each other at O

(i) Since ABC is an isosceles with AB = AC,
∴ ∠B = ∠C
⇒ 1/2∠B = 1/2∠C
⇒ ∠OBC = ∠OCB (Angle bisectors.)
⇒ OB = OC (Side opposite to the equal angles are equal.)

(ii) In ΔAOB and ΔAOC,
AB = AC (Given)
AO = AO (Common)
OB = OC (Proved above)
Therefore, ΔAOB ≅ ΔAOC by SSS congruence condition.
∠BAO = ∠CAO (by CPCT)
Thus, AO bisects ∠A.

2. In ΔABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that ΔABC is an isosceles triangle in which AB = AC.

Answer

Given,
AD is the perpendicular bisector of BC
To show,
AB = AC
Proof,
In ΔADB and ΔADC,
AD = AD (Common)
∠ADB = ∠ADC
BD = CD (AD is the perpendicular bisector)
Therefore, ΔADB ≅ ΔADC by SAS congruence condition.
AB = AC (by CPCT)

Page No: 124

3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.

Answer

Given,
BE and CF are altitudes.
AC = AB
To show,
BE = CF
Proof,
In ΔAEB and ΔAFC,
∠A = ∠A (Common)
∠AEB = ∠AFC (Right angles)
AB = AC (Given)
Therefore, ΔAEB ≅ ΔAFC by AAS congruence condition.
Thus, BE = CF by CPCT.

4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that
(i) ΔABE ≅ ΔACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.

Answer

Given,
BE = CF

(i) In ΔABE and ΔACF,
∠A = ∠A (Common)
∠AEB = ∠AFC (Right angles)
BE = CF (Given)
Therefore, ΔABE ≅ ΔACF by AAS congruence condition.

(ii) Thus, AB = AC by CPCT and therefore ABC is an isosceles triangle.

Answer

Given,
ABC and DBC are two isosceles triangles.
To show,
∠ABD = ∠ACD
Proof,
In ΔABD and ΔACD,
AD = AD (Common)
AB = AC (ABC is an isosceles triangle.)
BD = CD (BCD is an isosceles triangle.)
Therefore, ΔABD ≅ ΔACD by SSS congruence condition. Thus, ∠ABD = ∠ACD by CPCT.

6. ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠BCD is a right angle.

Answer

Given,
AB = AC and AD = AB
To show,
∠BCD is a right angle.
Proof,
In ΔABC,
AB = AC (Given)
⇒ ∠ACB = ∠ABC (Angles opposite to the equal sides are equal.)
In ΔACD,
AD = AB
⇒ ∠ADC = ∠ACD (Angles opposite to the equal sides are equal.)
Now,
In ΔABC,
∠CAB + ∠ACB + ∠ABC = 180°
⇒ ∠CAB + 2∠ACB = 180°
⇒ ∠CAB = 180° – 2∠ACB — (i)
Similarly in ΔADC,
∠CAD = 180° – 2∠ACD — (ii)
also,
∠CAB + ∠CAD = 180° (BD is a straight line.)
Adding (i) and (ii)
∠CAB + ∠CAD = 180° – 2∠ACB + 180° – 2∠ACD
⇒ 180° = 360° – 2∠ACB – 2∠ACD
⇒ 2(∠ACB + ∠ACD) = 180°
⇒ ∠BCD = 90°

7. ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Answer

Given,
∠A = 90° and AB = AC
A/q,
AB = AC
⇒ ∠B = ∠C (Angles opposite to the equal sides are equal.)
Now,
∠A + ∠B + ∠C = 180° (Sum of the interior angles of the triangle.)
⇒ 90° + 2∠B = 180°
⇒ 2∠B = 90°
⇒ ∠B = 45°
Thus, ∠B = ∠C = 45°

8. Show that the angles of an equilateral triangle are 60° each.

Answer

Let ABC be an equilateral triangle.
BC = AC = AB (Length of all sides is same)
⇒ ∠A = ∠B = ∠C (Sides opposite to the equal angles are equal.)
Also,
∠A + ∠B + ∠C = 180°
⇒ 3∠A = 180°
⇒ ∠A = 60°
Therefore, ∠A = ∠B = ∠C = 60°
Thus, the angles of an equilateral triangle are 60° each.

Exercise 7.4

1. Show that in a right angled triangle, the hypotenuse is the longest side.

Answer

ABC is a triangle right angled at B.
Now,
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠C = 90° and ∠B is 90°.
Since, B is the largest angle of the triangle, the side opposite to it must be the largest.
So, BC is the hypotenuse which is the largest side of the right angled triangle ABC.

2. In Fig. 7.48, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Answer

Given,
∠PBC < ∠QCB
Now,
∠ABC + ∠PBC = 180°
⇒ ∠ABC = 180° – ∠PBC
also,
∠ACB + ∠QCB = 180°
⇒ ∠ACB = 180° – ∠QCB
Since,
∠PBC < ∠QCB therefore, ∠ABC > ∠ACB
Thus, AC > AB as sides opposite to the larger angle is larger.

3. In Fig. 7.49, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Answer

Given,
∠B < ∠A and ∠C < ∠D
Now,
AO <  BO — (i) (Side opposite to the smaller angle is smaller)
OD < OC —(ii) (Side opposite to the smaller angle is smaller)
Adding (i) and (ii)
AO + OD < BO + OC
⇒ AD < BC

4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50).
Show that ∠A > ∠C and ∠B > ∠D.

Answer


In ΔABD,
AB < AD < BD
∴ ∠ADB < ∠ABD — (i) (Angle opposite to longer side is larger.)
Now,
In ΔBCD,
BC < DC < BD
∴ ∠BDC < ∠CBD — (ii)
Adding (i) and (ii) we get,
∠ADB + ∠BDC < ∠ABD + ∠CBD
⇒ ∠ADC < ∠ABC
⇒ ∠B > ∠D
Similarly,
In ΔABC,
∠ACB < ∠BAC — (iii) (Angle opposite to longer side is larger.)
Now,
In ΔADC,
∠DCA < ∠DAC — (iv)
Adding (iii) and (iv) we get,
∠ACB + ∠DCA < ∠BAC + ∠DAC
⇒ ∠BCD < ∠BAD
⇒ ∠A > ∠C

5. In Fig 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Answer

Given,
PR > PQ and PS bisects ∠QPR
To prove,
∠PSR > ∠PSQ
Proof,
∠PQR > ∠PRQ — (i) (PR > PQ as angle opposite to larger side is larger.)
∠QPS = ∠RPS — (ii) (PS bisects ∠QPR)
∠PSR = ∠PQR + ∠QPS — (iii) (exterior angle of a triangle equals to the sum of opposite interior angles)
∠PSQ = ∠PRQ + ∠RPS — (iv) (exterior angle of a triangle equals to the sum of opposite interior angles)
Adding (i) and (ii)
∠PQR + ∠QPS > ∠PRQ + ∠RPS
⇒ ∠PSR > ∠PSQ [from (i), (ii), (iii) and (iv)]

Page No: 133

6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Answer

Let l is a line segment and B is a point lying o it. We drew a line AB perpendicular to l. Let C be any other point on l.
To prove,
AB < AC
Proof,
In ΔABC,
∠B = 90°
Now,
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠C = 90°
∴ ∠C must be acute angle. or ∠C < ∠B
⇒ AB < AC (Side opposite to the larger angle is larger.)

https://youtu.be/8zyFiuaWVec

Exercise 6.1

1. In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Answer
Given,
∠AOC + ∠BOE = 70° and ∠BOD = 40°
A/q,
∠AOC + ∠BOE +∠COE = 180° (Forms a straight line)
⇒ 70° +∠COE = 180°
⇒ ∠COE = 110°
also,
∠COE +∠BOD + ∠BOE = 180° (Forms a straight line)
⇒ 110° +40° + ∠BOE = 180°
⇒ 150° + ∠BOE = 180°
⇒ ∠BOE = 30°

Page No: 97

2. In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.

Answer

Given,
∠POY = 90° and a : b = 2 : 3
A/q,
∠POY + a + b = 180°
⇒ 90° + a + b = 180°
⇒ a + b = 90°
Let a be 2x then will be 3x
2x + 3x = 90°
⇒ 5x = 90°
⇒ x = 18°
∴ a = 2×18° = 36°
and b = 3×18° = 54°
also,
b + c = 180° (Linear Pair)
⇒ 54° + c = 180°
⇒ c = 126°

3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Answer

Given,
∠PQR = ∠PRQ
To prove,
∠PQS = ∠PRT
A/q,
∠PQR +∠PQS = 180° (Linear Pair)
⇒ ∠PQS = 180° – ∠PQR — (i)
also,
∠PRQ +∠PRT = 180° (Linear Pair)
⇒ ∠PRT = 180° – ∠PRQ
⇒ ∠PRQ = 180° – ∠PQR — (ii) (∠PQR = ∠PRQ)
From (i) and (ii)
∠PQS = ∠PRT = 180° – ∠PQR
Therefore,  ∠PQS = ∠PRT

4. In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.

Answer

Given,
x + y = w + z
To Prove,
AOB is a line or x + y = 180° (linear pair.)
A/q,
x + y + w + z = 360° (Angles around a point.)
⇒ (x + y) +  (w + z) = 360°
⇒ (x + y) +  (x + y) = 360° (Given x + y = w + z)
⇒ 2(x + y) = 360°
⇒ (x + y) = 180°
Hence, x + y makes a linear pair. Therefore, AOB is a staright line.

5. In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 1/2(∠QOS – ∠POS).

Answer

Given,
OR is perpendicular to line PQ
To prove,
∠ROS = 1/2(∠QOS – ∠POS)
A/q,
∠POR = ∠ROQ = 90° (Perpendicular)
∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS — (i)
∠POS = ∠POR – ∠ROS = 90° – ∠ROS — (ii)
Subtracting (ii) from (i)
∠QOS – ∠POS = 90° + ∠ROS – (90° – ∠ROS)
⇒ ∠QOS – ∠POS = 90° + ∠ROS – 90° + ∠ROS
⇒ ∠QOS – ∠POS = 2∠ROS
⇒ ∠ROS = 1/2(∠QOS – ∠POS)
Hence, Proved.

6. It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.

Answer

Given,
∠XYZ = 64°
YQ bisects ∠ZYP

∠XYZ +∠ZYP = 180° (Linear Pair)
⇒ 64° +∠ZYP = 180°
⇒ ∠ZYP = 116°
also, ∠ZYP = ∠ZYQ + ∠QYP
∠ZYQ = ∠QYP (YQ bisects ∠ZYP)
⇒ ∠ZYP = 2∠ZYQ
⇒ 2∠ZYQ = 116°
⇒ ∠ZYQ = 58° = ∠QYP
Now,
∠XYQ = ∠XYZ + ∠ZYQ
⇒ ∠XYQ = 64° + 58°
⇒ ∠XYQ = 122°
also,
reflex ∠QYP = 180° + ∠XYQ
∠QYP = 180° + 122°
⇒ ∠QYP = 302°

https://youtu.be/jum4L0ChTLw

Exercise 6.2

1. In Fig. 6.28, find the values of x and y and then show that AB || CD.

 

Answer

x + 50° = 180° (Linear pair)
⇒ x = 130°
also,
y = 130° (Vertically opposite)
Now,
x = y = 130° (Alternate interior angles)
Alternate interior angles are equal.
Therefore, AB || CD.

Page No: 104

2. In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Answer

Given,
AB || CD and CD || EF
y : z = 3 : 7
Now,
x + y = 180° (Angles on the same side of transversal.)
also,
∠O = z (Corresponding angles)
and, y + ∠O = 180° (Linear pair)
⇒ y + z = 180°
A/q,
y = 3w and z = 7w
3w + 7w = 180°
⇒ 10 w = 180°
⇒ w = 18°
∴ y = 3×18° = 54°
and, z = 7×18° = 126°
Now,
x + y = 180°
⇒ x + 54° = 180°
⇒ x = 126°

3. In Fig. 6.30, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Answer

Given,
AB || CD
EF ⊥ CD
∠GED = 126°
A/q,
∠FED = 90° (EF ⊥ CD)
Now,
∠AGE = ∠GED (Since, AB || CD and GE is transversal. Alternate interior angles.)
∴ ∠AGE = 126°
Also, ∠GEF = ∠GED – ∠FED
⇒ ∠GEF = 126° – 90°
⇒ ∠GEF = 36°
Now,
∠FGE +∠AGE = 180° (Linear pair)
⇒ ∠FGE = 180° – 126°
⇒ ∠FGE = 54°

4. In Fig. 6.31, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
[Hint : Draw a line parallel to ST through point R.]

 

∠PQR + ∠QRX = 180° (Angles on the same side of transversal.)
⇒ 110° + ∠QRX = 180°
⇒ ∠QRX = 70°
Also,
∠RST + ∠SRY = 180° (Angles on the same side of transversal.)
⇒ 130° + ∠SRY = 180°
⇒ ∠SRY = 50°
Now,
∠QRX +∠SRY + ∠QRS = 180°
⇒ 70° + 50° + ∠QRS = 180°
⇒ ∠QRS = 60°

5. In Fig. 6.32, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

 Answer

Given,
AB || CD, ∠APQ = 50° and ∠PRD = 127°
A/q,
x = 50° (Alternate interior angles.)
∠PRD + ∠RPB = 180° (Angles on the same side of transversal.)
⇒ 127° + ∠RPB = 180°
⇒ ∠RPB = 53°
Now,
y + 50° + ∠RPB = 180° (AB is a straight line.)
⇒ y + 50° + 53° = 180°
⇒ y + 103° = 180°
⇒ y = 77°

6. In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.

 

 As PQ || RS
So, BE || CF

From (i) and (ii),

⇒ ∠ABC = ∠DCB

https://youtu.be/rKakJ64X8Nc

Exercise 6.3

1. In Fig. 6.39, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Answer

Given,
∠SPR = 135° and ∠PQT = 110°
A/q,
∠SPR +∠QPR = 180° (SQ is a straight line.)
⇒ 135° +∠QPR = 180°
⇒ ∠QPR = 45°
also,

∠PQT +∠PQR = 180° (TR is a straight line.)
⇒ 110° +∠PQR = 180°
⇒ ∠PQR = 70°
Now,
∠PQR +∠QPR + ∠PRQ = 180° (Sum of the interior angles of the triangle.)
⇒ 70° + 45° + ∠PRQ = 180°
⇒ 115° + ∠PRQ = 180°
⇒ ∠PRQ = 65°

2. In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of Δ XYZ, find ∠OZY and ∠YOZ.

Answer

Given,
∠X = 62°, ∠XYZ = 54°
YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively.
A/q,
∠X +∠XYZ + ∠XZY = 180° (Sum of the interior angles of the triangle.)
⇒ 62° + 54° + ∠XZY = 180°
⇒ 116° + ∠XZY = 180°
⇒ ∠XZY = 64°
Now,
∠OZY = 1/2∠XZY (ZO is the bisector.)
⇒ ∠OZY = 32°
also,
∠OYZ = 1/2∠XYZ (YO is the bisector.)
⇒ ∠OYZ = 27°
Now,
∠OZY +∠OYZ + ∠O = 180° (Sum of the interior angles of the triangle.)
⇒ 32° + 27° + ∠O = 180°
⇒ 59° + ∠O = 180°
⇒ ∠O = 121°

3. In Fig. 6.41, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Answer

Given,
AB || DE, ∠BAC = 35° and ∠ CDE = 53°
A/q,
∠BAC = ∠CED (Alternate interior angles.)
∴ ∠CED = 35°
Now,
∠DCE +∠CED + ∠CDE = 180° (Sum of the interior angles of the triangle.)
⇒ ∠DCE + 35° + 53° = 180°
⇒ ∠DCE + 88° = 180°
⇒ ∠DCE = 92°

4. In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Answer

Given,
∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°
A/q,
∠PRT +∠RPT + ∠PTR = 180° (Sum of the interior angles of the triangle.)
⇒ 40° + 95° + ∠PTR = 180°
⇒ 40° + 95° + ∠PTR = 180°
⇒ 135° + ∠PTR = 180°
⇒ ∠PTR = 45°
∠PTR = ∠STQ = 45° (Vertically opposite angles.)
Now,
∠TSQ +∠PTR + ∠SQT = 180° (Sum of the interior angles of the triangle.)
⇒ 75° + 45° + ∠SQT = 180°
⇒ 120° + ∠SQT = 180°
⇒ ∠SQT = 60°

Page No: 108

5. In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Answer

Given,
PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°
A/q,
x +∠SQR = ∠QRT (Alternate angles  as QR is transveersal.)
⇒ x + 28° = 65°
⇒ x = 37°
also,
∠QSR = x
⇒ ∠QSR = 37°
also,
∠QRS +∠QRT = 180° (Linea pair)
⇒ ∠QRS + 65° = 180°
⇒ ∠QRS = 115°
Now,
∠P + ∠Q+ ∠R +∠S = 360° (Sum of the angles in a quadrilateral.)
⇒ 90° + 65° + 115° + ∠S = 360°
⇒ 270° + y + ∠QSR = 360°
⇒ 270° + y + 37° = 360°
⇒ 307° + y = 360°
⇒ y = 53°

6. In Fig. 6.44, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = 1/2∠QPR.

Answer

Given,
Bisectors of ∠PQR and ∠PRS meet at point T.
To prove,
∠QTR = 1/2∠QPR.
Proof,
∠TRS = ∠TQR +∠QTR (Exterior angle of a triangle equals to the sum of the two interior angles.)
⇒ ∠QTR = ∠TRS – ∠TQR — (i)
also,
∠SRP = ∠QPR + ∠PQR
⇒ 2∠TRS = ∠QPR + 2∠TQR
⇒ ∠QPR =  2∠TRS – 2∠TQR
⇒ 1/2∠QPR =  ∠TRS – ∠TQR — (ii)
Equating (i) and (ii)
∠QTR – ∠TQR = 1/2∠QPR
Hence proved.

https://youtu.be/UBrZiKewNmQ

Exercise 5.1

1. Which of the following statements are true and which are false? Give reasons for your
answers.
(i) Only one line can pass through a single point.
(ii) There are an infinite number of lines which pass through two distinct points.
(iii) A terminated line can be produced indefinitely on both the sides.
(iv) If two circles are equal, then their radii are equal.
(v) In Fig. 5.9, if AB = PQ and PQ = XY, then AB = XY.

Answer

(i) False. There can be infinite line drawn passing through a single point.

(ii) False. Only one line can be drawn which passes through two distinct points.

(iii) True. A terminated line can be produced indefinitely on both the sides.
In geometry, a line can be extended in both direction. A line means infinite long length.

(iv) True. If two circles are equal, then their radii are equal.
By superposition, we will find that the centre and circumference of the both circles coincide. Hence, their radius must be equal.

(v) True. By Euclid’s first axiom things which are equal to the same thing, are equal to one another.

2. Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they, and how might you define them?
(i) parallel lines                 (ii) perpendicular lines                (iii) line segment  
(iv) radius of a circle         (v) square

Answer

Yes, other terms need to be defined first which are:
Plane: A plane is flat surface on which geometric figures are drawn.
Point: A point is a dot drawn on a plane surface and is dimensionless.
Line: A line is collection of points which can extends in both direction and has only length not breadth.

(i) Parallel lines: When two or more never intersect each other in a plane and perpendicular distance between them is always constant then they are said to be parallel lines.

(ii) Perpendicular lines: When two lines intersect each other at right angle in a plane then they are said to be perpendicular to each other.

(iii) Line segment: A line segment is a part of a line with two end points and cannot be extended further.

(iv) Radius of circle: The fixed distance between the centre and the circumference of the circle is called the radius of the circle.

(v) Square: A square is a quadrilateral in which all the four sides are equal and each internal angle is a right angle.

3. Consider two ‘postulates’ given below:
(i) Given any two distinct points A and B, there exists a third point C which is in between A and B.
(ii) There exist at least three points that are not on the same line.
Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow from Euclid’s postulates? Explain.

Answer

Undefined terms in the postulates:
→ Many points lie in a plane. But here it is not given about the position of the point C whether it lies on the line segment joining AB or not.
→ Also, there is no information about the plane whether the points are in same plane or not.
Yes, these postulates are consistent when we deal with these two situation:
(i) Point C is lying in between and on the line segment joining A and B.
(ii) Point C not lies on the line segment joining A and B.

No, they don’t follow from Euclid’s postulates. They follow the axioms.

Page No: 86

4. If a point C lies between two points A and B such that AC = BC, then prove that AC = 1/2 AB. Explain by drawing the figure.

Answer

Here, AC = BC
Now, adding AC both sides.
AC + AC = BC + AC
also, BC +AC  = AB (as it coincides with line segment AB)
∴ 2 AC = AB (If equals are added to equals, the wholes are equal.)
⇒ AC = 1/2 AB.

5. In Question 4, point C is called a mid-point of line segment AB. Prove that every line segment has one and only one mid-point.

Answer
 

Let A and B be the line segment and points P and Q be two different mid points of AB.
Now,
∴ P and Q are midpoints of AB.
Therefore AP=PB and also AQ = QB.
also, PB + AP = AB (as it coincides with line segment AB)
Similarly, QB + AQ = AB.
Now,
AP + AP = PB + AP (If equals are added to equals, the wholes are equal.)
⇒ 2 AP = AB — (i)
Similarly, 
2 AQ = AB — (ii)
From (i) and (ii)
2 AP  = 2 AQ (Things which are equal to the same thing are equal to one another.)
⇒ AP = AQ (Things which are double of the same things are equal to one another.)
Thus, P and Q are the same points. This contradicts the fact that P and Q are two different mid points of AB. Thus, it is proved hat every line segment has one and only one mid-point.

6. In Fig. 5.10, if AC = BD, then prove that AB = CD.

 
Answer

Given, AC = BD
From the figure,
AC = AB + BC
BD = BC + CD   
⇒ AB + BC = BC + CD
According to Euclid’s axiom, when equals are subtracted from equals, remainders are also equal.
Subtracting BC both sides,
AB + BC – BC = BC + CD – BC
AB = CD

7. Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that the question is not about the fifth postulate.)

Answer

Axiom 5 : The whole is always greater than the part.
Take an example of a cake. When it is whole it will measures 2 pound but when we took out a part from it and measures its weigh it will came out lower than the previous one. So, the fifth axiom of Euclid is true for all the universal things. That is why it is considered a ‘universal truth’.

https://youtu.be/VuFtU6PVj9U

Exercise 5.2

1. How would you rewrite Euclid’s fifth postulate so that it would be easier to understand?

Answer

The fifth postulates is about parallel lines.
When two or more never intersect each other in a plane and perpendicular distance between them is always constant then they are said to be parallel lines.
Two facts of the postulates:
(i) If P doesn’t lie on l then we can draw a line through P which will be parallel to the line l.
(ii) There will be only one line can be drawn through P which is parallel to the line l.

2. Does Euclid’s fifth postulate imply the existence of parallel lines? Explain.

Answer

Yes, Euclid’s fifth postulate imply the existence of parallel lines.
If the sum of the interior angles will be equal to sum of the two right angles then two lines will not meet each other on either sides and therefore they will be parallel to each other.

m and n will be parallel if
∠1 + ∠3 = 180°
Or  ∠3 + ∠4 = 180°

https://youtu.be/MxNJ_K6ybLs

Exercise 4.1

1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
(Take the cost of a notebook to be x and that of a pen to be y).

Answer

Let the cost of pen be y and the cost of notebook be x.
A/q,
Cost  of a notebook = twice the pen = 2y.
∴2y = x
⇒ x – 2y = 0
This is a linear equation in two variables to represent this statement.

2. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:

(i) 2x + 3y = 9.35               (ii) x – y/5 – 10 = 0                  (iii) -2x + 3y = 6                (iv) x = 3y

(v) 2x = -5y                        (vi) 3x + 2 = 0                        (vii) y – 2 = 0                     (viii) 5 = 2x

Answer

(i) 2x + 3y = 9.35
⇒ 2x + 3y – 9.35 = 0
On comparing this equation with ax + by + c = 0, we get
a = 2x, b = 3 and c = -9.35

(ii) x – y/5 – 10 = 0
On comparing this equation with ax + by + c = 0, we get
a = 1, b = -1/5 and c = -10


(iii) -2x + 3y = 6
⇒ -2x + 3y – 6 = 0
On comparing this equation with ax + by + c = 0, we get
a = -2, b = 3 and c = -6

(iv) x = 3y
⇒ x – 3y = 0
On comparing this equation with ax + by + c = 0, we get
a = 1, b = -3 and c = 0

(v) 2x = -5y
⇒ 2x + 5y = 0
On comparing this equation with ax + by + c = 0, we get
a = 2, b = 5 and c = 0

(vi) 3x + 2 = 0
⇒ 3x + 0y + 2 = 0
On comparing this equation with ax + by + c = 0, we get
a = 3, b = 0 and c = 2

(vii) y – 2 = 0
⇒ 0x + y – 2 = 0
On comparing this equation with ax + by + c = 0, we get
a = 0, b = 1 and c = -2

(viii) 5 = 2x
⇒ -2x + 0y + 5 = 0
On comparing this equation with ax + by + c = 0, we get
a = -2, b = 0 and c = 5

https://youtu.be/_31qm2RCxmw

Exercise 4.2

1. Which one of the following options is true, and why?
y = 3x + 5 has
(i) a unique solution,               (ii) only two solutions,             (iii) infinitely many solutions

Answer

Since the equation, y = 3x + 5 is a linear equation in two variables. It will have (iii) infinitely many solutions.

2. Write four solutions for each of the following equations:

    (i) 2x + y = 7             (ii) πx + y = 9                (iii) x = 4y

Answer

(i) 2x + y = 7
⇒ y = 7 – 2x
→ Put x = 0,
y = 7 – 2 × 0 ⇒ y = 7
(0, 7) is the solution.
→ Now, put x = 1
y = 7 – 2 × 1 ⇒ y = 5
(1, 5) is the solution.
→ Now, put x = 2
y = 7 – 2 × 2 ⇒ y = 3
(2, 3) is the solution.
→ Now, put x = -1
y = 7 – 2 × -1 ⇒ y = 9
(-1, 9) is the solution.
The four solutions of the equation 2x + y = 7 are (0, 7), (1, 5), (2, 3) and (-1, 9).

(ii) πx + y = 9
⇒ y = 9 – πx
→ Put x = 0,
y = 9 – π×0 ⇒ y = 9
(0, 9) is the solution.
→ Now, put x = 1
y = 9 – π×1 ⇒ y = 9-π
(1, 9-π) is the solution.
→ Now, put x = 2
y = 9 – π×2 ⇒ y = 9-2π
(2, 9-2π) is the solution.
→ Now, put x = -1
y = 9 – π× -1 ⇒ y = 9+π
(-1, 9+π) is the solution.
The four solutions of the equation πx + y = 9 are (0, 9), (1, 9-π), (2, 9-2π) and (-1, 9+π).

(iii) x = 4y
→ Put x = 0,
0 = 4y ⇒ y = 0
(0, 0) is the solution.
→ Now, put x = 1
1 = 4y ⇒ y = 1/4
(1, 1/4) is the solution.
→ Now, put x = 4
4 = 4y ⇒ y = 1
(4, 1) is the solution.
→ Now, put x = 8
8 = 4y ⇒ y = 2
(8, 2) is the solution.
The four solutions of the equation πx + y = 9 are (0, 0), (1, 1/4), (4, 1) and (8, 2).

3. Check which of the following are solutions of the equation x – 2y = 4 and which are not:

    (i) (0, 2)              (ii) (2, 0)             (iii) (4, 0)            (iv) (√2, 4√2)              (v) (1, 1)

Answer

(i) Put x = 0 and y = 2 in the equation x – 2y = 4.
0 – 2×2 = 4
⇒ -4 ≠ 4
∴ (0, 2) is not a solution of the given equation.

(ii) Put x = 2 and y = 0 in the equation x – 2y = 4.
2 – 2×0 = 4
⇒ 2 ≠ 4
∴ (2, 0) is not a solution of the given equation.

(iii) Put x = 4 and y = 0 in the equation x – 2y = 4.
4 – 2×0 = 4
⇒ 4 = 4
∴ (4, 0) is a solution of the given equation.

(iv) Put x = √2 and y = 4√2 in the equation x – 2y = 4.
√2 – 2×4√2 = 4 ⇒ √2 – 8√2 = 4 ⇒ √2(1 – 8) = 4
⇒ -7√2  ≠ 4
∴ (√2, 4√2) is not a solution of the given equation.

(v) Put x = 1 and y = 1 in the equation x – 2y = 4.
1 – 2×1 = 4
⇒ -1 ≠ 4
∴ (1, 1) is not a solution of the given equation.

4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Answer

Given equation = 2x + 3y = k
x = 2, y = 1 is the solution of the given equation.
A/q,
Putting the value of x and y in the equation, we get
2×2 + 3×1 = k
⇒ k = 4 + 3
⇒ k = 7

https://youtu.be/RzvMsJVBsvo

Exercise 4.3

1. Draw the graph of each of the following linear equations in two variables:
   (i) x + y = 4                  (ii) x – y = 2               (iii) y = 3x              (iv) 3 = 2x + y

Answer

(i) x + y = 4
Put x = 0 then y = 4
Put x = 4 then y = 0

x	0	4
y	4	0

(ii) x – y = 2
Put x = 0 then y = -2
Put x = 2 then y = 0

x	0	2
y	-2	0

(iii) y = 3x
Put x = 0 then y = 0
Put x = 1 then y = 3

x	0	1
y	0	3

(iv) 3 = 2x + y
Put x = 0 then y = 3
Put x = 1 then y = 1

x	0	1
y	3	1

2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

Answer

Here, x = 2 and y =14.
Thus, x + y = 1
also, y = 7x ⇒ y – 7x = 0
∴ The equations of two lines passing through (2, 14) are
x + y = 1 and y – 7x = 0.
There will be infinite such lines because infinite number of lines can pass through a given point.

3. If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.

Answer

The point (3, 4) lies on the graph of the equation.
∴ Putting x = 3 and y = 4 in the equation 3y = ax + 7, we get
3×4 = a×3 + 7
⇒ 12 = 3a + 7
⇒ 3a = 12 – 7
⇒ a = 5/3

4. The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this information, and draw its graph.

Answer

Total fare = y
Total distance covered = x
Fair for the subsequent distance after 1st kilometre = Rs 5
Fair for 1st kilometre = Rs 8
A/q
y = 8 + 5(x-1)
⇒ y = 8 + 5x – 5
⇒ y = 5x + 3

x	0	-3/5
y	3	0

5. From the choices given below, choose the equation whose graphs are given in Fig. 4.6 and Fig. 4.7.
   For Fig. 4. 6                           For Fig. 4.7
   (i) y = x                                  (i) y = x + 2
   (ii) x + y = 0                           (ii) y = x – 2
   (iii) y = 2x                              (iii) y = –x + 2
   (iv) 2 + 3y = 7x                     (iv) x + 2y = 6

Answer

In fig. 4.6, Points are (0, 0), (-1, 1) and (1, -1).
∴ Equation (ii) x + y = 0 is correct as it satisfies all the value of the points.

In fig. 4.7, Points are (-1, 3), (0, 2) and (2, 0).
∴ Equation (iii) y = –x + 2 is correct as it satisfies all the value of the points.

Page No: 75

6. If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is
                     (i) 2 units               (ii) 0 unit

Answer

Let the distance traveled by the body be x and y be the work done by the force.
y ∝ x (Given)
⇒ y = 5x (To equate the proportional, we need a constant. Here, it was given 5)
A/q,
(i) When x = 2 units then y = 10 units
(ii) When x = 0 unit then y = 0 unit

x	2	0
y	10	0

7. Yamini and Fatima, two students of Class IX of a school, together contributed Rs 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs x and Rs y.) Draw the graph of the same.

Answer

Let the contribution amount by Yamini be x and contribution amount by Fatima be y.
A/q,
x + y = 100
When x = 0 then y = 100
When x = 50 then y = 50
When x = 100 then y = 0

x	0	50	100
y	100	50	0

 

 

8. In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:

                                           F = (9/5)C + 32

(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.

(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95°F, what is the temperature in Celsius?
(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

 

Answer

 

(i) F = (9/5)C + 32

When C = 0 then F = 32

also, when C = -10 then F = 14

C	0	-10
F	32	14

(ii) Putting the value of C = 30 in F = (9/5)C + 32, we get
F = (9/5)×30  + 32
⇒ F = 54 + 32
⇒ F = 86

(iii) Putting the value of F = 95 in F = (9/5)C + 32, we get
95 = (9/5)C  + 32
⇒ (9/5)C = 95 – 32
⇒ C = 63 × 5/9
⇒ C = 35

(iv) Putting the value of F = 0 in F = (9/5)C + 32, we get
0 = (9/5)C  + 32
⇒ (9/5)C = -32
⇒ C = -32 × 5/9
⇒ C = -160/9

Putting the value of C = 0 in F = (9/5)C + 32, we get
F = (9/5)× 0  + 32
⇒ F = 32

(v) Here, we have to find when F = C.
Therefore, Putting F = C in F = (9/5)C + 32, we get
F = (9/5)F + 32
⇒ F – 9/5 F = 32
⇒ -4/5 F = 32
⇒ F = -40
Therefore at -40, both Fahrenheit and Celsius numerically the same.

https://youtu.be/wnegZVDCOzo

Exercise 4.4

1. Give the geometric representations of y = 3 as an equation
(i) in one variable
(ii) in two variables

Answer

(i) in one variable, it is represented as
y = 3

 

2. Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable
(ii) in two variables

Answer

(i) in one variable, it is represented as
x = -9/2