Category: 10th mathematics exercise questions with answers

# Mensuration Exercise 15.6 – Class 10

### Mensuration Exercise 15.6 – Questions

1. Draw a plan and calculate the area of a level ground using the information given below:
 Metre to C To D 120To E 180 22021012080 200 to B From A

1. Plan out and find the area of the field from the date given from the Surveyor’s field book.
 Metre to C To D 100To C 75To B 50 35030025015050 150 to F100 to G From A

1. Sketch a rough plan and calculate the area of the field ABCDEFG from the following data:
 Metre to D To E 90To F 60To G 15 2251751251008060 20 to C70 to B From A

1. Calculate the area of the field shown in the diagram below: {measures in Meters}

## Mensuration Exercise 15.6 – Solutions

1. Draw a plan and calculate the area of a level ground using the information given below:
 Metre to C To D 120To E 180 22021012080 200 to B From A

Solution:

Take a scale 20m = 1cm

area of ABCDE = area of ∆APE + trapezium PEDR + area of ∆CDR + area of ∆CQB + area of ∆AQB

= 1/2 bh + 1/2 h(a+b) + 1/2 bh + 1/2 bh + 1/2 bh

= 1/2 x 80 x 180 + 1/2 x 130(180+120) + 1/2 x 10 x 120 + 1/2 x 100 x 200 + 1/2 x 200 x 120

area of ABCDE = 49,300 m2

1. Plan out and find the area of the field from the date given from the Surveyor’s field book.
 Metre to E To D 100To C 75To B 50 35030025015050 150 to F100 to G From A

Solution:

Scale 20 m = 1 cm

Area of ABCDEFG = area of ∆ABP + area of trapezium PRBC + area of trapezium SRCD + area of ∆DES + area of ∆EFS + area of trapezium SFGQ + area of ∆AQG

= 1/2 bh + 1/2 h(a+b) + 1/2 h(a+b) + 1/2 bh + 1/2 bh + 1/2 h(a+b) + 1/2 bh

= 1/2 x 50 x 50 + 1/2 x 200(50 + 75) + 1/2 x 50(100 + 75) + 1/2 x100 x 50 + 1/2 x 50 x 150 + 1/2 x 150 x (100 + 150) + 1/2 x 100 x 150

= 50,625 m2

1. Sketch a rough plan and calculate the area of the field ABCDEFG from the following data:
 Metre to D To E 90To F 60To G 15 2251751251008060 20 to C70 to B From A

Solution:

Scale 20 m = 1cm

Area of ABCDEFG = area of ∆AGQ + area of trapezium QRFG + area of trapezium RTEF + area of ∆TED + area of ∆DSC + area of trapezium SCBP + area of ∆APB

= 1/2 bh + 1/2 h(a+b) + 1/2 h(a+b) + 1/2 bh + 1/2 bh + 1/2 h(a+b) + 1/2 bh

= 1/2 x 80 x 15 + 1/2 x20(15+60) + 1/2 x75(60+ 90) + 1/2 x 50 x 90 + 1/2 x 100 x 20 + 1/2 x 65 x(70 + 20) + 1/2 x 60 x 70

= 15250 m2

1. Calculate the area of the field shown in the diagram below: {measures in Meters}

Solution:

 Metre To C 40 to C30 to E 1309570 30 to B From A

Scale 20 m = 1cm

Area of ABCDE = area of ∆AFE + area of trapezium CFEG + area of ∆BCG + area of ∆ABG

= 1/2 bh + 1/2 h(a+b) + 1/2 bh + 1/2 bh

= 1/2 x 95 x 30 + 1/2 x 35(30+40) + 1/2 x 60 x 30 + 1/2 x 70 x 30

= 4,600 m2

# Mensuration Exercise 15.5 – Class 10 – Solutions

### Mensuration Exercise 15.5 – Questions

1. A petrol tank is in the shape of a cylinder with hemisphere of same radius attached to both ends. If the total length of the tank is 6m and the radius is 1m, what is the capacity of the tank in litres.
2. A rocket is in the shape of a cylinder with a cone attached to one end and a hemisphere attached to the other. All of them are of the same radius of 1.5m. The total length of the rocket is 7m and height of the cup is 2m. Find the volume of the rocket.
3. A cup is in the form of a hemisphere surmounted by a cylinder. The height of the cylindrical portion is 8cm and the total height of the cup is 11.5cm. Find the TSA of the cup.
4. A storage tank consists of a circular cylinder with a hemisphere adjoined on either ends. The external diameter of the cylinder is 1.4 m and length is 8m, find the cost of painting it on the outside at the rate of Rs. 10 per m2.
5. A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 16 cm and its height is 15 cm. Find the cost of painting the toy at 7 per Rs. 100 cm2.
6. A circus tent is cylindrical up-to a height of 3m and conical above it. If the diameter of the base is 105m and the slant height of the conical part is 53m, find the total cost of canvas used to make the tent if the cost of the canvas per m2 is Rs. 100.

## Mensuration Exercise 15.5 – Solutions

1. A petrol tank is in the shape of a cylinder with hemisphere of same radius attached to both ends. If the total length of the tank is 6m and the radius is 1m, what is the capacity of the tank in litres.

Solution:

r = 1m

total length of the tank = 6m

h = 6 – (1 + 1) = 6 – 2 = 4 m

Volume of the tank = volume of the hemisphere + volume of the cylinder + volume of the hemisphere

= 2/3πr3 + πr2h + 2/3πr3

= 2/3 x 22/7­ x 13 + 22/7 x 12 x 4 + 2/3 x 22/7 x 13

= 16.76 m

Capacity of the water tank in litres = 16.76 x 1000 = 16761.9 l

1. A rocket is in the shape of a cylinder with a cone attached to one end and a hemisphere attached to the other. All of them are of the same radius of 1.5m. The total length of the rocket is 7m and height of the cup is 2m. Find the volume of the rocket.

Solution:

r = 1.5 m

Height of the cylinder, h = 7 – (1.5 + 2) = 7 – 3.5 = 3.5 m

height of the cone = 2m

Volume of the rocket = Volume of the cone + volume of the cylinder + volume of hemisphere

= 1/3πr2h+ πr2h + 2/3πr3

= 1/3 x 22/7 x (1.5)2 x 2 + 22/7 x (1.5)2 x 3.5 + 2/3 x  22/7 x (1.5)3

= 36.53 m3

1. A cup is in the form of a hemisphere surmounted by a cylinder. The height of the cylindrical portion is 8cm and the total height of the cup is 11.5cm. Find the TSA of the cup.

Solution:

total height of the cup = 11.5 cm

Height of the cylindrical portion = 8 cm

Height of the hemisphere = 11.5 – 8 = 3.5 cm

r = 3.5 cm

TSA of the cup = TSA of cylinder + TSA of hemisphere

= 2πrh + 2πr2

= 2 x 22/7 x 3.5 x 8 + 2 x 22/7 x 3.52

= 253 cm2

1. A storage tank consists of a circular cylinder with a hemisphere adjoined on either ends. The external diameter of the cylinder is 1.4 m and length is 8m, find the cost of painting it on the outside at the rate of Rs. 10 per m2.

Solution:

External diameter = 1.4 m

Radius = d/2 = 1.4/2 = 0.7 m

length = 8m

length of the hemisphere = 8 – (0.7+0.7) = 6.6 m

Total surface area of the storage tank = TSA of the cylindrical part + 2x TSA of  hemispherical part

= 2πr(r +h) + 2 x 3 πr2

= 2x22/7x(0.7)(0.7+6.6)+2x3x22/7 x (0.7)2

= 41.36 m2

Cost of polishing at the rate of  Rs. 10 per m2 = 41.34 x 10 = Rs. 413.6

1. A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 16 cm and its height is 15 cm. Find the cost of painting the toy at 7 per Rs. 100 cm2.

Solution:

diameter of the base of the cone = 16 cm

radius of the base of the cone = 8 cm

height = 15 cm

l2 = r2 + h2

= 82 + 152

= 64 + 25

= 289

l = 17 cm

Surface area of the toy = CSA of hemisphere + CSA of cone

= 2πr2 + πrl

= 2 x 22/7 x 82 + 22/7 x 8 x 17

= 829.71 cm2

Cost of painting the toy at Rs. 7 per 100 cm2 = 829.71×7/100 = Rs. 58.08

1. A circus tent is cylindrical up-to a height of 3m and conical above it. If the diameter of the base is 105m and the slant height of the conical part is 53m, find the total cost of canvas used to make the tent if the cost of the canvas per m2 is Rs. 100.

Solution:

d = 105 m

r = d/2 = 105/2 = 52.5 m

h = 3m

l = 53 m

l2 = r2 + h2

532 = 52.52 + h2

532 – 52.52 = h2

52.75 = h2

h = 7.26 m

Total canvas used = CSA of cylindrical part + CSA of conical part

= 2πrh + πrl

= 2 x 22/7 x 52.5 x 3 + 22/7 x 52.5 x 53

= 9735 m2

The total cost of canvas used to make the tent if the cost of the canvas per m2 is Rs. 100 = 9735 x 100 = Rs. 97,35,000

# Mensuration Exercise15.4 – Class 10 – Solutions

### Mensuration Exercise15.4 – Questions

1. Find the surface area of a sphere of radius 14 cm.
2. Find the TSA of a hemisphere of radius 5 cm.
3. A hemisphere bowl made of wood has inner diameter of 10.5 cm. Find the cost of painting it on the inside at the rate of Rs. 12 per 100 cm2.
4. Calculate the surface area of the largest sphere that can be cut out of a cube of side 15 cm.
5. Find the volume of the sphere whose radius is 7 cm.
6. Find the volume of a sphere whose surface area is 154 cm2.
7. The volume of a solid hemisphere is 1152πcm3. Find its curved surface area.
8. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine is needed to fill this capsule?
9. A right circular cone of height 20 cm and base radius 5 cm is melted and recast into a sphere. Find the radius of the sphere.
10. The diameter of a metallic sphere is 18 cm. It is melted and drawn into a wire having diameter of cross section 0.4 cm. Find the length of the wire.

## Mensuration Exercise15.4 – Solutions

1. Find the surface area of a sphere of radius 14 cm.

Solution:

Surface area of the sphere = 4πr2

= 4 x 22/7 x 142

= 2464 cm2

1. Find the TSA of a hemisphere of radius 5 cm.

Solution:

TSA of hemisphere = 3πr2

= 3 x 22/7 x 52

= 235.71 cm2

1. A hemisphere bowl made of wood has inner diameter of 10.5 cm. Find the cost of painting it on the inside at the rate of Rs. 12 per 100 cm2.

Solution:

diameter of the hemisphere = 10.5 cm

radius of the hemisphere = 5.25 cm

CSA of hemisphere = 2πr2

= 2 x 22/7 x 5.252

= 173.25 cm2

The cost of painting it on the inside at the rate of Rs. 12 per 100 cm2. Then the cost of painting the 173.25cm2 = 173.25×12/100 = Rs. 20.79

1. Calculate the surface area of the largest sphere that can be cut out of a cube of side 15 cm.

Solution:

d = 15 cm

r = d/2 = 15/2 = 7.5 cm

Surface area of the sphere = 4πr2

= 4 x 22/7 x 7.52

= 707.14 cm2

1. Find the volume of the sphere whose radius is 7 cm.

Solution:

Volume of the sphere = 4/3 πr3

= 4/3 x 22/7 x 73

= 1437.33 cm3

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

Solution:

We know, surface area of the sphere = 4πr2

154 = 4πr2

154/ = r2

r2 = 12.54

r = 3.5 cm

Volume of the sphere = 4/3 πr3

= 4/3 x 22/7 x 3.53

= 179.67 cm3

1. The volume of a solid hemisphere is 1152πcm3. Find its curved surface area.

Solution:

Volume of the hemisphere = 2/3 πr3

1152π = 2/3 πr3

1152π x 3/ = r3

1728 = r3

r = 12 cm

CSA of hemisphere = 2πr2

= 2 x 22/7 x (12)2

=  905.14 cm2

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

Solution:

d = 3.5 mm

r = d/2­ = 3.5/2 = 1.75 mm

Volume of the sphere = 4/3πr3

= 4/3 x 22/7 x (1.75)3

= 25.45 mm3

Therefore, 25.25mm3 medicine is needed to fill the capsule.

1. A right circular cone of height 20 cm and base radius 5 cm is melted and recast into a sphere. Find the radius of the sphere.

Solution:

h = 20 cm

r = 5 cm

Volume of the sphere = Volume of the cone

4/3 πrs3 = 1/3 πrc2hc

4/3 πrs3 = 1/3 π x 52 x 20

4rs3 = 52 x 20

rs3 = 25×20/4 = 125

= 5 cm

1. The diameter of a metallic sphere is 18 cm. It is melted and drawn into a wire having diameter of cross section 0.4 cm. Find the length of the wire.

Solution:

d = 18 cm

r = 9 cm

Volume of sphere = Volume of wire

4/3πrs3 = πrw2hw

4/3rs3 = rw2hw

4/3 x 93 = 0.22hw

hw = (4/3 x 93)/0.22

hw = 24300 cm

The length of the wire = 243 m

# Mensuration Exercise 15.3 – Class 10

### Mensuration Exercise 15.3 – Questions

1.Flower vase is in the form of a frustum of a cone. The perimeter of the ends are 44 cm and 8.4π cm. If the depth is 14 cm, fond how much water it can hold?

1. A bucket is in the shape of a frustum with the top and bottom circles of radii 15 cm and 10 cm. Its depth is 12 cm. Find its curved surface area and total surface area.(express your answer in terms of π)
2. From the top of a cone of base radius 24 cm and height 45 cm, a cone of slant height 17 cm is cut off What is the volume of the remaining frustum of the cone?
3. A vessel is in the form of a frustum of a cone. Its radius at one end is 8 cm and the height is 14 cm. If its volume is 5676/3 cm3, find the radius of the other hand.

## Mensuration Exercise 15.3 – Solutions

1.Flower vase is in the form of a frustum of a cone. The perimeter of the ends are 44 cm and 8.4π cm. If the depth is 14 cm, fond how much water it can hold?

Solution:

Given, Perimeter, P1 = 2πr1 =  44 cm

Then, r1 = 44/ = 22×7/22 = 7 cm

Perimeter, P2 = 2πr2 =  8.4π cm

Then, r2 = 8.4π/ = 4.2 cm

Given: h = 14 cm

Volume of the flower vase = 1/3 πh(r12 + r22 + r1r2)

= 1/3 x π x 14 [(7)2 + (4.2)2 + (7 x 4.2)]

= 1/3 x 22/7 x 14(49 + 17.64 + 29.4)

= 1408.58 cm3

1. A bucket is in the shape of a frustum with the top and bottom circles of radii 15 cm and 10 cm. Its depth is 12 cm. Find its curved surface area and total surface area.(express your answer in terms of π)

Solution:

r1 = 15 cm

r2 = 10 cm

h = 12 cm

l2 = h2 + (r2 – r1)2

l2 = 122 + (15 – 10)2

= 122 + 52

= 144 + 25

= 169

l = 13 cm

Curved surface area of the frustum = π(r1 + r2)l

= 22/7 x (15 + 10) x 13

= 1042.43 cm2

1. From the top of a cone of base radius 24 cm and height 45 cm, a cone of slant height 17 cm is cut off. What is the volume of the remaining frustum of the cone?

Solution:

L2 = R2 + H2

= 242 + 452

= 576 + 2025

= 2601

l = √2601

l = 51 cm

We know,

r/R = l/L and h/H = l/L

r/24 = 17/51 and h/45 = 17/51

r = 17/51 x 24 and h = 17/51 x 45

r = 8 cm and h = 15 cm

Height of the frustum = 45 – 15 = 30 cm

Volume of the frustum = 1/3 πh(r12 + r22 + r1r2)

= 1/3 x 22/7 x 30 x [82 + 242 + 8 x 24]

= 26148.57 cm3

1. A vessel is in the form of a frustum of a cone. Its radius at one end is 8 cm and the height is 14 cm. If its volume is 5676/3 cm3, find the radius of the other hand.

Solution:

r1 = 8 cm

h = 14 cm

Volume of the frustum = 5676/3 cm3

Volume of the frustum = 1/3 πh(r12 + r22 + r1r2)

5676/3 = 1/3x 22/7 x 14(82 + r22 + 8xr2)

5676/3 x 3×7/22×14 = (64 + r22 + 8r2)

5676/3 x 3×7/22×14 = 64 + r22 + 8r2

129 – 64 = r22 + 8r2

r22 + 8r2 = 65

r22 + 8r2 – 65 = 0

r22 + 13r2 – 5r2 – 65 = 0

r2(r2 + 13) – 5 (r2 + 13) = 0

(r2 + 13)(r2 – 5) = 0

r2 = -13 and r2 = 5

Therefore, radius of the other hand, r2 = 5 cm.

# Mensuration Exercise 15.2 – Class 10

### Mensuration Exercise 15.2 – Questions

1. Find the curved surface area of a cone, if its slant height is 60 cm and the radius of its base is 21 cm.
2. The radius of a cone is 7 cm and area of curved surface is 176 cm2. Find its slant height.
3. The area of the curved surface of a cone is 60πcm2. If the slant height of the cone is 8 cm, find the radius of the base.
4. Curved surface area of a cone is 308cm2 and its slant height is 14 cm. Find the radius of the base and total surface area of the cone.
5. A clown’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
6. Find the ratio of the curved surface areas of two cones if their diameters of the bases are equal and slant heights are in the ratio 4:3.
7. Find the volume of a right circular cone with radius 5 cm and height 7 cm.
8. Two cones have their heights in the ratio 1:3 and the radii of their bases in the ratio 3:1. Find the ratio of their volumes.
9. A right angled triangle of which the sides containing the right angle are 6.3 cm and 10 cm in length, is made to turn around on the longer side. Find the volume of the solid thus generated.
10. A tent is of the shape of a right circular cylinder up to a height of 3m and then becomes a right circular cone with a maximum height of 13.5 m above the ground. Calculate the cost of painting the inner side of the tent at the rate of Rs. 2 per sq. m., if the radius of the base is 14m.

## Mensuration Exercise 15.2 – Solutions

1. Find the curved surface area of a cone, if its slant height is 60 cm and the radius of its base is 21 cm.

Solution:

Slant height, l = 60 cm

Curved surface area of the cone = πrl

= 22/7 x 21 x 60

= 3960 cm2

1. The radius of a cone is 7 cm and area of curved surface is 176 cm2. Find its slant height.

Solution:

Curved surface area of the cone = 176 cm2.

We have to find the slant height, l.

We know, curved surface area of the cone = πrl.

176 = 22/x 7 x l

l = 176 x 7/7 x 21

l = 8 cm

1. The area of the curved surface of a cone is 60πcm2. If the slant height of the cone is 8 cm, find the radius of the base.

Solution:

Curved surface area of the cone = 60π cm2

Slant height, l = 8 cm

We have to find the radius of the cone. Then,

Curved surface area = πrl

60π = π x r x 8

r = 60π/ = 7.5 cm

1. Curved surface area of a cone is 308cm2 and its slant height is 14 cm. Find the radius of the base and total surface area of the cone.

Solution:

Curved surface area of the cone = 308 cm2

Slant height = 14 cm

To find the radius of the cone, we have,

Curved surface area = πrl

308 = π x r x 14

r = 308×7/22×14

r = 7 cm.

Total surface area of the cone = πr (r + l)

= 22/7 x 7(7 + 14)

= 22/7 x 7(21)

= 462 cm2

Therefore, radius of the cone is 7 cm and total surface area of the cone is 462 cm2.

1. A clown’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Solution:

Height = 24 cm

We know that, l2 = r2 + h2

l2 = 72 + 242 = 625

l = √625 = 25 cm

Curved surface area = πrl

= 22/7 x 7 x 25

= 550 cm2

The sheet required to make 1 cap is 550 cm2 .Then the area of the sheet required to make 10 such caps = 550 x 10 = 5500 cm2

1. Find the ratio of the curved surface areas of two cones if their diameters of the bases are equal and slant heights are in the ratio 4:3.

Solution:

Let diameters of the cone be d1 and d2.

Slant heights be l1 and l2.

Given, diameters of two cones are equal then their radius are also equal.Their slant heights of the cone 4:3 i.e., l1 = 4 and l2 = 3

We have to find the curved surface areas of two cones. let its be A1 and A2.

We know, curved surface area = πrl

Curved surface area of the cone with A1 = πrl1 = 4πr

Curved surface area of the cone with A2 = πrl2 = 3πr

A1/A2 = 4πr/3πr = 4/3

Ratio of the curved surface area of two cones = 4:3

1. Find the volume of a right circular cone with radius 5 cm and height 7 cm.

Solution:

height = 7cm

Volume of the cone = 1/3 πr2h

= 1/3 x 22/7 x 52 x 7

= 183.33 cm3

1. Two cones have their heights in the ratio 1:3 and the radii of their bases in the ratio 3:1. Find the ratio of their volumes.

Solution:

Height of the cones be h1 = 1 and h2 = 3.

Radii of two cones be r1 = 3 and r2 = 1

Volume of the cone with h1 is 1 and r1 is 3 = 1/3 πr2h

= 1/3 π x (3)2 x 1

= 3π

Volume of the cone with h2 is 3 and r2 is 1 = 1/3 πr2h

= 1/3 π x (1)2 x 3

= π

Ratio of  volumes of two cones = v1/v2 = /π = 3/1

Then, Ratio of  volumes of two cones is 3:1

1. A right angled triangle of which the sides containing the right angle are 6.3 cm and 10 cm in length, is made to turn around on the longer side. Find the volume of the solid thus generated.

Solution:

height = 10 cm

Volume = 1/3 πr2h

= 1/3 x 22/7 x (6.3)2 x (10)

= 415.8 cm3

1. A tent is of the shape of a right circular cylinder up to a height of 3m and then becomes a right circular cone with a maximum height of 13.5 m above the ground. Calculate the cost of painting the inner side of the tent at the rate of Rs. 2 per sq. m., if the radius of the base is 14m.

Solution:

Total height of the tent = 13.5 m

Height of the cylinder = 3 m

Therefore, height of the cone = total height of the tent – height of the cone

= 13.5 – 3

= 10.5 m

We know, slant height of the cone , l2 = r2 +h2

= 142 + 10.52

= 196 + 110.25

l2 = 306.25

l = 17.5 m

Curved surface area of the cone = πrl

= 22/7 x 14 x 17.5

= 770 m2

Curved surface area of the cylinder = 2πrh

= 2 x 22/7 x 14 x 10.5

= 264 m2

Total curved surface area = curved surface area of the cone + curved surface area of the cylinder

= 770 + 264

= 1034 m2

Therefore, cost of painting inter side of the tent at the rate of Rs. per sq. m = 2 x 1034 = Rs. 2068

# Mensuration Exercise 15.1 – Class 10

### Mensuration Exercise 15.1 – Questions:

1. The height of a right circular cylinder is 14 cm and the radius of its base is 2 cm. Find its (i) CSA (ii) TSA
2. An iron pipe 0 cm long has external radius equal to 12.5 cm and external radius equal to 11.5 cm. Find the TSA of the pipe.
3. The radii of two circular cylinders are in the ratio 2:# and the ratio of their curved surface areas is 5:6. Find the ratio of their heights.
4. The inner diameter of a circular well is 2.8 cm. It is 10 m deep. Find its inner curved surface area. Also find the cost of plastering this curved surface at the rate of Rs. 42 per m2?
5. Craft teacher of the school taught the students to prepare cylindrical pen holders out of card board. In a class of strength 42, if each child prepared a pen holder of radius 5 cm and height 14 cm, how much cardboard was consumed?
6. The diameter of a garden roller is 1.4 m and 2 m long. How much area will it cover in 5 revolutions?
7. Find the volume of right circular cylinder whose radius is 10.5 cm and height is 16 cm.
8. The inner diameter of a cylinder wooden pipe is 24 cm and its outer diameter is 28 cm. The pipe is 35 cm long. Find the mass of the pipe if 1cm3 of wood has a mass of 0.6 gm.
9. Two circular cylinder of equal volumes have their heights in the ratio 1:2. Find the ratio of their radii.
10. A rectangular sheet of paper, 44 cm x 20 cm is rolled along its length to form a cylinder. Find the volume of the cylinder so formed.

## Mensuration Exercise 15.1 – Solutions:

1. The height of a right circular cylinder is 14 cm and the radius of its base is 2 cm. Find its (i) CSA (ii) TSA

Solution:

Given:

Height of a right circular cylinder, h = 14 cm.

(i) CSA of cylinder = 2πrh

= 2 x 22/7 x 2 x 14

= 176 cm2

(ii) TSA of cylinder = 2πr(h + r)

= 2 x 22/7 x 2 x (14 + 2)

= 201.14 cm2

1. An iron pipe 20 cm long has external radius equal to 12.5 cm and external radius equal to 11.5 cm. Find the TSA of the pipe.

Solution:

Given:

h = 20 cm

External radius = R = 12.5 cm

Inner radius = r = 11.5 cm

(a) Inner CSA = S1 = 2πrh = 2 x 22/7 x 11.5 x 20 = 1445.71 cm2

(b) Outer CSA = S2 = 2πRh = 2 x 22/7 x 12.5 x 20 = 1571.43 cm2

(c) TSA = S1 + S2 + area of two bases

= S1 + S2 + 2(πR2 – πr2)

= 1445.71 + 1571.43 + 2[22/(­12.5)222/7(11.5)2]

= 3167.98 cm2

1. The radii of two circular cylinders are in the ratio 2:3 and the ratio of their curved surface areas is 5:6. Find the ratio of their heights.

Solution:

Let the radii of 2 cylinders be 2r and 3r respectively and their curved surface areas be 5cm2 and 6cm2 respectively.

We know, Curved surface area S = 2πrh

Then, h = S/2πr

Let h1 be the height of the cylinder of radii 2r and curved surface area 5 cm2 = 5/2πx2r  = 5/4πr

Let h2 be the height of the cylinder of radii 3r and curved surface area 6 cm2 = 6/2πx3r  = 1/πr

Therefore, h1/h2 = (5/4πr)/(1/πr) = 5/4

h1: h2 = 5: 4

1. The inner diameter of a circular well is 2.8 cm. It is 10 m deep. Find its inner curved surface area. Also find the cost of plastering this curved surface at the rate of Rs. 42 per m2?

Solution:

Inner diameter of a circular well = 2.8 cm r = 2.8/2 = 1.4 cm and h = 10 m

Inner curved surface area = 2πrh = 2 x 22/7 x 1.4 x 10 = 88 cm2

The cost of plastering for curved surface area 1m2 is 42. Then, the cost of plastering 88cm2 =Rs. 3696

1. Craft teacher of the school taught the students to prepare cylindrical pen holders out of card board. In a class of strength 42, if each child prepared a pen holder of radius 5 cm and height 14 cm, how much cardboard was consumed?

Solution:

No. of students = 42

radius of the pen holder, r = 5 cm

height of the pen holder, h = 14 cm

The card board consumed for 1 pen holder = curved surface area + area of the base = 2πrh + πr2

= 2 x 22/7 x 5 x 14 + 22/7 x 52

= 440 +78.57

= 518.57 cm2

Card board consumed to make 1 pen holder is 518.57cm2. Then the card board consumed to make 42 pen holder = 42×518.57 = 21780 cm2

1. The diameter of a garden roller is 1.4 m and 2 m long. How much area will it cover in 5 revolutions?

Solution:

The diameter of the garden roller, d = 1.4 m

Then, radius, r = 1.4/2 = 0.7m

Height of the garden roller, h = 2m

Curved surface area = 2πrh

= 2 x 22/7 x 0.7 x 2

= 8.8 m2

No. of revolutions garden roller takes is 5. Then the area covered in 5 revolutions = 8.8 x 5 = 44m2

1. Find the volume of right circular cylinder whose radius is 10.5 cm and height is 16 cm.

Solution:

Radius of the right circular cylinder, r = 10.5 cm

Height of the right circular cylinder, h = 16 cm

Volume of a cylinder = πr2h

= 22/7 x 10.52 x 16

= 5544 cm3

1. The inner diameter of a cylinder wooden pipe is 24 cm and its outer diameter is 28 cm. The pipe is 35 cm long. Find the mass of the pipe if 1cm3 of wood has a mass of 0.6 gm.

Solution:

Inner diameter of a cylinder wooden pipe, d = 24 cm, r = 24/2 = 12 cm

Outer diameter of a cylinder wooden pipe, D = 28 cm, d = 28/2 = 14 cm

Height of a cylinder wooden pipe, h = 35 cm

Volume of the of a cylinder wooden pipe, V = πR2h – πr2h

= 22/7 x 35x(142 – 122)

= 5720 cm3

The mass of the pipe if 1cm3 of wood has  a mass of 0.6 gm = 5720 x 0.6 = 3432 gm.

1. Two circular cylinder of equal volumes have their heights in the ratio 1:2. Find the ratio of their radii.

Solution:

Ratio of heights of two circular cylinders of equal volumes = 1:2

Therefore, h1 = 1 and h2 = 2

Volume of the circular cylinder = πr2h

Given, V1 = V2 = V

Radius of the circular cylinder of h1 = 1, πr12h = V, r12 = V/πh1

Radius of the circular cylinder of  h2 = 2, πr22h = V, r22 = V/πh2

Ratio of their radii = r12/r22 = (V/πh1)/ (V/πh2) = h2/h1 = 2/1

r1/r2 = √(2/1) = √2/1

Therefore, ratio of their radii = √2:1

1. A rectangular sheet of paper, 44 cm x 20 cm is rolled along its length to form a cylinder. Find the volume of the cylinder so formed.

Solution:

Area of rectangular sheet = Curved surface area of the cylinder =

⇒ l x b = 2πr x h

⇒ 44cm x 20cm = l x b

Thus, h = 20 cm and 2πr = 44 cm

⇒ 2πr  = 44 cm

⇒ πr = 22

⇒ r = 22×7/22 = 7 cm

Volume of the cylinder = πr2h

= 22/7 x 72 x 20

= 3080 cm3

# Coordinate Geometry Exercise 14.4 – Class 10

### Coordinate Geometry Exercise 14.4 – Questions:

1. In what ratio does the point (-2, 3) divide the line segment joining the points (-3, 5) and (4, -9)?
2. In the point C(1, 1) divides the line segment joining A(-2, 7) and B in the ratio 3:2, find the coordinates of B.
3. Find the ratio in which the point (-1, k) divides the line joining the points (-3, 10) and (6, -8) and find the value of k.
4. Find the coordinates of the midpoint of the line joining the points (-3, 10) and (6, -8).
5. Three consecutive vertices of a parallelogram are A(1, 2), B(2, 3) and C(8, 5). Find the fourth vertex. [Hint: diagonals of a parallelogram bisect each other]

## Coordinate Geometry Exercise 14.4 – Solutions:

1. In what ratio does the point (-2, 3) divide the line segment joining the points (-3, 5) and (4, -9)?

Solution:

Let the required ratio be m:n

Given: (x1, y1­) = (-2, 3), (x2, y2) = (4, -9) and (x , y) = (-2, 3)

By section formula,

1. In the point C(1, 1) divides the line segment joining A(-2, 7) and B in the ratio 3:2, find the coordinates of B.

Solution:

Let the required ratio be m:n

Given: (x1, y1­) = (-2, 7) , m:n = 3:2 and (x , y) = (1, 1) . We have to find (x2, y2).

By section formula,

To find x2,

To find y2,

Therefore, (x2, y2) = (3, -3) .Hence B(3, -3).

1. Find the ratio in which the point (-1, k) divides the line joining the points (-3, 10) and (6, -8) and find the value of k.

Solution:

Given, (x, y)= (-1, k), (x1, y1) = (-3, 10) and (x2, y2) = (6, -8)

By section formula,

To find m:n,

To find y2,

Therefore, m:n = 2:7  and k = 6.

1. Find the coordinates of the midpoint of the line joining the points (-3, 10) and (6, -8).

Solution:

Given, (x1, y1) = (-3, 10) and (x2, y2) = (6, -8).

1. Three consecutive vertices of a parallelogram are A(1, 2), B(2, 3) and C(8, 5). Find the fourth vertex. [Hint: diagonals of a parallelogram bisect each other]

Solution:

Given, A(1, 2), B(2, 3) and C(8, 5)

Diagonal of a parallelogram bisect each other. Therefore, diagonal of the parallelogram is the midpoint.

We know, diagonals of the parallelogram are equal.

Therefore, we have, (9/2, 7/2) = (2+x/2, 3+y/2)

Then,

Also

# Coordinate Geometry Exercise 14.3 – Class 10

### Coordinate Geometry Exercise 14.3 – Questions:

1. Find the distance between the following pairs of points

(i) (8, 3) and (8,-7)

(ii) (1,-3) and (-4, 7)

(iii) (-4, 5) and (-12, 3)

(iv) 6, 5) and (4, 4)

(v) (2,0) and (0, 3)

(vi) (2, 8) and (6, 8)

(vii) (a, b) and (c, b)

(viii) (cosθ, -sinθ) and (sinθ, -cosθ)

1. Find the distance between the origin and the point

(i) (-6, 8)

(ii) (5, 12)

(iii) (-8, 15)

(i)The distance between the points (3, 1) and (0, x) is 5 units. Find x.

(ii) A point P(2, 1) is equidistant from the points (a, 7) and (-3, a). Find a.

(iii) Find a point on y-axis which is equidistant from the points (5, 2) and (-4, 3)

1. Find the perimeter of the triangles whose vertices have the following coordinates

(i)(-2, 1), (4, 6) and (6, -3)

(ii) (3, 10), (5, 2), (14, 12)

1. Prove that the points A(1, -3), B(-3, 0) and C(4, 1) are the vertices of a right isosceles triangle.
2. Find the radius of the circle whose centre is (-5, 4) and which passes through the point (-7, 1).
3. Prove that each of the set of coordinates are the vertices of parallelogram.

(i) (-5, -3), (1, -11), (7, -6), (1, 2)

(ii) (4, 0), (-2, -3), (3, 2), (-3, -1)

1. The coordinates of vertices of triangles are given. Identify the types of the triangle.

(i) (2, 1), (10, 1), (6, 9)

(ii) (1, 6), (3, 2), (10,8)

(iii) (3, 5), (-1, 1), (6, 2)

(iv) (3, -3), (3, 5), (11, -3)

## Coordinate Geometry Exercise 14.3 – Solutions:

1. Find the distance between the following pairs of points

(i) (8, 3) and (8,-7)

Solution:

(ii) (1,-3) and (-4, 7)

Solution:

(iii) (-4, 5) and (-12, 3)

Solution:

(iv) 6, 5) and (4, 4)

Solution:

(v) (2,0) and (0, 3)

Solution:

(vi) (2, 8) and (6, 8)

Solution:

(vii) (a, b) and (c, b)

Solution:

(viii) (cos θ, -sinθ) and (sinθ, -cosθ)

Solution:

1. Find the distance between the origin and the point

(i) (-6, 8)

Solution:

Distance between the origin and (x, y) = √(x2 + y2)

Here, (x, y) = (-6, 8)

d = √[(-6)2 + (8)2]

d = √[36+64]

d = √100

d = 10 units

(ii) (5, 12)

Solution:

Distance between the origin and (x, y) = √(x2 + y2)

Here, (x, y) = (5, 12)

d = √[(5)2 + (12)2]

d = √[25+144]

d = √169

d = 13 units

(iii) (-8, 15)

Solution:

Distance between the origin and (x, y) = √(x2 + y2)

Here, (x, y) = (-8, 15)

d = √[(-8)2 + (15)2]

d = √[64+225]

d = √289

d =  17 units

3. (i)The distance between the points (3, 1) and (0, x) is 5 units. Find x.

Solution:

We know,

52 = x2 – 2x +  10

x2 – 2x + 10  – 25 = 0

x2 – 2x – 15  = 0

x2 + 3x – 5x – 15 = 0

x(x + 3) – 5(x + 3) = 0

(x + 3)(x – 5) = 0

x + 3 = 0 or x – 5 = 0

x = -3 or x = 5

(ii) A point P(2, 1) is equidistant from the points (a, 7) and (-3, a). Find a.

Solution:

Let P(2, 1), Q(a , 7) and R(-3, a)

a2 – 4a + 40 = a2 – 2a + 26

-4a + 2a + 40 – 26 = 0

-2a – 14 = 0

2a = 14

a = 7 units

(iii) Find a point on y-axis which is equidistant from the points (5, 2) and (-4, 3)

Solution:

Let A(5, 2),  B(4, 3) and the point on y-axis be Y(0, y)

y2 – 4y + 29 = y2 – 6y + 25

-4y + 6y + 29 – 25 = 0

2y + 4 = 0

2y = -4

y = -2 units

Therefore, a point on y-axis which is equidistant from the points (5, 2) and (-4, 3) is (0, -2)

1. Find the perimeter of the triangles whose vertices have the following coordinates

(i)(-2, 1), (4, 6) and (6, -3)

Solution:

Perimeter of the triangle = sum of 3 sides of the triangle

P(∆ABC) = AB + BC + CA

Let A(-2, 1), B(4, 6) and C(6, -3)

Therefore, Perimeter of triangle ABC = AB + BC + CA = √61+√85+√80

(ii) (3, 10), (5, 2), (14, 12)

Solution:

Perimeter of the triangle = sum of 3 sides of the triangle

P(∆ABC) = AB + BC + CA

Let A(3, 10), B(5, 2) and C(14, 12)

Therefore, Perimeter of triangle ABC = AB + BC + CA = √68+√181+√125

1. Prove that the points A(1, -3), B(-3, 0) and C(4, 1) are the vertices of a right isosceles triangle.

Solution:

To show that the triangle is isosceles, we must show that the length of 2 sides of the triangle are equal.

Let A(1, -3), B(-3, 0) and C(4, 1)

Since AB = CA , the triangle is isosceles.

1. Find the radius of the circle whose centre is (-5, 4) and which passes through the point (-7, 1).

Solution:

Let O(-5, 4) and P(-7, 1)

1. Prove that each of the set of coordinates is the vertices of parallelogram.

(i) (-5, -3), (1, -11), (7, -6), (1, 2)

Solution:

A quadrilateral is a parallelogram if it’s opposite sides is equal.

Let A(-5, -3), B(1, -11), C(7, -6), D(1, 2)

Therefore, AB = CD and BC = DA

(ii) (4, 0), (-2, -3), (3, 2), (-3, -1)

Solution:

A(4, 0), B(-2, -3), C(3, 2), D(-3, -1)

Therefore, AB = CD and BC = DA

1. The coordinates of vertices of triangles are given. Identify the types of the triangle.

(i) (2, 1), (10, 1), (6, 9)

Solution:

Let A(2, 1), B(10, 1) and  C(6, 9)

Since BC = CA, the triangle is isosceles.

(ii) (1, 6), (3, 2), (10,8)

Solution:

Let A(1, 6), B(3, 2), C(10,8)

Since AB ≠ BC ≠ CA, the triangle isosceles.

(iii) (3, 5), (-1, 1), (6, 2)

Solution:

Let A(3, 5), B(-1, 1), and C(6, 2)

Since AB ≠ BC ≠ CA, the triangle scalene.

(iv) (3, -3), (3, 5), (11, -3)

Solution:

Let A(3, -3), B(3, 5), C(11, -3)

Since AB ≠ BC ≠ CA, the triangle isosceles.

# Coordinate Geometry Exercise 14.2 – Class 10

### Coordinate Geometry Exercise 14.2 – Questions:

1. Find the equation of the line whose angle of inclination and y-intercept are given.

(i) θ = 60˚, y-intercept is -2.

(ii) θ = 45˚, y-intercept is 3.

1. Find the equation of the line whose slope and y-intercept are given.

(i) Slope = 2, y-intercept = -4

(ii) Slope = –2/3, y-intercept = 1/2

(iii) Slope = -2, y-intercept = 3

1. Find the slope and y-intercept of the lines

(i) 2x + 3y = 4

(ii) 3x = y

(iii) x – y + 5 = 0

(iv) 3x – 4y = 5

1. Is the line x = 2y parallel to 2x – 4y + 7 = 0[Hint: Parallel lines have some slopes]
2. Show that the line 3x + 4y + 7 = 0 and 28x – 21y + 50 = 0 are perpendicular to each other.[Hint: For perpendicular lines, m1.m2 = -1].

## Coordinate Geometry Exercise 14.2 – Solutions:

1. Find the equation of the line whose angle of inclination and y-intercept are given.

(i) θ = 60˚, y-intercept is -2.

Solution:

m = tan θ = tan 60˚ = √3

y – intercept = c = -2

Equation: y = mx + c

y = √3x – 2

(ii) θ = 45˚, y-intercept is 3.

Solution:

m = tan θ = tan 45˚ = 1

y – intercept = c = 3

Equation: y = mx + c

y = x + 3

1. Find the equation of the line whose slope and y-intercept are given.

(i) Slope = 2, y-intercept = -4

Solution:

m = slope = 2

y – intercept = c = -4

Equation: y = mx + c

y = 2x – 4

(ii) Slope = –2/3, y-intercept = 1/2

Solution:

m = slope = –2/3

y – intercept = c = 1/2

Equation: y = mx + c

y = –2/3 x + 1/2

(iii) Slope = -2, y-intercept = 3

Solution:

m = slope = -2

y – intercept = c = 3

Equation: y = mx + c

y = -2x + 3

1. Find the slope and y-intercept of the lines

(i) 2x + 3y = 4

Solution:

2x + 3y = 4

3y =  4 – 2x

y = – 2/3x + 4/3

Standard equation is y = mx + c

Therefore, m = –2/3 and c = 4/3

(ii) 3x = y

Solution:

Given, 3x = y

y = 3x + 0

Standard equation is y = mx + c

Therefore, m = 3 and c = 0

(iii) x – y + 5 = 0

Solution:

x – y + 5 = 0

-y = – x – 5

y = x + 5

Standard equation is y = mx + c

Therefore, m = 1 and c = 5

(iv) 3x – 4y = 5

Solution:

3x – 4y = 5

-4y = 5 – 3x

4y = 3x – 5

y = 3/4 x – 5/4

Standard equation is y = mx + c

Therefore, m = 3/4 and c = – 5/4

1. Is the line x = 2y parallel to 2x – 4y + 7 = 0[Hint: Parallel lines have same slopes]

Solution:

line 1:

x = 2y

y = 1/2 x + 0

Standard equation is y = mx + c

Therefore, m1 = 1/2 and c1 = 0

line 2:

2x – 4y + 7 = 0

-4y = 7 – 2x

4y = 2x – 7

y = 2/4 x – 7/4

Standard equation is y = mx + c

Therefore, m2 = 2/4 = 1/2 and c2 = –7/4

Since slope of the line x = 2y and the line 2x – 4y + 7 = 0 are same i.e., m1 = m2 = 1/2.

Therefore, the line x = 2y parallel to 2x – 4y + 7 = 0.

1. Show that the line 3x + 4y + 7 = 0 and 28x – 21y + 50 = 0 are perpendicular to each other.[Hint: For perpendicular lines, m1.m2 = -1].

Solution:

line 1:

3x + 4y + 7 = 0

4y = – 3x – 7

y = – 3/4 x – 7/4

Standard equation is y = mx + c

Therefore, m1 = – 3/4 and c1 = –7/4

line 2:

28x – 21y + 50 = 0

-21y = – 28x – 50

21y = 50 + 28x

y = 28/21 x + 50/21

Standard equation is y = mx + c

Therefore, m2 = 28/21 and c2 = 50/21

If slope of the line 3x + 4y + 7 = 0 and 28x – 21y + 50 = 0 are perpendicular then we must get m1.m2 = -1

i.e.,

3/4 . 28/21 = -1

Therefore, the line 3x + 4y + 7 = 0 is perpendicular to 28x – 21y + 50 = 0.

# Coordinate Geometry Exercise 14.1 – Class 10

### Coordinate Geometry Exercise 14.1 – Questions:

1. Find the slope of the curve whose inclination is

(i) 90˚

(ii) 45˚

(iii) 30˚

(iv) 0˚

1. Find the angles of inclination of straight lines whose slopes are

(i)√3

(ii) 1

(iii) 1/√3

1. Find the slope of the line joining the points

(i) (-4, 1) and (-5, 2)

(ii) 4, -8) and (5, -2)

(iii) (0, 0) and (√3, 3)

(iv) (-5, 0) and (0, -7)

(v) (2a, 3b) and (a, -b)

1. Find whether the lines drawn through the two pairs of points are parallel or perpendicular

(i)(5, 2), (0, 5) and (0, 0), (-5, 3)

(ii) (3, 3), (4, 6) and (4, 1), (6, 7)

(iii) (4, 7), (3, 5) and (-1, 7), (1, 6)

(iv) (-1, -2), (1, 6) and (-1, 1), (-2, -3)

1. Find the slope of the line perpendicular to the line joining the points

(i)(1, 7) and (-4, 3)

(ii) (2, -3) and (1, 4)

1. Find the slope of the line parallel to the line joining the points

(i) (-4, 3) and (2, 5)

(ii) (1, -5) and (7, 1)

1. A line passing through the points (2, 7) and (3, 6) is parallel to a line joining (9, a) and (11, 3). Find a.
2. A line joining through the points (1, 0) and (4, 3) is perpendicular to the line joining (-2, -1) and (m, 0). Find the value of m.

## Coordinate Geometry Exercise 14.1 – Solutions:

1. Find the slope of the curve whose inclination is

(i) 90˚

Solution:

θ = 90˚

Slope = m = tan θ

m = tan 90˚

m = undefined

(ii) 45˚

Solution:

θ = 45˚

Slope = m = tan θ

m = tan 45˚

m = 1

(iii) 30˚

Solution:

θ = 30˚

Slope = m = tan θ

m = tan 30˚

m = 1/√3

(iv) 0˚

Solution:

θ = 0˚

Slope = m = tan θ

m = tan 0˚

m = 0

1. Find the angles of inclination of straight lines whose slopes are

(i)√3

Solution:

Slope = tan θ

√3 = tanθ

We know that, tan 60˚ = √3

θ = 60˚

(ii) 1

Solution:

Slope = tan θ

1 = tan θ

We know that, tan 45˚ = 1

θ = 45˚

(iii) 1/√3

Solution:

Slope = tan θ

1/√3 = tanθ

We know that, tan 30˚ = 1/√3

θ = 30˚

1. Find the slope of the line joining the points

(i) (-4, 1) and (-5, 2)

Solution:

Let (x1 ,y1) = (-4, 1) and (x2 ,y2) (-5, 2)

Slope = y2-y1/x2-x1 = 2-1/-5-(-4) = 1/-5+4 = -1

(ii) 4, -8) and (5, -2)

Solution:

Let (x1 ,y1) = (4, -8) and (x2 ,y2) = (5, -2)

Slope = y2-y1/x2-x1 = -2-(-8)/5-4 = -2+8/1 = 6

(iii) (0, 0) and (√3, 3)

Solution:

Let (x1 ,y1) = (0, 0) and (x2 ,y2) = (√3, 3)

Slope = y2-y1/x2-x1 = 3-0/√3-0 = 3/√3 = √3

(iv) (-5, 0) and (0, -7)

Solution:

Let (x1 ,y1) = (-5, 0) and (x2 ,y2) = (0, -7)

Slope = y2-y1/x2-x1 = -7-0/0-(-5) = -7/5

(v) (2a, 3b) and (a, -b)

Solution:

Let (x1 ,y1) = (2a, 3b) and (x2 ,y2) = (a, -b)

Slope = y2-y1/x2-x1 = -b-3b/a-2a = -4b/-a = 4b/a

1. Find whether the lines drawn through the two pairs of points are parallel or perpendicular

(i)(5, 2), (0, 5) and (0, 0), (-5, 3)

Solution:

For line 1:

Let (x1 ,y1) = (5, 2) and (x2 ,y2)= (0, 5)

Slope = m1 = y2-y1/x2-x1 = 5-2/0-5 = 3/-5

For line 2:

Let (x1 ,y1) = (0, 0) and (x2 ,y2)= (-5, 3)

Slope = m2 = y2-y1/x2-x1 = 3-0/-5-0 = 3/-5

We have m1 = m2

Therefore, (5, 2), (0, 5) and (0, 0), (-5, 3) lines are parallel.

(ii) (3, 3), (4, 6) and (4, 1), (6, 7)

Solution:

For line 1:

Let (x1 ,y1) = (3, 3) and (x2 ,y2)= (4, 6)

Slope = m1 = y2-y1/x2-x1 = 6-3/4-3 = 3/1 = 3

For line 2:

Let (x1 ,y1) = (4, 1) and (x2 ,y2)= (6, 7)

Slope = m2 = y2-y1/x2-x1 = 7-1/6-4 = 6/2 = 3

We have m1 = m2

Therefore, (3, 3), (4, 6) and (4, 1), (6, 7) parallel.

(iii) (4, 7), (3, 5) and (-1, 7), (1, 6)

Solution:

For line 1:

Let (x1 ,y1) = (4, 7) and (x2 ,y2)= (3, 5)

Slope = m1 = y2-y1/x2-x1 = 5-7/3-4 = -2/-1 = 2

For line 2:

Let (x1 ,y1) = (-1, 7) and (x2 ,y2)= (1, 6)

Slope = m2 = y2-y1/x2-x1 = 6-7/1-(-1) = -1/2

We have m1 ≠ m2

m1.m2 = 2 x (1/-2) = – 1

Therefore, (4, 7), (3, 5) and (-1, 7), (1, 6) are perpendicular.

(iv) (-1, -2), (1, 6) and (-1, 1), (-2, -3)

Solution:

For line 1:

Let (x1 ,y1) = (-1, -2) and (x2 ,y2)= (1, 6)

Slope = m1 = y2-y1/x2-x1 = 6-(-2)/1-(-1) = 8/2 = 4

For line 2:

Let (x1 ,y1) = (-1, 1) and (x2 ,y2)= (-2, -3)

Slope = m2 = y2-y1/x2-x1 = -3-1/-2-(-1) = -4/-1 = 4

We have m1 = m2

Therefore, (3, 3), (4, 6) and (4, 1), (6, 7) are parallel.

1. Find the slope of the line perpendicular to the line joining the points

(i)(1, 7) and (-4, 3)

Solution:

Let (x1 ,y1) = (1, 7) and (x2 ,y2)= (-4, 3)

Slope = m1 = y2-y1/x2-x1 = 3-7/-4-1 = -4/-5 = 4/5

The slope of the line perpendicular to the line joining points (1, 7) and (-4, 3) is  -5/4.[since m1.m2 = -1]

(ii) (2, -3) and (1, 4)

Solution:

Let (x1 ,y1) = (2, 3) and (x2 ,y2)= (1, 4)

Slope = m1 = y2-y1/x2-x1 = 4-3/1-2 = 1/-1 = -1

The slope of the line perpendicular to the line joining points (2, -3) and (1, 4) is 1. [Since m1.m2 = -1]

1. Find the slope of the line parallel to the line joining the points

(i) (-4, 3) and (2, 5)

Solution:

Let (x1 ,y1) = (-4, 3) and (x2 ,y2)= (2, 5)

Slope = m1 = y2-y1/x2-x1 = 5-3/2-(-4) = 2/6 = 1/3

The slope of the line parallel to the line joining points (-4, 3) and (2, 5) is 1/3.

(ii) (1, -5) and (7, 1)

Solution:

Let (x1 ,y1) = (1, -5) and (x2 ,y2)= (7, 1)

Slope = m1 = y2-y1/x2-x1 = 1-(-5)/7-1 = 6/6 = 1

The slope of the line parallel to the line joining points (1, -5) and (7, 1) is 1.

1. A line passing through the points (2, 7) and (3, 6) is parallel to a line joining (9, a) and (11, 3). Find a.

Solution:

Line 1:

Let (x1 ,y1) = (2, 7) and (x2 ,y2)= (3, 6)

Slope = m1 = y2-y1/x2-x1 = 6-7/3-2 = -1/1 = -1

Line 2:

Let (x1 ,y1) = (9, a) and (x2 ,y2)= (11, 3)

Slope = m2 = y2-y1/x2-x1 = 3-a/11-9 = 3-a/2

Since m1 = m2 as line (2, 7) and (3, 6) is parallel to (9, a) and (11, 3).

We have, -1 = 3-a/2

-2 = 3 – a

-2 – 3 = – a

– 5 = -a

a = 5

1. A line joining through the points (1, 0) and (4, 3) is perpendicular to the line joining (-2, -1) and (m, 0). Find the value of m.

Solution:

Line 1:

Let (x1 ,y1) = (1, 0) and (x2 ,y2)= (4, 3)

Slope = m1 = y2-y1/x2-x1 = 3-0/4-1 = 3/3 = 1

Line 2:

Let (x1 ,y1) = (-2, -1) and (x2 ,y2)= (m, 0)

Slope = m2 = y2-y1/x2-x1 = 0-(-1)/m-(-2) = 1/m+2

Since m1.m2 = -1 as A line joining through the points (1, 0) and (4, 3) is perpendicular to the line joining (-2, -1) and (m, 0).

We have, m1.m2 = -1

1.(1/m+2) = -1

1/m+2 = -1

1 = -(m+2)

1 = -m – 2

-m = 1 + 2

m = -3