Surface Area and Volumes – Exercise 13.1 – Class IX

  1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine:

(i) The area of the sheet required for making the box.

(ii) The cost of sheet for it, if a sheet measuring 1m2 costs Rs 20.

Solution:

Given: l = 1.5 m ; b = 1.25m ; h = 65cm = 0.65 m.

It is given that the box is open at the top, so it has only 5 faces.

Surface  area of the box = lb + 2(bh + hl)

= 1.5*1.25 + 2(1.25*0.65 + 0.65*1.5)

= 1.875 + 2(1.7875)

= 5.45 m2

Hence 5.45 m2 is the required sheet.

Cost of 1m2 costs Rs. 20, then cost of 5.45m2 sheet = Rs. 20 * 5.45m2 = Rs. 109.


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

Solution:

Given: l = 5m , b = 4m and h = 3m

Surface area of the wall = lb + 2(bh + hl)

= 5*4 + 2(4*3 + 3*5)

= 20 + 2(12 + 15)

= 20 + 2(27)

= 20 + 54

= 74 m2

The cost for 1m2 is Rs. 7.50 , then , the cost of 74m2 = 74m2*Rs.7.50 = Rs. 555


  1. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs 10 per m2 is Rs 15000, find the height of the hall.

[Hint : Area of the four walls = Lateral surface area.]

Solution:

Perimater of the rectangular hall 250m

i.e., 2(l +b) = 250 m

Area of the four walls of the hall = 2h(l + b)…………….(1)

It is given that, the cost of painting the four walls at the rate of Rs 10 per m2 is Rs 15000.

The area of the four walls of the hall = 15000/10 m2 = 1500m2………………..(2)

From (1) and (2), we have,

2h(l + b) = 1500

h*250 = 1500

h = 1500/250

h = 6

Hence the height o the wall is 6 m


  1. 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?

Solution:

Since, l = 22.5 cm ; b =  10 cm ; h = 7.5 cm

Total surface area of 1 brick = 2(lb + bh + lh)

= 2(22.5*10 + 10*7.5 + 7.5*22.5)

=2(225 + 75 + 168.75)

= 937.5 cm2

= 0.09375 m2

Therefore, no. of bricks can be painted = 9.375/0.09375 = 100


  1. 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?

Solution:

(i)

Latteral surface area of the cube = 4a2 = 4(10)2 = 4*100 = 400 cm2

Lateral surface area of the cuboidal box = 2(lh + bh)

= 2(12.5*8 + 8*10)

= 2*8(22.5)

= 360 cm2

The difference in the lateral surface area of cube and cuboidal box =(400 – 360)cm2 = 40 cm2

Therefore, Cube has the greater lateral surface area.

(ii)

Total surface area of the cube = 6a2 = 6(10)2  = 6(100) = 600cm2

Total surface area of the cuboidal box = 2(lb + bh + lh)

= 2(12.5*10 + 10*8 + 8*12.5)

= 2(125 + 80 + 100)

= 2(305)

= 610 cm2

Difference in the total surface are of the cube and cuboidal box = (610 – 600)cm2 = 10cm2

Therefore, cube has the smaller total surface area.


  1. 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?

Solution:

Here, l = 30 cm , b = 25 cm and h = 25 cm.

Total surface area of the glass = 2(lb + bh + lh)

= 2(30*25 + 25*25 + 25*30)

= 2*25(30 + 25 + 30)

= 50(85)

= 4250 cm2

(ii)

A cuboid has 12 edges i.e., 4lengths, 4 breadths and 4 heights.

Length of the tape required = 4(l+b+h)

= 4(30+25+25)

=4(80)

= 320 cm


  1. 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 Rs 4 for 1000 cm2 , find the cost of cardboard required for supplying 250 boxes of each kind.

Solution:

Total surface area of the bigger box = 2(lb+hb+ lh)

= 2(25*20 + 20*5 + 5*25)

= 2*(500 + 100+ 125)

= 2(725)

= 1450 cm2

Area required for the overlaps is 5% of 1450cm2 = 1450*5/100 cm2 = 72.5 cm2

Total area of cardboard required for 1 bigger box = (1450+72.5)cm2 = 1522.5 cm2

Total area of cardboard needed for 250 bigger boxes = 1522.5 * 250 cm2 = 380625 cm2

 

Total surface area of the smaller box = 2(lb+hb+ lh)

= 2(15*12 + 12*5 + 5*15)

= 2(180 + 60 + 75)

= 630 cm2

Area required for the overlaps is 5% of 630cm2 = 630*5/100 cm2 = 31.5 cm2

Total area of cardboard required for 1 smaller boxes = (630+31.5)cm2 = 661.5 cm2

Total area of cardboard needed for 250 bigger boxes = 661.5 * 250 cm2 = 165375 cm2

Now, the total area of cardboard required for 250bigger boxes and 250 smaller boxes  = (380625+165375) cm2 = 546000 cm2

The cost of 1000 cm2 of cardboard = Rs. 4

Cost of 546000 cm2 of cardboard = Rs. 4/1000 *546000 = Rs. 2184


  1. Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m?

Solution:

h = 2.5 m , l = 4m and b = 3m

The tarpaulin required only 5 faces as excluding the floor area.

Surface area of the shelter for the car = lb + 2(bh + lh)

= 4*3 + 2(3*2.5 + 4*2.5)

= 12 + 2(7.5 + 10)

= 12 + 2(17.5)

= 12 + 35

= 47 m2


 

Heron’s Formula – Exercise 12.2 – Class IX

 

  1. A park, in the shape of a quadrilateral ABCD, has ∠ C = 90, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

Solution:

Heron's Formula - exercise 12.2. - CLass IX

Given: a park, in the shape of a quadrilateral ABCD, has ∠ C = 90, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m

Construction: Join BD

In ∆DBC, by  Pythagoras theorem, we have,

DB2 = BC2 + DC2

DB2 = (12)2 + (5)2

DB2 = 144 + 25

DB2 = 169

DB = √169 = 13 m

Area  of ∆DBC = 1/2 x base x height

= 1/2 x 12 x 5

= 6 x 5

= 30 m2

In ∆ADB, a = 9m,b  = 8m , c = 13m

Therefore, s = a+b+c/2 = 9+8+13/2 = 15 m

area of ∆ABD = √[s(s-a)(s-b)(s-c)

= √[15(15-9)(15-8)(15-13) m2

= √[15x6x7x2] m2

=√1260 m2

= 35.5 m2

Area of the park = area of ∆DBC + area of ∆ABD

=(30 + 35.5) m2

= 65.5 m2


  1. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Solution:

Heron's Formula - exercise 12.2. - CLass IXIn ∆ABC, we have,

a = 3 cm, b = 4 cm , c = 5 cm

s = a+b+c/2 = 3+4+5/2 = 12/2 = 6 cm

Area of the triangle = √[s(s-a)(s-b)(s-c)

=√[6(6-3)(6-4)(6-5)

= √[6x3x2x1]

=√[6×6]

= 6 cm2

In ∆ADC, we have

a = 5 cm , b = 5 cm, c = 4 cm

s = a+b+c/2 = 5+5+4/2 = 14/2 = 7 cm

Area of the triangle = √[s(s-a)(s-b)(s-c)

= √[7(7-5)(7-5)(7-4)]

= √[7x2x2x3]

= √84

= 9.2 cm2

Area of the quadrilateral = area of ∆ABC + area of ∆ADC

= (6 + 9.2) cm2

= 15.2 cm2


  1. Herons formula - exercise 12.2 - class IXRadha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.

 

Solution:

For the triangle marked as I:

a = b = 5 cm, c = 1 cm

s = a+b+c/2 = 5+5+1/2 = 11/2 = 5.5 cm

Area of triangle1 = √[s(s-a)(s-b)(s-c)

= √[5.5(5.5-5)(5.5-5)(5.5-1)]

= √[5.5×0.5×0.5×4.5]

= √6.1875 m2

≅ 2.5 m2

For the rectangle marked as II:

Area of the rectangle = Length x breadth [Length = 6.5 cm and breadth = 1 cm]

= 6.5 x 1

= 6.5 m2

For the trapezium marked as III:

Draw AF||DC and AE ⊥ BC

Therefore, AD = FC = DC = AF = 1 cm

⸫BF = BC – FC = (2 – 1) cm = 1 cm

Hence, ∆ABF is equilateral

Also, E is the midpoint of BF

⸫ BE = 1/2 cm = 0.5 cm

Also, AB2 = AE2 + BE2

AE2 = 12 – (0.5)2 = 0.75

AE ≅ 0.9 cm

Area of trapezium = 1/2 (sum of the parallel sides) x (distance between them)

= 1/2 x(BC + AD) X AE

= 1/2 X(2 + 1) X 0.9 cm2

= 1.4 cm2

For triangle marked as IV and V:

[Triangle IV and V are congruent to one another as these are the wings of the aeroplane which needs to same for proper figure]

These are right angled triangles, therefore, by Pythagoras theorem, we have,

Area of triangle = 1/2 x base x height

= 1/2 x 6 x 1.5

= 3 x 1.5

= 4.5 cm2

Area of the triangle IV and V are 4.5cm2 and 4.5cm2.

Total area of the paper used = area of I + area of II + area of III + area of IV + area of V

= (2.5 + 6.5 + 1.4 + 4.5 + 4.5) cm2

= 19.4 cm2


  1. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

Solution:

HErons formula exercise 12.2 class IXLet ABCD is a parallelogram and CDE is a triangle with same base CD = 28 cm.

Given area of the triangle = area of the parallelogram.

In triangle ABC,

a = 26 cm , b = 28 cm, c = 30 cm

s = 26+28+30/2 = 84/2 = 42

Area of the triangle = √[s(s-a)(s-b)(s-c)]

= √[42(42-26)(42-28)(42 – 30)

= √[42x16x14x12]

= √112896

=336m2

Since, area of the triangle = area of the parallelogram

Area of parallelogram = base x height

336 = 28 x height

height = 336/28 = 12 cm

Therefore, height of the parallelogram = 12 cm


  1. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

Solution:

HErons formula - exercise 12.2. class IXGiven: . If each side of the rhombus is 30 m and its longer diagonal is 48 m.

In rhombus ABCD, we have 2 equal triangles i.e., ∆ABC and ∆ADC

Therefore, area of ∆ABC = area of ∆ADC ————(1)

area of the rhombus = area of ∆ABC + area of  ∆ADC

= 2 x area of ∆ABC [from (1)]

In ∆ABC and ∆ADC,

a = b = 30 m ; c = 48 m

s = a+b+c/2 = 30+30+48/2 = 108/2 = 54 m

area of ∆ABC = √[s(s-a)(s-b)(s-c)]

=√[54(54-30)(54-30)(54-48)]

=√[54*24*24*6]

=√1,86,624

=432 m2

Therefore, area of rhombus = 2 x area of ∆ABC

= 2 x 432

= 864 m2

Area of the grass field each cow gets = area of rhombus/no. of cows

= 864/18

= 48 m2


  1. Herons Formula Exercise 12.2. class IXAn umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig. 12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Solution:

Given: An umbrella is made by stitching 10 triangular pieces of cloth of two different colours, each piece measuring 20 cm, 50 cm and 50 cm.

Sides of each triangle are 20cm, 50cm and 50cm i.e., a = 20 cm, b = 50 cm, c = 50 cm

s = 20+50+50/2 = 120/2 = 60

Area of each triangle = √[s(s-a)(s-b)(s-c)]

= √[60(60-20)(60-50)(60-50)]

= √[60*40*10*10]

= 10√[20x3x20x2]

=200 √6 m2

Therefore, area of 10 triangular pieces = 200 √6 x 10 = 2000 √6 m2

Hence, cloth required for each color = 2000 √6/2 = 1000 √6 m2


  1. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much paper of each shade has been used in it?A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much paper of each shade has been used in it?

 

 

Solution:

Area of a kite = diagonal^2/2 = 32^2/2 = 1024/2 = 512 cm2

Area of  shades in a kite are equal . Therefore area of two equal shades of a kite = area of  a kite/2 = 512/2 = 256 cm2

Given an isosceles triangle with base 8 cm and 6cm. Therefore, a = 8 cm, b = 6 cm and c = 6 cm

s = a+b+c/2= 8+6+6/2 = 20/2 = 10 cm

Area of an equilateral triangle = √[s(s-a)(s-b)(s-c)

= √[10(10-8)(10-6)(10-6)

= √[10*2*4*4]

= √[2*5*2*4*4]

= 8√5 cm2

Therefore, paper needed for shade I = 256cm2, paper needed for shade II = 256cm2, paper needed for shade III = 8√5 cm2


  1. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm2.A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm2.

Solution:

Given: A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm.

Therefore, in each triangle a = 9 cm , b = 28 cm and c = 35 cm

s = a+b+c/2 = 9+28+35/2 = 36 cm

Area of each triangle = √[s(s-a)(s-b)(s-c)

= √[36(36-9)(36-28)(36-35)]

= √[36*27*8*1]

= √7776

= 88.18 cm2

Area of 16 such tiles = 88.18 x 16 = 1410.88 cm2

Given: the cost of polishing the tiles at the rate of 50p per cm2 i.e., the cost of polishing the tiles for 1 cm2 is 50p

Then, the cost of polishing the tiles for 1410.88 cm2 will be = 1410.88 x .50 = Rs. 705.44


  1. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Solution:

A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm2.Let ABCD be trapezium.

Draw a line segment from B parallel to AD which intersects DC at F and BE parallel to DC.

In ∆BFC,

a = 13m , b = 14m , c = 15m

s = 13+14+15/2 = 42/2 = 21 m

area of triangle BFC = √[s(s-a)(s-b)(s-c)

=√[21(21-13)(21-14)(21-15)]

=√[21*8*7*6]

=√7056

= 84 m2

In ∆BFC,

Area of ∆BFC = 1/2 x base x height

84 =  1/2 X 15 x BE

84 x 2/15 = BE

BE = 11.2 m

Area of trapezium = 1/2 (sum of parallel sides)x(distance between them)

= 1/2 x (25 + 10) x (11.2)

= 196 m2

Hence area  if the field = 196 m2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Heron’s Formula – Exercise 12.1 – Class IX

  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?

Solution:

Each side of a triangle = a

Perimeter of a triangle = 3a

s = 3a/2

Area of the signal traingel signal board = √[s(s-a)(s-b)(s-c)] [sign board is an equilateral triangle a = b = c ]

= √[s(s-a)(s-a)(s-a)

= (s – a) √[s(s-a)]

= (3a/2 – a) √[3a/2(3a/2 – a)]

= a/2 √(3a^2/4)

=a/2 . a/2 .√3

= a^2/4 . √3

Thus, the area of the signal sign board = a^2/4. √3 sq. units

Therefore, perimeter = 180cm (given)

⇒each side of the triangle = 180/3 = 60 cm

⇒ area of the triangle = (60)^2/4 x √3 cm2 = 900√3 cm2


  1. 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 Rs 5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?Heron's Formula - Exercise 12.1 - Class IX

Solution:

Given sides of the triangular walls are 122m, 22m , 120m i.e., a = 122m , b = 22m , c = 120m

Therefore, s = 3a/2 = 122+22+120/2 = 132 m

Area of the triangular side wall = √[s(s-a)(s-b)(s-c)

= √[132(132 – 122)(132 – 22)(132 – 120)] m2

= √[132(10)(110)(12) m2

= 1320 m2

Given: The advertisements yield an earning of Rs 5000 per m2 per year. i.e., the advertisement earning of 1m2 of the wall for one year is Rs. 5000

⸫ advertisement earning of 1m2 wall for 1month = Rs. 5000/12

⸫ advertisement earning of 1m2 of the wall for 3 months = Rs. 5000/12 x 3

⸫ advertisement earning of complete wall for 3 months = Rs. 5000/12 x 3 x 1320 = Rs. 15, 50, 000


  1. 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. Fig. 12.10Heron's Formula - Exercise 12.1 - Class IX

Solution:

The sides of the walls are 15m , 11m and 6m

Therefore, s = 15+11+6/2 = 32/2 = 16m

Area of the triangular walls = √[s(s-a)(s-b)(s-c)] [a= 15m, b = 11m, c = 6m]

=√[16(16-15)(16-11)(16-6)]

=√[16x1x5x10]

=4×5√2

= 20√2


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

Solution:

Perimeter = 42m

Two sides of the triangle are 18cm and 10 cm , i.e., a = 18cm, b = 10cm

Perimeter = a + b + c

42 = 18 + 10 + c

c = 42 – 18 – 10 = 14m

Therefore, s = a+b+c/2 = 18+10+14/2 = 42/2 = 21cm

Area of the triangle = √[s(s-a)(s-b)(s-c)

= √[21(21-18)(21-10)(21-14)] cm2

=√[21x3x11x7] cm2

=√[7x3x3x11x7] cm2

= 21√11cm2


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

Solution:

Sides of the triangle are in the ratio 12:17:25 i.e., a = 12x , b = 17x, c = 25x

Perimeter = 540cm = 12x + 17x + 25x

54x = 540

x = 540/54 = 10cm

Thus, a = 120 , b = 170 and c = 250

s =a+b+c/2 = 540/2 = 270 cm

Area of the triangle = √[s(s-a)(s-b)(s-c)

= √[270(270-120)(270-170)(270-250)

=√[270x150x100x20]

= √30x9x30x5x10x10x5x4]

=30x3x5x10x2

= 9000cm2


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

Solution:

Perimeter of an isosceles triangle = 30 cm

Perimeter = a+b+c = 30cm

Each of the equal side is 12cm i.e., a = b = 12 cm

12+12+c = 30

12+12+c  = 60

24 + c = 30

c = 30 – 24 = 6 cm

s = 30/2 = 15 cm

Area of the triangle = √[s(s-a)(s-b)(s-c)

=√[15(15-12)(15-12)(15-6)]

=√[15x3x3x9]

=√[3x5x9x9]

=9√15 cm2


 

Construction – Exercise 11.2 – Class IX

  1. Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13 cm.

Solution:

  • Construction - Exercise 11.2 – Class IXDraw a line segment BC = 7 cm
  • Make an ∠B = 75° i.e., ∠CBX = 75°
  • Cut a line  segment BX at D such that BD = AB + AC = 13 cm
  • Join DC
  • Make an ∠DCY = ∠BDC
  • Let CY intersect BD at A .
  • Then the required triangle is ABC

 


  1. Construct a triangle ABC in which BC = 8cm, ∠B = 45° and AB – AC = 3.5 cm.

Solution:

  • Construction - Exercise 11.2 – Class  IXDraw a line segment BC = 8c
  • Make an angle ∠B = 45° i.e., ∠CBX = 45°
  • Cut the line segment BD = 3.5 cm from the ray BX
  • Join DC
  • Draw perpendicular bisector MN of DC and intersect it to BX at A
  • Join AC
  • Then ABC is the required triangle.

  1. Construct a triangle PQR in which QR = 6cm, ∠Q = 60° and PR – PQ = 2cm.

Construction - Exercise 11.2 – Class IX

Solution:

  • Draw a line segment QR = 6cm
  • Make an angle ∠Q = 60° i.e., ∠RQX = 60°
  • Cut the line segment QX such that, QS = 2cm = PR – PQ which is extended in opposite side
  • Join SR . Draw perpendicular bisector of MN of SR which intersects QZ at P
  • Join RP then PQR is the required triangle.

 

 

 


  1. Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.

Solution:

  • Let AB = XY + ZX + XY = 11 cm
  • Then ∠BAP = 30° and ∠ABR = 90°
  • Bisect ∠LBC and ∠MCB . Let these bisector meet at X
  • Draw perpendicular bisectors DE and FG to XB and XC respectively.
  • Let DE intersect BC and FC intersect BC at Y and Z respectively.
  • Join XY and XZ . Then XYZ is the required triangle.

Construction - Exercise 11.2 – Class IX


  1. Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18 cm.

Solution:

  • Construction - Exercise 11.2 – Class IXLet the base of an right angled triangle is BC = 12 cm
  • At B make an angle XBC = 90°
  • Cut the line segment BX at D i.e., at 18cm = AB+AC
  • Draw a perpendicular bisector MN to DC which cut the line segment BX at A
  • Join AC
  • Therefore triangle ABC is the required right angled triangle.

 

 

 


 

 

Construction – Exercise 11.1 – Class IX

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

Solution:

Construction - Exercise 11.1 - Class IX

Construction:

  • Draw a line ray AB with initial point A
  • Taking A as a centre draw an arc XY with any radius intersecting the ray AB at P.
  • Draw another arc with same radius by taking C as centre which intersects the arc XY at Q. Draw a ray joining AQ.
  • Taking Q as centre with same radius draw another arc which intersects the arc XY at R. Draw a ray joining AR.
  • Now, taking Q and R as centres with any radius draw two arc intersecting one another at S.
  • Join AS. Then angle SAB = 90˚

Justification:

By construction, we have AP = AQ = PQ, therefore, triangle APQ is an equilateral triangle. Thus, ∠APQ = ∠AQP = ∠QAP = 60˚

Also, we have, AQ = AR =RQ , therefore, triangle ARQ is also an equilateral triangle.  Thus, ∠ARQ = ∠AQR = ∠QAR = 60˚

Since, AS bisects ∠QAR  then we have ∠QAS = ∠SAR = 1/2∠QAR = 1/2 x 60˚  = 30˚

Thus, ∠PAS = ∠PAQ + ∠QAS = 60˚  + 30˚  = 90˚


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

Solution:

Construction:

  • Construction - Exercise 11.1 - Class IXDraw a line ray AB with initial point A
  • Taking A as a centre draw an arc XY with any radius intersecting the ray AB at P.
  • Draw another arc with same radius by taking C as centre which intersects the arc XY at Q. Draw a ray joining AQ.
  • Taking Q as centre with same radius draw another arc which intersects the arc XY at R. Draw a ray joining AR.
  • Now, taking Q and R as centres with any radius draw two arc intersecting one another at S.
  • Join AS. Then angle SAB = 90˚
  • Again taking S and P as centres with any radius draw two arcs intersecting each other at L.
  • Join AL . Therefore, ∠PAL = 45 ˚

 

Justification:

Join GH and CH

In ∆AHG and ∆AHC,  we have [arcs of equal radii]

HG = HC [radii of the same arc]

AG = AC [common]

⸫∆AHG ≅ ∆AHC [SSS congruence]
⇒∠HAG = ∠HAC [CPCT] ……………(i)

But ∠HAG + ∠HAC = 90˚ [By construction] ……………….(ii)

⇒∠HAG = ∠HAC = 45˚ [from (i) and (ii)]


  1. Construct the angles of the following measurements:

(i) 30° (ii) 22 1/2° (iii) 15°

Solution:

(i) 30° = 60 ˚/2

Construction:

  • Construction - Exercise 11.1 - Class IXDraw a ray AB with initial point A
  • Draw an arc XY with any radius by taking A as centre which cuts the ray AB at C.
  • Draw an arc by taking C as centre with same radius which intersects arc XY at D. Since AC = AD = DC then, ∆ADC is an equilateral triangle.
  • ∠ADC = ∠ACD = ∠DAC = 60°
  • Taking C and D as centres with any radius draw two arcs intersecting one another at E. Join AE.
  • Thus, 1/2∠EAC = ∠DAC ; 1/2∠EAC = 60˚ ⇒∠EAC = 30˚

 

(ii)221/2˚ = 90˚/2 = 45˚

Construction:

  • Construction - Exercise 11.1 - Class IXDraw a line ray AB with initial point A
  • Taking A as a centre draw an arc XY with any radius intersecting the ray AB at P.
  • Draw another arc with same radius by taking C as centre which intersects the arc XY at Q. Draw a ray joining AQ.
  • Taking Q as centre with same radius draw another arc which intersects the arc XY at R. Draw a ray joining AR.
  • Now, taking Q and R as centres with any radius draw two arc intersecting one another at S.
  • Join AS. Then angle SAB = 90˚
  • Taking S and P as centres draw two arcs intersecting each other at T. Join AT. i.e., ∠TAP = 1/2 ∠SAP

∠TAP = 1/2 x 90 ˚ = 45 ˚ = 221/2˚

 

(iii) 15°

Construction:

  • Construction - Exercise 11.1 - Class IXDraw a ray AB with initial point A
  • Draw an arc XY with any radius by taking A as centre which cuts the ray AB at C.
  • Draw an arc by taking C as centre with same radius which intersects arc XY at D. Since AC = AD = DC then, ∆ADC is an equilateral triangle.
  • ∠ADC = ∠ACD = ∠DAC = 60°
  • Taking C and D as centres with any radius draw two arcs intersecting one another at E. Join AE.
  • Thus, 1/2∠EAC = ∠DAC ; 1/2∠EAC = 60˚ ⇒∠EAC = 30˚
  • Taking E and C as centres draw two arcs with any radius intersecting each other at F. Join AF
  • Thus, 1/2∠FAC = ∠EAC ; 1/2∠FAC = 30˚ ⇒∠FAC = 15˚

  1. Construct the following angles and verify by measuring them by a protractor:

(i) 75° (ii) 105° (iii) 135°

Solution:

(i) 75°

Construction:

  • Construction - Exercise 11.1 - Class IXDraw a line ray AB with initial point A
  • Taking A as a centre draw an arc XY with any radius intersecting the ray AB at P.
  • Draw another arc with same radius by taking C as centre which intersects the arc XY at Q. Draw a ray joining AQ.
  • Since AQ = AP = PQ, ∆APQ is an equilateral triangle. Therefore, ∠PAQ = 60 ˚
  • Taking Q as centre with same radius draw another arc which intersects the arc XY at R. Draw a ray joining AR.
  • Now, taking Q and R as centres with any radius draw two arc intersecting one another at S.
  • Join AS. Then angle SAB = 90˚
  • ∠SAB = ∠PAQ + ∠QAS ⇒90 ˚ = 60 ˚+ ∠QAS ⇒∠QAS = 30˚
  • Taking S and Q as centres draw two arcs intersecting each other at M.
  • Join Then ∠MAQ = 1/2∠QAS = 1/2x 30˚ = 15˚
  • Therefore, ∠PAM = ∠MAQ + ∠PAQ = 60˚ + 15˚  = 75˚

 

(ii) 105°

Construction:

  • Construction - Exercise 11.1 - Class IXDraw a line ray AB with initial point A
  • Taking A as a centre draw an arc XY with any radius intersecting the ray AB at P.
  • Draw another arc with same radius by taking C as centre which intersects the arc XY at Q. Draw a ray joining AQ.
  • Since AQ = AP = PQ, ∆APQ is an equilateral triangle. Therefore, ∠PAQ = 60 ˚
  • Taking Q as centre with same radius draw another arc which intersects the arc XY at R. Draw a ray joining AR.
  • Now, taking Q and R as centres with any radius draw two arc intersecting one another at S.
  • Join AS. Then angle SAB = 90˚
  • Taking S and R as centre with any radius draw two arcs intersecting each other at T.
  • Join AT. Then, ∠SAT = 1/2 ∠SAR = 1/2 x 30 ˚ = 15 ˚
  • Thus, ∠PAT = ∠PAS + ∠SAT = 90 ˚ + 15 ˚  = 105 ˚

 

(iii) 135°

Construction:

  • Construction - Exercise 11.1 - Class IXDraw a line ray AB with a point O.
  • Taking O as a centre draw an arc XY with any radius intersecting the ray AB at C.
  • Draw another arc with same radius by taking C as centre which intersects the arc XY at D.
  • Taking D as centre with same radius draw another arc which intersects the arc XY at E. Draw a ray joining OF.
  • Now, taking D and E as centres with any radius draw two arc intersecting one another at F.
  • Join OF. Then angle ∟FOC = 90˚
  • Taking O as radius draw an arc interesting the ray AB at G. ∟FOG = 90˚
  • Now, taking G and F as centres with any radius draw two arc intersecting one another at H.
  • Join OH. ∟HOF = 1/2∟FOG =1/2 x 90˚ = 45˚
  • ∠HOC = ∠FOC + ∠HOF = 90˚ + 45˚ = 135˚

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

Solution:

Construction - Exercise 11.1 - Class IX

Steps of Construction:

  • Draw a line segment AB  of the given length
  • Taking A and B as centres with radius same as that of the length of AB draw two arcs intersecting each other at C
  • Join AC and BC.
  • Then triangle ABC is an equilateral triangle.

Justification:

  • By construction AB= BC = AC
  • Then triangle ABC is an equilateral triangle.

Circles – Exercise 10.5 – Class IX

  1. Circles – Exercise 10.5  – Class IXIn Fig. 10.36, A,B and C are three points on a circle with centre O such that ∠ BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.

Solution:

Since ABC is an arc of the circle with centre O, which makes an angle AOC = angle AOB + angle BOC

i.e., ∠AOC = ∠AOB + ∠BO = 60° + 30° =  90°

We know ∠AOC = 2 ∠ADC [angle at the centre of a circle is twice the angle at the circumference]

 Therefore ∠ADC = 1/2 (∠AOC ) = 1/2 x 90° = 45°


  1. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Solution:

Circles – Exercise 10.5  – Class IXSince OA = OB = AB

Therefore, ∆AOB is an equilateral triangle.

⇒∠AOB = 60°

We know that the angle at the centre of a circle is twice the angle at the circumference.

Therefore, ∠AOB  = 2 ∠ACB

⇒∠ACB = 1/2 ∠AOB = 1/2 x 60° = 30°

Also, we have , ∠ADB = 1/2reflex ∠AOB = 1/2(360° – 60°) = 1/2 (300°) = 150°

Hence, the angle subtended by the chord at a point on the minor arc and also at a point on the major arc is 30 °


  1. Circles – Exercise 10.5  – Class IXIn Fig. 10.37, ∠ PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠ OPR.

Solution:

Since, angle at the centre of a circle is twice the angle at the circumference.

Therefore, ∠POR = 2 ∠PQR

⇒reflex ∠POR = 2 x 100 ° = 200 °

⇒ ∠POR = 360 ° – 200 ° = 160 °

In ∆OPR, OP = OR [Radii of the same circle]

⇒ ∠OPR = ∠ORP [Angles opposite to equal sides are equal]

and ∠POR = 160 ° [which is proved]

In ∆OPR ,

∠OPR + ∠ORP + ∠PQR = 180 ° [Sum of interior angles of a triangle are equal to 180 °]

⇒160 ° + 2 ∠OPR = 180 °

⇒ 2 ∠OPR = 180 ° – 160 ° = 20 °

⇒ ∠OPR = 10 °


  1. Circles – Exercise 10.5  – Class IXIn Fig. 10.38, ∠ ABC = 69°, ∠ ACB = 31°, find ∠ BDC.

Solution:

In ∆ABC,

∠BAC + ∠ABC + ∠BCA = 180 °

⇒ ∠BAC + 69 ° + 31 ° = 180 °

⇒ 180 ° – 100 ° = 80 °

Since the angles in the same segment are equal.

Therefore, ∠BDC = ∠BAC = 80 °


  1. Circles – Exercise 10.5  – Class IXIn Fig. 10.39, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.

Solution:

Given A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°

We have find ∠BAC

Since, ∠CED + ∠CEB = ∠ BED = 180 °  [linear pair]

⇒∠CED + 130 ° = 180 °

⇒∠CED = 180 ° – 130 ° = 50 °

In ∆ECD, we have,

∠EDC + ∠EDC + ∠CED = 180 °

⇒ ∠EDC + 50 ° + 20 ° = 180 °

⇒∠EDC = 180 ° – 50 ° – 20 °  = 110 °

⇒ ∠BDC = ∠EDC = 110 ° [angles in the same segment are equal]

Therefore, ∠BAC = ∠BDC = 110 °


  1. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠ DBC = 70°, ∠ BAC is 30°, find ∠ BCD. Further, if AB = BC, find ∠ ECD.

Solution:

Circles – Exercise 10.5  – Class IXGiven, ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠ DBC = 70°,  ∠ BAC is 30°.

∠CAD = ∠DBC= 70° [Angles in the same segment]

Therefore, ∠DAB = ∠CAD + ∠BAC = 70° + 30° = 100°

But, ∠DAB + ∠BCD = 180° [Opposite angles of a cyclic quadrilateral]

So, ∠BCD = 180° – 100° = 80°

Now, we have AB = BC

Therefore, ∠BCA = 30° [Opposite angles of an isosceles triangle]

Again, ∠DAB + ∠BCD = 180° [Opposite angles of a cyclic quadrilateral]

⇒ 100° + ∠BCA + ∠ECD = 180° [∵∠BCD = ∠BCA + ∠ECD]

⇒ 100° + 30° + ∠ECD = 180°

⇒ 130° + ∠ECD = 180°

⇒ ∠ECD = 180° – 130° = 50°

Hence, ∠BCD = 80° and ∠ECD = 50°


  1. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Solution:

Circles – Exercise 10.5  – Class IXDiagonals AC and BD of a cyclic quadrilateral are diameters of the circle through the vertices A, B, C and D of the quadrilateral ABCD.

We have to prove quadrilateral ABCD is a rectangle.

We know, all the radius of the same circle are equal.

Therefore, OA = OB = OC = OD

⇒OA = OC = 1/2 AC

⇒OB = OC = 1/2 BD

⇒AC = BD

Thus, the diagonals of the quadrilateral ABCD are equal and bisect each other.

Quadrilateral ABCD is rectangle.


  1. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Solution:

Circles – Exercise 10.5 – Class IX

Given : A trapezium ABCD in which AB || CD and AD = BC.

To Prove : ABCD is a cyclic trapezium.

Construction : Draw DE ⊥ AB and CF ⊥ AB.

Proof : In ∆DEA and ∆CFB, we have

AD = BC [Given]

∠DEA = ∠CFB = 90° [DE ⊥ AB and CF ⊥ AB]

DE = CF [Distance between parallel lines remains constant]

∴ ∆DEA ≅ ∆CFB [RHS axiom]

⇒ ∠A = ∠B …(i) [CPCT]

and, ∠ADE = ∠BCF ..(ii) [CPCT]

Since, ∠ADE = ∠BCF [From (ii)]

⇒ ∠ADE + 90° = ∠BCF + 90°

⇒ ∠ADE + ∠CDE = ∠BCF + ∠DCF

⇒ ∠D = ∠C ……………………………….(iii)

[∠ADE + ∠CDE = ∠D, ∠BCF + ∠DCF = ∠C]

∴ ∠A = ∠B and ∠C = ∠D [From (i) and (iii)] (iv)

∠A + ∠B + ∠C + ∠D = 360° [Sum of the angles of a quadrilateral is 360°]

⇒ 2(∠B + ∠D) = 360° [Using (iv)]

⇒ ∠B + ∠D = 180°

⇒ Sum of a pair of opposite angles of quadrilateral ABCD is 180°.

⇒ ABCD is a cyclic trapezium


  1. Circles – Exercise 10.5  – Class IXTwo circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 10.40). Prove that ∠ ACP = ∠ QCD.

Solution:

Since angles in the same segment are equal.

Therefore, ∠ACP = ∠ABP ……………….(1)

and ∠QCD = ∠QBD ……………………(2)

Also ∠ABP = ∠QBD ……………………..(3) [vertically opposite angles]

Therefore, from (1), (2) and (3), we have

∠ACP = ∠QCD


  1. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Solution:

Circles – Exercise 10.5  – Class IXGiven: Two circles are drawn with sides AB and AC of ∆ABC as diameters . The circle intersect at D

To prove: D lies on BC

Construction:

Join A and D

Proof: since AB and AC are the diameters of the two circles[given]

Therefore, ∠ADB = 90° [angles are in a semi circle]

and ∠ADC = 90° [angles in a semi circle]

On adding, we get,

⇒∠ADB + ∠ADC = 90 + 90 = 180

⇒BDC is a straight line

Hence D lies on BC


  1. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠ CAD = ∠ CBD.

Solution:

Circles – Exercise 10.5 – Class IX∆ABC and ∆ADC are right angles with common hypotenuse AC. Draw a circle AC as diameter passing through B and D . Join BD

Clearly, ∠CAD = ∠CBD [since angles in the same segment are equal]

 

 


  1. Prove that a cyclic parallelogram is a rectangle.

Solution:

Circles – Exercise 10.5  – Class IXGiven: ABCD is a parallelogram inscribed in a circle

To prove: ABCD is a rectangle

Proof: Since ABCD is a cyclic parallelogram

Therefore, ∠A + ∠C = 180° ………………..(1)

But ∠A = ∠C ………………..(2)

From (1) and (2), we have

∠A = ∠C = 90°

Similarly, ∠B = ∠D = 90°

Therefore, each angle of ABCD is of 90°

Hence, ABCD is a rectangle.


 

Circles – Exercise 10.4 – Class IX

  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.

Solution:

Circles – Exercise 10.4 – Class IX

In ∆AOO’,

AO2 = 52 = 25

AO’2 = 32 = 9

OO’2 = 42 = 16

AO’2 + OO’2 = 9 + 16 = 25 = AO2

⇒∠AO’O = 90˚ [By converse of Pythagoras theorem]

Similarly, ∠BO’O = 90˚

∠AO’B = 90˚ + 90˚ = 180˚

Therefore, ∠AO’B is a straight line whose midpoint is O’.

⇒ AB = (3 + 3)cm = 6 cm


  1. 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.

Solution:

Circles – Exercise 10.4 – Class IXGiven: Let AB and CD be the two equal chords intersecting at point X.

⇒AB = CD

To prove:

Corresponding segments are equal i.e., AX = DX

and BX = CX

Proof: We draw OM ⊥AB and ON ⊥CD

So, AM = BM = 1/2 AB

and DN = CN = 1/2 CD

We know, AB = CD

1/2 AB  = 1/2 CD

⸫AM = DN ………………………..….(1)

and MB = CN …………………………(2)

In ∆OMX and ∆ONX,

∠OMX = ∠ONX = 90˚

OX = OX (common)

OM = ON [AB and Cd are equal chords and equal chords are equidistant from the centre)

⸫∆OMX ≅ ∆ONX (RHS congruence rule)

⸫MX = NX (CPCT) …………….(3)

On adding (1) and (3) we get,

AM + MX = DN + NX

AX = DX

On subtracting (2) and (3), we get,

BM – MX = CN – MX

BX = CX

Therefore, AX = DX and  BX = CX


  1. 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.

Solution:

Circles – Exercise 10.4 – Class IX

Given AB and CD are two equal chords of a circle which meets at E within the centre and a line PQ joining the point of intersection to the centre

To prove:

∠AEQ = ∠DEQ

Construction: Draw OL⊥AB and OM⊥CD

Proof:

In ∆OLE and ∆OME, we have,

OL = OM (equal chords are equidistant]

OE = OE [common]

∠OLE = ∠OME = 90˚

⇒∠LEO = ∠MEO (CPCT)


  1. Circles – Exercise 10.4 – Class IXIf 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).

Solution:

Given: Two concentric circles with centre O and a line intersects the circles at A, B, C and D

To prove:

AB = CD

Proof:

Circles – Exercise 10.4 – Class IXLet two circles be C1 and C2 and line be l.

We draw Op perpendicular to line l.

In circle C1,

OP ⊥BC (As OP is perpendicular to line l)

So, OP bisects BC

i.e., BP = CP …………(1) (perpendicular drawn from centre of a circle to a chord bisects the chord)

In circle C2,

OP ⊥ AD (As Op is perpendicular to line l)

So, OP bisects AD

i.e., AP = DP …………(2) (perpendicular drawn from centre of a circle to a chord bisects the chord)

On subtracting (2) and (1)

AP – BP = DP – CP

⇒ AB = CD


  1. 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?

Solution:

Circles – Exercise 10.4 – Class IXLet Reshma, Salma and Mandip be represented by R, S and M respectively.

Draw OL ⊥ RS,

OL2 = OR2 – RL2

OL2 = 52 – 32 [RL = 3 m, because OL ⊥ RS]

= 25 – 9 = 16

OL = √(16) = 4

Now, area of triangle ORS = 1/2 × KR × 05

= 1/2 × KR × 05

Also, area of ∆ORS = 1/2 × RS × OL = 1/2 × 6 × 4 = 12 m2

1/2 × KR × 5 = 12

⇒ KR = 12×2/5 = 24/5 = 4.8 m

⇒ RM = 2KR

⇒ RM = 2 × 4.8 = 9.6 m

Hence, distance between Reshma and Mandip is 9.6 m


  1. 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.

Solution:

Circles – Exercise 10.4 – Class IX

Let Ankur, Syed and David be represented by A, S and D respectively (as shown on the figure)

Let PD = SP = SQ = QA = AR = RD = x

In ∆OPD,

OP2 = 400 – x2

⇒ OP = √(400 – x2)

⇒ AP = 2√(400 – x2) + √(400 – x2) [∵ AP divides the median in the ratio 2 : 1]

= 3√(400 – x2)

In ∆APD,

PD2 = AD2 – DP2

⇒ x2 = (2x)2 – (3√(400 – x2)2

⇒ x2 = 4x2 – 9(400 – x2)

⇒ x2 = 4x2 – 3600 + 9x2

⇒ 12x2 = 3600

⇒ x2 = 3600/12 = 300

⇒ x = 10√3

Thus, SD = 2x = 2 × 10√3 = 20√3

∴ ASD is an equilateral triangle.

⇒ SD = AS = AD = 20√3

Hence, length of the string of each phone is 20√3 m