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Question-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:
Given, the triangular trafic signal Board is an equilateral triangle,therefore the sides ‘a’ = ‘b’ = ‘c’ = a

⇒  s =  =

Area of the signal board

 

   =

Perimeter = 180 cm
‘a’ + ‘b’ + ‘c’ = 180 a + a + a = 180 3a = 180
a = a = 60 cm
Area of the signal board =

                                   = = 900cm2
Alternatively,
s = = (60) = 90cm

Area of the signal board      

                                     =

                                     =

                                     = 900cm2

Question-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.). 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? 


Solution:


a = 122 m

b = 22 m
c = 120 m
⇒  s = = m

      = = 132 m

Area of the wall =

                          =

                         

                          =

                          = (12)(11)(10)m2 = 1320 m2
1 year = 12 months
 Rent for 12 months per m2 `5000

Rent for 1 month per m2 `

Rent for 3 months per m2 `× 3 =  `1250

Rent for 3 months of 1320 m2 `(1250 × 1320) = 1650000.

Question-3

There is a slide in a park. One of its sidewalls has been painted in some colour with a message " keep the green and clean" (SEE FIGURE). If the sides of the wall are 15 m, 11 m and 6 m. Find the area painted in colour.
 
Solution:


a = 15 m

b = 11 m
c = 6 m
s =

     = m = 16 m

Area painted in colour

     =  

     =

     = 20m2.

Question-4

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

Solution:
 
a = 18 cm

b = 10 cm

Perimeter = 42 cm

a + b + c = 42 18 + 10 + c = 42
28 + c = 42 c = 42 – 28
c = 14 cm
s = = 21 cm

Area of the triangle =

=

= =

= (7)(3) = 21cm2.

Question-5

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

Solution:
Given, the ratio of the triangle = 12: 17 : 25
∴ length of the sides = 12k, 17k, 25k
Then,
Perimeter = 12k + 17k + 25 k = 54 k
According to the question,


54k = 540

k = k = 10

a = 12k = 12 × 10 = 120 cm
b = 17k = 17 × 10 = 170 cm
c = 25k = 25 × 10 = 250 cm

s =


= = = 270 cm

Area =


          =

          = =

          = (3)(30)(5)(20) cm2 = 9000 cm2.

Question-6

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

Solution:
 
a = 12 cm, 
b = 12 cm
Perimeter = 30 cm
a + b + c = 30 12 + 12 + c = 30
24 + c = 30 c = 30 – 24
c = 6 cm
s = cm = 15 cm
                                                                                               
Area of the triangle =  

                              =  

                              = = 9 cm2.

Question-7

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

Solution:
 
Given, ABCD is a quadrilateral,  = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m 
Join BD
∴ Area of right triangle BCD =
                                     =   = 30 m2
In right triangle BCD,
BD2 = BC2 + CD2
      = (12)2+(5)2 = 144 + 25 = 169

BD =

= (12)2 + (5)2 = 144 + 25 = 169


BD = = 13 m.

For Δ ABD
a = 13 m
b = 8 m
c = 9 m
s =
= = = 15 m

Area of the Δ ABD =

                            =

                            = =

                            = 3 × 2 = 6m2

                            = 6 × 5.916 = 35.5 m2 (approx.)

Area of the quadrilateral ABCD = Area of Δ BCD + Area of Δ ABD
                                             = 30 m2 + 35.5 m2
                                             = 65.5 m2(approx.)

Hence the park occupies the area 65.5 m2. (approx.)

Question-8

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:
 
For Δ ABC
AB(a) = 3 cm, BC(b) = 5 cm, AC(c) = 3 cm
 a2 + c2 = b2
Δ ABC is right angled with B = 90°.

Area of right triangle ABC
=
                                      = = 6 cm2
For Δ ACD
a = 4 cm, b = 5 cm, c = 5 cm
s =

     = = = 7 cm

Area of the Δ ACD =

                            =

                            = = 2cm2

                            = 2 × 4.6 cm2(approx.) = 9.2 cm2(approx.)

Area of the quadrilateral ABCD = Area of Δ BCD + Area of Δ ACD

                                            = 6 cm2 + 9.2 cm2 = 15.2 cm2(approx.)

Question-9

Radha made a picture of an aeroplane with coloured paper as shown in Figure. Find the total area of the paper used.
 

Solution:

For Triangular Area I
a = 5 cm, b = 5 cm, c = 1 cm
s =

      = = = 5.5 cm


Area I =

            =

            = = (.5)


            = 


            = 0.75 = 0.75 (3.3) (approx.)

            = 2.5 cm2 (approx.)

Area II = 6.5 × 1 = 6.5 cm2

For Area III
Area III = =

= =

= = 1.3 cm2(approx.)


Area IV = = 4.5 cm2

Area V = = 4.5 cm2
Total area of the paper used = Area I + Area II + Area III + Area IV + Area V

                                          = 2.5 cm2 +6.5 cm2 + 1.3 cm2 + 4.5 cm2 + 4.5 cm2
                                          = 19.3 cm2.

Question-10

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:

a = 26 cm, b = 28 cm, c = 30 cm
s =
      = =
      = 42 cm
Area of the triangle =

                              =

                              =

                              =

                              = 6 × 4 × 7 × 2 = 336 cm2

Let the height of the parallelogram be h cm.
Then, area of the parallelogram = Base × Height
                                           = 28 × h cm2
According to the question,
28h = 336 h = h = 12 cm
Hence the height of the parallelogram is 12 cm.

Question-11

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:


 
For Triangular Area I
a = 30 cm
b = 48 cm
c = 30 cm
s =
      = = = 54 cm

Area of Triangle I =

                           =

                           =

                           =

                           = 3 × 6 × 24 = 432
m2
Area of the rhombus = 2 Area of Δ ABC
                               = 2 × 432 = 864 m2
Area of grass for 18 cows = 864 m2

Area of grass for 1 cow = = 48 m2

Aliter: Draw BE AC. Then E is the mid-point of AC.

AE = CE = AC = (48) = 24 m.


In right triangle AEB,
AB2 = AE2 + BE2                | By Pythagoras Theorem
(30)2 = (24)2 + BE2
900 = 576 + BE2

BE2 = 900 – 576

BE2 = 324
BE = = 18 m
BD = 2 BE = 2 × 18 = 36 m
Area of rhombus ABCD
= product of diagonals
= = 864 m2  
Area of grass of 18 cows = 864 m2
Area of grass for 1 cow = = 48 m2
Area of right Δ AEB =
                          = = 216 m2
Area of rhombus ABCD = 4 Area of Δ AEB = 4 × 216 = 864 m2.

Question-12

An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig.), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umberalla?

Solution:
For one triangle piece
a = 20 cm
b = 50 cm
c = 50 cm
s = = = = 60 cm


 Area of one triangle =

                              =

                              = = 200cm2


∴ Area of five triangles of one colour = 5(200) cm2 = 1000cm2  
Hence 1000cm2 of each colour is required for the umbrella.

Question-13

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

Solution:
                                                                                                      
Area of paper of shade I = = 256 cm2
Area of paper of shade II = 256 cm2
For paper of shade III
a = 8 cm, b = 6 cm, c = 6 cm

s =

      = = 10 cm            
                                                                                                
Area of paper of shade III  =

                                       =

                                       = = 8 = 17.92 cm2.

Question-14

A Floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28cm and 35 cm ( see Figure). Find the cost of polishing the tiles at the rate of 50P Per cm2.

Solution:

For one tile
a = 9 cm, b = 28 cm, c = 35 cm

s = = = 36 cm


Area of one tile =

                        =

                        = =

                        = 6 × 3 × 2 = 36 cm2

Area of 16 tiles = 36× 16 = 576cm2
Cost of polishing the tiles at the rate of 50p per cm2= 576× 50 p

                                                                         =   

                                                                         =   288 र 705.60.

Question-15

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:

Let the given field be in the shape of a trapezium ABCD in which AB = 25 m, CD = 10 m,
BC = 13 m and AD = 14 m.

                                                                                                             
From D, draw DE || BC meeting AB at E. Also, draw DF AB.
DE = BC = 13 m
AE = AB – EB = AB – DC = 25 – 10 = 15 m
For Δ AED
a = 14 cm
b = 13 cm

c = 15 cm
s =

     = = = 21 m

Area of the ΔAED =

                           =

                           = =


                           = 7 × 3 × 2 × 2 = 84 m2

= 84 = 84

DF = DF =

Area of paralleogram EBCD = Base × Height

                                       = EB × DF = 10 × = 112 m2

Area of the field = Area of Δ AED + Area of parallelogram EBCD
                          = 84 m2 + 112 m2 = 196 m2.




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