1.    Let the Traffic signal board is Δ ABC.

According to question, Semi-perimeter of Δ ABC 

Using Heron’s Formula,   Area of triangle ABC

Now,   Perimeter of this triangle = 180 cm     ⇒  Side of triangle (a) =  = 60 cm

⇒ Semi-perimeter of this triangle =  = 90 cm

Using Heron’s Formula,           Area of this triangle

2.    Given: a = 122 m, b = 22 m and c = 120 m

Semi-perimeter of triangle (s) = 

Using Heron’s Formula,      Area of triangle

∵ Rent for advertisement on wall for 1 year = Rs. 5000 per m2

∴ Rent for advertisement on wall for 3 months for 1320 m2

    = Rs. 1650000

Hence rent paid by company = Rs. 16,50,000

3.    Since, sides of coloured triangular wall are 15 m, 11 m and 6 m.

∴ Semi-perimeter of coloured triangular wall = 

Now,     Using Heron’s formula,        Area of coloured triangular wall

Hence area painted in blue colour = 20√2 m2

4.    Given: a = 18 cm, b = 10 cm.

Since Perimeter = 42 cm               ⇒          a+b+c =42            ⇒           18 + 10 + c = 42

⇒ c = 42 – 28 = 14 cm

∴ Semi-perimeter of triangle =

∴ Area of triangle

5.    Let the sides of the triangle be 12x, 17x and 25x

Therefore,            12x + 17x + 15x = 540     ⇒       54x =540      ⇒           x = 10

∴ The sides are 120 cm, 170 cm and 250 cm.

Semi-perimeter of triangle (s) = 

Now,     Area of triangle

6.    Given: a = 12 cm, b = 12 cm

Since Perimeter = 30 cm             ⇒         a+b+c =30               ⇒          12 + 12 + c = 30

⇒ c = 30 – 24 = 6 cm

∴ Semi-perimeter of triangle = 

∴ Area of triangle

7.    Since BD divides quadrilateral ABCD in two triangles:

(i) Right triangle BCD and (ii) Δ ABD.

In right triangle BCD, right angled at C,

therefore Base = CD = 5 m and Altitude = BC = 12 m

In Δ ABD, AB = 9 m, AD = 8 m

∴ Area of quadrilateral ABCD = Area of Δ BCD + Area of Δ ABD

= 30 + 35.4

= 65.4 m2

8.    In quadrilateral ABCE, diagonal AC divides it in two triangles, Δ ABC and Δ ADC.

In Δ ABC,       Semi-perimeter of Δ ABC = 

Using Heron’s formula,

Again, In Δ ADC, Semi-perimeter of Δ ADC = 

Using Heron’s formula,           Area of Δ ABC

Now area of quadrilateral ABCD = Area of Δ ABC + Area of Δ ADC

= 6 + 9.2

= 15.2 cm2

9.    Area of triangular part I: Here,    Semi-perimeter  

Therefore, Area

Area of triangular part II = Length × Breadth = 6.5 × 1 = 6.5 cm2

Total area = 2.4825 + 6.2 + 1.299 + 9 = 19.28 cm2.

10.  Semi-perimeter of triangle 

Using Heron’s formula, Area of triangle

According to question, Area of parallelogram = Area of triangle

⇒ Base × Corresponding height = 336

⇒ 28 × Height = 336

⇒ Height = 12 cm

11.  Here, AB = BC = CD = DA = 30 m and Diagonal AC = 48 m which divides the rhombus ABCD in two congruent triangle.

∴ Area of A ABC = Area of Δ ACD (s) 

Now, Semi-perimeter of Δ ABC

Now Area of rhombus ABCD = Area of Δ ABC + Area of Δ ACD

∵ Field available for 18 cows to graze the grass = 864 m2

∴ Field available for 1 cow to graze the grass =  = 48 m2.

12.  Here, sides of each of 10 triangular pieces of two different colours are 20 cm, 50 cm and 50 cm.

Semi-perimeter of each triangle 

Now, Area of each triangle

 

According to question, there are 5 pieces of red colour and 5 pieces of green colour.

∴ Cloth required for 5 red pieces = 5 × 200√6 = 1000√6 cm2

And     Cloth required to 5 green pieces = 5 × 200√6 = 1000√6 cm2

13.  Let ABCD is a square of side a cm and diagonals AC = BD = 32 cm

In right triangle ABC, AB2 + BC2 = AC2         [Using Pythagoras theorem]

Area of square = 512 cm2                    [Area of square = side x side]

Diagonal BD divides the square in two equal triangular parts I and II.

Area of shaded part I = Area of shaded part II =  × 512 = 256 cm2

Now, semi-perimeter of shaded part III

Area of shaded part III

14.  Here,  Sides of a triangular shaped tile area 9 cm, 28 cm and 35 cm.

Semi-perimeter of tile 

Area of triangular shaped tile

∴ Area of 16 such tiles = 16 × 88.2 = 1411.2 cm2 (Approx.)

∵ Cost of polishing 1 cm2 of tile = Rs. 0.50

∴ Cost of polishing 1411.2 cm2 of tile = Rs. 0.50 × 1411.2 = Rs. 705.60 (Approx.)

15.  Let ABCD be a field in the shape of trapezium and parallel side AB = 10 m & CD = 25 m And Non-parallel sides AD = 13 m and BC = 14 m

Draw BM ⊥ DC and BE || AD so that ABED is a parallelogram.

∴ BE = AD = 13 m and DE = AB = 10 m