**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 m^{2}

∴ Rent for advertisement on wall for 3 months for 1320 m^{2} =

= 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 m^{2}

**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 m^{2}

**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 cm^{2}

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

Therefore, Area

**Area of triangular part II** = Length × Breadth = 6.5 × 1 = 6.5 cm^{2}

**Total area = **2.4825 + 6.2 + 1.299 + 9 = 19.28 cm^{2}.

**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 m^{2}

∴ Field available for 1 cow to graze the grass = ^{ }= 48 m^{2}.

**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 cm^{2}

And Cloth required to 5 green pieces = 5 × 200√6 = 1000√6 cm^{2}

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

In right triangle ABC, AB^{2} + BC^{2} = AC^{2} [Using Pythagoras theorem]

Area of square = 512 cm^{2} [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 cm^{2}

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.

Area of triangular shaped tile

∴ Area of 16 such tiles = 16 × 88.2 = 1411.2 cm^{2} (Approx.)

∵ Cost of polishing 1 cm^{2} of tile = Rs. 0.50

∴ Cost of polishing 1411.2 cm^{2} 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