Heron'S Formula - Class 9 Maths - Extra Questions

Area of equilateral triangle $$=(16\times \sqrt{3})\,cm^2$$

Length of sides = $$6,8,10$$

In $$\triangle{ABC}$$ ,

Let $$a,b$$ and $$c$$ be the sides of a triangle.
Apply Heron's Formula to find the area of triangle.
Area $$= \sqrt{S(S-a)(S-b)(S-c)}$$
Whee $$S = \dfrac{a+b+c}{2}$$
Here $$a = 42\,cm,b = 34\,cm$$ and $$c = 20\,cm$$
$$S = \dfrac{(42+34+20)}{2} = 48$$
Area $$= \sqrt{(48(48-42)(48-34)(48-20))}$$
          $$= \sqrt{(48\times 6\times 14\times 28)}$$
          $$= 336$$
Area of triangle is $$336\,cm^{2}$$

Clearly,
Length of longest side $$= 42\,cm$$
Also we know,
Area of a triangle $$= \dfrac{1}{2}\times Base \times Height$$
$$\Rightarrow 336 = \dfrac{1}{2}\times 42\times Height$$
$$\Rightarrow 336 = 21\times Height$$
or Height $$= 16$$
The height corresponding to the longest side is $$16\,cm$$

Area of tiles = 16 x area of 1 $$\Delta$$

$$S = \dfrac{a + b + c} {2} = \dfrac{9 + 28 + 35} {2} = \dfrac{72} {2} = 36$$

$$Area = \sqrt {36(36 - 9)(36 - 28)(36 - 35)}$$

$$\sqrt {36 \times 27 \times 8 \times 1}$$

$$36\sqrt 6 c{m^2}$$

Area of tiles = 16 $$\times 36\sqrt 6$$

= $$1410.20.$$

Cost of polishing per $$c{m^2}$$ = 50.

Cost of polishing = $$1410.90$$ $$c{m^2}$$

$$1410.90$$ x $${1 \over 2}$$

$$705.45$$

Given:-

Ratio of sides
7: 12:8

let them be
7x,12x, 8x respectively

perimeter of a triangle = sum of all sides

270 = 7x+12x, 8x
270 = 27x
x = 10

all sides measure
7x = 7×10 = 70
12x = 12× 10 = 120
8x= 8 × 10 = 80

it's semi perimeter = 270/2
= 135

using heron's formula area of the triangle =

root {(s)(s-a)(s-b)(s-c)}
where s is the semi perimeter and a,b,c are the sides of the triangle.

root {( 135)(135-70)(135-120)(135-80)}

= 2690.6cm^2.

Heron's formula is used for calculating the area of a triangle when the length of all 3 sides are known.

$$a=78m$$ , $$b=112m$$ , $$c=50m$$
$$s=\cfrac { 1 }{ 2 } \left( 78+112+50 \right) =120m\quad$$
$$s-a=42m,s-b=8m,s-c=70m$$
$$\therefore$$ Area $$=\sqrt { 120\times 70\times 42\times 8 }$$
$$=1680$$ sq.m

Consider the given sides.

Let $$ABC$$ be the given triangle.

Semiperimeter $$=s=\dfrac{a+b+c}{2}=\dfrac{6+8+10}{2}=\dfrac{24}{2}=12\,cm$$

Given $$,$$ the sides of a triangle are $$5cm,12cm$$ and $$13cm$$

Let the side be $$5x,3x$$ and $$4x$$

We have,

$$According\quad to\quad Heron's\quad formula\quad the\quad area\quad (A)\quad of\quad triangle\quad with\quad sides\quad a,\quad b\quad \& \quad c\quad is\quad given\quad as\\ \quad A\quad =\quad \sqrt [ 2 ]{ \left[ s(s-a)(s-b)(s-c) \right]  } where\quad 2s\quad =\quad (a+b+c).\\ The\quad triangular\quad side\quad walls\quad have\quad the\quad following\quad dimensions\quad :\quad a=13m,\quad b=14m,\quad c=15m.\\ Hence\quad 2s\quad =\quad (13+14+15)\quad =\quad 42\quad \\ So,\quad the\quad area\quad of\quad one\quad triangular\quad side\quad wall\quad \\ =\quad \sqrt [ 2 ]{ \left[ 21(21-13)(21-14)(21-15) \right]  } \quad \\ =\quad \sqrt [ 2 ]{ \left[ 21(8)(7)(6) \right]  } \quad \\ =\quad \sqrt [ 2 ]{ \left[ 3\times 7\times 2\times 2\times 2\times 7\times 2\times 3 \right]  } \\ =\quad \sqrt [ 2 ]{ \left[ (2\times 2\times 3\times 7)(2\times 2\times 3\times 7) \right]  } \quad \\ =\quad (2\times 2\times 3\times 7)\quad \\ =\quad 84\quad \\ The\quad advertisements\quad yield\quad =\quad Rs.\quad 2000\quad per\quad \quad for\quad 1\quad year.\\ Since\quad the\quad company\quad hired\quad 1\quad such\quad wall\quad for\quad 6\quad months,\quad so\quad the\quad amount\quad of\quad rent\quad to\quad be\quad paid\quad at\quad this\quad rate\quad for\quad 6\quad months\quad \\ =\quad \frac { 84\times 2000 }{ 2 } \quad \\ =\quad 84\times 1000\\ =\quad Rs.\quad 84000$$

$$We\quad know\quad that\quad the\quad diagonal\quad of\quad a\quad parallelogram\quad ({ \parallel  }^{ gm })\quad divides\quad it\quad into\quad two\quad congruent\quad triangles.\\ So\quad Area\quad of\quad { \parallel  }^{ gm }\quad ABCD\quad =\quad 2\quad \times \quad Area\quad of\quad \triangle BCD.\\ According\quad to\quad Heron's\quad formula\quad the\quad area\quad (A)\quad of\quad triangle\quad with\quad sides\quad a,\quad b\quad \& \quad c\quad is\quad given\quad as\\ A\quad =\quad \sqrt [ 2 ]{ \left[ s(s-a)(s-b)(s-c) \right]  } where\quad 2s\quad =\quad (a+b+c).\\ Here\quad a\quad =\quad 12\quad ,\quad b\quad =\quad 17\quad ,\quad c\quad =\quad 25\quad ,\quad s\quad =\quad \frac { \left( 12\quad +\quad 17\quad +\quad 25 \right)  }{ 2 } \quad =\quad \frac { 54 }{ 2 } \quad =\quad 27\quad \quad \\ Area\quad of\quad { \parallel  }^{ gm }\quad ABCD\quad \\ =\quad 2\quad \times \quad \sqrt [ 2 ]{ \left[ 27\quad \times \quad \left( 27\quad -\quad 12 \right) \left( 27\quad -\quad 17 \right) \left( 27\quad -\quad 25 \right)  \right]  } \quad \quad \quad \\ =\quad 2\quad \times \quad \sqrt [ 2 ]{ \left( 27\quad \times \quad 15\quad \times \quad 10\quad \times \quad 2 \right)  } \quad \quad \quad \quad \quad \quad \\ =\quad 2\quad \times \quad \sqrt [ 2 ]{ \left( 3\quad \times \quad 3\quad \times \quad 3\quad \times \quad 3\quad \times \quad 5\quad \times \quad 5\quad \times \quad 2\quad \times \quad 2 \right)  } \quad \quad \\ =\quad 2\quad \times \quad \sqrt [ 2 ]{ \left[ (3\quad \times \quad 3\quad \times \quad 5\quad \times \quad 2)\quad (3\quad \times \quad 3\quad \times \quad 5\quad \times \quad 2) \right]  } \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ =\quad 2\quad \times \quad (3\quad \times \quad 3\quad \times \quad 5\quad \times \quad 2)\\ =\quad 2\quad \times \quad 90\\ =\quad 180\quad \\ Area\quad of\quad { \parallel  }^{ gm }\quad =\quad base\quad \times \quad height\\ Height\quad of\quad altitude\quad from\quad vertex\quad A\quad on\quad side\quad CD\quad of\quad the\quad of\quad { \parallel  }^{ gm }\\ =\quad \frac { area\quad of\quad { \parallel  }^{ gm }\quad ABCD }{ base\quad CD } \quad \\ =\quad \frac { 180 }{ 12 } \\ =\quad 15 cm$$

$$Let\  the\  vertices\  of\  the\  \triangle \  be\  A,B,C\  \& \  the\  length\  of\  smaller\  side\  BC\  is\  a.\\ Then\  side\  AB\  =\  a+4\  and\  side\  AC\  =\  2a-6.\\ Given\  the\  perimeter\  of\  \triangle \  ABC\  is\  50cm.\\ So\  we\  have\\ a\  +\  \left( a+4 \right) \  +\left( 2a-6 \right) \  =\  50\\ \Rightarrow \  4a\  -\  2\  =\  50\\ \Rightarrow \  4a\  =\  52\\ \Rightarrow \  a\  =\  13.\\ \therefore \  The\  sides\  are\  BC\  =\  a\  =\  13cm,\  AB\  =\  \left( a+4 \right) \  =\  17cm\  and\  AC\  =\  \left( 2a-6 \right) \  =\  20cm.\\ Now\  according\  to\  Heron's\  formula\  the\  area\  of\  triangle\  ABC\  with\  sides\  a,\  b\  \& \  c\  is\  given\  as\\ A\  =\  \sqrt [ 2 ]{ \left[ s\left( s-a \right) \left( s-b \right) \left( s-c \right)  \right]  } where\  2s\  =\  \left( a+b+c \right) .\\ Here\  a\  =\  13cm,\  b\  =\  17\  ,\  c\  =\  20\  ,\  s\  =\  \frac { \left( 13\  +\  17\  +\  20 \right)  }{ 2 } \  =\  \frac { 50 }{ 2 } \  =\  25\  \\ Area\  of\  \triangle \  ABC\  \\ =\  \sqrt [ 2 ]{ \left[ 25\  \times \  \left( 25\  -\  13 \right) \left( 25\  -\  17 \right) \left( 25\  -\  20 \right)  \right]  } \  \  \  \\ =\  \sqrt [ 2 ]{ \left( 25\  \times \  12\  \times \  8\  \times \  5 \right)  } \  \  \  \  \  \  \\ =\  \sqrt [ 2 ]{ \left( 5\  \times \  5\  \times \  6\  \times \  2\  \times \  8\  \times \  5 \right)  } \\ =\  \sqrt [ 2 ]{ \left( 5\  \times \  5\  \times \  6\  \times \  16\  \times \  5 \right)  } \  \  \\ =\  5\  \times \  4\  \times \  \sqrt [ 2 ]{ \left( 6\  \times \  5 \right)  } \\ =\  20\  \sqrt [ 2 ]{ 30 }$$

Area of triangle with sides a, b, and c and s $$= \dfrac { a+b+c }{ 2 }$$ is $$\sqrt { s(s-a)(s-b)(s-c) } \ $$ .

For triangle with sides 18 mm, 24 mm and 30 mm, $$s = \dfrac { 18+24+30 }{ 2 } =36\   mm\ $$
Area of the triangle with sides 18 mm, 24 mm and 30 \ mm 

Area of triangle with $$a,b,c$$ as sides and  $$s= \dfrac { a+b+c }{ 2 }$$ is $$\sqrt { s(s-a)(s-b)(s-c) } \ $$ .

Area of triangle with $$a,b,c$$ as sides and  $$s= \dfrac { a+b+c }{ 2 }$$ is $$\sqrt { s(s-a)(s-b)(s-c) } \ $$ .
For triangle with sides $$16 cm, 12 cm$$ and $$20 cm$$ , $$s = \dfrac { 16+12+20 }{ 2 } =24  cm\ $$
Area of the triangle with sides 16 cm, 12 cm and 20 cm  $$=\sqrt { 24(24-16)(24-12)(24-20) }$$

$$=\sqrt { 24\times 8\times 12\times 4 }$$

$$=\sqrt { 8\times 3\times 8\times 4\times 3\times 4 }$$

$$= 8\times 3\times 4$$

$$= 96  {cm}^{2}$$

Let the sides of the triangle be $$4a, 5a, 3a$$ .
Perimeter of the triangle $$=$$ Sum of all sides $$= 4a + 5a + 3a = 12a$$

Let the sides of the triangular field be $$5a, 3a, 4a$$ . 

Perimeter of the triangle $$=$$ Sum of all sides $$= 5a + 3a + 4a = 12a$$

Given, perimeter of the triangular fied $$= 180\ m$$  

$$\Rightarrow  12a = 180 \ m$$   

$$\Rightarrow  a = 15 \ m$$  

 
So, the sides of the triangular field are $$5a = 75\ m$$  

                                                                  $$3a = 45\ m$$

                                                                  $$4a  = 60\ m$$ .

$$\Rightarrow$$ Area of triangle with sides $$a,b,c$$ is $$\sqrt { s(s-a)(s-b)(s-c) } \ $$ . 

where  $$s= \dfrac { a+b+c }{ 2 }$$

$$\Rightarrow$$ For the triangular field with sides $$75\ m$$ , $$45\ m$$   and $$60\ m$$ , 

       $$s = \cfrac { 75 + 45 + 60 }{ 2 }$$

          $$= 90\ m$$

$$\Rightarrow$$ Area of the triangular field with sides $$75$$  m, $$45$$ m and $$60$$ m 

$$=\sqrt { 90(90-75)(90-45)(90-60) }$$

$$=\sqrt { 90\times 15\times 45\times 30 }$$

$$=\sqrt { 15\times 6\times 15\times 15\times 3\times 15 \times 2 }$$

$$= 15\times 15\times 6 = 1350\   {m}^{2}$$

Also, Area of triangle $$= \dfrac { 1 }{ 2 } \times$$  base $$\times$$ height 

Considering the height on the largest side of the triangle $$75\ m$$ as $$h$$ ,

$$\Rightarrow  \dfrac { 1 }{ 2 } \times 75\times h = 1350\ sq.m$$  

$$\Rightarrow  h = 36\ m$$  

Let the sides of the triangular field be $$5a, 3a, 4a$$ . 

Perimeter of the triangle $$=$$ Sum of all sides $$= 5a + 3a + 4a = 12a$$

Given, perimeter of the triangular fied $$= 180$$ m

$$\implies12a = 180$$ m 

$$a = 15$$ m 

So, the sides of the triangular field are $$5a = 75$$ m, $$3a = 45$$ m, $$4a  = 60$$ m.

Area of triangle with sides $$a,b,c$$  and $$s$$ $$= \dfrac { a+b+c }{ 2 }$$ is $$\sqrt { s(s-a)(s-b)(s-c) } \ $$ . 

For the triangular field with sides $$75$$ m, $$45$$ m and $$60$$ m, $$s = \dfrac { 75 + 45 + 60 }{ 2 } = 90$$ m

Area of the triangular field with sides $$75$$ m, $$45$$ m and $$60$$ m $$=\sqrt { 90(90-75)(90-45)(90-60) }$$

$$=\sqrt { 90\times 15\times 45\times 30 }$$

$$=\sqrt { 15\times 6\times 15\times 15\times 3\times 15 \times 2 }$$

$$= 15\times 15\times 6$$

$$= 1350 {m}^{2}$$

Cost of levelling the field at the rate of Rs. $$10$$ per square metre $$= 1350 \times 10 =$$ Rs. $$13500$$ .

Area of triangle with sides a, b, and c 

s $$= \dfrac { a+b+c }{ 2 }$$ is $$\sqrt { s(s-a)(s-b)(s-c) } \ $$ .

For triangle with sides 21 cm, 28 cm and 35 cm, 

$$s = \dfrac { 21+28+35 }{ 2 } =42  cm\ $$


Area of the triangle $$=\sqrt { 42(42-21)(42-28)(42-35) }$$

                                  $$=\sqrt { 42\times 21\times 14\times 7 }$$

                                  $$=\sqrt { 21\times 2\times 21\times 7\times 2\times 7 }$$  

                                  $$= 21\times 7\times 2 = 294 {cm}^{2}$$

Let the sides of the triangle be $$3a, 4a, 5a$$ .
Perimeter of the triangle $$=$$ Sum of all sides $$= 3a + 4a + 5a = 12a$$
Given, perimeter of the triangle $$= 144\ cm$$
$$\Rightarrow  12a = 144\ cm$$   
$$\Rightarrow  a = 12\ cm$$  

So, the sides of the triangle are $$3a = 36\ cm$$ , $$4a = 48\ cm$$ , $$5a  = 60\ cm$$ .

Area of triangle with sides $$a,b,c$$   is $$\sqrt {s(s-a)(s-b)(s-c)}$$

where, $$s = \dfrac { a+b+c }{ 2 }$$

Here, $$s=\dfrac {36+48+60}{2}\ cm=72\ cm$$ .

Area of the triangle  $$=\sqrt { 72(72-36)(72-48)(72-60) } \ cm^2$$

                                  $$=\sqrt { 72\times 36\times 24\times 12 }\ cm^2$$

                                  $$=\sqrt { 36\times 2\times 36\times 12\times 2\times 12 }\ cm^2$$

                                  $$= 36\times 12\times 2 \ cm^2$$

                                  $$= 864\  {cm}^{2}$$

Hence, the required area is  $$864\  {cm}^{2}$$ .

Area of region $$I:$$  

$$2S=5+4+3$$ $$\Rightarrow$$ $$S=6$$

Join $$BD$$ in $$\Delta {BCD}$$ ,   $$BC$$ and $$DC$$ are given. 

Let the third side of the triangle be $$x$$

Two sides length are given, we need to figure out the length of this side.

Given: 

Given: 

$$\Rightarrow$$ Let $$x$$ be common ratio

$$\Rightarrow$$ Sides:  $$a=122\ m$$ , $$b=22\ m$$ and $$c=120\ m$$

The umbrella's figure consists of $$5$$ triangle pieces

$$\Rightarrow$$ Area of square 

$$\Rightarrow$$ Semi perimeter of tile $$=\cfrac{35+28+9}{2}=36cm$$

$$2S=26+28+30=84$$

Given in triangle $$ABC, BC=11\ cm ,CA=13\ cm$$ and the median to side $$AB =10$$ cm

Let $$a=13$$ , $$b=14$$ , $$c=15$$ then  $$s=\dfrac { 1 }{ 2 } \left( a+b+c \right) =21$$

We know in triangle if the three sides are $$a,b,c$$ then the area of the triangle is

To prove:

Given, diagonal of the square $$=40\ \text{cm}$$

Let the sides of triangles are $$3x,4x$$ and $$5x.$$

Points:  $$( - 5, - 1)(, - 5)(5,2)$$

The area of the triangle between the co-ordinate axis of line $$ax + by + c = 0$$ is $$\left| \dfrac{c^2}{ab} \right|$$ sq.units.

Given 

Given that the sides of a triangle are $$3 \ cm, 7\ cm\ and \ 8 \ cm$$ .

Heron's formula

The sides of a quadrilateral field taken order as 

Using the formula 

Heron's Formula =

Assume the Triangle has vertices $$O,A,B$$

Here $$2s$$ $$=$$ $$(61+60+11)cm$$

In $$\triangle{BCD}$$ , using Pythagoras theorem,

Let the third side be $$x\ cm$$

Semi perimeter of triangle $$s= \cfrac{5 + 12 + 13}{2} = 15 \; cm$$

In $$\Delta COD$$ , we have

Equal sides of measure $$= 10\,cm$$
Here, base and perpendicular are equal sides.
Area of a triangle $$= \dfrac{1}{2} \times Base \times Height$$  
                              $$= \dfrac{1}{2} \times 10 \times 10$$
                              $$= 50$$

From Pythagorean Theorem:
Hypotenuse $$^2 =$$ Base $$^2 +$$ Perpendicular $$^2$$
                        $$= 10^2 + 10^2$$
                        $$= 200$$
or Hypotenuse $$= 10\,\sqrt2\,cm = 10 \times 1.41\,cm = 14.1\,cm$$
Perimeter $$=$$ Sum of all the sides $$= 10 + 10 + 14.1 = 34.1$$
Therefore, Area of required triangle is $$50\,cm^2$$ and perimeter $$= 34.1\,cm$$

Let $$a,b$$ and $$c$$ be the sides of a triangle.
Apply Heron's Formula to find the area of triangle.
Area $$= \sqrt{S(S-a)(S-b)(S-c)}$$
Where $$S = \dfrac{a+b+c}{2}$$
Here $$a = 18\,cm,b = 24\,cm,c = 30\,cm$$
Now, 
$$S=\dfrac 12(18+24+30) = 36$$
Area $$= \sqrt{(36(36-18)(36-24)(36-30))}$$
          $$= \sqrt{(36\times 18\times 12\times 6)}$$
          $$= 216$$
Area is $$216\,cm^{2}$$

From given, Length of smallest side $$= 18\,cm$$
Area of a triangle $$= \dfrac{1}{2}\times Base\times Height$$
$$\Rightarrow  216 = \dfrac{1}{2}\times 18\times height$$
Height $$= 24$$
Therefore, the height corresponding to the smallest side is $$24\,cm$$

Let h be the height, then
Each of the equal sides of an isosceles triangle $$= (h + 2)\,cm$$ and
Base $$= 12\,cm$$

Area of a triangle $$= \dfrac{1}{2} \times  Base \times  Height$$
$$= \dfrac{1}{2} \times  12 \times  h ....(1)$$

Area of isosceles triangle $$= \dfrac{1}{4}\times b\times \sqrt{(4a^{2}-b^{2})}$$
$$= \dfrac{1}{4}\times 12\times \sqrt{(4(h+2)^{2}-144)}...(2)$$

From (1) and (2), we get
$$\Rightarrow \dfrac{1}{2} \times 12\times h = \dfrac{1}{4}\times 12\sqrt{(4(h+2)^{2}-144)}$$
$$\Rightarrow  6h = 3\sqrt{(4h^2+16h+16-144)}$$
$$\Rightarrow  2h = \sqrt{(4h^{2}+16h-128)}$$
$$\Rightarrow  4h^{2} = 4h^{2}+16h-128$$
$$\Rightarrow  16h = 128$$
or $$h = 8$$
Height is 8 cm.
Now, 
Area of a triangle $$= \dfrac{1}{2}\times Base \times Height$$
$$= \dfrac{1}{2}\times 12\,cm\times 8\,cm$$
$$= \dfrac{1}{2}\times 96\,cm^{2}$$
$$= 48\,cm^{2}.$$  

Height of an equilateral triangle $$= 6\,cm$$
Let x be a side of triangle.
Height of the equilateral triangle $$= \dfrac{\sqrt{3}}{2}(side) = \dfrac{\sqrt{3}}{2}x.$$
$$6 = \dfrac{\sqrt{3}}{2}x$$
$$x = \dfrac{12}{\sqrt{3}}$$
$$x = \dfrac{12\sqrt{3}}{3} = 6.92\,cm$$
Now,
Area of a triangle $$= \dfrac{\sqrt{3}}{4}(side)^{2} = \dfrac{\sqrt{3}}{4}\times 6.92\times 6.92 = 20.76\,cm^{2}$$

Let each side of an equilateral triangle be "a" cm
Area of an equilateral triangle $$= 81\sqrt{3}cm^{2}$$ (Given)
Area of an equilateral triangle $$=\dfrac{\sqrt{3}}{4}(side)^{2}$$
$$\Rightarrow 81\sqrt{3} = \dfrac{\sqrt{3}}{4}\times a^{2}$$
$$\Rightarrow a^{2} = 324$$
or $$a = 18\,cm$$
Now,
Height of an equilateral triangle $$= \dfrac{\sqrt{3}}{2}(side) = \dfrac{\sqrt{3}}{4}(18) = 9\sqrt{3}cm.$$
Perimeter $$= 3 (side) = 3\times 12 = 36\,cm$$

We are given with a right-angled triangle whose measures are
Base $$= 48 \,cm$$
Hypotenuse $$= 50\, cm$$

Using Pythagoras Theorem:
Hypotenuse $$^2 =$$ Base $$^2 +$$ Perpendicular $$^2$$
$$50^2 = 48^2 +$$ Perpendicular $$^2$$
or Perpendicular $$^2 = 2500 - 2304=196$$
or Perpendicular $$= 14\, cm$$

Now,
Area of a triangle $$= \dfrac{1}{2} \times Base \times  Height$$
                              $$= \dfrac{1}{2} \times  48 \,cm \times 14\, cm$$
                              $$= 336\, cm^2$$

Given, $$\Delta ABC$$ is an equilateral triangle the length of whose side is equal to 10 cm, and $$\Delta DBC$$ is right-angled at D 
and $$BD = 8\,cm.$$

Side of an equilateral triangle $$= 10\,cm$$
(i) Area of an equilateral  triangle $$=\dfrac{\sqrt{3}}{4}(side)^{2} =\dfrac{ \sqrt{3}}{4}\times 10\times 10 = 43.3$$
Area of a triangle is $$43.3\,cm^{2}$$

(ii) Height of the equilateral triangle $$= \dfrac{\sqrt{3}}{2}(side) = \dfrac{\sqrt{3}}{2}\times 10 = 8.66$$
Height of the triangle is $$8.66\,cm$$

$$\textbf{Step - 1: Writing sides by using  proportionality constant.}$$

                    $$\text{Let the proportionality constant be x.}$$

                    $${\text{So we'll remove the proportionality sign and multiply each of the sides with x}}{\text{.}}$$

                    $${\text{As we know that the ratio between the sides is given as }}5:12:13.$$

                    $${\text{We get the sides to be 5x,12x and 13x}}{\text{.}}$$

$${\textbf{Step - 2: Finding perimeter.}}$$

                    $$\text{We know that the perimeter is the sum of all the sides of a triangle.}$$

                    $$\text{Let the perimeter be P.}$$

                    $$\therefore \text{P} = 5x + 12x + 13x$$

                    $$\Rightarrow P = 30x$$

$${\textbf{Step - 3: Equate the given perimeter and find the value of sides}}{\text{.}}$$

                    $$\text{We know that the perimeter is }$$ $$150\ m.$$

                    $$\Rightarrow 30x = 150$$

                    $$\Rightarrow x = 5$$

                    $$\therefore \text{The sides of triangle are of length will be}$$   $$5\times5\ \text{m},12\times5\ \text{m}, 13\times5\ \text{m}$$

                    $$\Rightarrow \text{The sides of triangle will be } 25\ \text{m},60\ \text{m},65\ \text{m}.$$

$${\textbf{Step  - 4: Apply Heron's formula to find the area of the triangle}}{\text{.}}$$

                    $$\text{We know from Heron's formula area } A= \sqrt {s(s - a)(s - b)(s - c)}.$$

                    $${\text{where s is the semi perimeter and  }}s= \dfrac{{a + b + c}}{2}.$$

                    $$\therefore s= \dfrac{{25 + 60 + 65}}{2}$$   

                            $$=\dfrac{{150}}{2}$$

                            $$= 75\ m.$$

                    $${\text{Now substituting all the values in the Heron's formula, we get,}}$$

                    $$A = \sqrt {75(75 - 25)(75 - 60)(75 - 65)}$$

                        $$= \sqrt {75 \times 50 \times 15 \times 10}$$

                        $$= \sqrt {562500}  = 750\ {\text{m}^2}$$

$${\textbf{Hence, the area of the triangle is 750 }}{{\mathbf{m}}^2}.$$

Consider  a=29 cm , b = 20 cm and c = 21 cm
We know that 
S = $$\left ( a + b + c \right )/2$$  
Substituting the values 
  $$\left ( 29 +20+ 21 \right )/2$$
 = 70/2
= 35 cm

We know that 
Perimeter of an equilateral triangle = $$3 \times side$$
Substituting the values 
  $$36 = 3 \times side$$
By further calculation 
side $$= 36/3 = 12 cm$$
So $$AB =BC = CA = 12 cm$$

It is given that 
Side of equilateral triangle $$= 8 \ cm$$
We know that, 
Area of an equilateral triangle with side $$a$$ is $$\dfrac{\sqrt3 a^2}{4}$$
Substituting value of $$a$$ , we get, 
Area $$=\dfrac{\sqrt{3}}4 \times 8\times 8$$
$$= 16\sqrt3$$

$$The\quad kite\quad is\quad divided\quad in\quad two\quad parts\quad -\quad upper\quad square\quad and\quad lower\quad triangle.\\ The\quad square\quad ABCD\quad is\quad having\quad diagonals\quad of\quad length\quad 44.\\ The\quad area\quad of\quad the\quad square\quad ABCD\quad =\quad \frac { 1 }{ 2 } \times \quad \left( { diagonal }^{ 2 } \right) \quad =\quad \frac { 1 }{ 2 } \times \quad \left( { 44 }^{ 2 } \right) \quad =\quad 22\quad \times \quad 44\quad =\quad 968\\ Since\quad diagonals\quad of\quad a\quad square\quad always\quad divide\quad the\quad square\quad into\quad 4\quad equal\quad triangles.\\ \therefore \quad Area\quad of\quad \triangle OAB\quad =\quad area\quad of\quad \triangle OBC\quad =\quad area\quad of\quad \triangle OCD\quad =\quad area\quad of\quad \triangle OAD\quad =\quad \frac { 1 }{ 4 } \times \quad 968\quad =\quad 242\\ Now\quad in\quad the\quad lower\quad triangle\quad of\quad kite,\quad using\quad Heron's\quad formula\quad the\quad area\quad of\quad triangle\quad with\quad sides\quad a,\quad b\quad \& \quad c\quad is\quad given\quad as\\ A\quad =\quad \sqrt [ 2 ]{ \left[ s\left( s-a \right) \left( s-b \right) \left( s-c \right)  \right]  } where\quad 2s\quad =\quad \left( a+b+c \right) .\\ Here\quad a\quad =\quad c\quad =\quad 20\quad ,\quad b\quad =\quad 14\quad ,\quad s\quad =\quad \frac { \left( 20\quad +\quad 20\quad +\quad 14 \right)  }{ 2 } \quad =\quad \frac { 54 }{ 2 } \quad =\quad 27\quad \\ Area\quad of\quad lower\quad \triangle \quad =\quad \sqrt [ 2 ]{ \left[ 27\quad \times \quad \left( 27\quad -\quad 14 \right) \left( 27\quad -\quad 20 \right) \left( 27\quad -\quad 20 \right)  \right]  } \quad \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \sqrt [ 2 ]{ \left( 27\quad \times \quad 13\quad \times \quad 7\quad \times \quad 7 \right)  } \quad \quad \quad \quad \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \sqrt [ 2 ]{ \left( 3\quad \times \quad 3\quad \times \quad 3\quad \times \quad 13\quad \times \quad 7\quad \times \quad 7 \right)  } \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 3\quad \times \quad 7\quad \times \quad \sqrt [ 2 ]{ \left( 3\quad \times \quad 13 \right)  } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 21\quad \times \quad \sqrt [ 2 ]{ 39 } \quad \left\{ Use\quad division\quad method\quad to\quad find\quad square\quad root \right\} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 21\quad \times \quad 6.24\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 131.04\\ Total\quad yellow\quad paper\quad required\quad =\quad area\quad of\quad \triangle OAB\quad +\quad area\quad of\quad \triangle OCD\quad =\quad 2\quad \times \quad 242\quad =\quad 484\\ Total\quad red\quad paper\quad required\quad =\quad area\quad of\quad \triangle OBC\quad =\quad 242\\ Total\quad green\quad paper\quad required\quad =\quad area\quad of\quad \triangle AOD\quad +\quad area\quad of\quad lower\quad triangle\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 242\quad +\quad 131.04\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 373.04\\ $$

For the first triangle, we have 

The dimension of the triangular park is $$30\;{\rm{m}} \times 26\;{\rm{m}} \times 28\;{\rm{m}}$$ .

$$s = \dfrac{{a + b + c}}{2}$$

$$= \dfrac{{30 + 26 + 28}}{2}$$

$$= 42\;{\rm{m}}$$

The area of the triangular park is,

$$A = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$$

$$= \sqrt {42\left( {42 - 30} \right)\left( {42 - 26} \right)\left( {42 - 28} \right)}$$

$$= 336\;{{\rm{m}}^2}$$

Therefore, the amount paid to the gardener is,

$$336 \times 1.50 = {\rm{Rs}}\;504$$

Here, P = 8+15 +17= 21 cm

$$S=\dfrac { a+b+c }{ 2 }$$

So  $$S=\dfrac { 40 }{ 2 } =20cm$$

$$s - a = 12, s - b = 5, s - c = 3$$

Area of triangle= $$\sqrt { S(S-a)(S-b)(S-c) }$$

                        $$=\sqrt { 20(12)(5)(3) }$$

                        $$=\sqrt { 40(12)(5)(3) }$$

                        $$=60$$

$$\Box ABED$$ is parallelogram,
$$\Rightarrow$$ $$AD=BE=13\ \ m$$
$$\Rightarrow$$ $$AB=DE=10\ \ m$$
$$\Rightarrow$$ $$BC=14\ \ m$$

$$Let\, \, a\, \, triangle\, \, whose\, \, sides\, \, are\, \, e\left( a \right) . \\ New\, \, triangle\, whose\, \, area\, \, is\, \, =\frac { { \sqrt { 3 }  } }{ 4 } { \left( { 2a } \right) ^{ 2 } } \\ Let\, \, a\, \, triangle\, \, is\, \, equilateral\, \,  \\ Then,\, \, its\, \, area\, =\frac { { \sqrt { 3 }  } }{ 4 } { a^{ 2 } } \\ Ratio\, \, of\, area\, \, of\, \, new\, \, triangle\, \, to\, \, gine\, \, triangle\, \, is \\ =\frac { { \frac { { \sqrt { 3 }  } }{ 4 } \times 4{ a^{ 2 } } } }{ { \frac { { \sqrt { 3 }  } }{ 4 } \times { a^{ 2 } } } } =\frac { 4 }{ 1 } =4:1\, \, \,$$

Area of $$\triangle{BCD}=\dfrac{1}{2}\times12\times5$$

Given sides are $$18cm,10cm$$ and $$14cm$$

Using basic proportionality theorem, we can say that :