Herons Formula Test - 22

Area of triangle with sides a, b and c is given by: $$ \sqrt { s(s-a)(s-b)(s-c) }  $$.
where $$s = \displaystyle \frac { a+b+c }{ 2 } $$ 

For triangle with sides 13 cm, 14 cm and 15 cm:

$$ s = \displaystyle \frac { 13+14+15 }{ 2 } \ \text{cm}=21\ \text{  cm} $$
$$\therefore $$ Area of the triangle with sides 13 cm, 14 cm and 15 cm $$ =\sqrt { 21(21-13)(21-14)(21-15) }$$

$$ =\sqrt { 21\times 8\times 7\times 6 }$$

$$ =\sqrt { 7\times 3\times 2\times 4\times 7\times 2 \times 3 } $$

$$= 7\times 3\times 2 \times 2 = 84\  \text{cm}^{2}$$

Let, 

Let the sides of the triangle be $$3x cm, 4x cm$$ and 5x cm Then 
$$\displaystyle s=\frac{3x+4x+5x}{2}=\frac{12x}{2}=6x$$
$$\displaystyle \therefore$$  Area of  $$\displaystyle \Delta=\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}$$
$$\displaystyle = \sqrt{6x\left ( 6x - 3 \right )\left ( 6x - 4x \right )\left ( 6x - 5x \right )}$$
$$\displaystyle= \sqrt{6x\times 3x\times 2x\times x=6x^{2}}$$
Given $$\displaystyle 6x^{2}=216\Rightarrow x^{2}=\frac{216}{6}=36\Rightarrow x=6$$
$$\displaystyle \therefore $$ The sides of the triangle are $$(3\times 6)cm, (4\times 6)\ cm$$ and $$(5\times 6)\ cm$$ i.e $$18\ cm, 24\ cm$$ and $$30\ cm$$
Perimeter of the $$\displaystyle \Delta  $$ =$$ 18 cm + 24 cm +30 cm$$
= 72 cm

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

                                         $$= 12a$$
Given, perimeter of the triangle $$ = 144  m $$

$$ \Rightarrow 12a = 144  m $$

$$ a = 12   m $$
So, the sides of the triangle are

 $$ 3a = 36  m,\\ 4a = 48  m, \\5a  = 60  m $$.

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

Where

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

For triangle with sides $$36  m, 48  m \ and\  60  m,$$

 $$ s = \dfrac { 36+48+60 }{ 2 } \\= 72  m  $$
Thus 

$$A =\sqrt { 72(72-36)(72-48)(72-60) } \\=\sqrt { 72\times 36\times 24\times 12 }$$

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

$$= 36\times 12\times 2 \\= 864 {m}^{2}$$

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 13  m, 14  m and 15  m, $$ s = \dfrac { 13+14+15 }{ 2 } = 21\  m  $$

Area of the triangle with $$13\  m$$, $$14\  m$$ and $$15\  m$$ $$ =\sqrt { 21(21-13)(21-14)(21-15) } $$

$$=\sqrt { 21\times 8\times 7\times 6 }$$

$$ =\sqrt { 3\times 7\times 2\times 4\times 7\times 3 \times 2}$$

$$ = 3\times 7\times 2 \times 2 = 88\  {m}^{2}$$

$$\therefore $$ cost of painting it at the rate of Rs $$ 8.75 $$ per $$ { m }^{ 2 } = 84 \times 8.75 = $$ Rs. 735 

Let $$a, b$$ and $$c$$ denotes the length of the sides of the triangle. 
Area of the triangle, $${ A }_{ 1 }$$ $$=$$ $$\sqrt{s(s-a)(s-b)(s-c)}$$, where s is the semi-perimeter of the triangle.
So, semiperimeter $$\displaystyle s = \frac{a+b+c}{2}$$
When the sides of the triangle are doubled, we get
$${ s }^{ \prime  }=\dfrac { 2a+2b+2c }{ 2 } =a+b+c=2s$$, where $${ s }^{ \prime  }$$ is the semi-perimeter of the new triangle

For a quadrilateral with $$4$$ sides given and none of the angle is $$90°$$, then area of quadrilateral area can  found by using Heron's formula only and sum of all the angles is $$360°$$

Since $$16^2 + 30^2 = 34^2$$, clearly, given triangle is a right angled triangle. Hence
Area $$=\, \displaystyle \frac {1}{2}\, \times\, base\, \times\, height$$
$$=\, \displaystyle \frac {1}{2}\, \times\, 16\, \times\, 30\, =\, 240\, cm^2$$