CBSE Questions for Class 9 Maths Heron'S Formula Quiz 2 - MCQExams.com

Let the sides of the triangle be $$ 13a, 14a, 15a $$.
Perimeter of the triangle $$ = $$ Sum of all sides $$ = 13a + 14a + 15a = 42a$$
Given, perimeter of the triangle $$ = 84  cm $$

$$ \therefore  42a = 84  cm $$

$$ a = 2   cm $$
So, the sides of the triangle are $$ 13a = 26  cm, 14a = 28  cm, 15a  = 30  cm $$.

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

For triangle with sides 26  cm, 28  cm and 30  cm, $$ s = \displaystyle \frac { 26+28+30 }{ 2 }= 42  cm  $$
Area of the triangle with 26  cm, 28  cm and 30  cm $$ =\sqrt { 42(42-26)(42-28)(42-30) }$$

$$ =\sqrt { 42\times 16\times 14\times 12 } $$

$$=\sqrt { 6\times 7\times 4\times 4\times 7\times 2 \times 6 \times 2} $$

$$= 6\times 7\times 4 \times 2 = 336  {cm}^{2}$$

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 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 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