Question

Quality Needed...✔️


There is slide in a park .One of its side walls has been painted in some colour with a message " Keep the park green and clean " . lf the side of the wall are 15 m, 11 m, and 6 m , find the area painted in colour ?​
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Answer 1

Answer :-

• Area if painted wall = 20√2 cm²

Given :-

• Length of first side = 15 metres
• Length of second side = 11 metres
• Length of third side = 6 metres

To Find :-

• The area of Painted wall.

Explanation :-

As we know, by using Heron's Formulae we need semi-perimeter of triangle or triangular object.

Finding Semi-perimeter of wall

$\implies \rm \dfrac{Side \: 1 + Side \: 2 + Side \: 3}{2}$

$\implies \rm \dfrac{15 + 11 + 6}{2}$

$\implies \rm \dfrac{32}{2}$

$\pink{\boxed { \implies \rm \bold { 16 \: cm}}}$

Now, Substituting the values

$\blue { \boxed { \bf \bigstar \: Heron's \: Formulae = \sqrt{s (s - a) (s - b) (s - c) \: \bigstar }}}$

$\implies \rm \sqrt{16 (16 - 15) (16 - 11) (16 - 6)}$

$\implies \rm \sqrt{16 (1) (5) (10)}$

$\implies \rm \sqrt{16 \times 1 \times 5 \times 10}$

$\implies \rm \sqrt{800}$

$\red{ \underline { \boxed { \implies \rm \bold { 20 \sqrt{2} \: cm^2 }}}}$

Hence, the area of painted wall is 20√2 cm²

Answer 2

Answer:

Answer :-

Area if painted wall = 20√2 cm²

Given :-

Length of first side = 15 metres

Length of second side = 11 metres

Length of third side = 6 metres

To Find :-

The area of Painted wall.

Explanation :-

As we know, by using Heron's Formulae we need semi-perimeter of triangle or triangular object.

Finding Semi-perimeter of wall

\implies \rm \dfrac{Side \: 1 + Side \: 2 + Side \: 3}{2}⟹

2

Side1+Side2+Side3

\implies \rm \dfrac{15 + 11 + 6}{2}⟹

2

15+11+6

\implies \rm \dfrac{32}{2}⟹

2

32

\pink{\boxed { \implies \rm \bold { 16 \: cm}}}

⟹16cm

Now, Substituting the values

\blue { \boxed { \bf \bigstar \: Heron's \: Formulae = \sqrt{s (s - a) (s - b) (s - c) \: \bigstar }}}

★Heron

′

sFormulae=

s(s−a)(s−b)(s−c)★

\implies \rm \sqrt{16 (16 - 15) (16 - 11) (16 - 6)}⟹

16(16−15)(16−11)(16−6)

\implies \rm \sqrt{16 (1) (5) (10)}⟹

16(1)(5)(10)

\implies \rm \sqrt{16 \times 1 \times 5 \times 10}⟹

16×1×5×10

\implies \rm \sqrt{800}⟹

800

\red{ \underline { \boxed { \implies \rm \bold { 20 \sqrt{2} \: cm^2 }}}}

⟹20

2

cm

2

Hence, the area of painted wall is 20√2 cm²