Question

Find the area of a triangular field with sides 55m, 60m and 65m. Find the cost of

laying the grass in the same field at the rate of Rs.8 per m2

.​

Answer 1

Concept Used

• Heron's formula

Let us suppose that three sides be represented as

$\begin{gathered}\begin{gathered}\bf \: sides \: are \: - \begin{cases} &\sf{a \: = \: 55} \\ &\sf{b \: = \: 60}\\ &\sf{c \: = \: 65} \end{cases}\end{gathered}\end{gathered}$

Now,

• Semi perimeter is given by

$\bigstar \: \: \underline{\boxed{\sf Semi \ perimeter \: (s)=\dfrac{1}{2}(a \: + \: b \: + \: c)}}$

$\rm : \implies \:s \: = \: \dfrac{55 + 60 + 65}{2}$

$\rm : \implies \:s \: = \: \dfrac{180}{2}$

$\boxed{ \pink{\rm : \implies \:s \: = \: 90 \:m \: }}$

Now,

• Area of triangle is given by

$\underline{\boxed{\bf Area \ of \ triangle=\sqrt{s(s-a)(s-b)(s-c)} }}$

$\rm : \implies \:Area = \sqrt{90(90 - 55)(90 - 60)(90 - 65)}$

$\rm : \implies \:Area = \sqrt{90 \times 35 \times 30 \times 25}$

$\rm : \implies \:Area = \sqrt{3 \times 3 \times 10 \times 3 \times 10 \times 5 \times 5 \times 7 \times 5}$

$\rm : \implies \:Area = 3 \times 10 \times 5 \sqrt{3 \times 5 \times 7}$

$\rm : \implies \:Area = 150 \sqrt{105} \: {m}^{2}$

$\boxed{ \pink{ \rm : \implies \:Area \: = \: 1537.04 \: {m}^{2} }}$

Now to find the Cost of laying grass in the field.

$\rm : \implies \: Cost \: of \: laying \: grass \: 1 \: {m}^{2} \: is \: Rs 8$

$\rm : \implies \: Cost \: of \: laying \: grass \: 1537.04 \: {m}^{2} \: = \: Rs \: 8 \times 1537.04$

$\boxed{ \pink{ \ \rm : \implies \:Cost \: is \: Rs \: 12296 \: .32}}$

Answer 2

Answer:

Heron's formula

Let us suppose that three sides be represented as

\begin{gathered}\begin{gathered}\begin{gathered}\bf \: sides \: are \: - \begin{cases} &\sf{a \: = \: 55} \\ &\sf{b \: = \: 60}\\ &\sf{c \: = \: 65} \end{cases}\end{gathered}\end{gathered}\end{gathered}

sidesare−

⎩

⎪

⎪

⎨

⎪

⎪

⎧

a=55

b=60

c=65

Now,

Semi perimeter is given by

\bigstar \: \: \underline{\boxed{\sf Semi \ perimeter \: (s)=\dfrac{1}{2}(a \: + \: b \: + \: c)}}★

Semi perimeter(s)=

2

1

(a+b+c)

\rm : \implies \:s \: = \: \dfrac{55 + 60 + 65}{2}:⟹s=

2

55+60+65

\rm : \implies \:s \: = \: \dfrac{180}{2}:⟹s=

2

180

\boxed{ \pink{\rm : \implies \:s \: = \: 90 \:m \: }}

:⟹s=90m

Now,

Area of triangle is given by

\underline{\boxed{\bf Area \ of \ triangle=\sqrt{s(s-a)(s-b)(s-c)} }}

Area of triangle=

s(s−a)(s−b)(s−c)

\rm : \implies \:Area = \sqrt{90(90 - 55)(90 - 60)(90 - 65)}:⟹Area=

90(90−55)(90−60)(90−65)

\rm : \implies \:Area = \sqrt{90 \times 35 \times 30 \times 25}:⟹Area=

90×35×30×25

\rm : \implies \:Area = \sqrt{3 \times 3 \times 10 \times 3 \times 10 \times 5 \times 5 \times 7 \times 5}:⟹Area=

3×3×10×3×10×5×5×7×5

\rm : \implies \:Area = 3 \times 10 \times 5 \sqrt{3 \times 5 \times 7}:⟹Area=3×10×5

3×5×7

\rm : \implies \:Area = 150 \sqrt{105} \: {m}^{2}:⟹Area=150

105

m

2

\boxed{ \pink{ \rm : \implies \:Area \: = \: 1537.04 \: {m}^{2} }}

:⟹Area=1537.04m

2

Now to find the Cost of laying grass in the field.

\rm : \implies \: Cost \: of \: laying \: grass \: 1 \: {m}^{2} \: is \: Rs 8:⟹Costoflayinggrass1m

2

isRs8

\rm : \implies \: Cost \: of \: laying \: grass \: 1537.04 \: {m}^{2} \: = \: Rs \: 8 \times 1537.04:⟹Costoflayinggrass1537.04m

2

=Rs8×1537.04

\boxed{ \pink{ \ \rm : \implies \:Cost \: is \: Rs \: 12296 \: .32}}

:⟹CostisRs12296.32