Question

PLEASE ANSWER THIS....!! ​

Answer 1

Question :

If the perimeter of a triangular field is 144 m and ratio of the sides is 3:4:5, find the area of the field.

Given :

• Ratio of the sides = 3:4:5
• Perimeter of triangular field = 144 m

To find :

• Area of the field

According to the question,

Let,

• The side be 3x,4x and 5x.

We know,

• Perimeter of triangle = Sum of all its side

⇒ Perimeter of triangle = Sum of all its side

⇒ 144 m = 3x + 4x + 5x

⇒ 144 m = 12x

⇒ 144 m ÷ 12 = x

⇒ 12 m = x

★ First side of triangle :

⇒ 3x

⇒ 3(12 m)

⇒ 36 m

★ Second side of triangle :

⇒ 4x

⇒ 4(12 m)

⇒ 48 m

★ Third side of triangle :

⇒ 5x

⇒ 5(12 m)

⇒ 60 m

• So, the sides of triangle is 36 m,48 m and 60 m.

Now,

$\sf : \implies{Semi \: perimeter,s = \dfrac{a + b + c}{2} }$

$\sf : \implies{ Semi \: perimeter,s = \dfrac{36 m+ 48 m+ 60m}{2} }$

$\sf: \implies{Semi \: perimeter ,s= \dfrac{144m}{2} \: \leadsto { \underline{ \boxed{ \sf{ \red{72m }}}}}}$

Now,

$\sf: \implies{Area \: of \: triangle = \sqrt{s(s - a)(s - b)(s - c)} }$

$\sf: \implies{ \sqrt{72(72 -36)(72 - 48)(72 - 60)} {m}^{2} }$

$\sf: \implies{ \sqrt{72(36)(24)(12)} \: {m}^{2} }$

$\sf: \implies{12 \times 6 \times 6 \times 2 \: {m}^{2} }$

${ \underline{ \boxed{ \bf{ \orange{: \implies{ {864 \: m}^{2} }}}}}} \: \bigstar$

$\therefore{ \underline{ \sf{So, \: the \: area \: of \: the \: field \: is \: {864 m}^{2} }}}$

Answer 2

Answer:

Given :-

• Perimeter = 144
• Sides = 3:4:5

To Find :-

Area

SoluTion :-

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

3x + 4x + 5x = 144

12x = 144

x = 144/12

x = 12

3x = 3 × 12 = 36

4x = 4 × 12 = 48

5x = 5 × 12 =60

$\rule{375}{3.5}$

Now,

Semiperimeter = Perimeter/2

Semiperimeter = 144/2

Semiperimeter = 72

Now,

Finding Area

Area = √s(s-a)(s-b)(s-c)

Area = √72(72 - 36)(72-48)(72-60)

Area = √72(36)(24)(12)

Area = √746,496

Area = 864 m