Question

The perimeter of a triangular field is 120 m and the sides are in the ratio of 25::15::20. find its area​

Answer 1

Answer :

600 m²

____________

As per the provided information in the given question, we have :

• Perimeter of the triangular field = 120 m
• Sides are in the ratio of 25:15:20

We are asked to calculate,

• Area of the field.

In order to calculate the area of triangle. We need to find its side.

Finding the sides of the field :

Let us assume the sides of the field as 25x, 15x and 20x.

Now, as we know that,

$\longmapsto \bf {Perimeter_{(\Delta)} = Sum_{(All \; sides)} }$

According to the question,

$\longmapsto \rm { 120 = 25x + 15x + 20x}$

Performing addition in R.H.S.

$\longmapsto \rm { 120 = 60x}$

Transposing 60 from R.H.S to L.H.S. Its arithmetic operator will get changed.

$\longmapsto \rm { \cancel{\dfrac{120}{60}} = x}$

$\longmapsto \rm { 2 = x}$

So, the sides are :

First side :

$\longmapsto \rm { 1st \; side = 25x \; m}$

Substituting the value of x.

$\longmapsto \rm { 1st \; side = 25(2) \; m}$

Performing multiplication.

$\longmapsto \bf { 1st \; side = 50 \; m}$

Second side :

$\longmapsto \rm { 2nd \; side = 15x \; m}$

Substituting the value of x.

$\longmapsto \rm {2nd \; side = 15(2) \; m}$

Performing multiplication.

$\longmapsto \bf { 2nd \; side = 30 \; m}$

Third side :

$\longmapsto \rm { 3rd \; side = 20x \; m}$

Substituting the value of x.

$\longmapsto \rm {3rd \; side = 20(2) \; m}$

Performing multiplication.

$\longmapsto \bf { 3rd \; side = 40 \; m}$

Now, finding the area of the field.

By using Heron's formula,

$\bigstar \; \underline{\boxed{\bf Area_{(\Delta)} = \sqrt{s(s -a)(s -b)(s -c)} }}$

• a,b,c are sides
• s is Semi-perimeter

Finding Semi-perimeter :

$\longmapsto \rm {s = \dfrac{Perimeter}{2} }$

Substituting the value of Perimeter.

$\longmapsto \rm {s = \dfrac{120}{2} m}$

Dividing 120 with 2.

$\longmapsto \bf {s = 60 \; m}$

Substituting all the value in the formula :-

$\longmapsto \rm{Area_{(\Delta)} = \sqrt{s(s -a)(s -b)(s -c)}}$

$\longmapsto \rm{Area_{(\Delta)} = \sqrt{60(60-50)(60 -30)(60 -40)}\; m^2}$

Performing subtraction in the brackets.

$\longmapsto \rm{Area_{(\Delta)} = \sqrt{60(10)(30)(20)}\; m^2}$

$\longmapsto \rm{Area_{(\Delta)} = \sqrt{360000}\; m^2}$

$\longmapsto \rm{Area_{(\Delta)} = \sqrt{6 \times 6 \times 10 \times 10 \times 10 \times 10 }\; m^2}$

$\longmapsto \rm{Area_{(\Delta)} = 6 \times 10 \times 10 \; m^2}$

$\longmapsto \bf{Area_{(\Delta)} = 600 \; m^2}$

∴ Area of the field is 600 m².

Answer 2

Answer:

$\huge\mathcal{\fcolorbox{aqua}{azure}{\red{✿Question ♡❖}}}$

The perimeter of a triangular field is 120 m and the sides are in the ratio of 25::15::20. find its area

$\huge\mathcal{\fcolorbox{aqua}{azure}{\red{✿Yøur-Answer♡❖}}}$

The perimeter of a triangular field =120m

Let the side =25x,15x,20x

Perimeter of a triangle =Sum of three side

$25x + 15x + 20x = 120 \\ 60x =\begin{gathered} \rm { \cancel{\dfrac{120}{60}} = x}\\ \end{gathered}⟼=x = 2$

1st side (a)=25×2=50m

2nd side(b)=15×2=30m

3rd side (c)=20×2=40

Semi--perimeter(S)=

$\begin{gathered} \longmapsto \rm { {\dfrac{a+b+c}{2}} }\\ \end{gathered}$

$\begin{gathered} \longmapsto \rm { {\dfrac{(50+30+40)}{2}} = }\\ \end{gathered}\begin{gathered} \longmapsto \rm { \cancel{\dfrac{120}{2}} = 60m}\\ \end{gathered}$

Area of the triangle =

$\sqrt{S}(s - a)(s - b)(s - c)$

[By Heron's Formula]

$★Area(Δ)=s(s−50)(s−30)(s−40)$

$\sqrt{60 = (60 - 50)(60 - 30(60 - 40} \\ \sqrt{60 \times 10 \times 30 \times 20} \\ = 360000$

$★Area \: \: of \: \: (Δ)=600m {}^{2}$

$600m {}^{2}$

$⟹ \large\bf{\underline{\red{VERIFIED✔}}}$

$\bf\colorbox{lavender}{{\color{b}{✿Hope It Helps You ♡❖ }}}$

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