Math, asked by yashjumale, 8 months ago

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Answered by atahrv
1

Answer:

\red\large\boxed{3)\:Area\:of\:Equilateral\:Triangle=\frac{\sqrt{3} }{4} \times49}

Step-by-step explanation:

Given:-

   Sides of an Equilateral Triangle are (2a-b+5), (a+b) and (2b-a+2).

To Find:-

   Area of the Equilateral Triangle.

We Know That:-

  All sides of an Equilateral Triangle are Equal.

∴(2a-b+5)=(a+b)=(2b-a+2)

Formula Applied:-

  • Area of Equilateral=\frac{\sqrt{3} }{4} \times(side)^2

Solution:-

→(2a-b+5)=(a+b)=(2b-a+2)

⇒(2a-b+5)=a+b

⇒2a-a-b-b+5=0

⇒a-2b+5=0

⇒a=2b-5 ---------(1)

(2b-a+2)=a+b --------------(2)

Put equation(1) in equation(2):-

⇒[2b-(2b-5)+2]=2b-5+b

⇒2b-2b+5+2=3b-5

⇒3b-5=7

⇒3b=7+5=12

⇒b=\frac{12}{3}

⇒b=4

2a-b+5=a+b

Put value of b=4,

⇒2a-4+5=a+4

⇒2a-a=4-1

⇒a=3

Now, we know that side of Equilateral Triangle=a+b

⇒Side of Equilateral Triangle=a+b, where a=3 and b=4.

⇒Side of Equilateral Triangle=3+4

⇒Side of Equilateral Triangle=7 units.

Area of Equilateral Triangle=\frac{\sqrt{3} }{4} \times(side)^2, where side=7 units.

Area of Equilateral Triangle=\frac{\sqrt{3} }{4} \times(7)^2

\blue\boxed{Area\:of\:Equilateral\:Triangle=\frac{\sqrt{3} }{4} \times49}

Answered by silentlover45
3

\underline\mathfrak{Given:-}

  • \: \: \: \: \: The \: \: sides \: \: of \: \: an \: \: equilateral \: \: triangle \: \: are \: \: {(2a \: - \: b \: + \: 5),} \: \: {(a \: + \: b),} :\ \: and \: \: {(2b \: - \: a \: + \: 2)}

\underline\mathfrak{To \: \: Find:-}

  • \: \: \: \: \: The \: \: area \: \: of \: \: the \: \: triangle.?

\huge\underline\mathfrak{Solutions:-}

  • \: \: \: \: \: \: \: In \: \: \triangle \: \: ABC

\: \: \: \: \: \fbox{AB \: \: = \: \: BC \: \: = \: \: AC \: \: is \: \: an \: \: Equilateral \: \: Triangle.}

  • \: \: \: \: \: AB \: \: = \: \: (2a \: - \: b \: + \: 5)

  • \: \: \: \: \: BC \: \: = \: \: (a \: + \: b)

  • \: \: \: \: \: AC \: \:= \: \: (2b \: - \: a \: + \: 2)

\: \: \: \: \: \underline{\angle{AB} \: \: = \: \: \angle{BC}}

\: \: \: \: \: \leadsto {(2a \: - \: b \: + \: 5)} \: \: = \: \: {(a \: + \: b)}

\: \: \: \: \: \leadsto {a \: - \: 2b} \: \: = \: \: {-5} \: \: .......(1)

\: \: \: \: \: \underline{\angle{BC} \: \: = \: \: \angle{AC}}

\: \: \: \: \: \leadsto {(2b \: - \: a \: + \: 2)} \: \: = \: \: {(a \: + \: b)}

\: \: \: \: \: \leadsto {b \: - \: 2a} \: \: = \: \: {-2}

\: \: \: \: \: \leadsto {2a \: - \: b} \: \: = \: \: {2} \: \: ......(2)

\: \: \: \: \: Now, \: \: the \: \: multiplying \: \: {2} \: \: in \: \: Eq. \: \: (2).

\: \: \: \: \: \leadsto {4a \: - \: 2b} \: \: = \: \: {4} \: \: ......(2)

\: \: \: \: \: \underline{By \: \: Elimination \: \: Mathod}

 4a \:  \:  -  \:  \: 2b \:  \:  =  \:  \:  \:  \: 4

\:  \:  a \:  \:  -  \:  \: 2b \:  \:  =  \:  - 5

  \: \: -  \:  \:  \:  \:  \:  +  \:  \:  \:  \:  \:  \: =  \:  \:  +

___________________

  \:  \:  \:  \:  \:  \:  \:  \:  \: 3a \:  \:  =  \:  \: 9

\: \: \: \: \: \leadsto {a} \: \: = \: \: {3}

\: \: \: \: \: Now, \: \: putting \: \: the \: \: value \: \: of \: \: in \: \: Eq. \: (1).

\: \: \: \: \: \leadsto {3 \: - \: 2b} \: \: = \: \: {-5}

\: \: \: \: \: \leadsto {3 \: + \: 5} \: \: = \: \: {2b}

\: \: \: \: \: \leadsto {8} \: \: = \: \: {2b}

\: \: \: \: \: \leadsto {b} \: \: = \: \: {4}

\: \: \: \: \: So, \: \: putting \: \: the \: \: value \: \: of  \: \: a \: and \: b \: \: all \: \: sides.

  • \: \: \: \: \: \angle{AB} \: \: = \: \: (2a \: - \: b \: + \: 5)

\: \: \: \: \: \leadsto \: \: {(2 \: \times \: 3 \: - \: 4 \: + \: 5)}

\: \: \: \: \: \leadsto \: \: {7} \: \: unit.

  • \: \: \: \: \: \angle{BC} \: \: = \: \: (a \: + \: b)

\: \: \: \: \: \leadsto \: \: {3 \: + \: 4}

\: \: \: \: \: \leadsto \: \: {7} \: \: unit.

  • \: \: \: \: \: \angle{AC} \: \:= \: \: (2b \: - \: a \: + \: 2)

\: \: \: \: \: \leadsto \: \: {2 \: \times \: 4 \: - \: 3 \: + \: 2}

\: \: \: \: \: \leadsto \: \: {7} \: \: unit.

\: \: \: \: \: Area \: \: of \: \: Equilateral \: \: triangle \: \: \frac{\sqrt{3}}{4} \: {(side)}^{2}

\: \: \: \: \: \leadsto \: \: \frac{\sqrt{3}}{4} \: {(7)}^{2}

\: \: \: \: \: \leadsto \: \: \frac{{49}\sqrt{3}}{4}

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