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Engineering
5 months, 3 weeks ago

Can someone explain If an equilateral triangle, having vertexat the (2,-1), has a side along the line, x+y=2, then the area (in sq. units) of this triangle is :

If an equilateral triangle, having vertex at the (2,-1), has a side along the line, x+y=2, then the area (in sq. units) of this triangle is :

 

  • Option 1)

    3\sqrt{6}

  • Option 2)

    6

  • Option 3)

    6\sqrt{3}

  • Option 4)

    \frac{9}{2}\sqrt{3}

 
Answers (2)
60 Views
A arun shelke
Answered 5 months ago

ACCORDING TO QUESTION ,WE JUST HAVE TO FIND THE PERPENDICULAR DISTANCE OF POINT (2,-1) FORM THE LINE

x + y = 2,THIS PERPENDICULAR DISTANCE IS EQUAL TO THE HEIGHT OF THE TRIANGLE FORMED,AND THEN WE CAN CALCULATE THE SIDE OF TRIANGLE,AND THEN BY APPLING FORMULA OF AREA OF TRIANGLE  {(1/2)*b*h}

SO,

DISTANCE OF POINT FRON THE LINE (h) = \left |x_1+y_1-c \right |/\left | \sqrt{x_1^{2}+y_1^{2}} \right |  

AND ALSO WE KNOW THAT HEIGHT OF THE TRIANGLE = (\sqrt{3}*a)/2

NOW, h = 1/\sqrt{2 } AND ALSO h = (\sqrt{3}*a)/2

FROM HERE WE GET THE VALUE OF a(BASE OF TRIANGLE(b)) = \sqrt{2/3} 

NOW BY FORMULA {(1/2)*b*h} = (1/2) *  \sqrt{2/3} *  1/\sqrt{2 } = 1/(\sqrt{3})*2 sq.units.

 BUT UNFORTUNATELY NO OPTION IS MATCHING OUR ANSWER.SO, THE ANSWER IS 1/2\sqrt{3}

 

S solutionqc
Answered 5 months, 3 weeks ago


Option 1)

3\sqrt{6}

Option 2)

6

Option 3)

6\sqrt{3}

Option 4)

\frac{9}{2}\sqrt{3}

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