Question

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Show that the point O(0,0) , A(3,root3 ) and B(3,-root3 ) are the vertices of an equilateral triangle. Find the area of this triangle.

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Answer 1

SOLUTION:-

Given

• O(0,0)
• A( 3, √3)

• B(3, -√3)

To find

• Show that the points 0(0,0) , A(3,√3) and ⠀⠀⠀⠀B(3,-√3) are the vertices of equilateral tri.

⠀⠀• Find area of equilateral triangle

Explanation

By distance formula

$\sqrt{(x1 - x2) {}^{2} + (y1 - y2) {}^{2} }$

For AO⤵️

$= \sqrt{(0 - 3) {}^{2} + (0 - \sqrt{3}) {}^{2} }$

$= \sqrt{( - 3) {}^{2} + ( - \sqrt{3}) {}^{2} }$

$= \sqrt{9 + 3}$

$= \sqrt{12}$

By distance formula

$\sqrt{(x1 - x2) {}^{2} + (y1 - y2) {}^{2} }$

For AB⤵️

$= \sqrt{(3 - 3) {}^{2} + ( \sqrt{3} - \sqrt{3}) {}^{2} }$

$= \sqrt{(0) {}^{2} + ( - 2 \sqrt{3}) {}^{2} }$

$= \sqrt{12}$

By distance formula

$\sqrt{(x1 - x2) {}^{2} + (y1 - y2) {}^{2} }$

For OB⤵️

$\sqrt{(3 - 0) {}^{2} + (0 - ( \sqrt{3}) {}^{}) {}^{2} }$

$= \sqrt{(3) {}^{2} + ( \sqrt{3}) {}^{2} }$

$= \sqrt{9 + 3}$

$= \sqrt{1} 2$

_____________________

$\sqrt{12} = \sqrt{12} = \sqrt{12}$

All Sides are equal

So, It is an equilateral triangle and OAB are the vertices of equilateral triangle.

➡ Area for equilateral triangle = √3/4×(side)²

Area = √3/4×(side)²

➡ √3/4×(√12)²

➡ √3/4 ×12

➡ √3 ×3

Area = 3√3 sq. unit

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