A circle is inscribed in an equilateral triangle of side a. What is th

1.	A circle is inscribed in an equilateral triangle of side a. What is the area of any square inscribed in this circle?1). a2/32). a2/43). a2/64). a2/8
Answer» Solution: Area of an equilateral triangle with side $a = [ \sqrt{3} a ^2 ] / 4$ Perimeter of an equilateral triangle $= 3 a$ Semi-perimeter of an equilateral triangle $= 3 a /2$ Radius of the incircle = Area / Semi-perimeter $r =[ \sqrt{3} a ^2 ] × 2 / 4 × 3a$ $r =[ \sqrt{3} a ] / 6$ Diameter of the incircle = DIAGONAL of the SQUARE inscribed in the incircle $2r =$ Diagonal of the square Diagonal of the square $= 2 × [ \sqrt{3} a] / 6$ Diagonal of the square $= [ \sqrt{3}) a] / 3$ Area of the square $=(diagonal )^2 / 2$ $= [ \sqrt{3} a]^2 / 3^2 ×2$ $= 3 a^2 / ( 9 × 2)$ $= 3 a^2 / 18$ $= a^2 / 6$ Therefore the area of the square inscribed in the circle is $a^2 / 6$. The correct option is 3). $a^2 / 6$

A circle is inscribed in an equilateral triangle of side a. What is the area of any square inscribed in this circle?1). a2/32). a2/43). a2/64). a2/8

Answer»

Solution:

Area of an equilateral triangle with side $a = [ \sqrt{3} a ^2 ] / 4$

Perimeter of an equilateral triangle $= 3 a$

Semi-perimeter of an equilateral triangle $= 3 a /2$

Radius of the incircle = Area / Semi-perimeter

$r =[ \sqrt{3} a ^2 ] × 2 / 4 × 3a$

$r =[ \sqrt{3} a ] / 6$

Diameter of the incircle

= DIAGONAL of the SQUARE inscribed in the incircle

$2r =$ Diagonal of the square

Diagonal of the square $= 2 × [ \sqrt{3} a] / 6$

Diagonal of the square $= [ \sqrt{3}) a] / 3$

Area of the square $=(diagonal )^2 / 2$

$= [ \sqrt{3} a]^2 / 3^2 ×2$

$= 3 a^2 / ( 9 × 2)$

$= 3 a^2 / 18$

$= a^2 / 6$

Therefore the area of the square inscribed in the circle is $a^2 / 6$.

The correct option is 3). $a^2 / 6$

Discussion

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