Question

An equilateral triangle is inscribed in a circle which is incribed in a square. If the side of the

equilateral triangle is 8
$\sqrt{3}$
cm. then find the area of square?​

Answer 1

Answer:

Required area of square is 256 unit^2.

Step-by-step explanation:

As given in the diagram, O is the center of the center of the circle. Since OE is the radius of the circle( according to the figure ), BD or AC should be equal to the diameter of the circle.

Let, radius of circle be a units.  

Thus, side of square is 2a units.  Draw a perpendicular from E on FH( on side of triangle ).

Let the length of OG

By Pythagoras Theorem : -

= > EG^2 + GH^2 = EH^2

= > EG^2 + ( 8√3 / 2 )^2 = ( 8√3 )^2        { Since FG = GH , by properties of equilateral Δ }

= > EG^2 = ( 8√3 )^2 - ( 4√3 )^2

= > EG^2 = 192 - 48

= > EG^2 = 144

= > EG = 12        { Since length can't be negative }

From above,

= > EG = radius of circle + OG

= > 12 = a + OG

= > 12 - a = OG

Again by Pythagoras theorem : -

= > OG^2 + GH^2 = OH^2

= > ( 12 - a )^2 + ( 8√3 / 2 )^2 = a^2              { since OH is radius }

= > 144 + a^2 - 24a + ( 4√3 )^2 = a^2

= > 144 - 24a + 48 = 0

= > 192 = 24a

= > 8 = a

Therefore,

Radius of circle = a = 8 units

Then,

Diameter of circle = side of square = 2 x radius of circle

= > side of square =2 x 8 units

= > side of square = 16 units

From the properties of square : -

• Area of square = side^2

Thus,

= > Area of this square = ( 16 units )^2

= > Area of this square = 256 unit^2

Hence the required area of square is 256 unit^2.