An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle.
Answers
Given: An equilateral triangle of side 9 cm is inscribed in a circle.
To find : The radius of the circle.
Solution :
Let ∆ ABC be an equilateral triangle of side 9 cm and let, AD be one of its medians and G be the centroid of the ∆ABC. Then, AG : GD = 2 : 1
[Centroid divides the median in the ratio 2 : 1]
We know that, in an equilateral triangle centroid coincides with the circumcentre.
Therefore, G is the centre of the circumcircle with circumradius GA.
Also, G is the centre and GD ⊥ BC.
∴ BD = CD = 4.5 cm
In right ∆ ADB,by using Pythagoras theorem ,
AB² = AD² + DB²
9² = AD² + 4.5²
81 = AD² + (45/10)²
81 = AD² + (9/2)²
81 = AD² + 81/4
AD² = 81 - 81/4
AD² = (4 × 81 - 81)/4
AD² = (324 - 81)/4
AD² = 243/4
AD² = (81× 3)/4
AD = √(81× 3)/4
AD = 9√3/2 cm
∴ Radius = AG = ⅔ AD
Radius = AG = ⅔ × 9√3/2
Radius = 3√3 cm
Hence, the radius of the circle is 3√3 cm.
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Answer:
Step-by-step explanation:
In right ∆ ADB,by using Pythagoras theorem ,
AB² = AD² + DB²
9² = AD² + 4.5²
81 = AD² + (45/10)²
81 = AD² + (9/2)²
81 = AD² + 81/4
AD² = 81 - 81/4
AD² = (4 × 81 - 81)/4
AD² = (324 - 81)/4
AD² = 243/4
AD² = (81× 3)/4
AD = √(81× 3)/4
AD = 9√3/2 cm