## Quant Question Of The Day: 247

**Find the ratio of Area of Circumcircle to the are of triangle having side 3, 4, 5. (circumcircle of the same triangle)**

**275/84****3/1****121/60****225/125**

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**Find the ratio of Area of Circumcircle to the are of triangle having side 3, 4, 5. (circumcircle of the same triangle)**

**275/84****3/1****121/60****225/125**

**Answer: (1) 275/84**

**Solution**

**Area of triangle by hero’s formula = [s(s-a)(s-b)(s-c)]^0.5**

**s= semiperimeter = (3+4+5)/2 = 6**

**so area = [6(3)(2)(1)]^0.5 = 6 units**

**now area = abc /4R = 6 units , where R is circumradius**

**3*4*5/4R = 6 ; R = 5/2 units**

**so area of circumcircle = 22/7 (25/4) = 275/14 units**

**So ratio of Area = (275/14)/6 = 275/84**

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option 1) 275/84

The right angled triangle with sides 3K 4K 5K has inradius of K and circumradius of 2.5K.

Remember it and get the answer.