In a Δ A B C ,   2 Δ 2 = a 2 b 2 c 2 a 2 + b 2 + c 2 , then the triangle is

In a ΔABC, 2Δ2=a2b2c2a2+b2+c2, then the triangle is
  1. Equilateral
  2. Isosceles
  3. Right angled
  4. Acute angled triangle

Solution

We have, 2Δ2(a2+b2+c2)=a2b2c2 

(a2+b2+c2)=(abcΔ)2.12=8R2 

sin2A+sin2B+sin2C=2 

cos2A+cos2B+cos2C=1 

1 4cosAcosBcosC=1 

cosAcosBcosC=0 

cosa=0 or cosB=0 or cosC=0 

Therefore, any one of the angle is 90°, hence, it is a right angled triangle

Asked in: AP EAMCET 2021 (20 Aug Shift 1)

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