Let A(α,−2),B(α,6) and Cα4,−2 be vertices of a △ABC. If 5,α4 is isthe circumcentre of △ABC , then which of the following is NOT correct about △ABC=

A(α,−2):B(α,6):Cα4,−2Since AC is perpendicular to AB.So, △ABC is right angled at A.Circumcentre = mid point of BC.=5α8,2∴5α8=5&α4=2Area =12(6)(8)=24Perimeter = 24Circumradius = 5Inradius =Δs=2412=2

Let A(α,−2),B(α,6) and Cα4,−2 be vertices of a △ABC. If 5,α4 is isthe circumcentre of △ABC , then which of the following is NOT correct about △ABC=

A(α,−2):B(α,6):Cα4,−2Since AC is perpendicular to AB.So, △ABC is right angled at A.Circumcentre = mid point of BC.=5α8,2∴5α8=5&α4=2Area =12(6)(8)=24Perimeter = 24Circumradius = 5Inradius =Δs=2412=2

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Let $A (α, - 2), B (α, 6) and C (\frac{α}{4}, - 2)$ be vertices of a $△ ABC$ . If $(5, \frac{α}{4})$ is isthe circumcentre of $△ ABC$ , then which of the following is NOT correct about $△ ABC =$

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perimeter is 25

circumradius is 5

area is 24

inradius is 2

answer is B.

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Detailed Solution

$A (α, - 2) : B (α, 6) : C (\frac{α}{4}, - 2)$
Since AC is perpendicular to AB.
So, $△ ABC$ is right angled at A.
Circumcentre = mid point of $BC . = (\frac{5 α}{8}, 2)$
$∴ \frac{5 α}{8} = 5 & \frac{α}{4} = 2$
Question Image
Area $= \frac{1}{2} (6) (8) = 24$
Perimeter = 24
Circumradius = 5
Inradius $= \frac{Δ}{s} = \frac{24}{12} = 2$