What is the Circumradius of the triangle with vertices (22,0) ,(2a, 2b), (0,2b)
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2
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7
a
2
+b
2
Let A(2a,0)B(2a,2b)C(0,2b)
Slope of AB=
2a−2a
2b−0
=
0
2b
=∞
Slope of BC=
0−2a
2b−2b
=0
AB⊥BC(∵AB∣∣y axis and BC∣∣ x axis)
The triangle is right at the vertex B. Hence AC is the hypotenuse, whose mid-point is the circumcentre S.
S=(
2
2a+0
,
2
0+2b
)=(a,b)
Circum radius=SA=SB=SC
SB=
(2a−a)
2
+(2b−b)
2
=
a
2
+b
2
.
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