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Question
Show that each of the triangles whose vertices are given below are isosceles :
(i) (8, 2), (5,-3) and (0,0)
(ii) (0,6), (-5, 3) and (3,1).
Solution
(i) Let A (8, 2) B (5, -3) and C (0, 0) be the given point
Then AB = `sqrt((8 - 5)^2 + (2 - 3)^2)`
= `sqrt(9 + 25) = sqrt(34)"units"`.
BC = `sqrt((5 - 10)^2 + (-3 - 0)^2)`
= `sqrt(25 + 9) = sqrt(34)"units"`.
AC = `sqrt((8 - 0)^2 + (2 - 0)^2)`
= `sqrt(64 + 4) = sqrt(68)."units"`.
Here AB = BC = `sqrt(34)`.
Hence, the triangle is isosceles.
(ii) The given points are A(0, 6), B(-5, 3) and C(3, 1)
Then AB = `sqrt((0 + 5)^2 + (6 - 3^2)`
= `sqrt(25 + 9) = sqrt(34)"units"`.
BC = `sqrt((-5 - 3)^2 + (3 - 1)^2)`
= `sqrt(64 + 4) = sqrt(68)"units"`.
Also AC = `sqrt((0 - 3)^2 + (6 - 1)^2)`
= `sqrt(9 + 25) = sqrt(34)"units"`.
Since AB = AC = `sqrt(34)`
Hence, the triangle is an isosceles.
Hence proved.
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