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प्रश्न
Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 3.
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उत्तर १
The distance d between two points `(x_1,y_1)` and `(x_2, y_2)` is given by the formula
`d = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)`
In an isosceles triangle two sides will be of equal length.
Here two vertices of the triangle is given as A (2, 0) and B (2, 5). Let the third side of the triangle be C(x, y)
It is given that the length of the equal sides is 3 units.
Let us now find the length of the side in which both the vertices are known.
`AB = sqrt((2 - 2)^2 + (0 - 5)^2)`
`= sqrt((0)^2 + (-5)^2)`
`= sqrt(0 + 25)`
`= sqrt25`
AB = 5
So, now we know that the side ‘AB’ is not one of the equal sides of the isosceles triangle.
So, we have AC = BC
`AC = sqrt((2 - x)^2 + (0 - y)^2)`
`BC = sqrt((2 - x)^2 + (5 - y)^2)`
Equating these two equations we have,
`sqrt((2 - x)^2 + (0 - y)^2) = sqrt((2 - x)^2 + (5 - y)^2)`
Squaring on both sides of the equation we have,
`(2 - x)^2 + (0 - y)^2 = (2 - x)^2 + (5 - y)^2`
`4 + x^2 - 4x + y^2 = 4 + x^2 - 4x + 25 + y^2 - 10y`
10y = 25
`y = 5/2`
y = 2.5
We know that the length of the equal sides is 3 units. So substituting the value of ‘y’ in equation for either ‘AC’ or ‘BC’ we can get the value of ‘x’.
`AC = sqrt((2 - x)^2 + (0 - y)^2)`
`3 = sqrt((2 -x)^2 + (-5/2)^2)`
Squaring on both sides,
`9 = 4 + x^2 - 4x + 25/4`
`5 = x^2 - 4x + 25/4`
`20 = 4x^2 - 16x + 25`
`-5 = 4x^2 - 16x`
We have a quadratic equation for ‘x’. Solving for roots of the above equation we have,
`4x^2 - 16x + 5 = 0
`x = (16+-sqrt(256 - 4(4)(5)))/8`
` = (16 +- sqrt176)/8`
`= (16 +- 4sqrt11)/8`
`x = 2 +- sqrt(11)/2`
Hence the possible co−ordinates of the third vertex of the isosceles triangle are `(2 + sqrt11/2, 5/2)` or `(2 - sqrt11/2, 5/2)`
उत्तर २
The distance d between two points `(x_1,y_1)` and `(x_2, y_2)` is given by the formula
`d = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)`
In an isosceles triangle two sides will be of equal length.
Here two vertices of the triangle is given as A (2, 0) and B (2, 5). Let the third side of the triangle be C(x, y)
It is given that the length of the equal sides is 3 units.
Let us now find the length of the side in which both the vertices are known.
`AB = sqrt((2 - 2)^2 + (0 - 5)^2)`
`= sqrt((0)^2 + (-5)^2)`
`= sqrt(0 + 25)`
`= sqrt25`
AB = 5
So, now we know that the side ‘AB’ is not one of the equal sides of the isosceles triangle.
So, we have AC = BC
`AC = sqrt((2 - x)^2 + (0 - y)^2)`
`BC = sqrt((2 - x)^2 + (5 - y)^2)`
Equating these two equations we have,
`sqrt((2 - x)^2 + (0 - y)^2) = sqrt((2 - x)^2 + (5 - y)^2)`
Squaring on both sides of the equation we have,
`(2 - x)^2 + (0 - y)^2 = (2 - x)^2 + (5 - y)^2`
`4 + x^2 - 4x + y^2 = 4 + x^2 - 4x + 25 + y^2 - 10y`
10y = 25
`y = 5/2`
y = 2.5
We know that the length of the equal sides is 3 units. So substituting the value of ‘y’ in equation for either ‘AC’ or ‘BC’ we can get the value of ‘x’.
`AC = sqrt((2 - x)^2 + (0 - y)^2)`
`3 = sqrt((2 -x)^2 + (-5/2)^2)`
Squaring on both sides,
`9 = 4 + x^2 - 4x + 25/4`
`5 = x^2 - 4x + 25/4`
`20 = 4x^2 - 16x + 25`
`-5 = 4x^2 - 16x`
We have a quadratic equation for ‘x’. Solving for roots of the above equation we have,
`4x^2 - 16x + 5 = 0
`x = (16+-sqrt(256 - 4(4)(5)))/8`
` = (16 +- sqrt176)/8`
`= (16 +- 4sqrt11)/8`
`x = 2 +- sqrt(11)/2`
Hence the possible co−ordinates of the third vertex of the isosceles triangle are `(2 + sqrt11/2, 5/2)` or `(2 - sqrt11/2, 5/2)`
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