Two corners of an isosceles triangle are at $(2, 6)$ $(2 ,6 )$ and $(3, 2)$ $(3 ,2 )$ . If the triangle's area is $48$ $48$ , what are the lengths of the triangle's sides?

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Two corners of an isosceles triangle are at $(2, 6)$ $(2 ,6 )$ and $(3, 2)$ $(3 ,2 )$ . If the triangle's area is $48$ $48$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-02-10T10:40:58

The length of three sides of triangle are $4.12, 23.37, 23.37$ $4.12, 23.37 ,23.37$ unit

Explanation:

The base of the isosceles triangle, $b = \sqrt{{(x_{1} - x_{2})}_{1}^{2} + {(y_{1} - y_{2})}_{1}^{2}} = \sqrt{{(2 - 3)}^{2} + {(6 - 2)}^{2}} = \sqrt{17} = 4.12 (2 d p) u n i t$ $b=sqrt( (x_1-x_2)^2+(y_1-y_2)^2) = sqrt( (2-3)^2+(6-2)^2) =sqrt17=4.12(2dp)unit$

The area of an isosceles triangle is $A_t = 1/2 * b *h =1/2*4.12 *h ; A_t=48 :. h = (2* A_t )/b = (2*48)/4.12=96/4.12= 23.28(2dp) unit$ . Where $h$ is the altitude of triangle.

The legs of the isosceles triangle are $l_1=l_2= sqrt(h^2+(b/2)^2)=sqrt(23.28^2+(4.12/2)^2) =23.37(2dp) unit$

Hence the length of three sides of triangle are $4.12(2dp), 23.37(2dp) ,23.37(2dp)$ unit [Ans]

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Two corners of an isosceles triangle are at $(2, 6)$ $(2 ,6 )$ and $(3, 2)$ $(3 ,2 )$ . If the triangle's area is $48$ $48$ , what are the lengths of the triangle's sides?

1 Answer

Explanation:

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Science

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Humanities

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Two corners of an isosceles triangle are at (2 ,6 )(2,6)(2 ,6 ) and (3 ,2 )(3,2)(3 ,2 ). If the triangle's area is 48 4848 , what are the lengths of the triangle's sides?

1 Answer

Explanation:

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Two corners of an isosceles triangle are at $(2, 6)$ $(2 ,6 )$ and $(3, 2)$ $(3 ,2 )$ . If the triangle's area is $48$ $48$ , what are the lengths of the triangle's sides?