Two corners of an isosceles triangle are at $(5, 2)$ $(5 ,2 )$ and $(2, 1)$ $(2 ,1 )$ . If the triangle's area is $7$ $7$ , what are the lengths of the triangle's sides?

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Two corners of an isosceles triangle are at $(5, 2)$ $(5 ,2 )$ and $(2, 1)$ $(2 ,1 )$ . If the triangle's area is $7$ $7$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-02-14T10:39:56

Lengths of three sides of triangle are $3.16, 4.70, 4.70$ $3.16, 4.70,4.70$ unit

Explanation:

The base of the isosceles triangle, $b = \sqrt{{(x_{1} - x_{2})}_{1}^{2} + {(y_{1} - y_{2})}_{1}^{2}} = \sqrt{{(5 - 2)}^{2} + {(2 - 1)}^{2}} = \sqrt{10} = 3.16 (2 d p) u n i t$ $b=sqrt( (x_1-x_2)^2+(y_1-y_2)^2) = sqrt( (5-2)^2+(2-1)^2) =sqrt10=3.16(2dp)unit$

The area of the isosceles triangle is $A_t = 1/2 * b *h =1/2*3.16 *h ; A_t=7 :. h = (2* A_t )/b = (2*7)/3.16=14/3.16= 4.43(2dp) unit$ . Where $h$ is the altitude of triangle.

The legs of the isosceles triangle are $l_1=l_2= sqrt(h^2+(b/2)^2)=sqrt(4.43^2+(3.16/2)^2) =4.70(2dp) unit$

Hence the length of three sides of triangle are $3.16(2dp), 4.70(2dp) ,4.70(2dp)$ unit [Ans]

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Two corners of an isosceles triangle are at $(5, 2)$ $(5 ,2 )$ and $(2, 1)$ $(2 ,1 )$ . If the triangle's area is $7$ $7$ , what are the lengths of the triangle's sides?

1 Answer

Explanation:

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Two corners of an isosceles triangle are at (5 ,2 )(5,2)(5 ,2 ) and (2 ,1 )(2,1)(2 ,1 ). If the triangle's area is 7 77 , what are the lengths of the triangle's sides?

1 Answer

Explanation:

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Two corners of an isosceles triangle are at $(5, 2)$ $(5 ,2 )$ and $(2, 1)$ $(2 ,1 )$ . If the triangle's area is $7$ $7$ , what are the lengths of the triangle's sides?