Two corners of an isosceles triangle are at $(2, 1)$ $(2 ,1 )$ and $(8, 5)$ $(8 ,5 )$ . If the triangle's area is $4$ $4$ , what are the lengths of the triangle's sides?

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Two corners of an isosceles triangle are at $(2, 1)$ $(2 ,1 )$ and $(8, 5)$ $(8 ,5 )$ . If the triangle's area is $4$ $4$ , what are the lengths of the triangle's sides?

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Answer 1 · 2018-02-02T02:28:28

Measure of the triangle's sides $7.2111, 3.7724, 3.7724$ $color(violet)(7.2111, 3.7724, 3.7724)$

Explanation:

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Length of the base (b) is the distance between the given two points (2,1) , (8,5).

Using distance formula,

$B C = a = \sqrt{{(x 2 - x 1)}^{2} + {(y 2 - y 1)}^{2}}$ $BC = a = sqrt((x2-x1)^2 + (y2-y1)^2)$

$a = \sqrt{{(8 - 2)}^{2} + {(5 - 1)}^{2}} = 7.2111$ $a = sqrt((8-2)^2 + (5-1)^2) = color(green)(7.2111)$

Area of triangle $A = (\frac{1}{2}) a h$ $A = (1/2) a h$

$4 = (\frac{1}{2}) 7.2111 \cdot h$ $4 = (1/2) 7.2111 * h$

$A N = h = \frac{2 \cdot 4}{7.2111} = 1.1094$ $AN = h = (2 * 4) / 7.2111 = color(purple)(1.1094)$

$A B = A C = b = c = \sqrt{{(A N)}^{2} + {(B N)}^{2}}$ $AB = AC = b = c = sqrt((AN)^2 + (BN)^2)$

$b = c = \sqrt{h^{2} + {(\frac{a}{2})}^{2}} = \sqrt{{1.1094}^{2} + {(\frac{7.2111}{2})}^{2}} = 3.7724$ $b = c = sqrt(h^2 + (a/2)^2) = sqrt(1.1094^2 + (7.2111/2)^2) = color(red)(3.7724)$

Measure of the triangle's sides $7.2111, 3.7724, 3.7724$ $color(violet)(7.2111, 3.7724, 3.7724)$

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Two corners of an isosceles triangle are at $(2, 1)$ $(2 ,1 )$ and $(8, 5)$ $(8 ,5 )$ . If the triangle's area is $4$ $4$ , 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 ,1 )(2,1)(2 ,1 ) and (8 ,5 )(8,5)(8 ,5 ). If the triangle's area is 4 44 , what are the lengths of the triangle's sides?

1 Answer

Explanation:

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Two corners of an isosceles triangle are at $(2, 1)$ $(2 ,1 )$ and $(8, 5)$ $(8 ,5 )$ . If the triangle's area is $4$ $4$ , what are the lengths of the triangle's sides?