Two corners of an isosceles triangle are at $(8, 2)$ $(8 ,2 )$ and $(7, 5)$ $(7 ,5 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?

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Answer 1 · 2018-05-12T04:10:33

$Length of triangle sides 3.16, 40.51, 40.51$ $color(brown)("Length of triangle sides " 3.16, 40.51, 40.51$

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$A = (8, 2), C = (7, 5) A_{t} = 64$ $A = (8,2), C = (7,5) A_t = 64$

$¯ ¯¯¯¯ ¯ A C = b = \sqrt{{(8 - 7)}^{2} + {(2 - 5)}^{2}} = \sqrt{10} = 3.16$ $bar (AC) = b = sqrt((8-7)^2 + (2-5)^2) = sqrt10 = 3.16$

$A_{t} = 64 = (\frac{1}{2}) \cdot b \cdot h = (\frac{1}{2}) \cdot \sqrt{10} \cdot h$ $A_t = 64 = (1/2) * b * h = (1/2) * sqrt10 * h$

$h = \frac{2 \cdot 64}{\sqrt{10}} = \frac{128}{\sqrt{10}}$ $h = (2 * 64) / sqrt(10) = 128 / sqrt10$

$¯ ¯¯¯¯ ¯ A B = ¯ ¯¯¯¯ ¯ A C = a = \sqrt{{(\frac{b}{2})}^{2} + h^{2}}$ $bar (AB) = bar (AC) = a = sqrt((b/2)^2 + h^2)$

$a = \sqrt{{(\frac{\sqrt{10}}{2})}^{2} + {(\frac{128}{\sqrt{10}})}^{2}}$ $a = sqrt((sqrt10/2)^2 + (128/sqrt10)^2)$

$a = \sqrt{(\frac{10}{4}) + (\frac{16384}{10})} = 40.51 units$ $a = sqrt((10/4) + (16384/10)) = 40.51 " units"$

Two corners of an isosceles triangle are at (8,2)(8 ,2 ) and (7,5)(7 ,5 ). If the triangle's area is 6464 , what are the lengths of the triangle's sides?