Two corners of an isosceles triangle are at $(2, 4)$ $(2 ,4 )$ and $(8, 5)$ $(8 ,5 )$ . If the triangle's area is $9$ $9$ , what are the lengths of the triangle's sides?

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Answer 1 · 2018-05-20T12:10:24

Lengths of three sides are $6.08, 4.24, 4.24$ $color(purple)(6.08, 4.24, 4.24$

Given : $A (2, 4), B (8, 5), A r e a = 9$ $A(2,4), B(8,5), Area = 9$ and it’s an isosceles triangle. To find the sides of the triangle.

$A B = c = \sqrt{{(8 - 2)}^{2} + {(5 - 4)}^{2}} = \sqrt{37} = 6.08$ $AB = c = sqrt((8-2)^2 + (5-4)^2) = sqrt37 = 6.08$ , using distance formula.

$A r e a = A_{t} = 9 = (\frac{1}{2}) \cdot c \cdot h$ $Area = A_t = 9 = (1/2) * c * h$

$h = \frac{9 \cdot 2}{\sqrt{37}} = \frac{18}{\sqrt{37}}$ $h = (9*2) / sqrt37 = 18 / sqrt37$

Side $a = b = \sqrt{{(\frac{c}{2})}^{2} + h^{2}}$ $a = b = sqrt((c/2)^2 + h^2)$ , using Pythagoras theorem

$a = b = \sqrt{{(\frac{\sqrt{37}}{2})}^{2} + {(\frac{18}{\sqrt{37}})}^{2}}$ $a = b = sqrt((sqrt37/2)^2 + (18/(sqrt37))^2)$

$\Rightarrow \sqrt{(\frac{37}{4}) + (\frac{324}{37})}$ $=> sqrt((37/4) + (324/37))$

$a = b = 4.24$ $a = b = 4.24$

Two corners of an isosceles triangle are at (2 ,4 )(2,4)(2 ,4 ) and (8 ,5 )(8,5)(8 ,5 ). If the triangle's area is 9 99 , what are the lengths of the triangle's sides?