Two corners of an isosceles triangle are at #(2 ,4 )# and #(3 ,8 )#. If the triangle's area is #48 #, what are the lengths of the triangle's sides?

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1 Answer
sankarankalyanam
Apr 19, 2018

#color(maroon)("Lengths of the sides of the triangle are "#

#color(indigo)(a = b = 23.4, c = 4.12#

Explanation:

![http://www.algebra.com/algebra/homework/Geometry-proofs/Geometry_proofshttp://.faq.question.209023.html](https://useruploads.socratic.org/zwonrgLRSdCBPBNldBqq_isosceles%20triangle.png)

#A (2,4), B (3,8), " Area "A_t = 48, " To find AC, BC"#

#vec(AB) = c = sqrt((2-3)^2 + (4-8)^2) = 4.12#

#A_t = (1/2) (AB) * (CD) #

#vec(CD) = h = (2 * 48) / 4.12 = 23.3#

#color(crimson)("Applying Pythagoras Theorem,"#

#vec(AC) = vec(BC) = b = sqrt(h^2 + (c/2)^2)#

#b = sqrt(23.3^2 + (4.12/2)^2) = 23.4#

#color(indigo)(a = b = 23.4, c = 4.12#

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