SOLUTION: 2 isosceles triangles have legs that are 26 and areas that are equal. If the base of one triangle is 48. Find the base of the other triangle ? *i know that in an isosceles triangl

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Question 855917: 2 isosceles triangles have legs that are 26 and areas that are equal. If the base of one triangle is 48. Find the base of the other triangle ?
*i know that in an isosceles triangle the sides that are connected to the vertex angle is congruent and the base angles are also congruent* i have tried using the Pythagorean theorem but its not working*(Pythagorean theorem= a^2+b^2=C^2
please do help me and thank you

Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 
Hi,
Perhaps this picture will help:
the 'h' is a 'leg', other leg = 24 (h bisects the base)
hypotenuse = 26

line segment of Height to the base bisects the base (base = 48)
Therefore: 26^2 - 24^2 = h^2 = 100, h = 10
A = = 240
Find the base of the other triangle
A = 240 =
480=
480^2/b^2 = 26^2 - (b^2/4
b^4 -4(26^2)b^2 + 4(480^2) Let x = b^2
x^2 - 2704x + - 921600
x = 400 0r x = 2304
b = 20 0r b = 48
the base of the other triangle is 20
CHECKING our answer***
26^2 -10^2 = 576= h^2, h = 24 AND
Wish You the Best in your Studies.

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