First, we need to find the length of the line segment making up the base of the isosceles triangle. The formula for calculating the distance between two points is:
d = sqrt((color(red)(x_2) - color(blue)(x_1))^2 + (color(red)(y_2) - color(blue)(y_1))^2)
Substituting the values from the points in the problem gives:
d = sqrt((color(red)(5) - color(blue)(8))^2 + (color(red)(9) - color(blue)(3))^2)
d = sqrt((-3)^2 + 6^2)
d = sqrt(9 + 36)
d = sqrt(45)
d = sqrt(9 * 5)
d = sqrt(9)sqrt(5)
d = 3sqrt(5)
he formula for area of a triangle is:
A= (bh_b)/2
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Substituting the Area from the problem and the length of the base we calculated and solving for h_b gives:
4= (3sqrt(5)h_b)/2
2/(3sqrt(5)) xx 4= 2/(3sqrt(5)) xx (3sqrt(5)h_b)/2
8/(3sqrt(5)) = cancel(2/(3sqrt(5))) xx cancel((3sqrt(5))/2)h_b
h_b = 8/(3sqrt(5))
From an isosceles triangle we know the base and h_b are at right angles. Therefore we can use the Pythagorean Theorem to find the length of the sides.
c^2 = a^2 + b^2
c is what we are solving for.
a is the side of the triangle made up of 1/2 the base or:
1/2 xx 3sqrt(5) = (3sqrt(5))/2
b is h_b = 8/(3sqrt(5))
Substituting and solving for c gives:
c^2 = ((3sqrt(5))/2)^2 + (8/(3sqrt(5)))^2
c^2 = (9 * 5)/4 + 64/(9 * 5)
c^2 = 45/4 + 64/45
c^2 = (45/45 xx 45/4) + (4/4 xx 64/45)
c^2 = 2025/180 + 256/180
c^2 = 2281/180
sqrt(c^2) = sqrt(2281/180)
c = sqrt(2281)/sqrt(180)
c = sqrt(2281)/sqrt(36 * 5)
c = sqrt(2281)/(sqrt(36)sqrt(5))
c = sqrt(2281)/(6sqrt(5))
