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?

Question

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?

Sophia Alfred · Accepted Answer

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

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

#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#

Aaliyah Anderson · Answer

undefined The distance between the two given points is the base of the triangle. Using the distance formula, the distance between (2, 4) and (3, 8) is √((3-2)^2 + (8-4)^2) = √(1 + 16) = √17.

Since the triangle is isosceles, the altitude from the top vertex bisects the base. The length of the altitude can be found using the formula for the area of a triangle: area = 1/2 * base * height. Thus, height = 2 * area / base = 2 * 48 / √17.

The length of the other two sides of the triangle (the congruent sides) can be calculated using the Pythagorean theorem, where one leg is the altitude we just found and the other leg is half the base length.

So, the lengths of the sides are √((1/2 * base)^2 + (height)^2) and √((1/2 * base)^2 + (height)^2).