The perimeter of an isoceles triangle is 16. The length of each of the equal sides is 8-x.?

perimeter of triangle = sum of sides = 16

an isosceles triangle has two equal sides and they give that each of those is 8 - x

so the third side = 16 - 2(8 - x)

= 16 - 16 + 2x

= 2x

An altitude (height) of the isosceles triangle divides it in half making 2 right triangles

the length of the base of each RIGHT triangle is 2x / 2 = x

Using one of the right triangle and the Pythagorean Theorem:

(8 - x)² = h² + x²

h² = (8 - x)² - x²

h² = 64 - 16x + x² - x²

h² = 64 - 16x

h² = 16(4 - x)

h = √[16(4 - x)]

h = √16 * √(4 - x)

h = 4 √(4 - x)

and that can also be written as h = 4 (4 - x)^(1/2)

Area of triangle = (1/2) * Base * height

The base of the isosceles triangle is 2x so:

Area of triangle = (1/2) * 2x * 4 √(4 - x)

= 4x √(4 - x)

and that can also be written as 4x (4 - x)^(1/2)

Use x because of the fact the fringe of the triangle and six - x because of the fact the fringe of the sq.. subsequently, the portion of the triangle is x/3 and the portion of the sq. is (6-x)/4. as a result, the sum of their aspects is: A(x) = (6-x)² / sixteen + [sqrt (3) / 4]* x²/9 A(x) = (6-x)² / sixteen + [sqrt (3) / 36]* x² A' = (-a million/8)(6-x) + [sqrt (3) / 18] x A' = -3/4 + x/8 + [sqrt (3) / 18] x 0 = -3/4 + x/8 + [sqrt (3) / 18] x 3/4 =x/8 + [sqrt (3) / 18] x 3/4 = x((a million/8 + sqrt (3) / 18 ) 3/4 = x (9 + 4sqrt (3))/seventy two fifty 4/(9 + 4sqrt (3)) = x x= [18(9-4sqrt3)/11] it is approximately 3.6 so because it is your purely serious fee, you are able to attempt if it is a optimal, yet this x fee represents the fringe of the triangle, it is the place you may decrease the twine and use something for the sq.. A(3.6) is the section.