jugoslav

 ABC is an equilateral triangle such that vertices B,C lie on two parallel lines at a distance of 6 units. A lies between the parallel lines at a distance of 4 units from one of them.  Find the side length of the equilateral triangle.

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The side length of a triangle ABC is ≈ 6.11 units

Good job, Phill!

AB = 4 * cos^-1(3/4) = 3.2

AB and CD are parallel. The length of a chord AB is 7 cm. AD = BC = 3 cm. DC is the diameter of a circle O. Find radius DO.

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Let AF be the altitude of a trapezoid ABCD.

AO = DO => r          AB = 7       AD = BC = 3

FO = AB / 2  ==>    FO = 3.5         

AD² - (r - FO)² = r² - FO²        ==>        3² - (r - 3.5)² = r² - 3.5²       ==>  r = 4.5 cm

I "borrowed" Dragan's diagram and added a couple of red lines to make my answer even clearer. 

A vertical pole AB measuring 5 meters snaps at point C. The pole remains in contact at C and the top of the pole touches the ground at point T, a distance of 3 meters from A. Find the length AC, in meters, the point where the pole snapped.

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AB = 5       AT = 3       AC => x

3² + x² = (5 - x)²          x = 1.6           CT = 3.4

Find the area of the triangle with sides sqrt(13), sqrt(61), sqrt(80).

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The triangle area, using Heron's formula, is 14 square units. 

Perimeter of ΔABC = 18 units and the area = 12 square units.

AB = 8

AP = 8.4641

BP = 1.5359

AB + AP + BP = 8 + 8.4641 + 1.5359 = 18

Area of ΔABP = 6 u²

CP ≈ 4.582456 units

Square side is a

2(a / tan45º) + a = √72           a = √8           A = 8 cm²

White circle:        diameter D = 2     area  A_c = pi

Square:               side b = √2          area   A_s = 2

Semicircles:       diameter d = √2     total area A_sc = pi

A_s + A_sc - A_c       ==>        2 + pi - pi = 2        shaded area = 2 u²