Angle MQS = 55 degrees
Using this diagram, you should be able to find the area of the equilateral triangle QRP.
Area of a triangle QRP = 3√3
This may help you a bit
A = (LX * XY) / 2 = 16
Let WXYZ be a trapezoid with bases XY and WZ. In this trapezoid, ZXW=105 degrees, XWZ=43 degrees, and XYZ=141 degrees. Find YXZ, in degrees.
Angle YXZ = 32 degrees
AB cannot be shorter than BY!!!
https://www.youtube.com/watch?v=LlXFrwRBozc
ΔBCY and ΔACX are similar. Let CX be an N (CY = 2N)
N / 4 = 18 / 2N
N * 2N = 18 * 4
2N2 = 72
N = 6
XY = 3 * N = 18
AB = sqrt [ (XY)2 + (BY - AX)2 ]
An equilateral triangle of side length 2 units is inscribed in a circle. Find the length of a chord of this circle which passes through the midpoints of two sides of the triangle.
BC = 2 BQ = 1 ∠B = 60º
Height AQ = tan(B) * BQ = 1.732050808
AO = 2/3 * AQ = 1.154700537 AO = XO
AP = AQ / 2 = 0.866025402
PO = AO - AP = 0.288675134
XP = sqrt( XO² - PO² ) = 1.118033987
XY = 2 * XP = 2.236067975
https://web2.0calc.com/questions/i-really-need-help_17#r3
∠PTO = 25º
∠TOP = ∠OPQ = ∠OQP = 65º
∠QOP = 50º
∠POQ = ∠QOP
Find the area of the quadrilateral with vertices (0,3), (3,0), (0,9), and (9,0).
A(0,3) B(3,0) C(9,0) D(0,9)
BC = 6 angle DCB = 45º
Area of the quadrilateral (ABCD) = (92 - 32) / 2
The lengths of the sides of a triangle are 5, 6, and 10 cm respectively. Find the square of the length of the median of the greatest side.
a = 5 b = 6 c = 10
(mc)2 = {[ sqrt( 2a2 + 2b2 - c2 )] / 2}2 = 5.5 cm2