1)
In , we know that and Find
2)
In quadrilateral and We know that and What is
3)
Two diagonals of quadrilateral are perpendicular to each other at We know and What is
4)
In , we have , , and . Point is on such that bisects . If , then what is
5)
Points and are on side of rectangle such that and trisect . If and , then what is the area of ?
1) Draw altitude BD.....this will bisect angle ABC
Then angle BCA = 30°
And since triangle BDC is a 30 - 60- 90 triangle...then CD = [√3 / 2] [ 6 √3] = 9
And CD = (1/2) AC
So AC = 2CD = 18
3) AO = BO and CO = DO
Then the angles opposite these sides = 45°
And since triangles ABO and DCO are 45 - 45 - 90 triangles
Then DO = DC/√2 and BO = AB / √2
So DO + BO = DC/√2 + AB/√2 = [ DC + AB ] / √2
And DO + BO = BD = [ DC + AB ] / √2 = [ 5 √2 ] / √2 = 5
4) Since QX bisects PQR, then angles PQX = RQX = QPX = 30°
Then angle QXP = 120°...so angle QXR = 60°
And triangles QRP and QRX are 30 - 60 - 90 triangles
So.........since QR is opposite a 30° angle, it is (1/2)PQ = (1/2) 4√3 = 2√3
So.......QX = [ 2 / √3 ] * [ 2√3 ] = 4
5) Since BAD is trisected, angle SAD = 30°
And triangle SAD is a 30 - 60 - 90 triangle
So.......AD = √3SD = √3 * 1 = √3
And angle TAD = 60° so triangle ATD is also a 30 - 60 - 90 triangle
And TD = √3 AD = √3 √3 = 3
So CD = CT + TD = 2√3 - 3 + 3 = 2√3
So......the area of ABCD = CD * AD = 2√3 * √3 = 6 units^2