1. Point X is on AC such that AX=3CX=12 . If $\angle ABC = \angle BXA = 90, then what is BX?
2. Find the length of the shortest altitude of a triangle with sides of lengths 10, 24, and 26.
3.In the diagram below, we have DE=2EC and AB=DC=20 . Find the length of FG.
4.In the diagram below, we have $\angleABC= \angleACB=\angleDEC=\angleCDE$, BC=8, and DB=2. Find AB.
5.Right triangle ABC has , AB=3,BC=4 and AC=5. Square XYZW is inscribed in triangle ABC with X and Y on AC , W on AB , and Z on BC. What is the side length of the square?
6. In triangle ABC, we have\angleBAC= 60 and \angle ABC = 45. The bisector of \angle A intersects BC at point T, and AT=24. What is the area of triangle ABC?
+25 1: 4√3
2: 120/13
3: 12
4: 6*(√3)/3
5: 60/37
6: 216 + 72√3
PS. I am not sure about all of these answers
+128474 2. Find the length of the shortest altitude of a triangle with sides of lengths 10, 24, and 26.
The area of the triangle will be (1/2)(10) (24) = 120 units^2
So....if we let 10 be one altitude....the shortest altitude will be drawn from the intersection of the legs to the hypotenuse
And since the areas are equal....we have that
(1/2) (length of the shortest altitude) (26) = 120
(13) (length of shortest altitude) = 120 divide both sides by 13
length of the shortest altitude = 120 / 13 units .....just as CluelesssPersonn found !!!!
+128474 3.
Since DE = 2EC....then
EC + 2EC = DC
3EC = 20
EC = 20/3
And DE = 2EC = 2(20/3) = 40/3
Note that triangle AFB is similar to triangle EFD
So
AB/ED = FB / FD
So
20/(40/3) = FB / FD
60/40 = FB / FD
6/4 = FB /FD
3/2 = FB / FD
Then BF consists of 5 equal parts.....FB is 3 of them and FD is 2 of them
And in triangle BDC, FG is drawn parallel to DC....and we have the following relationship :
BF / BD = FG / DC
3 / 5 = FG / 20 multiply both sides by 20
20 * 3 / 5 = FG
60 / 5 = FG
12 = FG
+128474 4.In the diagram below, we have \(angleABC= angleACB=angleDEC=angleCDE\), BC=8, and DB=2. Find AB.
Since angle DEC = angle BEC, and angle DEC = angle ABC = angle ECB.....then angles BEC and ECB are equal....then in triangle BEC, BC = BE
Then
BC = BD + DE
8 = 2 + DE
6 = DE
And angle ABC = angle ACB...so...AC = AB
Note that since angles ACB, ACB, DEC and CDE are all equal, we have the following similar triangles :
ΔABC ~ Δ BEC ~ ΔCED and because angles DEC and CDE are equal, then CD = EC
So
AC/BC = BC /EC
AC / 8 = 8 / EC
AC * EC = 8*8
AC * EC = 64
AC = 64/EC (1)
And
AC/BC = CD/ED
AC/8 = CD / 6
AC/8 = EC/ 6
AC = (8/6)EC
AC = (4/3)EC (2)
Equating (1) and (2), we have that
64/EC = (4/3)EC
(64)(3/4) = EC^2
48 = EC^2
√48 = EC
4√3 = EC
And AC = (4/3)EC = (4/3) * 4√3 = 16/√3 = AB
