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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?

Guest Jul 11, 2018

edited by
Guest
Jul 11, 2018

#1**+3 **

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

CluelesssPersonnn Jul 11, 2018

#3**+2 **

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 !!!!

CPhill Jul 11, 2018

#4**+2 **

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

CPhill Jul 11, 2018

#5**+1 **

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

CPhill Jul 12, 2018